2026-09-10
Differential equations are equations involving an unknown function and its derivatives, like \[\tag{0.1.1} m \frac { d ^ 2 x } { d t ^ 2 } + k x = 0 ,\] \[\tag{0.1.2} \frac { d ^ 2 x } { d t ^ 2 } - \mu ( 1 - x ^ 2 ) \frac { d x } { d t } + x = 0\] \[\tag{0.1.3} \frac { \partial ^ 2 u } { \partial x ^ 2 } + \frac { \partial ^ 2 u } { \partial y ^ 2 } = 0\] \[\tag{0.1.4} \frac { \partial u } { \partial t } - k \frac { \partial ^ 2 u } { \partial x ^ 2 } = 0\] \[\tag{0.1.5} \frac { \partial ^ 2 u } { \partial t ^ 2 } - c ^ 2 \frac { \partial ^ 2 u } { \partial x ^ 2 } = 0\] \[\tag{0.1.6} \frac { \partial u } { \partial t } + u \frac { \partial u } { \partial x } = 0\] \[\tag{0.1.7} \frac { \partial u } { \partial t } + \frac { \partial ^ 3 u } { \partial x ^ 3 } + 6 u \frac { \partial u } { \partial x } = 0\] \[\tag{0.1.8} \frac { \partial v } { \partial \tau } + \frac 1 2 \sigma ^ 2 s ^ 2 \frac { \partial ^ 2 v } { \partial s ^ 2 } + r s \frac { \partial v } { \partial s } - r v = 0\]
All of these equations have names. Equation (0.1.1) is the simple harmonic oscillator equation. It appears in mechanics and many other places. Equation (0.1.2) is the Van der Pol equation. It first appeared in electrical engineering. Equation (0.1.3) is the Laplace equation. It appears in the study of gravitational and electrostatic fields. Equation (0.1.4) is the diffusion equation, also known as the heat equation. It appears in study of diffusion, originally diffusion of heat but it also applies to, for example, diffusion of chemical solutions. Equation (0.1.5) is the wave equation. It appears in the study of various kinds of waves, e.g. electromagnetic, acoustic, etc., but not generally water waves. Equation (0.1.6) is Burgers' equation. Unlike the wave equation Burgers' equation is often used as a simple model for water waves. Equation (0.1.7) is the Korteweg–De Vries equation. It is a more refined model of water waves, but still simpler than real water waves. Equation (0.1.8) is the Black-Scholes equation. It appears in mathematical finance.
Some terminology is useful for describing these. We say that variables which are differentiated are dependent variables, variables with respect to which we differentiate them are independent variables, and variables which don't appear in derivatives at all are parameters. Note that which variables play which role varies from equation to equation. $x$, for example, is a dependent variable in the first two equations, an independent variable in the next five equations, and doesn't appear in the last equation. If there's only one independent variable, i.e. if we only differentiate with respect to one variable, then the derivatives are ordinary derivatives and so the equation is called an ordinary differential equation. If there is more than one then the derivatives are partial derivatives and so the equation is called a partial differential equation, which is the subject of these notes. In the list above equations (0.1.1) and (0.1.2) are ordinary, while the others are all partial differential equations. Some knowledge of ordinary differential equations is useful for studying partial differential equations, but for what we'll do here it's not essential. The order of a differential equation is the order of the highest derivative appearing in it. Equation (0.1.6) is first order and equation (0.1.7) is third order while all the other equations above are second order. Second order equations seem to be pervasive in mathematical physics.
The most important distinction in the theory of differential equations is between linear and nonlinear equations. The distinction is somewhat subtle though. A linear differential equation is a linear equation in the unknown function, i.e. the dependent variable, and its derivatives, the coefficients of which are allowed to be functions of the independent variables and parameters, but not of the dependent variable. It's this last bit which tends to cause confusion. In the list above equations (0.1.1), (0.1.3), (0.1.4), (0.1.5) and (0.1.8) are linear, while (0.1.2), (0.1.6) and (0.1.7) are nonlinear. Equation (0.1.8), for example, is linear because the coefficients of the derivatives, $\partial v / \partial t$, $\partial ^ 2 v / \partial s ^ 2$, $\partial v / \partial s$ and $v$, the last of these being considered as a zeroeth order derivative of the unknown function $v$, are $1$, $\frac 1 2 \sigma ^ 2 s ^ 2$, $r s$ and $- r$, all of which are functions of the independent variables $t$ and $s$ and the parameters $r$ and $\sigma$, but don't depend on the independent variable $v$. Equation (0.1.7), by contrast, is nonlinear because if we try to write it as linear equation for the derivatives $ \partial u / \partial t $, $ \partial ^ 3 u / \partial x ^ 3 $ and $ \partial u / \partial x $ with coefficients $ 1 $, $ 1 $ and $ 6 u $ then the first two are okay but $ 6 u $ is a function of the dependent variable $u$, which is not allowed. Note that constant functions are functions, so $ 1 $ is a function of the independent variables $t$ and $x$, it's just a constant function. $ 6 u $, by contrast, is not a function of $t$ and $x$.
We met the wave equation \[\tag{1.0.1} \frac { \partial ^ 2 u } { \partial t ^ 2 } - c ^ 2 \frac { \partial ^ 2 } { \partial x ^ 2 } = 0\] in the introduction. Although it wasn't stated there, we always assume that $ c $ is positive. That's natural for physical applications and doesn't really limit the generality since replacing $ c $ by $ - c $ leaves the equation unchanged and the case $ c = 0 $ is both easy to analyse and uninteresting.
Suppose that $u$ is a twice continuously differentiable function on $\mathbf R ^ 2$. For this chapter we'll label the coordinates on $\mathbf R ^ 2$ as $t$ and $x$, written in that order, and in diagrams the $t$ will be vertical and the $x$ axis will be horizontal. This is slightly awkward since we are used to coordinate systems in the plane where the first coordinate corresponds to the horizontal axis and the second one corresponds to the vertical. The letters $ t $ and $ x $ for time and space coordinates are far too well established to consider changing. Similarly the convention that in space-time diagrams time corresponds to the vertical direction is fairly universal. The only other option to avoid listing the vertical coordinate before the horizontal one would be to reverse the order of $ t $ and $ x $. Some authors do this, but listing coordinates in alphabetical order is standard nearly everywhere else and physicists largely switched from the convention of listing time last to listing time last more than half a century ago so the conventions we're using here are probably the least bad option.
It's helpful to introduce the auxiliary functions \[\tag{1.1.1} v = \frac { \partial u } { \partial t } + c \frac { \partial u } { \partial x } , \quad w = \frac { \partial u } { \partial t } - c \frac { \partial u } { \partial x }.\] Since $u$ was twice continuously differentiable $v$ and $w$ are once continuously differentiable.
Suppose that $ ( t _ 1 , x _ 1 ) $ and $ ( t _ 2 , x _ 2 ) $ are points such that \[\tag{1.1.2} x _ 1 - c t _ 1 = x _ 2 - c t _ 2 .\] and set \[\tag{1.1.3} \tau ( r ) = t _ 1 + r ( t _ 2 - t _ 1 ) , \quad \xi ( r ) = x _ 1 + r ( x _ 2 - x _ 1 ) .\] Then, by the chain rule \[\tag{1.1.4} \begin{split} \frac d { dr } z ( \tau ( r ) , \xi ( r ) ) & = \left ( t _ 2 - t _ 1 \right ) \frac { \partial z } { \partial t } ( \tau ( r ) , \xi ( r ) ) \\ & \quad {} + \left ( x _ 2 - x _ 1 \right ) \frac { \partial z } { \partial x } ( \tau ( r ) , \xi ( r ) ) \\ & = \left ( t _ 2 - t _ 1 \right ) \left ( \frac { \partial z } { \partial t } + c \frac { \partial z } { \partial x } \right ) ( \tau ( r ) , \xi ( r ) ) \end{split}\] for any function $z$ which is at least once continuously differentiable. Integrating from $r = 0$ to $r = 1$ and using the fundamental theorem of calculus we see that \[\tag{1.1.5} \begin{split} & z ( t _ 2 , x _ 2 ) = z ( t _ 1 , x _ 1 ) \\ & \qquad {} + ( t _ 2 - t _ 1 ) \int _ 0 ^ 1 \left ( \frac { \partial z } { \partial t } + c \frac { \partial z } { \partial x } \right ) \left ( t _ 1 + r ( t _ 2 - t _ 1 ) , x _ 1 + r ( x _ 2 - x _ 1 ) \right ) \, d r . \end{split}\] This holds in particular for $z = u$ and $z = w$. In the latter case note that \[\tag{1.1.6} \frac { \partial w } { \partial t } + c \frac { \partial w } { \partial x } = \frac { \partial ^ 2 u } { \partial t ^ 2 } - c ^ 2 \frac { \partial ^ 2 u } { \partial x ^ 2 }\] so if, as we will assume from now until the end of this section, $u$ satisfies the wave equation, the integrand vanishes throughout the interval of integration. We then conclude that \[\tag{1.1.7} u ( t _ 2 , x _ 2 ) = u ( t _ 1 , x _ 1 ) + ( t _ 2 - t _ 1 ) \int _ 0 ^ 1 v \left ( t _ 1 + r ( t _ 2 - t _ 1 ) , x _ 1 + r ( x _ 2 - x _ 1 ) \right ) \, d r .\] and \[\tag{1.1.8} w ( t _ 2 , x _ 2 ) = w ( t _ 1 , x _ 1 ) .\] The preceding calculation was carried out under the assumption that $ x _ 1 - c t _ 1 = x _ 2 - c t _ 2 $, which is certainly true if \[\tag{1.1.9} x _ 1 = x _ 2 + c t _ 1 - c t _ 2 ,\] so we can rewrite the preceding equations as \[\tag{1.1.10} \begin{split} u ( t _ 2 , x _ 2 ) & = u ( t _ 1 , x _ 2 + c t _ 1 - c t _ 2 ) \\ & \quad {} + ( t _ 2 - t _ 1 ) \int _ 0 ^ 1 v \left ( t _ 1 + r ( t _ 2 - t _ 1 ) , x _ 2 - ( 1 - r ) c ( t _ 2 - t _ 1 ) \right ) \, d r . \end{split}\] and \[\tag{1.1.11} w ( t _ 2 , x _ 2 ) = w ( t _ 1 , x _ 2 + c t _ 1 - c t _ 2 ) .\] If instead we assume that \[\tag{1.1.12} x _ 1 + c t _ 1 = x _ 2 + c t _ 2\] then a very similar calculation leads to \[\tag{1.1.13} \begin{split} u ( t _ 2 , x _ 2 ) & = u ( t _ 1 , x _ 2 - c t _ 1 + c t _ 2 ) \\ & \quad {} + ( t _ 2 - t _ 1 ) \int _ 0 ^ 1 w \left ( t _ 1 + r ( t _ 2 - t _ 1 ) , x _ 2 + ( 1 - r ) c ( t _ 2 - t _ 1 ) \right ) \, d r . \end{split}\] and \[\tag{1.1.14} v ( t _ 2 , x _ 2 ) = v ( t _ 1 , x _ 2 - c t _ 1 + c t _ 2 ) .\]
Since $ ( t _ 2 , x _ 2 ) $ is an arbitrary point in the equations above we can substitute any other point for it. In particular we can substitute \[\tag{1.1.15} ( t _ 1 + r ( t _ 2 - t _ 1 ) , x _ 2 + ( r - 1 ) c ( t _ 2 - t _ 1 ) )\] for it in (1.1.14), which gives \[\tag{1.1.16} v ( t _ 1 + r ( t _ 2 - t _ 1 ) , x _ 1 + r c ( t _ 2 - t _ 1 ) ) = v ( t _ 1 , x _ 2 + ( 2 r - 1 ) c ( t _ 2 - t _ 1 ) ) ,\] which we can substitute into (1.1.10) to obtain \[\tag{1.1.17} \begin{split} u ( t _ 2 , x _ 2 ) & = u ( t _ 1 , x _ 2 + c t _ 1 - c t _ 2 ) \\ & \quad {} + ( t _ 2 - t _ 1 ) \int _ 0 ^ 1 v \left ( t _ 1 , x _ 2 + ( 2 r - 1 ) c ( t _ 2 - t _ 1 ) \right ) \, d r . \end{split}\] The substitution \[\tag{1.1.18} y = x _ 2 + ( 2 r - 1 ) c ( t _ 2 - t _ 1 )\] converts this into \[\tag{1.1.19} \begin{split} u ( t _ 2 , x _ 2 ) & = u ( t _ 1 , x _ 2 + c t _ 1 - c t _ 2 ) \\ & \quad {} + \frac 1 { 2 c } \int _ { x _ 2 - c ( t _ 2 - t _ 1 ) } ^ { x _ 2 + c ( t _ 2 - t _ 1 ) } v ( t _ 1 , y ) \, d y . \end{split}\] Similarly, we could substitute \[\tag{1.1.20} ( t _ 1 + r ( t _ 2 - t _ 1 ) , x _ 2 + ( 1 - r ) c ( t _ 2 - t _ 1 ) )\] for $ ( t _ 2 , x _ 2 ) $ in (1.1.11), substitute the result into (1.1.13), and make the substitution \[\tag{1.1.21} y = x _ 2 + ( 1 - 2 r ) c ( t _ 2 - t _ 1 )\] into the resulting integral to obtain \[\tag{1.1.22} \begin{split} u ( t _ 2 , x _ 2 ) & = u ( t _ 1 , x _ 2 - c t _ 1 + c t _ 2 ) \\ & \quad {} + \frac 1 { 2 c } \int _ { x _ 2 - c ( t _ 2 - t _ 1 ) } ^ { x _ 2 + c ( t _ 2 - t _ 1 ) } w ( t _ 1 , y ) \, d y . \end{split}\] Averaging (1.1.19) and (1.1.22) and noting that \[\tag{1.1.23} v + w = 2 \frac { \partial u } { \partial t }\] gives the equation \[\tag{1.1.24} \begin{split} u ( t _ 2 , x _ 2 ) & = \frac 1 2 u ( t _ 1 , x _ 2 + c t _ 1 - c t _ 2 ) \\ & \quad {} + \frac 1 2 u ( t _ 1 , x _ 2 - c t _ 1 + c t _ 2 ) \\ & \quad {} + \frac 1 { 2 c } \int _ { x _ 2 - c ( t _ 2 - t _ 1 ) } ^ { x _ 2 + c ( t _ 2 - t _ 1 ) } \frac { \partial u } { \partial t } ( t _ 1 , y ) \, d y . \end{split}\] Relabeling the variables gives D'Alembert's formula \[\tag{1.1.25} u ( t , x ) = \frac 1 2 f ( x + c s - c t ) + \frac 1 2 f ( x - c s + c t ) + \frac 1 { 2 c } \int _ { x + c s - c t } ^ { x - c s + c t } g ( y ) \, d y ,\] where \[\tag{1.1.26} f ( y ) = u ( s , y ) , \quad g ( y ) = \frac { \partial u } { \partial t } ( s , y ) .\] This gives the solution at time $ t $ in terms of its values and the values of its first derivatives at time $ s $. Note that we haven't assumed $ s < t $, although the formula is usually applied in this case. We haven't even assumed $ s \neq t $, although the formula doesn't give us any useful information when $ s = t $.
D'Alembert's formula has a natural interpretation in terms of the initial value problem for the wave equation, i.e. the problem of finding a classical solution to the wave equation with initial conditions at time $s$ given by \[\tag{1.2.1} u ( s , y ) = f ( y ) , \quad \frac { \partial u } { \partial t } ( s , y ) = g ( y ) .\] A classical solution is just a twice continuously differentiable function. It's natural to assume two derivatives because the equation is of second order. In other words, second derivatives are the highest ones which appear. The equation doesn't have any obvious interpretation if we assume much less differentiability than this. It is possible, and indeed useful, to give it less obvious interpretations which assume less differentiability, but that would be a topic for a more advanced text. Here we'll only consider classical solutions. Since $u$ should be twice continuously differentiable the initial conditions force $ f $ to be twice continuously differentiable as well and $ g $ to be continuously differentiable. So a more precise formulation of the initial value problem is, given a twice continuously differentiable function $ f $ and a continuously differentiable function $ g $, to find a classical solution $ u $ of the wave equation such that the initial conditions (1.2.1) are satisfied, i.e a twice continuously differentiable function $ u $ satisfying \[\tag{1.2.2} \frac { \partial ^ 2 u } { \partial t ^ 2 } - c ^ 2 \frac { \partial ^ 2 u } { \partial x ^ 2 } = 0 .\]
In terms of the initial value problem of the preceding paragraph the calculation of the preceding section provides a proof of the following theorem.
Both theorems above are pure uniqueness theorems. They assert that there is at most one solution of the initial value problem, without guaranteeing that there is at least one. And indeed there is no way to turn the calculation of the preceding section into an existence proof, since we assumed at a very early stage that we had a classical solution $ u $. The nice thing about an explicit formula though is that it naturally suggests a way of proving existence: we just need to take the formula and verify that what it gives is indeed a solution. Fortunately this works for D'Alembert's formula. The most naive way of doing this is rather awkward though, for two reasons. First, we'd need to differentiate under the integral, and the variables with respect to which we want to differentiate appear in the limits of the integral. Second, we need to take two derivatives and the function $ g $ appearing in the integrand is only known to have one derivative. Neither of these problems is insurmountable but it's possible to avoid facing either of them directly.
We start by choosing some point $ p $ in the interval in which the functions $ f $ and $ g $ are defined and set \[\tag{1.2.3} \begin{split} \varphi ( z ) & = \frac 1 2 f ( z ) + \frac 1 { 2 c } \int _ z ^ p g ( y ) \, d y , \\ \psi ( z ) & = \frac 1 2 f ( z ) + \frac 1 { 2 c } \int _ p ^ z g ( y ) \, d y \end{split}\] with the usual convention that if the lower limit of an integral is greater than the upper limit then the limits should be swapped and the sign should be changed. Then $ \varphi $ and $ \psi $ are twice continuously differentiable functions, defined on the same interval that $ f $ and $ g $ were. Indeed the fundamental theorem of calculus gives \[\tag{1.2.4} \begin{split} \varphi ' ( z ) & = \frac 1 2 f ' ( z ) - \frac 1 { 2 c } g ( z ) , \\ \psi ' ( z ) & = \frac 1 2 f ' ( z ) + \frac 1 { 2 c } g ( z ) \end{split}\] and then taking an additional derivative gives \[\tag{1.2.5} \begin{split} \varphi '' ( z ) & = \frac 1 2 f '' ( z ) - \frac 1 { 2 c } g ' ( z ) , \\ \psi '' ( z ) & = \frac 1 2 f '' ( z ) + \frac 1 { 2 c } g ' ( z ) . \end{split}\] By assumption $ f $ is twice continuously differentiable and $ g $ is continuously differentiable in the initial value problem so the right hand sides are continuous. Let \[\tag{1.2.6} u ( t , x ) = \varphi ( x + c s - c t ) + \psi ( x - c s + c t )\] and note that $ u $ is twice continuously differentiable. Then \[\tag{1.2.7} u ( s , x ) = \varphi ( x ) + \psi ( x ) = f ( x ) .\] Also, \[\tag{1.2.8} \frac { \partial u } { \partial t } ( t , x ) = - c \varphi ' ( x + c s - c t ) + c \psi ' ( x - c s + c t )\] so \[\tag{1.2.9} \frac { \partial u } { \partial t } ( s , x ) = - c \varphi ' ( x ) + c \psi ' ( x ) = g ( s ) .\] Thus $ u $ satisfies the initial conditions. Does it also satisfy the wave equation? Taking another derivative, \[\tag{1.2.10} \frac { \partial ^ 2 u } { \partial t ^ 2 } ( t , x ) = c ^ 2 \varphi '' ( x + c s - c t ) + c ^ 2 \psi '' ( x - c s + c t ) .\] Similarly, \[\tag{1.2.11} \frac { \partial u } { \partial x } ( t , x ) = \varphi ' ( x + c s - c t ) + \psi ' ( x - c s + c t )\] and \[\tag{1.2.12} \frac { \partial ^ 2 u } { \partial x ^ 2 } ( t , x ) = \varphi '' ( x + c s - c t ) + \psi '' ( x - c s + c t )\] so \[\tag{1.2.13} \frac { \partial ^ 2 u } { \partial t ^ 2 } ( t , x ) - c ^ 2 \frac { \partial ^ 2 u } { \partial x ^ 2 } ( t , x ) = 0\] and so $ u $ is a classical solution of the wave equation. In particular, since we've already checked that the initial conditions are satisfied, the initial value problem has at least one solution. We've already seen that it has at most one solution so there is exactly one solution. The uniqueness theorems tell us that this solution must be given by D'Alembert's formula so there is no need to check separately that the expression for $ u $ given above is equal to the one in D'Alembert's formula, although it is not difficult to do so.
We've now proved existence theorems, complementary to our earlier uniqueness theorems:
We saw in section 1.1 that when $u$ is a classical solution to the wave equation the functions \[\tag{1.3.1} v = \frac { \partial u } { \partial t } + c \frac { \partial u } { \partial x } , \quad w = \frac { \partial u } { \partial t } - c \frac { \partial u } { \partial x } \] satisfy the relations \[\tag{1.3.2} v ( t , x ) = v ( s , x - c s + c t ) , \quad w ( t , x ) = w ( s , x + c s - c t ) . \] The quantity \[\tag{1.3.3} e = \frac 1 4 v ^ 2 + \frac 1 4 w ^ 2 = \frac 1 2 \left ( \frac { \partial u } { \partial t } \right ) ^ 2 + \frac { c ^ 2 } 2 \left ( \frac { \partial u } { \partial x } \right ) ^ 2 \] has a physical interpretation as the energy density, so the integral \[\tag{1.3.4} I = \frac 1 4 \int _ a ^ b \left ( v ( t , x ) ^ 2 + w ( t , x ) ^ 2 \right) \, d x \] represents the energy present in the interval $ [ a , b ] $ at time $t$. We can split this into two parts, \[\tag{1.3.5} I = \frac 1 4 \int _ a ^ b v ( t , x ) ^ 2 \, d x + \frac 1 4 \int _ a ^ b w ( t , x ) ^ 2 \, d x \] and use the relations above to get \[\tag{1.3.6} I = \frac 1 4 \int _ a ^ b v ( s , x - c s + c t ) ^ 2 \, d x + \frac 1 4 \int _ a ^ b w ( s , x + c s - c t ) ^ 2 \, d x \] or, changing variables in the integrals, \[\tag{1.3.7} I = \frac 1 4 \int _ { a + c s - c t } ^ { b + c s - c t } v ( s , x ) ^ 2 \, d x + \frac 1 4 \int _ { a - c s + c t } ^ { b - c s + c t } w ( s , x ) ^ 2 \, d x . \] Now the integral of a non-negative integrand over an interval is at least as large as the integral over a smaller interval and at most as large as the integral over a larger integral so we see that \[\tag{1.3.8} I \ge \frac 1 4 \int _ { \max ( a + c s - c t , a - c s + c t ) } ^ { \min ( b + c s - c t , b - c s + c t ) } v ( s , x ) ^ 2 \, d x + \frac 1 4 \int _ { \max ( a + c s - c t , a - c s + c t ) } ^ { \min ( b + c s - c t , b - c s + c t ) } w ( s , x ) ^ 2 \, d x \] \[\tag{1.3.9} I \le \frac 1 4 \int _ { \min ( a + c s - c t , a - c s + c t ) } ^ { \max ( b + c s - c t , b - c s + c t ) } v ( s , x ) ^ 2 \, d x + \frac 1 4 \int _ { \min ( a + c s - c t , a - c s + c t ) } ^ { \max ( b + c s - c t , b - c s + c t ) } w ( s , x ) ^ 2 \, d x . \] Combining the integrals, and writing the limits in a slightly cleaner form, we find \[\tag{1.3.10} \int _ { a + c | s - t | } ^ { b - c | s - t | } e ( s , x ) \, d x \le I \le \int _ { a - c | s - t | } ^ { b + c | s - t | } e ( s , x ) \, d x . \] If $ 2 c | s - t | > b - a $ then the lower limit of the integral on the left is greater than its upper limit but that's okay. We continue to follow the convention that in such cases the limits are to be swapped and the sign is to be changed and we still get a valid inequality in that case.
Suppose that the integral \[\tag{1.3.11} \int _ { - \infty } ^ { + \infty } e ( s , x ) \, d x\] is finite, i.e. that the total energy in all of space at time $s$ is finite. This is true if and only if the limit of the integral over a finite interval, \[\tag{1.3.12} \int _ { \alpha } ^ \beta e ( s , x ) \, d x\] tends to a finite limit as $\alpha$ tends to $- \infty$ and $\beta$ tends to $+ \infty$, in which case the integrals in the upper and lower bounds of the inequality above tend to that same limit as $a$ tends to $- \infty$ and $b$ tends to $+ \infty$. It then follows from the squeeze principle from real analysis that the limit of $I$ as $a$ tends to $- \infty$ and $b$ tends to $+ \infty$ exists and is equal to the other limits considered. In other words, \[\tag{1.3.13} \int _ { - \infty } ^ { + \infty } e ( t , x ) \, d x = \int _ { - \infty } ^ { + \infty } e ( s , x ) \, d x \] in the sense that if the integral on the right is finite then so is the integral on the left and both are equal. Since $s$ and $t$ are arbitrary we therefore have the following theorem.
Symmetries of a differential equation are transformations of a function with the property that the transformed function satisfies the differential equation if and only if the original function does. Symmetries often depend on one or more parameters. For example, the wave equation has the scaling symmetry \[\tag{1.4.1} ( S _ \alpha u ) ( t , x ) = u ( t / \alpha , x / \alpha ) ,\] where $\alpha$ is non-zero. To verify that this is indeed a symmetry we need to check that if $ \tilde u = S _ \alpha u $ for some non-zero value of $\alpha$ then $ u $ is a solution of the wave equation if and only if $ \tilde u $ is. This is completely straightforward because the chain rule gives \[\tag{1.4.2} \frac { \partial \tilde u } { \partial t } ( t , x ) = \frac 1 \alpha \frac { \partial u } { \partial t } ( t / \alpha , x / \alpha ) ,\] \[\tag{1.4.3} \frac { \partial \tilde u } { \partial x } ( t , x ) = \frac 1 \alpha \frac { \partial u } { \partial x } ( t / \alpha , x / \alpha ) ,\] \[\tag{1.4.4} \frac { \partial ^ 2 \tilde u } { \partial t ^ 2 } ( t , x ) = \frac 1 { \alpha ^ 2 } \frac { \partial ^ 2 u } { \partial t ^ 2 } ( t / \alpha , x / \alpha ) ,\] and \[\tag{1.4.5} \frac { \partial ^ 2 \tilde u } { \partial x ^ 2 } ( t , x ) = \frac 1 { \alpha ^ 2 } \frac { \partial ^ 2 u } { \partial x ^ 2 } ( t / \alpha , x / \alpha )\] so \[\tag{1.4.6} \frac { \partial ^ 2 \tilde u } { \partial t ^ 2 } ( t , x ) - c ^ 2 \frac { \partial ^ 2 \tilde u } { \partial x ^ 2 } ( t , x ) = \frac 1 { \alpha ^ 2 } \left [ \frac { \partial ^ 2 u } { \partial t ^ 2 } ( t / \alpha , x / \alpha ) , - c ^ 2 \frac { \partial ^ 2 u } { \partial x ^ 2 } ( t / \alpha , x / \alpha ) \right ]\] and hence $u$ satisfies \[\tag{1.4.7} \frac { \partial ^ 2 \tilde u } { \partial t ^ 2 } - c ^ 2 \frac { \partial ^ 2 \tilde u } { \partial x ^ 2 } = 0\] wherever it's defined if and only if $ \tilde u $ satisfies \[\tag{1.4.8} \frac { \partial ^ 2 u } { \partial t ^ 2 } - c ^ 2 \frac { \partial ^ 2 u } { \partial x ^ 2 } = 0\] wherever it's defined. The domains of definition need not be the same, although they could be.
Although the preceding proof is straightforward there is one point where the notation could be confusing. In (1.4.2) the expression \[\tag{1.4.9} \frac { \partial u } { \partial t } ( t / \alpha , x / \alpha )\] means that we take the function $ u $, differentiate it with respect to its first argument, and then evaluate the resulting function at the point $ ( t / \alpha , x / \alpha ) $, not that we differentiate the function obtained by mapping $ ( t , x ) $ to $ u ( t / \alpha , x / \alpha ) $ with respect to its first argument. Either of these would be a plausible interpretation of the expression but they are unfortunately not equal in general. In these notes we'll be consistent in interpreting partial derivatives as in this example.
Another symmetry of the wave equation is the spatial reflection symmetry \[\tag{1.4.10} ( R u ) ( t , x ) = u ( t , - x ) .\] The proof that this is indeed a symmetry is straightforward. Temporal reflection is also a symmetry, but don't need to check this separately, since we can write a temporal reflection as the composition of a spatial reflection and a scaling by a factor of $- 1$, in either order: \[\tag{1.4.11} ( R S _ { - 1 } u ) ( t , x ) = ( S _ { - 1 } u ) ( t , - x ) = u ( - t , x )\] and \[\tag{1.4.12} ( S _ { - 1 } R u ) ( t , x ) = ( R u ) ( - t , - x ) = u ( - t , x ) .\] Although scaling and spatial reflection happen to commute this isn't true of symmetries in general. For example, we also have spacetime translational symmetries \[\tag{1.4.13} ( T _ { \tau , \xi } u ) ( t , x ) = u ( t - \tau , x - \xi ) .\] Again, the proof that this is a symmetry for any real numbers $ \tau $ and $ \xi $ is straightforward. Note that \[\tag{1.4.14} ( T _ { \tau , \xi } R u ) ( t , x ) = ( R u ) ( t - \tau , x - \xi ) = u ( t - \tau , \xi - x )\] while \[\tag{1.4.15} ( R T _ { \tau , \xi } u ) ( t , x ) = ( T _ { \tau , \xi } R u ) ( t , - x ) = u ( t - \tau , - x - \xi ) ,\] so although the compositions $ T _ { \tau , \xi } R $ and $ R T _ { \tau , \xi } $ are both symmetries they are not the same symmetry unless $ \xi = 0 $.
Not all symmetries of the wave equation are as easy to verify as the ones above. Another important class of symmetries are the Lorentz transformations \[\tag{1.4.16} ( L _ v u ) = u \left ( \frac { t + v x / c ^ 2 } { \sqrt { 1 - v ^ 2 / c ^ 2 } } , \frac { v t + x } { \sqrt { 1 - v ^ 2 / c ^ 2 } } \right )\] for $ v \in ( - c , c ) $. If $ \tilde u = L _ v u $ then \[\tag{1.4.17} \frac { \partial \tilde u } { \partial t } ( t , x ) = \frac 1 { \sqrt { 1 - v ^ 2 / c ^ 2 } } \left ( \frac { \partial u } { \partial t } + v \frac { \partial u } { \partial x } \right ) \left ( \frac { t + v x / c ^ 2 } { \sqrt { 1 - v ^ 2 / c ^ 2 } } , \frac { v t + x } { \sqrt { 1 - v ^ 2 / c ^ 2 } } \right ) ,\] \[\tag{1.4.18} \frac { \partial \tilde u } { \partial x } ( t , x ) = \frac 1 { \sqrt { 1 - v ^ 2 / c ^ 2 } } \left ( \frac v { c ^ 2 } \frac { \partial u } { \partial t } + \frac { \partial u } { \partial x } \right ) \left ( \frac { t + v x / c ^ 2 } { \sqrt { 1 - v ^ 2 / c ^ 2 } } , \frac { v t + x } { \sqrt { 1 - v ^ 2 / c ^ 2 } } \right ) ,\] \[\tag{1.4.19} \frac { \partial ^ 2 \tilde u } { \partial t ^ 2 } ( t , x ) = \frac 1 { 1 - v ^ 2 / c ^ 2 } \left ( \frac { \partial ^ 2 u } { \partial t ^ 2 } + 2 v \frac { \partial u } { \partial t \partial x } + v ^ 2 \frac { \partial u } { \partial x ^ 2 } \right ) \left ( \frac { t + v x / c ^ 2 } { \sqrt { 1 - v ^ 2 / c ^ 2 } } , \frac { v t + x } { \sqrt { 1 - v ^ 2 / c ^ 2 } } \right ) ,\] \[\tag{1.4.20} \frac { \partial ^ 2 \tilde u } { \partial x ^ 2 } ( t , x ) = \frac 1 { 1 - v ^ 2 / c ^ 2 } \left ( \frac { v ^ 2 } { c ^ 4 } \frac { \partial ^ 2 u } { \partial t ^ 2 } + 2 \frac { v ^ 2 } { c ^ 2 } \frac { \partial u } { \partial t \partial x } + \frac { \partial u } { \partial x ^ 2 } \right ) \left ( \frac { t + v x / c ^ 2 } { \sqrt { 1 - v ^ 2 / c ^ 2 } } , \frac { v t + x } { \sqrt { 1 - v ^ 2 / c ^ 2 } } \right ) ,\] and so \[\tag{1.4.21} \left ( \frac { \partial ^ 2 \tilde u } { \partial t ^ 2 } - c ^ 2 \frac { \partial ^ 2 \tilde u } { \partial x ^ 2 } \right ) ( t , x ) = \left ( \frac { \partial ^ 2 u } { \partial t ^ 2 } - c ^ 2 \frac { \partial ^ 2 u } { \partial x ^ 2 } \right ) \left ( \frac { t + v x / c ^ 2 } { \sqrt { 1 - v ^ 2 / c ^ 2 } } , \frac { v t + x } { \sqrt { 1 - v ^ 2 / c ^ 2 } } \right ) .\] So $ \tilde u $ is a solution of \[\tag{1.4.22} \frac { \partial ^ 2 \tilde u } { \partial t ^ 2 } - c ^ 2 \frac { \partial ^ 2 \tilde u } { \partial x ^ 2 } = 0\] if and only if $ u $ is a solution of \[\tag{1.4.23} \frac { \partial ^ 2 u } { \partial t ^ 2 } - c ^ 2 \frac { \partial ^ 2 u } { \partial x ^ 2 } = 0 ,\] and so $ L _ v $ is indeed a symmetry.
As symmetries get more complicated it becomes annoying to check each one in the way we just used for scaling, translations and Lorentz transformations. It's better to do one rather complicated calculation covering a whole class of transformations and then apply the result in particular cases. All the symmetries considered so far are of the form \[\tag{1.5.1} \tilde u ( t , x ) = u ( p ( t , x ) , q ( t , x ) )\] for some functions $ p $ and $ q $. This form isn't general enough to cover all symmetries of the wave equation, but it covers a large class of them so we'll assume we have a transformation of this form and deduces sufficient conditions for it to be a symmetry.
We start by noting that symmetries need to be invertible so we at least need the Jacobian of the transformation to be non-singular, i.e. \[\tag{1.5.2} \det \left [ \begin{matrix} \partial p / \partial t & \partial p / \partial x \\ \partial q / \partial t & \partial q / \partial x \end{matrix} \right ] \neq 0 .\] Assuming that $ p $ and $ q $ are twice continuously differentiable we can compute derivatives using the chain rule for such a general transformation in the same way we did in specific examples. \[\tag{1.5.3} \frac { \partial \tilde u } { \partial t } ( t , x ) = \frac { \partial p } { \partial t } ( t , x ) \frac { \partial u } { \partial t } ( p ( t , x ) , q ( t , x ) ) + \frac { \partial q } { \partial t } ( t , x ) \frac { \partial u } { \partial x } ( p ( t , x ) , q ( t , x ) ) ,\] \[\tag{1.5.4} \frac { \partial \tilde u } { \partial x } ( t , x ) = \frac { \partial p } { \partial x } ( t , x ) \frac { \partial u } { \partial t } ( p ( t , x ) , q ( t , x ) ) + \frac { \partial q } { \partial x } ( t , x ) \frac { \partial u } { \partial x } ( p ( t , x ) , q ( t , x ) ) ,\] \[\tag{1.5.5} \begin{split} \frac { \partial ^ 2 \tilde u } { \partial t ^ 2 } ( t , x ) & = \left ( \frac { \partial p } { \partial t } ( t , x ) \right ) ^ 2 \frac { \partial ^ 2 u } { \partial t ^ 2 } ( p ( t , x ) , q ( t , x ) ) \\ & \quad {} + 2 \frac { \partial p } { \partial t } ( t , x ) \frac { \partial q } { \partial t } ( t , x ) \frac { \partial ^ 2 u } { \partial t \partial x } ( p ( t , x ) , q ( t , x ) ) \\ & \quad {} + \left ( \frac { \partial q } { \partial t } ( t , x ) \right ) ^ 2 \frac { \partial ^ 2 u } { \partial x ^ 2 } ( p ( t , x ) , q ( t , x ) ) \\ & \quad {} + \frac { \partial ^ 2 p } { \partial t ^ 2 } ( t , x ) \frac { \partial u } { \partial t } ( p ( t , x ) , q ( t , x ) ) \\ & \quad {} + \frac { \partial ^ 2 q } { \partial t ^ 2 } ( t , x ) \frac { \partial u } { \partial x } ( p ( t , x ) , q ( t , x ) ) , \end{split}\] and \[\tag{1.5.6} \begin{split} \frac { \partial ^ 2 \tilde u } { \partial x ^ 2 } ( t , x ) & = \left ( \frac { \partial p } { \partial x } ( t , x ) \right ) ^ 2 \frac { \partial ^ 2 u } { \partial t ^ 2 } ( p ( t , x ) , q ( t , x ) ) \\ & \quad {} + 2 \frac { \partial p } { \partial x } ( t , x ) \frac { \partial q } { \partial x } ( t , x ) \frac { \partial ^ 2 u } { \partial t \partial x } ( p ( t , x ) , q ( t , x ) ) \\ & \quad {} + \left ( \frac { \partial q } { \partial x } ( t , x ) \right ) ^ 2 \frac { \partial ^ 2 u } { \partial x ^ 2 } ( p ( t , x ) , q ( t , x ) ) \\ & \quad {} + \frac { \partial ^ 2 p } { \partial x ^ 2 } ( t , x ) \frac { \partial u } { \partial t } ( p ( t , x ) , q ( t , x ) ) \\ & \quad {} + \frac { \partial ^ 2 q } { \partial x ^ 2 } ( t , x ) \frac { \partial u } { \partial x } ( p ( t , x ) , q ( t , x ) ) . \end{split}\] Then \[\tag{1.5.7} \begin{split} \left ( \frac { \partial ^ 2 \tilde u } { \partial t ^ 2 } - c ^ 2 \frac { \partial ^ 2 \tilde u } { \partial x ^ 2 } \right ) ( t , x ) & = A ( t , x ) \frac { \partial ^ 2 u } { \partial t ^ 2 } ( p ( t , x ) , q ( t , x ) ) % \\ & \quad {} + B ( t , x ) \frac { \partial ^ 2 u } { \partial t \partial x } ( p ( t , x ) , q ( t , x ) ) \\ & \quad {} + C ( t , x ) \frac { \partial ^ 2 u } { \partial x ^ 2 } ( p ( t , x ) , q ( t , x ) ) % \\ & \quad {} + D ( t , x ) \frac { \partial u } { \partial t } ( p ( t , x ) , q ( t , x ) ) \\ & \quad {} + E ( t , x ) \frac { \partial u } { \partial x } ( p ( t , x ) , q ( t , x ) ) , \end{split}\] where \[\tag{1.5.8} A = \left ( \frac { \partial p } { \partial t } \right ) ^ 2 - c ^ 2 \left ( \frac { \partial p } { \partial x } \right ) ^ 2 , \quad B = 2 \frac { \partial p } { \partial t } \frac { \partial q } { \partial t } - 2 c ^ 2 \frac { \partial p } { \partial x } \frac { \partial q } { \partial x }\] \[\tag{1.5.9} C = \left ( \frac { \partial q } { \partial t } \right ) ^ 2 - c ^ 2 \left ( \frac { \partial q } { \partial x } \right ) ^ 2 , \quad D = \frac { \partial ^ 2 p } { \partial t ^ 2 } - c ^ 2 \frac { \partial ^ 2 p } { \partial x ^ 2 } , \quad E = \frac { \partial ^ 2 q } { \partial t ^ 2 } - c ^ 2 \frac { \partial ^ 2 q } { \partial x ^ 2 } .\]
For a symmetry we need this to be a multiple of \[\tag{1.5.10} \frac { \partial ^ 2 u } { \partial t ^ 2 } ( p ( t , x ) , q ( t , x ) ) - c ^ 2 \frac { \partial ^ 2 u } { \partial x ^ 2 } ( p ( t , x ) , q ( t , x ) )\] so we need $ c ^ 2 A + C $, $ B $, $ D $ and $ E $ to be zero. If this is true then $ c ^ 2 A + C \pm c B $ is zero. We can factor this for either choice of sign: \[\tag{1.5.11} c ^ 2 A + c B + C = F G , \qquad c ^ 2 A - c B + C = H I ,\] where \[\tag{1.5.12} \begin{split} F & = c \frac { \partial p } { \partial t } + c ^ 2 \frac { \partial p } { \partial x } + \frac { \partial q } { \partial t } + c \frac { \partial q } { \partial x } , \\ G & = c \frac { \partial p } { \partial t } - c ^ 2 \frac { \partial p } { \partial x } + \frac { \partial q } { \partial t } - c \frac { \partial q } { \partial x } , \\ H & = c \frac { \partial p } { \partial t } + c ^ 2 \frac { \partial p } { \partial x } - \frac { \partial q } { \partial t } - c \frac { \partial q } { \partial x } , \\ I & = c \frac { \partial p } { \partial t } - c ^ 2 \frac { \partial p } { \partial x } - \frac { \partial q } { \partial t } + c \frac { \partial q } { \partial x } . \end{split}\] So for this to be a symmetry we need $ F $ or $ G $ to be zero and we need $ H $ or $ I $ to be zero. Now \[\tag{1.5.13} F I - G H = 4 c ^ 2 \det \left [ \begin{matrix} \partial p / \partial t & \partial p / \partial x \\ \partial q / \partial t & \partial q / \partial x \end{matrix} \right ]\] and the matrix on the right is the Jacobian, which, as we already observed, must be non-singular. So $ F $ and $ H $ can't both be zero and $ G $ and $ I $ can't both be zero. The remaining possibilities are that $ F $ and $ I $ are zero or $ G $ and $ H $ are zero.
We've just seen that for this to be a symmetry it is necessary that either $ F $ and $ I $ are both zero or $ G $ and $ H $ are both zero. In fact these conditions are also sufficient. To see this, suppose one of these conditions is satisfied. It follows immediately from the factorisation of $ c ^ 2 A + c B + C $ and $ c ^ 2 A - c B + C $ given earlier that both these quantities are zero, from which we see that $ B = 0 $ and $ C = - A ^ 2 $. But we also have \[\tag{1.5.14} D = \frac { \partial } { \partial t } \frac { F + I } { 2 c } - \frac { \partial } { \partial x } \frac { F - I } { 2 }\] and \[\tag{1.5.15} D = \frac { \partial } { \partial t } \frac { G + H } { 2 c } + \frac { \partial } { \partial x } \frac { G - H } { 2 }\] In the case where $ F $ and $ I $ are zero we use the first of these and in the case where $ G $ and $ H $ are zero we use the second. In either case we find that $ D $ is zero. $ E $ is handled similarly, but now we use the equations \[\tag{1.5.16} E = \frac { \partial } { \partial t } \frac { F - I } { 2 } - c ^ 2 \frac { \partial } { \partial x } \frac { F + I } { 2 c }\] and \[\tag{1.5.17} E = \frac { \partial } { \partial t } \frac { G - H } { 2 } - c ^ 2 \frac { \partial } { \partial x } { G + H } { 2 c } .\] So in either case we have \[\tag{1.5.18} \left ( \frac { \partial ^ 2 \tilde u } { \partial t ^ 2 } - c ^ 2 \frac { \partial ^ 2 \tilde u } { \partial x ^ 2 } \right ) ( t , x ) = A ( t , x ) \left ( \frac { \partial ^ 2 u } { \partial t ^ 2 } - c ^ 2 \frac { \partial ^ 2 u } { \partial x ^ 2 } \right ) ( p ( t , x ) , q ( t , x ) ) ,\] so the is indeed a symmetry. In other words, the conditions which we previously saw were necessary are also sufficient.
We can put the equations above in a slightly nicer form. First of all, $ F $ and $ I $ being zero is equivalent to $ \frac { F + I } { 2 c } $ and $ \frac { F - I } 2 $ being zero, i.e. to the equations \[\tag{1.5.19} \frac { \partial p } { \partial t } + \frac { \partial q } { \partial x } = 0 , \quad \frac { \partial q } { \partial t } + c ^ 2 \frac { \partial p } { \partial x } = 0\] while First of all, $ G $ and $ H $ being zero is equivalent to $ \frac { G + H } { 2 c } $ and $ \frac { G - H } 2 $ being zero, i.e. to the equations \[\tag{1.5.20} \frac { \partial p } { \partial t } - \frac { \partial q } { \partial x } = 0 , \quad \frac { \partial q } { \partial t } - c ^ 2 \frac { \partial p } { \partial x } = 0 .\] So our necessary and sufficient condition for a symmetry is that either of the pairs of equations above holds. Also, we can use the equations above to replace a derivative of $ p $ with a derivative of $ x $ in the equation for $ A $, obtaining \[\tag{1.5.21} A = \pm \left ( \frac { \partial p } { \partial t } \frac { \partial q } { \partial x } - \frac { \partial p } { \partial x } \frac { \partial q } { \partial t } \right ) .\] The sign is negative in the case of the first pair of equations and positive in the case of the second. The quantity inside the parentheses is of course the determinant of the Jacobian, which we've already met several times in this calculation.
We can use the method above to show that spacetime inversion \[\tag{1.5.22} ( J u ) ( t , x ) = u \left ( - \frac { t } { c ^ 2 t ^ 2 - x ^ 2 } , \frac { x } { c ^ 2 t ^ 2 - x ^ 2 } \right )\] is a symmetry. \[\tag{1.5.23} \frac { \partial p } { \partial t } ( t , x ) = \frac { c ^ 2 t ^ 2 + x ^ 2 } { \left ( c ^ 2 t ^ 2 - x ^ 2 \right ) ^ 2 } ,\] \[\tag{1.5.24} \frac { \partial p } { \partial x } ( t , x ) = - \frac { 2 t x } { \left ( c ^ 2 t ^ 2 - x ^ 2 \right ) ^ 2 } ,\] \[\tag{1.5.25} \frac { \partial q } { \partial t } ( t , x ) = - \frac { 2 c ^ 2 t x } { \left ( c ^ 2 t ^ 2 - x ^ 2 \right ) ^ 2 } ,\] and \[\tag{1.5.26} \frac { \partial q } { \partial x } ( t , x ) = \frac { c ^ 2 t ^ 2 + x ^ 2 } { \left ( c ^ 2 t ^ 2 - x ^ 2 \right ) ^ 2 } .\] These satisfy the conditions given earlier so $ \tilde u $ satisfies the wave equation if and only if $ u $ does. In other words, $ J $ is a symmetry of the wave equation.
It follows immediately from the definition of a symmetry that the composition of two symmetries is a symmetry and the inverse of a symmetry is a symmetry. In other words the symmetries of a differential equation form a group. The problem of determining the full symmetry group of a differential equation is in general quite a difficult one and indeed the theory of Lie groups was originally developed as a technique for solving this problem.
There are two particular simple symmetries which are valid not just for the wave equation but for all linear differential equations. One is scaling of the dependent variable \[\tag{1.5.27} ( M _ \lambda u ) ( t , x ) = \lambda u ( t , x )\] and the other is the addition of another solution of the equation \[\tag{1.5.28} ( A _ \varphi u ) ( t , x ) = u ( t , x ) + \varphi ( t , x ) ,\] where $ \lambda $ is a non-zero real number and $ \varphi $ is a solution of the wave equation. Neither of these symmetries is of the type we just considered so we can't apply the criterion given there but it is straightforward to check directly that they are symmetries. The fact that $ M _ { - 1 } $ is a symmetry means that $ \varphi $ is a solution if and only if $ - \varphi $ is and so we could equally well have said that subtraction of another solution is a symmetry rather than addition.
In general, a symmetry will take a solution to the differential equation and give another, different, solution, but it may happen to give the same solution, in which case we say that it is a symmetry not just of the differential equation but of the solution. It is often useful to find the set of solutions with a given symmetry or group of symmetries. Of course if we ask for too many symmetries we are unlikely to get any solutions. For example, the only solutions of the wave equation symmetric under the full group of spacetime translations are the constant solutions. Interestingly, as long as we restrict our attention to classical solutions defined on all of $ \mathbf R ^ 2 $, the only scaling invariant solutions are also constant. Indeed, if $ u $ is scaling invariant and $ \ell $ is a line through the origin then $ u $ must take the same value at all points of $ \ell $, except possibly the origin itself. But classical solutions are continuous, so it must take the same value at the origin as well. Every point is on some line through the origin so the value at that point is therefore equal to the value at the origin. What's interesting about this argument is we never actually needed the fact that $ u $ satisfies the wave equation!
If we impose less symmetry then we get more solutions. Instead of requiring solutions to be invariant under all space and time translations we could, for example, choose a particular $ v $ and then require invariance under those of the form $ T _ { s , v s } $ for all $ s $. The solutions obtained in this way, i.e. the ones which satisfy \[\tag{1.6.1} u ( t - s , x - v s ) = u ( t . x )\] are called travelling wave solutions. Differentiating with respect to $ s $ and then setting $ s = 0 $ we get the relation \[\tag{1.6.2} \frac { \partial u } { \partial t } + v \frac { \partial u } { \partial x } = 0\] satisfied by all travelling wave solutions. Applying the differential operator \[\tag{1.6.3} \frac { \partial } { \partial t } - v \frac { \partial } { \partial x }\] to this equation gives \[\tag{1.6.4} \frac { \partial ^ 2 u } { \partial t ^ 2 } - v ^ 2 \frac { \partial ^ 2 u } { \partial x ^ 2 } = 0\] and subtracting this from the wave equation gives \[\tag{1.6.5} ( c ^ 2 - v ^ 2 ) \frac { \partial ^ 2 u } { \partial x ^ 2 } = 0 .\]
Unlike when we considered Lorentz transformations, I haven't put any restrictions on $ v $ here, so $ v = \pm c $ is allowed. There are therefore two cases to consider, where $ c ^ 2 - v ^ 2 = 0 $ or $ c ^ 2 - v ^ 2 \neq 0 $. We consider the second case first. Then the second $ x $ derivative is zero and hence, from the wave equation, so is the second time derivative. The solutions must therefore be linear functions of space and time the relation between the coefficients is fixed by the equation $ \partial u / \partial t + v \partial u / \partial x = 0 $.
The first case is more interesting. In this case the wave equation provides no further information beyond the relation \[\tag{1.6.6} \frac { \partial u } { \partial t } + v \frac { \partial u } { \partial x } = 0\] obtained earlier. The solutions of this equation are precisely the functions \[\tag{1.6.7} u ( t , x ) = \theta ( x - v t )\] so for $ v = c $ or $ v = - c $ these are travelling wave solutions of the wave equation, as we can easily check directly.
We could also consider the set of solutions invariant under spatial reflection, i.e. the ones satisfying $ R u = u $, i.e. \[\tag{1.6.8} u ( t , - x ) = u ( t , x ) .\] These are just the solutions which are even functions of the spatial variable for each fixed value of the temporal variable. Note that differentiating the equation above gives \[\tag{1.6.9} - \frac{ \partial u } { \partial x } ( t , - x ) = \frac { \partial u } { \partial x } ( t , x )\] and so \[\tag{1.6.10} - \frac{ \partial u } { \partial x } ( t , 0 ) = \frac { \partial u } { \partial x } ( t , 0 ) ,\] from which it follows that \[\tag{1.6.11} \frac{ \partial u } { \partial x } ( t , 0 ) = 0 .\] More generally, any derivative of odd degree in $x$, if it exists, is zero on the time axis.
Similarly we could look at the solutions symmetric under the symmetry $ M _ { - 1 } R $. These are the solutions which are odd, considered as a function of the spatial variable for fixed value of the temporal variable, and they have the property that any derivative of even degree in $x$, if it exists, is zero on the time axis. In particular the zeroeth and second derivatives, which both certainly exist for any classical solution, are zero: \[\tag{1.6.12} u ( t , 0 ) = 0 .\] and \[\tag{1.6.13} \frac{ \partial ^ 2 u } { \partial x ^ 2 } ( t , 0 ) = 0 .\]
Symmetries, linearity and uniqueness interact in interesting ways. Suppose, for example, that $ u $ is a solution to the initial value problem considered earlier, the one with initial conditions given by (1.2.1). If $ u $ has $ R $ as a symmetry, i.e. if it is even as a function of the spatial variable, then $ f $ and $ g $ must also both be even. More interestingly, suppose $ f $ and $ g $ are even. Since $ u $ is a solution and $ R $ is a symmetry it follows that $ R u $ is also a solution. From linearity it then follows that $ u - R u $ is a solution. But the initial data for $ u - R u $ are identically zero and we already know a solution with zero initial data, namely the zero solution, so the uniqueness theorem implies that $ u - R u $ is the zero solution. In other words, $ u = R u $, or $ R $ is a symmetry of $ u $. So not only does an even solution necessarily have even initial data but even initial data can only give rise to an even solution. Similar remarks apply to odd solutions and odd initial data.
Up to now we've been trying to solve the wave equation in the whole of $ \mathbf R ^ 2 $ with initial data given on the whole of $ \mathbf R $ but often one wants to solve the equation in a region where the spatial variable is restricted to an interval, either bounded or semi-infinite, and the initial data are given in this interval. This, by itself, is not a problem that admits unique solutions, but it becomes one if we impose appropriate boundary conditions at the endpoint or endpoints of the interval. The two most important boundary conditions are the Dirichlet condition, $ u = 0 $, and the Neumann condition, $ \partial u / \partial x = 0 $.
To start with, let's consider the case of a finite interval, $ [ a , b ] $, with a Dirichlet condition at the left endpoint and a Neumann condition at the right endpoint: \[\tag{1.7.1} u ( t , a ) = 0 , \quad \frac { \partial u } { \partial x } ( t , b ) = 0 .\] These equations are to hold for all values of $t$. The initial conditions will be specified as usual by (1.2.1), where the functions $f$ and $g$ are defined on the interval $ [ a , b ] $ and the solution $u$ should be defined and twice continuously differentiable on $ \mathbf R \times [ a , b ] $, and should of course satisfy the wave equation there.
We see immediately that some additional restrictions are needed on $f$ and $g$. If we set $t = s$ in the equations above we get \[\tag{1.7.2} f ( a ) = 0 , \quad f ' ( b ) = 0\] since taking a partial derivative with respect to $ x $ and then fixing $ t $ is the same as fixing $ t $ and then taking an ordinary derivative. We could also take a $ t $ derivative of the boundary conditions, obtaining \[\tag{1.7.3} \frac { \partial u } { \partial t } ( t , a ) = 0 , \quad \frac { \partial ^ 2 u } { \partial t \partial x } ( t , b ) = 0 .\] Since $ u $ is twice continuously differentiable the mixed partial derivatives are equal so we can replace $ \partial ^ 2 u / \partial t \partial x $ with $ \partial ^ 2 u / \partial x \partial t $. If we do so and then set $ t = s $ then we get \[\tag{1.7.4} g ( a ) = 0 , \quad g ' ( b ) = 0 .\] There is one more restriction. Differentiating the Dirichlet condition once again gives \[\tag{1.7.5} \frac { \partial ^ 2 u } { \partial t ^ 2 } ( t , a ) = 0 ,\] but $ u $ satisfies the wave equation, so \[\tag{1.7.6} c ^ 2 \frac { \partial ^ 2 u } { \partial x ^ 2 } ( t , a ) = 0 ,\] so $ c ^ 2 f '' ( x ) = 0 $ and therefore \[\tag{1.7.7} f '' ( a ) = 0 .\] There is no similar condition for $ g $ and there is no analogous condition at $b$. To summarise, we've found that there can be no solution to the initial value problem with these boundary conditions unless the initial data satisfy the constraints \[\tag{1.7.8} f ( a ) = 0 , \quad g ( a ) = 0 , \quad f '' ( a ) = 0 , \quad f ' ( b ) = 0 , \quad g ' ( b ) = 0 .\]
Before considering existence and uniqueness let's examine energy conservation. It's useful to prove the following lemma.
The proof of the lemma is very simple. By the fundamental theorem of calculus \[\tag{1.7.11} q ( t , x _ 2 ) - q ( t , x _ 1 ) = \int _ { x _ 1 } ^ { x _ 2 } \frac { \partial q } { \partial x } ( t , x ) \, d x .\] Integrating this equation over the interval $ [ x _ 1 , x _ 2 ] $ gives \[\tag{1.7.12} \int _ { t _ 1 } ^ { t _ 2 } \left [ q ( t , x _ 2 ) - q ( t , x _ 1 ) \right ] \, d t = \int _ { t _ 1 } ^ { t _ 2 } \int _ { x _ 1 } ^ { x _ 2 } \frac { \partial q } { \partial x } ( t , x ) \, d x \, d t .\] On the left hand side we write the integral of the difference as a difference of integrals and on the right hand side we write the repeated integral as an area integral, using Fubini's theorem. \[\tag{1.7.13} \int _ { t _ 1 } ^ { t _ 2 } q ( t , x _ 2 ) \, d t - \int _ { t _ 1 } ^ { t _ 2 } q ( t , x _ 1 ) \, d t = \iint _ { [ t _ 1 , t _ 2 ] \times [ x _ 1 , x _ 2 ] } \frac { \partial q } { \partial x } \, d A .\] Similarly, \[\tag{1.7.14} \int _ { x _ 1 } ^ { x _ 2 } p ( t _ 2 , x ) \, d x - \int _ { x _ 1 } ^ { x _ 2 } p ( t _ 1 , x ) \, d x = \iint _ { [ t _ 1 , t _ 2 ] \times [ x _ 1 , x _ 2 ] } \frac { \partial p } { \partial t } \, d A .\] Subtracting this from the previous equation and using again the fact that that integral of a difference is the difference of the integrals we get (1.7.10).
Now that we've proved the lemma we can apply it to $ [ x _ 1 , x _ 2 ] = [ a , b ] $, \[\tag{1.7.15} p = e = \frac 1 2 \left ( \frac { \partial u } { \partial t } \right ) ^ 2 + \frac { c ^ 2 } 2 \left ( \frac { \partial u } { \partial x } \right ) ^ 2 , \quad q = c ^ 2 \frac { \partial u } { \partial t } \frac { \partial u } { \partial x }\] and note that \[\tag{1.7.16} \frac { \partial q } { \partial x } - \frac { \partial p } { \partial t } = - \left ( \frac { \partial ^ 2 u } { \partial t ^ 2 } - c ^ 2 \frac { \partial ^ 2 u } { \partial x ^ 2 } \right ) \frac { \partial u } { \partial t } ,\] so the right hand side of the equation in the lemma is zero when $ u $ satisfies the wave equation. The terms on the left hand side with a $ q $ in them are also zero if $ u $ satisfies the boundary conditions. When $ x = a $ this happens because $ \partial u / \partial t = 0 $ there and when $ x = b $ this happens because $ \partial u / \partial x = 0 $ there. The only terms which are left then are the ones with a $ p $, which is the same as $ e $, on the right hand side, so we have \[\tag{1.7.17} \int _ a ^ b e ( d , x ) \, d x - \int _ a ^ b e ( c , x ) \, d x = 0 .\] In other words, the energy in the interval $ [ a , b ] $ at time $ t _ 2 $ is the same as the energy at time $ t _ 1 $. We therefore have an energy conservation theorem for solutions on a bounded interval, just as we did in the case of an infinite interval.
We can use this energy conservation theorem to get a uniqueness theorem even though we don't yet have an explicit solution. Suppose that $ u _ 1 $ and $ u _ 2 $ are solutions of the initial value problem which satisfy the same boundary conditions, either Dirichlet or Neumann at each endpoint. Then their difference $ u = u _ 1 - u _ 2 $ satisfies the initial value problem with the same boundary conditions and zero initial data. It therefore has zero energy density everywhere. Looking at the definition of the energy density we see this means that its partial derivatives are everywhere zero. It must therefore be locally constant. Since $ \mathbf R \times [ a , b ] $ is connected it follows that $ u $ is constant. It's zero initially, since $ u $ has zero initial data, and hence is zero everywhere. So we've proved the following uniqueness theorem.
We haven't yet proved the existence of solutions. For the moment we'll return to the setting where we have a Dirichlet condition at $ x = a $ and Neumann at $ x = b $. In this setting we saw that there can be no solution unless $ f $ is twice continuously differentiable, $ g $ is continuously differentiable, and $ f ( a ) $, $ g ( a ) $, $ f '' ( a ) $, $ f ' ( b ) $, and $ g ' ( b ) $ are all zero, so we'll assume those conditions are satisfied. We then extend $ f $ and $ g $ to functions on all of $ \mathbf R $ as follows. Any real number can be written uniquely as $ n + r $ where $ n $ is an integer and $ r $ belongs to the half-open interval $ [ 0 , 1 ) $ and every integer $ n $ can be written at $ 4 m + l $ where $ m $ is an integer and $ l $ is $ 0 $, $ 1 $, $ 2 $, or $ 3 $. We use this to write \[\tag{1.8.1} \frac { x - a } { b - a } = 4 m ( x ) + l ( x ) + r ( x )\] and then define \[\tag{1.8.2} f ( x ) = \begin{cases} f ( a + ( b - a ) r ( x ) ) & \mbox{ if ${ l ( x ) = 0 }$,} \\ f ( b - ( b - a ) r ( x ) ) & \mbox{ if ${ l ( x ) = 1 }$,} \\ - f ( a + ( b - a ) r ( x ) ) & \mbox{ if ${ l ( x ) = 2 }$,} \\ - f ( b - ( b - a ) r ( x ) ) & \mbox{ if ${ l ( x ) = 3 }$.} \\ \end{cases}\] The intended interpretation of this equation is that the $ f $'s on the right hand side refer to the function we were originally given, while the $ f $ on the left hand side is the new function we are defining. We'll also define an extension of $ g $ via the same equation, except with all $ f $'s replaced by $ g $'s. Note that $ m ( x ) $ does not appear in the equations above, which implies that the extended functions are periodic of period $ 4 ( b - a ) $.
There are a number of things to check. One is that these really are extensions of the original functions, i.e. that they give the same values when evaluated at a point in the original interval $ [ a , b ] $. Suppose $ x \in [ a , b ) $. Then $ m ( x ) = 0 $, $ l ( x ) = 0 $, and $ r ( x ) = ( b - x ) / ( b - a ) $ so the right hand side above gives the value \[\tag{1.8.3} f ( a + ( b - a ) r ( x ) ) = f \left ( a + ( b - a ) \frac { x - a } { b - a } \right ) = f ( x ) ,\] as it should. On the other hand, if $ x = b $ then $ m ( x ) = 0 $, $ l ( x ) = 1 $ and $ r ( x ) = 0 $ and the right hand side gives \[\tag{1.8.4} f ( b - ( b - a ) r ( x ) ) = f ( b ) = f ( x ) ,\] and we again get the correct value. So the new $ f $ is indeed an extension of the old one, which is fortunate because otherwise we wouldn't know whether $ f ( x ) $ for $ x $ in $ [ a , b ] $ referred to the old function or the new one. Of course the same argument works equally well for $ g $.
We'd like to know, in addition, that the extended functions have the same differentiability properties as the old functions, i.e. that the extended $ f $ is twice continuously differentiable and the extended $ g $ is continuously differentiable. This might seem obvious from the definition since the various pieces of which they are composed have this property, but there's clearly something wrong with that line of reasoning since it would imply that the absolute value function \[\tag{1.8.5} | x | = \begin{cases} x & \mbox{ if ${ x > 0 }$,} \\ 0 & \mbox{ if ${ x = 0 }$,} \\ - x & \mbox{ if ${ x < 0 }$,} \end{cases}\] which is also composed of continuously differentiable pieces, is continuously differentiable, when it is in fact not differentiable at $ x = 0 $. We need a criterion for a function defined by different expressions on the two different sides of a point to have some number of continuous derivatives. This is supplied by the following lemma.
To prove the lemma we first introduce the functions \[\tag{1.8.9} f _ j ( x ) = \begin{cases} p ^ { ( j ) } ( x ) & \mbox{ if ${ \alpha < x < \beta }$,} \\ c _ j & \mbox{ if ${ x = \beta }$,} \\ q ^ { ( j ) } ( x ) & \mbox{ if ${ \beta < x < \gamma }$,} \end{cases}\] where $ c _ j $ is the common value of $ \lim _ { x \to \beta ^ - } p ^ { ( j ) } ( x ) $ and $ \lim _ { x \to \beta ^ + } q ^ { ( j ) } ( x ) $. It follows immediately from the definition of continuity that $ f _ j $ is continuous. Also, \[\tag{1.8.10} f _ j ' ( x ) = f _ { j + 1 } ( x )\] if $ j < k $ and $ \alpha < x < \beta $ or $ \beta < x < \gamma $. We would like to know that this is true also when $ x = \beta $. If $ \beta < z < \gamma $ then \[\tag{1.8.11} \int _ \beta ^ z f _ { j + 1 } ( y ) \, d y = \int _ \beta ^ w f _ { j + 1 } ( y ) \, d y + \int _ w ^ z f _ { j + 1 } ( y ) \, d y\] for any $ w $ in the interval $ ( \beta , z ) $. Now \[\tag{1.8.12} \lim _ { w \to \beta ^ + } \int _ \beta ^ w f _ { j + 1 } ( y ) \, d y = 0\] because integrals of continuous functions depend continuously on their limits of integration and \[\tag{1.8.13} \int _ w ^ z f _ { j + 1 } ( y ) \, d y = f _ j ( z ) - f _ j ( w )\] by the fundamental theorem of calculus. So \[\tag{1.8.14} \int _ \beta ^ z f _ { j + 1 } ( y ) \, d y = \lim _ { w \to \beta ^ + } \left ( f _ j ( y ) - f _ j ( w ) \right ) = f _ j ( z ) - f _ j ( \beta ) ,\] where we've used the continuity of $ f _ j $ to evaluate $ \lim _ { w \to \beta ^ + } f _ j ( w ) $. Now the change of variable $ r = ( y - \beta ) / ( z - \beta ) $ gives \[\tag{1.8.15} \int _ \beta ^ z f _ { j + 1 } ( y ) \, d y = ( z - \beta ) \int _ 0 ^ 1 f _ { j + 1 } ( \beta + r ( z - \beta ) ) \, d r\] and so \[\tag{1.8.16} \frac { f _ j ( z ) - f _ j ( \beta ) } { z - \beta } = \int _ 0 ^ 1 f _ { j + 1 } ( \beta + r ( z - \beta ) ) \, d r .\] This was proved for $ z $ in the interval $ ( \beta , \gamma ) $ but an almost identical argument gives the same equation when $ z $ is in the interval $ ( \alpha , \beta ) $. Taking limits inside the integral and using the continuity of $ f _ { j + 1 } $ gives \[\tag{1.8.17} \begin{split} \lim _ { z \to \beta } \frac { f _ j ( z ) - f _ j ( \beta ) } { z - \beta } & = \int _ 0 ^ 1 \lim _ { z \to \beta } f _ { j + 1 } ( \beta + r ( z - \beta ) ) \, d r . \\ & = \int _ 0 ^ 1 f _ { j + 1 } \left ( \lim _ { z \to \beta } \left ( \beta + r ( z - \beta ) \right ) \right ) \, d r . \\ & = \int _ 0 ^ 1 f _ { j + 1 } ( \beta ) \, d r = f _ { j + 1 } ( \beta ) . \end{split}\] Taking the limit inside the integral is justified in this case because we have continuous integrands with uniform convergence on a bounded integral. The equation above just says that \[\tag{1.8.18} f _ j ' ( x ) = f _ { j + 1 } ( x )\] when $ x = \beta $ though. Since we already had the same equation for $ x $ in the intervals $ ( \alpha , \beta ) $ and $ ( \beta , \gamma ) $ we now have it in the full interval $ ( \alpha , \gamma ) $. Now $ f ^ { ( 0 ) } = f = f _ 0 $ so we see by induction on $ j $ that $ f ^ { ( j ) } = f _ j $ for $ j \le k $. The $ f _ j $'s are already known to be continuous, so $ f $ is $ k $ times continuously differentiable.
Now that we have the lemma it's straightforward, if slightly tedious, to check that the extended $ f $ and $ g $ defined previously are twice and once continuously differentiable, respectively. We'll do this just at the point $ b $ as an illustration. To the left of $ b $ we have $ m ( x ) = 0 $, $ l ( x ) = 0 $ and $ r ( x ) = ( x - a ) / ( b - a ) $ so \[\tag{1.8.19} f ( x ) = f ( a + ( b - a ) r ( x ) ) = f ( x )\] and to the right of $ b $ we have $ m ( x ) = 0 $, $ l ( x ) = 1 $ and $ r ( x ) = ( x - b ) / ( b - a ) $ so \[\tag{1.8.20} f ( x ) = f ( b - ( b - a ) r ( x ) ) = f ( 2 b - x ) .\] The first and second derivatives to the left and right of $ b $ are \[\tag{1.8.21} f ' ( x ) = \begin{cases} f ' ( x ) = f ' ( x ) & \mbox{if ${ a < x < b }$,} \\ f ' ( x ) = - f ' ( 2 b - x ) & \mbox{if ${ b < x < 2 b - a }$} \end{cases}\] and \[\tag{1.8.22} f '' ( x ) = \begin{cases} f '' ( x ) = f '' ( x ) & \mbox{if ${ a < x < b }$,} \\ f '' ( x ) = f '' ( 2 b - x ) & \mbox{if ${ b < x < 2 b - a }$.} \end{cases}\] It's clear that the second derivatives approach a common limit as $ x $ tends to $ b $ from either side. It's less clear that the first derivatives do but here we have to remember that our original $ f $ was assumed to satisfy $ f ' ( b ) = 0 $. Without this assumption the extended $ f $ would not be continuously differentiable. The argument for $ g $ at $ b $ is similar, but we only need to worry about the first derivative.
Let $ u $ be the solution to the initial value problem in $ \mathbf R $ to the wave equation with initial data given by the extended $ f $ and $ g $. We can check directly that \[\tag{1.8.23} ( O _ a u ) ( t , x ) = - u ( t , 2 a - x )\] and \[\tag{1.8.24} ( E _ b u ) ( t , x ) = u ( t , 2 b - x )\] are symmetries of the wave equation, or we can note that $ O _ a = M _ { - 1 } T _ { a , 0 } R T _ { - a , 0 } $ and $ E _ b = T _ { b , 0 } R T _ { - b , 0 } $ so each of these is a composition of symmetries and hence a symmetry. Indeed $ O _ a $ and $ E _ b $ are symmetries not just of the wave equation in general but of our particular solution. To see this it suffices to check that the initial data is unchanged by each of these and then apply the uniqueness theorem. Then we note that \[\tag{1.8.25} u ( t , a ) = ( O _ a u ) ( t , a ) = - u ( t , a )\] so $ u ( t , a ) = 0 $ and $ u $ satisfies the Dirichlet boundary condition at the left endpoint. Similarly \[\tag{1.8.26} \frac { \partial u } { \partial x } ( t , b ) = \frac { \partial O _ b u } { \partial x } ( t , b ) = \frac { \partial O _ b u } { \partial x } ( t , b ) = - \frac { \partial u } { \partial x } ( t , b )\] so $ \partial u / \partial x ( t , b ) = 0 $. and $ u $ satisfies the Neumann boundary condition at the left endpoint. We now have a solution to our initial value problem with the given boundary conditions. We did this with a Dirichlet condition at the left endpoint and a Neumann condition at the right endpoint but the same technique works for any of the other three combinations of boundary conditions--we just need to choose the appropriate extension of the initial data $ f $ and $ g $, which will be the one which has the correct symmetry properties on spatial reflection through $ a $ and $ b $ to ensure the boundary conditions are satisfied. We therefore have the following complement to our earlier uniqueness theorem.
The solution we found above has two interesting properties. First, it is periodic in the spatial variable, with period $ 4 ( b - a ) $. To see this, recall that that $ E _ a $ and $ O _ b $ are symmetries of $ u $ and note that \[\tag{1.8.27} ( E _ a O _ b u ) ( t , x ) = ( O _ b u ) ( t , 2 b - x ) = - u ( t , 2 a - ( 2 b - x ) = - u ( t , x - 2 ( b - a ) )\] so \[\tag{1.8.28} ( E _ a O _ b E _ a O _ b u ) ( t , x ) = u ( t , x - 4 ( b - a ) ) = T _ { 0 , 4 ( b - a ) } u ( t , x ) .\] Similar arguments apply to the other three combinations of boundary conditions. With a Neumann condition at the left endpoint and a Dirichlet condition at the right endpoint we again get a solution which is periodic with period $ 4 ( b - a ) $. If have Dirichlet conditions at both endpoints or Neumann conditions at both endpoint then we get solutions which are periodic with period $ 2 ( b - a ) $. Of course the original problem was to solve the initial value problem in $ \mathbf R \times [ a , b ] $, so when we say that the solution is spatially periodic what we really mean is that the natural extension of the solution to $ \mathbf R ^ 2 $ is periodic.
The second interesting property of our solution is that it is periodic in time, with period $ 4 ( b - a ) / c $. To see this, note that in terms of our extended initial data the solution is given by the D'Alembert formula (1.1.25). Substituting $ t + 4 ( b - a ) / c $ for $ t $ in this formula gives \[\tag{1.8.29} \begin{split} u ( t + 4 ( b - a ) / c , x ) & = \frac 1 2 f ( x + c s - c t - 4 ( b - a ) ) \\ & \quad {} + \frac 1 2 f ( x - c s + c t + 4 ( b - a ) ) \\ & \quad {} + \frac 1 { 2 c } \int _ { x + c s - c t - 4 ( b - a ) } ^ { x - c s + c t + 4 ( b - a ) } g ( y ) \, d y . \end{split}\] Adding or subtracting $ 4 ( b - a ) $ from the argument of $ f $ has no effect, as we just saw when we discussed spatial periodicity, so the first two terms are equal to the corresponding terms in the formula for $ u ( t , x ) $. For the last term we split the integral into five pieces, corresponding to the subintervals $ [ x + c s - c t - 4 ( b - a ) , x + c s - c t - 2 ( b - a ) ] $, $ [ x + c s - c t - 2 ( b - a ) , x + c s - c t ] $, $ [ x + c s - c t , x - c s + c t ] $, $ [ x - c s + c t , x - c s + c t + 2 ( b - a ) ] $, and $ [ x - c s + c t + 2 ( b - a ) , x - c s + c t + 4 ( b - a ) ] $. The middle one corresponds to the corresponding terms in the formula for $ u ( t , x ) $. The first two cancel each other out because translation by $ 2 ( b - a ) $ changes the sign of $ g $, and the last two cancel for the same reason. So we find that $ u ( t + 4 ( b - a ) / c , x ) $ and $ u ( t , x ) $ are equal, as claimed.
The argument above works without changes in the case of a Neumann left endpoint and Dirichlet right endpoint. It also works when both endpoints satisfy Dirichlet boundary conditions. In fact a somewhat more careful argument gives periodicity with period $ 2 ( b - a ) / c $ in that case. No variant of the argument above can prove periodicity in time in the case of Neumann conditions at both endpoints though, because this is not true in general. Indeed the wave equation on any interval with Neumann conditions at both endpoints and initial data $ f ( x ) = 0 $, $ g ( x ) = 1 $ has as its solution $ u ( t , x ) = t $, which is not periodic in $ t $.
A number of results in earlier sections of this chapter are proved by evaluating double integrals by repeated integration. A more systematic approach is to use Green's theorem, which is the following generalisation of our earlier result about rectangles.
In most applications of the theorem the functions $ p $ and $ q $ are chosen so as to make the integrand $ \partial p / \partial t - \partial q / \partial x $ on the right hand side equal to zero. As a particular example, consider \[\tag{1.9.2} p = \frac { \partial u } { \partial t } , \quad q = c ^ 2 \frac { \partial u } { \partial x }\] where $ u $ is a classical solution of the wave equation. Then \[\tag{1.9.3} \frac { \partial q } { \partial x } - \frac { \partial p } { \partial t } = 0\] is just the wave equation with the signs reversed. In addition to the functions $ p $ and $ q $ we also need to choose a region $ R $. In this case we'll choose the triangle with vertices $ ( t _ 2 , x _ 2 ) $, $ ( t _ 1 , x _ 2 - c ( t _ 2 - t _ 1 ) ) $, and $ ( t _ 1 , x _ 2 + c ( t _ 2 - t _ 1 ) ) $. Let $ C _ 1 $, $ C _ 2 $ and $ C _ 3 $ be the edges from $ ( t _ 1 , x _ 2 + c ( t _ 2 - t _ 1 ) ) $ to $ ( t _ 2 , x _ 2 ) $, from $ ( t _ 2 , x _ 2 ) $ to $ ( t _ 1 , x _ 2 - c ( t _ 2 - t _ 1 ) ) $, and from $ ( t _ 1 , x _ 2 - c ( t _ 2 - t _ 1 ) ) $ to $ ( t _ 1 , x _ 2 + c ( t _ 2 - t _ 1 ) ) $.
We can also get energy conservation from Green's theorem using the pair of functions \[\tag{1.9.15} p = \frac 1 2 \left ( \frac { \partial u } { \partial t } \right ) ^ 2 + \frac { c ^ 2 } 2 \left ( \frac { \partial u } { \partial x } \right ) ^ 2 , \quad q = - c ^ 2 \frac { \partial u } { \partial t } \frac { \partial u } { \partial x } .\] In fact this is essentially how we proved in the case of a bounded interval, with $ R $ chosen to be a rectangle. We could also reprove the original version by the same method, taking $ R $ to be a trapezoid. Here, instead, we'll prove energy conservation for a semi-infinite interval, which we haven't treated yet.
Suppose then that $ u $ is a classical solution of the wave equation on $ \mathbf R \times [ a , + \infty ) $ which satisfies either the Dirichlet or Neumann boundary condition at $ a $. The region $ R $ we will choose is a quadrilateral with four sides, $ C _ 1 $ from $ ( t _ 1 , a ) $ to $ ( t _ 1 , x _ 1 ) $, $ C _ 2 $ from $ ( t _ 1 , x _ 1 ) $ to $ ( t _ 2 , x _ 2 ) $, $ C _ 3 $ from $ ( t _ 2 , x _ 2 ) $ to $ ( t _ 2 , a ) $, and $ C _ 4 $ from $ ( t _ 2 , a ) $ back to $ ( t _ 1 , a ) $, where \[\tag{1.9.16} x _ 2 = x _ 1 \pm c ( t _ 2 - t _ 1 ) .\]
One reason for developing multiple techniques for proving results about the wave equation in one spatial dimension is that some generalise better to higher dimensions or related equations than others. We won't consider higher dimensions here but we will briefly consider two related equations, the Klein-Gordon and Sine-Gordon equations.
The Klein-Gordon equation \[\tag{1.10.1} \frac { \partial ^ 2 u } { \partial t ^ 2 } - c ^ 2 \frac { \partial ^ 2 u } { \partial x ^ 2 } + m ^ 2 u = 0\] plays a fundamental result in relativistic quantum mechanics. It shares some, but not all, of the symmetries of the wave equation. Which ones is left as an exercise for the reader.
We can still prove energy conservation, but with the energy density \[\tag{1.10.2} e = \frac 1 2 \left ( \frac { \partial u } { \partial t } \right ) ^ 2 + \frac { c ^ 2 } 2 \left ( \frac { \partial u } { \partial x } \right ) ^ 2 + \frac { m ^ 2 } 2 u .\] The first argument we used for the wave equation, using the auxiliary functions $ v $ and $ w $, does not generalise but the argument using Green's theorem does. A solution with zero initial data has zero initial energy and so has zero energy for all time. Looking at the form of the energy density this means the solution can only be the zero solution. The equation is linear, so the difference of two solutions is also a solution. If the two solutions have the same initial data then the difference has zero initial data and hence is zero, so the two solutions are the same. Thus we see that analogues of the uniqueness theorems for the wave equation also hold for the Klein-Gordon equation. The technique above is the one we used for boundary value problems for the wave equation, but not the one we originally used in $ \mathbf R ^ 2 $, which relied on having an explicit solution. Here we haven't used an explicit solution. There is one, but it involves special functions and isn't nearly as convenient to work with as the D'Alembert formula.
The sine-Gordon equation \[\tag{1.10.3} \frac { \partial ^ 2 u } { \partial t ^ 2 } - \frac { \partial ^ 2 u } { \partial x ^ 2 } + \sin u = 0\] arises in a variety of geometric and physical contexts. It is reasonable to guess that small solutions should behave like solutions of the Klein-Gordon equation with $ c = m = 1 $ because $ \sin u \approx u $ for small $ u $, and this is at least somewhat correct, but it has many peculiarities of its own.
The sine-Gordon equation shares most, but not all, of the symmetries of the Klein-Gordon equation. The main one which is missing is scaling symmetry in the dependent variable.
Energy conservation also holds for the sine-Gordon equation, with energy density \[\tag{1.10.4} e = \frac 1 2 \left ( \frac { \partial u } { \partial t } \right ) ^ 2 + \frac 1 2 \left ( \frac { \partial u } { \partial x } \right ) ^ 2 + 1 - \cos u .\] As for Klein-Gordon, any solution with zero initial data must be zero for all time, but sine-Gordon is non-linear, so the difference of two solutions is not, in general, a solution, and so we can't obtain a uniqueness theorem in the same way. Perhaps unsurprisingly there is also no explicit solution formula for sine-Gordon.
We met the diffusion equation (0.1.4) \[\tag{2.1.1} \frac { \partial u } { \partial t } - k \frac { \partial ^ 2 u } { \partial x ^ 2 } = 0\] earlier. We will always assume $ k > 0 $. The natural differentiability assumption is that $ u $ is continuously differentiable in $ t $ and twice continuously differentiable in $ x $.
Like the wave equation, the diffusion equation has a spatial reflection symmetry, but unlike the wave equation it does not have temporal reflection symmetry, and indeed behaves very differently in the positive and negative time directions. It's also symmetric under spatial and temporal translation. It also has a scaling symmetry, in both the dependent and independent variables, but the scaling in the independent variables is different from the one we had for the wave equation. Let \[\tag{2.1.2} ( S _ { \alpha , \lambda } u ) ( t , x ) = \lambda u ( t / \alpha ^ 2 , x / \alpha ) ,\] where $ \alpha $ and $ \lambda $ are non-zero. A simple calculation shows that \[\tag{2.1.3} \begin{split} & \frac { \partial S _ { \alpha , \lambda } u } { \partial t } ( t , x ) - k \frac { \partial ^ 2 S _ { \alpha , \lambda } u } { \partial x ^ 2 } ( t , x ) \\ \quad & = \frac { \lambda } { \alpha ^ 2 } \left [ \frac { \partial u } { \partial t } ( t / \alpha ^ 2 , x / \alpha ) - k \frac { \partial ^ 2 u } { \partial x ^ 2 } ( t / \alpha ^ 2 , x / \alpha ) \right ] \end{split}\] so $ S _ { \alpha , \lambda } u $ satisfies the diffusion equation if and only $ u $ does. The diffusion equation also has some less obvious symmetries. Let \[\tag{2.1.4} ( G _ v u ) ( t , x ) = \exp \left ( - \frac { v x } { 2 k } + \frac { v ^ 2 t } { 4 k } \right ) u ( t , x - v t ) .\] Then $ G _ v $ is a symmetry for any value of $ v $. To verify this we set $ \tilde u = G _ v u $ and compute partial derivatives. \[\tag{2.1.5} \frac { \partial \tilde u } { \partial t } ( t , x ) = \exp \left ( - \frac { v x } { 2 k } + \frac { v ^ 2 t } { 4 k } \right ) \left [ \frac { \partial u } { \partial t } - v \frac { \partial u } { \partial x } + \frac { v ^ 2 } { 4 k } u \right ] ,\] \[\tag{2.1.6} \frac { \partial \tilde u } { \partial x } ( t , x ) = \exp \left ( - \frac { v x } { 2 k } + \frac { v ^ 2 t } { 4 k } \right ) \left [ \frac { \partial u } { \partial x } - \frac { v } { 2 k } u \right ] ,\] and \[\tag{2.1.7} \frac { \partial ^ 2 \tilde u } { \partial x ^ 2 } = \exp \left ( - \frac { v x } { 2 k } + \frac { v ^ 2 t } { 4 k } \right ) \left [ \frac { \partial ^ 2 u } { \partial x ^ 2 } - \frac { v } { k } \frac { \partial u } { \partial x } + \frac { v ^ 2 } { 4 k ^ 2 } u \right ] .\] On the right hand side $ u $ and its various derivatives are always evaluated at $ ( t , x - v t ) $ but this has not been written explicitly to prevent the equations from becoming too long. The same will be done in the next equation, \[\tag{2.1.8} \frac { \partial \tilde u } { \partial t } ( t , x ) - k \frac { \partial ^ 2 \tilde u } { \partial x ^ 2 } ( t , x ) = \exp \left ( - \frac { v x } { 2 k } + \frac { v ^ 2 t } { 4 k } \right ) \left [ \frac { \partial u } { \partial t } - k \frac { \partial ^ 2 u } { \partial x ^ 2 } \right ] ,\] which follows immediately from the preceding equations. The left hand side is zero if and only if the right hand side is zero and the exponential factor is non-zero so $ \tilde u $ satisfies the diffusion equation if and only if $ u $ does.
One thing we can do with symmetries of a differential equation is to look for solutions invariant under the symmetry. We may or may not get something interesting, depending on the equation and the symmetry. Spatial and temporal translations don't give very interesting solutions of the diffusion equation, for example. A function which is invariant under spatial translations, for example, is just a $ u ( t , x ) $ which is independent of $ x $, but then $ \partial ^ 2 u / \partial x ^ 2 = 0 $ and so if $ u $ is a solution of the diffusion equation then $ \partial u / \partial t = 0 $ as well, so $ u $ is independent of $ t $ as well, and hence constant. So the only solutions of the diffusion equation invariant under spatial translations are the constant solutions. The solutions which are invariant under temporal translations are just the linear functions of $ x $.
The only solution to the diffusion equation which is invariant under scaling in the dependent variable is the zero solution, which is also not very interesting. Scaling in the dependent variables is more interesting. A scale invariant solution is one which satisfies $ S _ { \alpha , 0 } u = u $ for all $ \alpha $. In other words \[\tag{2.2.1} u ( t , x ) = u ( t / \alpha ^ 2 , x / \alpha ) .\] Since this holds for all $ \alpha $ it must hold in particular for $ \alpha = \sqrt { k t } $ , so \[\tag{2.2.2} u ( t , x ) = \varphi ( x / \sqrt { k t } ) ,\] where \[\tag{2.2.3} \varphi ( y ) = u ( 1 / k , y ) .\] Here we've implicitly assumed that $ t > 0 $, so we can only expect this procedure to give us an invariant solution defined there. Of course we also need $ u $ to satisfy the diffusion equation. Taking partial derivatives, \[\tag{2.2.4} \frac { \partial u } { \partial t } ( t , x ) = - \frac 1 2 \frac x { \sqrt { k t ^ 3 } } \varphi ' ( x / \sqrt { k t } ) ,\] \[\tag{2.2.5} \frac { \partial u } { \partial x } ( t , x ) = \frac 1 { \sqrt { k t } } \varphi ' ( x / \sqrt { k t } ) ,\] and \[\tag{2.2.6} \frac { \partial ^ 2 u } { \partial x ^ 2 } ( t , x ) = \frac 1 { k t } \varphi '' ( x / \sqrt { k t } ) ,\] so \[\tag{2.2.7} \frac { \partial u } { \partial t } ( t , x ) - k \frac { \partial ^ 2 u } { \partial x ^ 2 } ( t , x ) = - \frac 1 t \left [ \varphi '' ( x / \sqrt { k t } ) + \frac 1 2 \frac x { \sqrt { k t } } \varphi ' ( x / \sqrt { k t } ) \right ] .\] The right hand side is zero for all $ t > 0 $ and all $ x $ if and only if $ \varphi $ satisfies the ordinary differential equation \[\tag{2.2.8} \varphi '' ( y ) + \frac y 2 \varphi ' ( y ) = 0 .\] This can be solved as follows. Let \[\tag{2.2.9} \psi ( y ) = \exp ( y ^ 2 / 4 ) \varphi ' ( y ) .\] Then \[\tag{2.2.10} \psi ' ( y ) = \exp ( y ^ 2 / 4 ) \left [ \varphi '' ( y ) + \frac y 2 \varphi ' ( y ) \right ]\] so $ \varphi $ satisfies the differential equation above if and only if $ \psi $ is constant. Calling that constant $ c _ 1 $ we then have \[\tag{2.2.11} \varphi ' ( y ) = c _ 1 \exp ( - y ^ 2 / 4 )\] and so \[\tag{2.2.12} \varphi ( y ) = c _ 1 \int _ 0 ^ y \exp ( - z ^ 2 / 4 ) \, d z + c _ 2\] for some other constant $ c _ 2 $. The integral above can't be expressed in terms of elementary functions. It can be expressed in terms of what's called the error function, named because of its interpretation in probability theory, but the error function is defined in terms of this integral so that doesn't really provide any new information. In any case, we conclude that the scale invariant solutions to the diffusion are the two parameter family given by the equation above.
Next we look for solutions of the diffusion equation which are invariant under the transformations $ G _ v $ defined earlier, i.e. those that satisfy \[\tag{2.2.13} u ( t , x ) = \exp \left ( - \frac { v x } { 2 k } + \frac { v ^ 2 t } { 4 k } \right ) u ( t , x - v t )\] for all $ v $. Again we'll look for solutions valid for $ t > 0 $. Since the equation above holds for all $ v $ it holds in particular for $ v = x / t $, which gives \[\tag{2.2.14} u ( t , x ) = \exp \left ( - \frac { x ^ 2 } { 4 k t } \right ) \varphi ( t )\] where $ \varphi ( t ) = u ( t , 0 ) $. We still need to impose the condition that $ u $ satisfies the diffusion equation. Computing partial derivatives, \[\tag{2.2.15} \frac { \partial u } { \partial t } ( t , x ) = \exp \left ( - \frac { x ^ 2 } { 4 k t } \right ) \left [ \varphi ' ( t ) + \frac { x ^ 2 } { 4 k t ^ 2 } \varphi ( t ) \right ] ,\] \[\tag{2.2.16} \frac { \partial u } { \partial x } ( t , x ) = \exp \left ( - \frac { x ^ 2 } { 4 k t } \right ) \left [ - \frac { x } { 2 k t } \varphi ( t ) \right ] ,\] and \[\tag{2.2.17} \frac { \partial ^ 2 u } { \partial x ^ 2 } ( t , x ) = \exp \left ( - \frac { x ^ 2 } { 4 k t } \right ) \left [ \frac { x ^ 2 } { 4 k ^ 2 t ^ 2 } \varphi ( t ) - \frac 1 { 2 k t } \varphi ( t ) \right ] ,\] so \[\tag{2.2.18} \frac { \partial u } { \partial t } ( t , x ) - k \frac { \partial ^ 2 u } { \partial x ^ 2 } ( t , x ) = \exp \left ( - \frac { x ^ 2 } { 4 k t } \right ) \left [ \varphi ' ( t ) + \frac 1 { 2 t } \varphi ( t ) \right ] ,\] and $ u $ satisfies the diffusion equation if and only if $ \varphi $ satisfies the ordinary differential equation \[\tag{2.2.19} \varphi ' ( t ) + \frac 1 { 2 t } \varphi ( t ) = 0 .\] The solutions of this equation are precisely the constant multiples of $ t ^ { - 1 / 2 } $. In this way we see that every solution of the diffusion equation invariant under the transformations $ G _ v $ is a constant multiple of \[\tag{2.2.20} K ( t , x ) = \frac 1 { \sqrt { 4 \pi k t } } \exp \left ( - \frac { x ^ 2 } { 4 k t } \right ) .\] This solution turns out to be so important to the theory of the the diffusion equation that it is know as the fundamental solution. The extra factor $ 1 / \sqrt { 4 \pi k } $ is chosen to simplify the form of various equations which will appear later in the chapter.
For the diffusion equation we will consider the initial value problem \[\tag{2.3.1} u ( s , x ) = f ( s ) .\] Note that unlike the case of the wave equation, for the diffusion we only specify the initial value and not the initial time derivative as data. One other difference is that we will only look for a solution for $ t \ge s $.
The diffusion equation is called the diffusion equation because it describes various diffusion processes, for chemicals, heat, etc. The last of these also explains why it is often also called the heat equation. This physical origin suggests a conjecture about the initial value problem, namely that if $ f $ is positive, or non-negative, then $ u $ should be as well. This conjecture is, unfortunately, false but it is true for bounded solutions. In fact most things we want to prove about the diffusion equation turn out to be false in general but true when we restrict our attention to bounded solutions and initial data. In fact it's possible to replace boundedness with considerably weaker growth conditions but we won't bother doing this. Of course there is no hope of $ u $ being bounded unless $ f $ is so for the remainder of the chapter every time $ u $ and $ f $ appear there will be an implicit assumption that both are bounded. That assumption will be made explicit though in the statements of theorems or in those parts of their proofs where we use it.
Suppose $ u $ is a bounded solution of the initial value problem for the diffusion equation with non-negative initial data. For $ \epsilon > 0 $ define $ w $ by \[\tag{2.3.2} w ( t , x ) = u ( t , x ) + 3 k \epsilon t + \epsilon x ^ 2 .\] Note that $ w $ doesn't satisfy the diffusion equation, but instead satisfies the related equation \[\tag{2.3.3} \frac { \partial w } { \partial t } - k \frac { \partial ^ 2 w } { \partial x ^ 2 } = k \epsilon .\] Since $ w $ is a continuous function it must have a minimum on the rectangle $ [ s , T ] \times [ - L , L ] $ for any $ T > s $ and $ L > 0 $. It does not have a minimum in the interior $ ( s , T ) \times ( - L , L ) $ of the rectangle. If it did then the first partial derivatives would be zero there, which would then imply that $ \partial ^ 2 w / \partial x ^ 2 = - \epsilon $ there, but the second partial derivative can't be negative at an interior minimum. A similar, but more careful, argument shows that there is also no minimum on $ \{ T \} \times ( - L , L ) $, the interior of the top of the rectangle. At such a minimum the $ \partial w / \partial t $ would have to be non-negative, since otherwise we would have points just below it where the value is smaller. Similarly, $ \partial w / \partial x $ would have to be non-negative since otherwise we'd have points just to its left where the value is smaller. But $ \partial w / \partial x $ would also have to be non-positive or we'd have points to its right where the value is smaller, so in fact $ \partial w / \partial x $ must be zero. It then follows that $ \partial w ^ 2 / \partial x ^ 2 $ must be non-negative at the minimum, because otherwise we'd have points on either side where the value is smaller. So $ \partial w / \partial t - k \partial w ^ 2 / \partial x ^ 2 $ would be non-positive at the minimum. But we've already seen that this is equal to $ k \epsilon $, the product of two positive numbers, so the assumption that there is a minimum on the interior of the top of the rectangle leads to a contradiction.
On the sides $ [ s , T ] \times \{ - L \} $ and $ [ s , T ] \times \{ L \} $ we have \[\tag{2.3.4} w ( t , x ) = u ( t , x ) + 3 k \epsilon t + \epsilon L ^ 2 \ge \inf u + 3 k \epsilon s + \epsilon L ^ 2 > 3 k \epsilon s\] if \[\tag{2.3.5} L > \sqrt { \frac { \max ( 0 , - \inf u ) } \epsilon } .\] Here we've used our boundedness assumption $ u $.
The only possibility not considered so far is that the minimum of $ w $ is located on $ \{ s \} \times ( - L , L ) $, the interior of the bottom of the rectangle. If so then the definition of $ w $ and our assumption that $ f $ is non-negative imply that this minimum value of $ w $ is at least $ 3 k \epsilon s $.
What we have shown is that the minimum of $ w $ in the rectangle is attained at a point on the sides or bottom and is at least $ 3 k \epsilon s $, at least provided $ L $ is sufficiently large. In particular, \[\tag{2.3.6} w ( t , x ) \ge 3 k \epsilon s\] for any $ ( t , x ) $ in the rectangle, and therefore \[\tag{2.3.7} u ( t , x ) \ge - k \epsilon x ^ 2 - 3 k \epsilon ( T - s )\] For any $ t > s $ there is a $ T > t $ and an \[\tag{2.3.8} L > \max ( | x | , \sqrt { \frac { \max ( 0 , - \inf u ) } \epsilon } )\] No assumptions other than positivity were made on $ \epsilon $ so this inequality holds for all positive $ \epsilon $ and therefore, taking limits, \[\tag{2.3.9} u ( t , x ) \ge 0 .\] In other words we've proved the following theorem.
If $ u $ is a solution of the diffusion equation then so is $ - u $ so the positivity theorem at the end of the last section also shows that if the initial data for the initial problem are non-positive then the solution is non-positive. Combining this with the original version we see that if the initial data are zero then the solution is also zero. The equation is linear so the difference of two solutions is also a solution. Considering the difference of two solutions then we see that if the difference of their initial data is zero then the difference of the solutions is zero. Put more simply, if they have the same initial data then they are the same solution. In this way we obtain the following uniqueness theorem.
Do we need to the boundedness assumption in our uniqueness theorem? We certainly used that assumption in our proof, but that doesn't necessarily mean that the theorem would be false if we omitted omitted it, just that our proof would no longer work. In fact there is an old example of Tychonoff which shows that the assumption cannot simply be removed without invalidating the theorem.
The example is simple to write down, but requires a fair amount of work to show that it does what it claims. It is \[\tag{2.4.1} u ( t , x ) = \sum _ { j = 0 } ^ \infty \frac { \varphi _ j ( t ) x ^ { 2 j } } { ( 2 j ) ! k ^ j } ,\] where $ \varphi $ is defined inductively by \[\tag{2.4.2} \varphi _ 0 ( t ) = \begin{cases} 0 & \mbox{if } t = 0 \\ \exp ( - 1 / t ^ 2 ) & \mbox{if } t \neq 0 \end{cases}\] and \[\tag{2.4.3} \varphi _ { j + 1 } ( t ) = \varphi _ j ' ( t ) .\]
It's not even obvious that $ \varphi _ j $ is well defined, i.e. that $ \varphi _ 0 $ is infinitely differentiable at $ 0 $, let alone that the series above converges. If we assume that it does converge though, and also assume that differentiation within the sum is legitimate, then it is clear that $ u $ is a solution to the diffusion equation, since \[\tag{2.4.4} \frac { \partial u } { \partial t } ( t , x ) = \sum _ { j = 0 } ^ \infty \frac { \varphi _ j ' ( t ) x ^ { 2 j } } { ( 2 j ) ! k ^ j } = \sum _ { j = 0 } ^ \infty \frac { \varphi _ { j + 1 } ( t ) x ^ { 2 j } } { ( 2 j ) ! k ^ j }\] while \[\tag{2.4.5} \frac { \partial ^ 2 u } { \partial x ^ 2 } ( t , x ) = \sum _ { j = 1 } ^ \infty \frac { \varphi _ j ( t ) x ^ { 2 j - 2 } } { ( 2 j - 2 ) ! k ^ j } = \sum _ { j = 0 } ^ \infty \frac { \varphi _ { j + 1 } ( t ) x ^ { 2 j } } { ( 2 j ) ! k ^ { j + 1 } } .\]
Note that $ u $ is a non-zero solution of the diffusion equation because \[\tag{2.4.6} u ( t , 0 ) = \varphi _ 0 ( t ) > 0\] for $ t \neq 0 $. We will show that it has zero initial data though by showing that \[\tag{2.4.7} \varphi _ j ( 0 ) = 0\] for all $ j $. So if we can show that $ u $ is continuous for $ t \ge 0 $ and that the differentiation under the summation is valid then we have our counterexample to the uniqueness theorem without the boundedness assumption.
Our first step is to find useful bounds on $ \varphi _ j $. There are a variety of ways to do this but the easiest is to use a basic fact about holomorphic functions. If $ f $ is holomorphic in the the open disc of radius $ r $ about $ w $ and continuous on the closed disc then \[\tag{2.4.8} | f ^ { ( j ) } ( w ) | \le j ! r ^ j \max _ { | z - w | \le r } | f ( z ) | .\] We apply this to the holomorphic function \[\tag{2.4.9} f ( z ) = \exp ( - 1 / z ^ 2 )\] at the point $ w = t $ with $ r = t / 3 $. Here we're assuming that $ t $ is positive. The case of negative $ t $ can be treated similarly. We need to find bounds on the absolute value of $ f $ in the given disc. It's helpful to write $ p $ for the real part of $ z - w $ and $ q $ for its imaginary part, so that \[\tag{2.4.10} z = t + p + i q .\] Then \[\tag{2.4.11} \begin{split} | f ( z ) | & = \left | \exp \left ( - \frac 1 { ( t + p + i q ) ^ 2 } \right ) \right | \\ & = \left | \exp \left ( - \frac { ( t + p ) ^ 2 - q ^ 2 - 2 i ( t + p ) q } { \left [ ( t + p ) ^ 2 + q ^ 2 \right ] ^ 2 } \right ) \right | \\ & = \exp \left ( - \frac { ( t + p ) ^ 2 - q ^ 2 } { \left [ ( t + p ) ^ 2 + q ^ 2 \right ] ^ 2 } \right ) . \end{split}\] Now from $ | z - w | < r = t / 3 $ we get the inequalities \[\tag{2.4.12} - t / 3 \le p \le t / 3 , \quad - t / 3 \le q \le t / 3 , \quad 2 t / 3 \le t + p \le 4 t / 3 ,\] \[\tag{2.4.13} 4 t ^ 2 / 9 \le ( t + p ) ^ 2 \le 16 t ^ 2 / 9 , \quad q ^ 2 \le t ^ 2 / 9 ,\] \[\tag{2.4.14} t ^ 2 / 3 \le ( t + p ) ^ 2 - q ^ 2 \le 16 t ^ 2 / 9 , \quad ( t + p ) ^ 2 + q ^ 2 \le 17 t ^ 2 / 9 ,\] \[\tag{2.4.15} 27 / 289 t ^ 2 \le \frac { ( t + p ) ^ 2 + q ^ 2 } { \left [ ( t + p ) ^ 2 + q ^ 2 \right ] ^ 2 } ,\] \[\tag{2.4.16} \exp \left ( - \frac { ( t + p ) ^ 2 + q ^ 2 } { \left [ ( t + p ) ^ 2 + q ^ 2 \right ] ^ 2 } \right ) \le \exp \left ( - \frac { 27 } { 289 t ^ 2 } \right ) .\] So \[\tag{2.4.17} | \varphi _ j ( t ) | \le \frac { 3 ^ j j ! } { t ^ j } \exp \left ( - \frac { 27 } { 289 t ^ 2 } \right ) .\] Since \[\tag{2.4.18} \lim _ { t \to 0 } \frac { 3 ^ j j ! } { t ^ j } \exp \left ( - \frac { 27 } { 289 t ^ 2 } \right ) = 0\] it follows from our differentiability criterion that $ \varphi _ 0 $ is infinitely differentiable and all derivatives are zero at $ t = 0 $. Assuming, as we have been, that the series defining $ u $ converges we see that \[\tag{2.4.19} u ( 0 , x ) = 0 .\]
To show that the series does in fact converge we need some more inequalities. The first derivative test shows that the maximum of \[\tag{2.4.20} \frac { 3 ^ j j ! } { t ^ j } \exp \left ( - \frac { 27 } { 289 t ^ 2 } \right )\] occurs at \[\tag{2.4.21} t = \sqrt { \frac { 54 } { 289 j } }\] so \[\tag{2.4.22} | \varphi _ j ( t ) | \le \left ( \frac { 289 j } { 3 e } \right ) ^ { j / 2 } j !\] Using this and the easy estimate \[\tag{2.4.23} ( 2 j ) ! \le j ! j ^ j\] we get the bound \[\tag{2.4.24} \left | \frac { \varphi ) j ( t ) x ^ { 2 j } } { ( 2 j ) ! k ^ j } \right | \le \left ( \frac { 289 | x | ^ 4 } { 3 e j k ^ 2 } \right ) ^ { j / 2 }\] for the absolute value of the summands in the definition of $ u $. Since \[\tag{2.4.25} \lim _ { j \to \infty } \sqrt \frac { 289 | x | ^ 4 } { 3 e j k ^ 2 } = 0\] uniformly for $ x $ in any bounded interval it follows that the series for $ u $ converges absolutely uniformly in any closed interval and so converges to a continuous function.
The only assertion we have yet to verify is that term by term differentiation is permissible. This is not difficult, but it does require a theorem not usually covered in our compulsory modules. There will be several cases where we need theorems about differentiating limits or integrals, so we consider that problem in general in the next section.
The most important theorem about derivatives is the one called, entirely appropriately, the fundamental theorem of calculus. Almost every non-trivial theorem about derivatives is proved by using the fundamental theorem of calculus to turn it into a theorem about integrals, and then proving that theorem. That is, for example, the way we proved our differentiability criterion for functions defined by different expressions on different intervals.
A series is just a limit of partial sums so we can derive theorems about term by term differentiation of series from theorems about term by term differentiation of sequences and these, as we just observed, can be derived from theorems about term by term integration of sequences. What are the main theorems about integration of sequences? This depends on which theory of integration we are using. For Riemann integration the main theorem is the one which says that integral of a uniform limit of Riemann integrable functions on an interval is the limit of their integrals. For Lebesgue integration the main theorem is the Lebesgue dominated convergence theorem. For a more advanced module we would certainly want to use Lebesgue integration but for this module Riemann integration suffices and is generally simpler, so that's what we'll use.
The fundamental theorem of calculus for Riemann integration has two parts, the first of which tells us that if a function $ f $ is continuous on an interval $ [ a , b ] $ then \[\tag{2.5.1} F ( y ) = \int _ a ^ y f ( x ) \, d x\] is differentiable on $ [ a , b ] $ and \[\tag{2.5.2} F ' ( y ) = f ( x )\] for all $ x \in [ a , b ] $. The second part says that if a function $ F $ is differentiable on an interval $ [ a , b ] $ and if $ F ' $ is Riemann integrable there then \[\tag{2.5.3} \int _ a ^ b F ' ( x ) \, d x = F ( b ) - F ( a ) .\] Note that in the second part $ F ' $ is only assumed to be Riemann integrable, not continuous, but all continuous functions on a closed interval are Riemann integrable and in this module we will only ever apply the theorem in that special case so you can, if you prefer, think of the second part as having continuity rather than Riemann integrability in its hypotheses.
Suppose now that we have a sequence $ F _ j $ of continuously differentiable functions on an interval $ [ a , b ] $ with limit \[\tag{2.5.4} \lim _ { j \to \infty } F _ j ( x ) = G ( x ) .\] Suppose further that the derivatives converge uniformly. Let \[\tag{2.5.5} g ( x ) = \lim _ { j \to \infty } F ' _ j ( x )\] and observe that $ g $, being the uniform limit of a sequence of continuous functions, is continuous. Then for any $ y \in [ a , b ] $ the functions $ F ' _ j $ converge uniformly on the subinterval $ [ a , y ] $ so by our theorem about interchanging integrals and uniform limits we have \[\tag{2.5.6} \int _ a ^ y g ( x ) \, d x = \lim _ { j \to \infty } \int _ a ^ y F ' _ j ( x ) \, d x\] and hence, by the second part of the fundamental theorem of calculus, \[\tag{2.5.7} \int _ a ^ y g ( x ) \, d x = \lim _ { j \to \infty } \left [ F _ j ( y ) - F _ j ( a ) \right ] = G ( y ) - G ( a ) .\] It then follows from the first part of the fundamental theorem of calculus that $ G $ is differentiable and that its derivative is $ g $. In other words, $ \lim _ { j \to \infty } F _ j $ is a differentiable function and \[\tag{2.5.8} \frac { d } { d x } \lim _ { j \to \infty } F _ j = G ' ( x ) = g ( x ) = \lim _ { j \to \infty } F ' _ j ( x ) .\] In other words, term by term differentiation is permissible if the sequence of derivatives converges uniformly. We can summarise this in the following theorem.
What if the interval isn't closed or bounded? A function on an interval is continuous, differentiable, or continuously differentiable if and only if it is continuous, differentiable, or continuously differentiable on every closed bounded subinterval, so we can derive the following theorem from the preceding one.
As mentioned previously, series are just limits of partial sums, so we can immediately derive the following theorem for series.
As you can easily check, using the bounds we've already proved, the derivatives of the series defining $ u $ satisfy the hypotheses of the theorem, so term by term differentiation is permissible, and therefore $ u $ is a solution of the diffusion equation. In fact we can take not only the one time derivative and two space derivatives required by the equation but any number of time and space derivatives, so the function $ u $ is in fact an infinitely differentiable solution of the diffusion equation.
The limit defining $ u $ was the limit of a sequence, i.e. a limit with respect to a discrete parameter. We often want to differentiate functions given by limits with respect to a continuous parameter. There is a similar theorem for that case. The first step is to find a theorem for limits of integrals with respect to a continuous parameter. Fortunately there is one, although its usual statement doesn't explicity refer to limits at all.
Just as series are limits of finite sums, Riemann integrals are, in some sense, limits of Riemann sums, so we should expect that under certain conditions we can exchange derivatives and integrals. Here I mean derivatives and integrals of a function of several variables, the derivatives and integrals being taken with respect to different variables. The case where we differentiate and integrate with respect to the same variable is already covered by the fundamental theorem of calculus. Unfortunately the limit in the definition of Riemann sums is not the limit of a sequence, so we can't apply our earlier theorem. Instead of trying to make the intuition above precise it's more convenient to follow our usual procedure of using the fundamental theorem of calculus to convert theorems about integration to theorems about differentiation, so to prove a theorem about exchanging derivatives and integrals we need a theorem about changing the order of integration.
The most important theorem on the exchange of integrals in Lebesgue integration is Fubini's theorem. Its counterpart in Riemann integration doesn't have a name but is often also called Fubini's theorem, even though it was known long before Fubini proved his theorem. It is as follows.
The theorems of the preceding sections are all based on theorems about integrals over closed bounded intervals. Riemann integration also has a notion of improper integration, which applies to functions on intervals which are not necessarily closed or bounded. If $ f $ is continuous on the open interval $ [ \alpha , \beta ] $ then \[\tag{2.6.1} \int _ a ^ b f ( x ) \, d x = \lim _ { \alpha \to a ^ + , \beta \to b ^ - } \int _ \alpha ^ \beta f ( x ) \, d x ,\] provided this limit exists, where the integral on the right hand side is over the closed bounded interval $ [ \alpha , \beta ] $, and so is an ordinary Riemann integral. Note that $ a $ and $ b $ are allowed to be infinite, but not required to be. If both are finite and the integrand is Riemann integrable on the closed interval $ [ a , b ] $ then we have a notational conflict because $ \int _ a ^ b f ( x ) \, d x $ could refer either to the ordinary Riemann integral over that closed interval or to the improper integral over the open interval $ ( a , b ) $. Fortunately Riemann integrals depend continuously on their limits of integration so these two integrals, ordinary and improper, have the same value, so the notational ambiguity is mostly harmless.
Most theorems about ordinary Riemann integration require additional hypotheses for improper integrals. For example, it's not true in general that \[\tag{2.6.2} \int _ a ^ b \lim _ { j \to \infty } f _ j ( x ) \, d x = \lim _ { j \to \infty } \int _ a ^ b f _ j ( x ) \, d x ,\] even if both sides are well defined. The sequence of functions \[\tag{2.6.3} f _ j ( x ) = \frac 1 { 1 + ( x - j ) ^ 2 }\] mentioned earlier is a counter example, since \[\tag{2.6.4} \int _ { - \infty } ^ { + \infty } \lim _ { j \to \infty } f _ j ( x ) \, d x = \int _ { - \infty } ^ { + \infty } 0 \, d x = 0\] while \[\tag{2.6.5} \lim _ { j \to \infty } \int _ { - \infty } ^ { + \infty } f _ j ( x ) \, d x = \lim _ { j \to \infty } \pi = \pi .\] If you know the Lebesgue dominated convergence theorem then you know that it has a hypothesis that there is an integrable function $ g $ such that for all $ j $ one has \[\tag{2.6.6} | f _ j ( x ) | \le g ( x ) .\] It turns out that this also works in the case of Riemann integration with improper integrals and we have the following theorem.
The proof of the theorem, which is rather long, will be given in detail, since the basic ideas are the same as those for two further theorems, whose proofs will be skipped. Suppose $ \theta > 0 $. By the definition of improper integrals as a limit of ordinary integrals there are $ A , B \in ( a , b ) $ such that if $ \alpha \in ( a , A ) $ and $ \beta \in ( B , b ) $ such then \[\tag{2.6.10} \left | \int _ \alpha ^ \beta g ( x ) \, d x - \int _ a ^ b g ( x ) \, d x \right | < \theta\] It follows from the definitions of indefinite integrals and the continuity of Riemann integrals as functions of their limits of integration that \[\tag{2.6.11} \int _ a ^ b g ( x ) \, d x = \int _ a ^ \alpha g ( x ) \, d x + \int _ \alpha ^ \beta g ( x ) \, d x + \int _ \beta ^ b g ( x ) \, d x ,\] just as we would have if these were all ordinary integrals. It follows from $ | f _ j ( x ) | \le g ( x ) $ that $ g $ is non-negative and we know that Riemann integrals of non-negative functions are non-negative. Since limits of non-negative functions are non-negative the same also applies to improper integrals. From this we see that \[\tag{2.6.12} 0 \le \int _ a ^ \alpha g ( x ) \, d x + \int _ \beta ^ b g ( x ) \, d x < \theta .\] Then from the inequality for $ f _ j $ we get \[\tag{2.6.13} - \theta < \int _ a ^ \alpha f _ j ( x ) \, d x + \int _ \beta ^ b f _ j ( x ) \, d x < \theta .\] These inequalities for all $ \alpha \in ( a , A ) $ and $ \beta \in ( B , b ) $. Since limits preserve non-strict inequalities the same also holds if $ \alpha = A $ or $ \beta = B $, except that we may need to replace strict inequalities with non-strict ones.
We've assumed that the sequence $ f _ j $ converges uniformly on closed bound subintervals of $ ( a , b ) $ so there is an $ m $ such that if $ j > m $ then \[\tag{2.6.14} \left | f _ j ( x ) - \lim _ { k \to \infty } f _ k ( x ) \right | < \frac \theta { b - a }\] for all $ j > m $ and $ x \in [ A , B ] $, from which it follows that \[\tag{2.6.15} \left | \int _ A ^ B f _ j ( x ) \, d x - \int _ A ^ B \lim _ { k \to \infty } f _ k ( x ) \, d x \right | < \theta \frac { B - A } { b - a } < \theta .\]
Now if $ \alpha \in ( a , A ) $ and $ \beta \in ( B , b ) $ then \[\tag{2.6.16} \int _ \alpha ^ \beta f _ j ( x ) \, d x = \int _ \alpha ^ A f _ j ( x ) \, d x + \int _ A ^ B f _ j ( x ) \, d x + \int _ B ^ \beta f _ j ( x ) \, d x\] so \[\tag{2.6.17} \left | \int _ \alpha ^ \beta f _ j ( x ) \, d x - \int _ A ^ B f _ j ( x ) \, d x \right | < \theta .\] It follows that if $ \alpha ' , \alpha '' \in ( a , A ) $ and $ \beta ' , \beta '' \in ( B , b ) $ then \[\tag{2.6.18} \left | \int _ { \alpha ' } ^ { \beta ' } f _ j ( x ) \, d x - \int _ { \alpha '' } ^ { \beta '' } f _ j ( x ) \, d x \right | < 2 \theta .\] For every positive $ \epsilon $ we can choose a positive $ \theta $ with $ \theta \le \epsilon / 2 $ and then choose $ A $ and $ B $ as above, so if $ \alpha ' , \alpha '' \in ( a , A ) $ and $ \beta ' , \beta '' \in ( B , b ) $ then \[\tag{2.6.19} \left | \int _ { \alpha ' } ^ { \beta ' } f _ j ( x ) \, d x - \int _ { \alpha '' } ^ { \beta '' } f _ j ( x ) \, d x \right | < \epsilon .\] It follows from the definition of improper integrals as limits and the Cauchy criterion for convergence of limits that the improper integrals \[\tag{2.6.20} \int _ a ^ b f _ j ( x ) \, d x\] exist. We also get the inequality \[\tag{2.6.21} \left | \int _ a ^ b f _ j ( x ) \, d x - \int _ A ^ B f _ j ( x ) \, d x \right | \le 2 \theta\] by taking the limits $ ( \alpha ' , \beta ' ) \to ( a ^ + , b ^ - ) $ and $ ( ( \alpha '' , \beta '' ) \to ( A ^ - , B ^ + ) $ in the inequality above. As usual the inequality becomes non-strict in the process. So far we haven't needed $ j > m $ but if we now use this then we also have \[\tag{2.6.22} \left | \int _ A ^ B f _ j ( x ) \, d x - \int _ A ^ B \lim _ { k \to \infty } f _ k ( x ) \, d x \right | < \theta\] and hence \[\tag{2.6.23} \left | \int _ a ^ b f _ j ( x ) \, d x - \int _ A ^ B \lim _ { k \to \infty } f _ k ( x ) \, d x \right | < 3 \theta .\] If we have $ j ' , j '' > m $ then \[\tag{2.6.24} \left | \int _ a ^ b f _ { j ' } ( x ) \, d x - \int _ a ^ b f _ { j '' } ( x ) \, d x \right | < 6 \theta .\] For every positive $ \epsilon $ we can find a positive $ \theta \le \epsilon / 6 $ so the sequence of integrals $ \int _ a ^ b f _ j ( x ) \, d x $ satisfies the Cauchy criterion and hence converges. At this point we know that the right hand side of the inequality we're trying to prove is at least well defined. We also have the inequality \[\tag{2.6.25} \left | \lim _ { k \to \infty } \int _ a ^ b f _ k ( x ) \, d x - \int _ A ^ B \lim _ { k \to \infty } f _ k ( x ) \, d x \right | \le 3 \theta ,\] obtained by setting $ j '' = j $ and letting $ j ' $ tend to infinity in our earlier inequality.
Checking that the left hand side of the inequality is well defined is easier. Taking limits inside the inequality for $ f _ j $ gives \[\tag{2.6.26} \left | \lim _ { k \to \infty } f _ k ( x ) \right | \le g ( x )\] so if $ \alpha \in ( a , A ) $ and $ \beta \in ( B , b ) $ then \[\tag{2.6.27} \left | \int _ \alpha ^ A \lim _ { k \to \infty } f _ k ( x ) \, d x + \int _ B ^ \beta \lim _ { k \to \infty } f _ k ( x ) \, d x \right | < \theta .\] Then if $ \alpha ' , \alpha '' \in ( a , A ) $ and $ \beta ' , \beta '' \in ( B , b ) $ then \[\tag{2.6.28} \left | \int _ { \alpha ' } ^ { \beta ' } \lim _ { k \to \infty } f _ k ( x ) \, d x - \int _ { \alpha '' } ^ { \beta '' } \lim _ { k \to \infty } f _ k ( x ) \, d x \right | < 2 \theta .\] Using the definition of the improper integral and the Cauchy criterion yet again we get the existence of the improper integral \[\tag{2.6.29} \int _ a ^ b \lim _ { k \to \infty } f _ k ( x ) \, d x .\] By taking the limits $ ( \alpha ' , \beta ' ) \to ( a ^ + , b ^ - ) $ and $ ( ( \alpha '' , \beta '' ) \to ( A ^ - , B ^ + ) $ in the inequality above we also obtain the inequality \[\tag{2.6.30} \left | \int _ a ^ b \lim _ { k \to \infty } f _ k ( x ) \, d x - \int _ A ^ B \lim _ { k \to \infty } f _ k ( x ) \, d x \right | \le 2 \theta\] and therefore \[\tag{2.6.31} \left | \int _ a ^ b \lim _ { k \to \infty } f _ k ( x ) \, d x - \lim _ { k \to \infty } \int _ a ^ b f _ k ( x ) \, d x \right | \le 5 \theta .\] Since this holds for all positive $ \theta $ the difference must by zero, and so the equation from the conclusion of the theorem holds.
A very similar argument gives the following theorem, which is essentially the previous one but for limits with respect to a continuous parameter rather than a discrete one.
Our theorem on the interchange of integrals also requires an additional hypothesis when the intervals are open and the integrals are improper, as the function \[\tag{2.6.34} f ( x , y ) = \frac { x y ( x ^ 2 - y ^ 2 ) } { ( x ^ 2 + y ^ 2 ) ^ 3 }\] on the products of intervals $ ( 0 , 1 ) \times ( 0 , 1 ) $ shows, since \[\tag{2.6.35} \begin{split} \int _ 0 ^ 1 \int _ 0 ^ 1 f ( x , y ) \, d y \, d x & = \int _ 0 ^ 1 \int _ 0 ^ 1 \frac { \partial } { \partial y } \left ( \frac 1 2 \frac { x y ^ 2 } { ( x ^ 2 + y ^ 2 ) ^ 2 } \right ) \, d y \, d x \\ & = \int _ 0 ^ 1 \left ( \frac 1 2 \frac { x } { ( 1 + x ^ 2 ) ^ 2 } \right ) \, d x \\ & = \int _ 0 ^ 1 \frac { \partial } { \partial x } \left ( - \frac 1 8 \frac { 1 - x ^ 2 } { 1 + x ^ 2 } \right ) \, d y = - \frac 1 8 , \end{split}\] while \[\tag{2.6.36} \begin{split} \int _ 0 ^ 1 \int _ 0 ^ 1 f ( x , y ) \, d x \, d y & = \int _ 0 ^ 1 \int _ 0 ^ 1 \frac { \partial } { \partial x } \left ( - \frac 1 2 \frac { x ^ 2 y } { ( x ^ 2 + y ^ 2 ) ^ 2 } \right ) \, d x \, d y \\ & = \int _ 0 ^ 1 \left ( - \frac 1 2 \frac { y } { ( 1 + y ^ 2 ) ^ 2 } \right ) \, d y \\ & = \int _ 0 ^ 1 \frac { \partial } { \partial y } \left ( \frac 1 8 \frac { 1 - y ^ 2 } { 1 + y ^ 2 } \right ) \, d y = \frac 1 8 . \end{split}\] The additional hypothesis needed is similar to the one for limits, that the absolute value of the integrand should be bounded by something integrable.
The theorem above has an analogue for derivatives and integrals.
For simplicity we have assumed in this section that integrals are with respect to a single variable, that limits are with respect to a single variable, and that derivatives are all first derivatives, and thus also with respect to a single variable. The same method of proof works, however, when we have more variables, when the functions in question are defined on products of intervals. We've also assumed for simplicity that the intervals for improper integrals are over open intervals, so that we take limits on both sides. In fact any interval is fine, so we can also apply them to intervals like $ [ 0 , + \infty ) $.
We saw earlier that there is at most one bounded classical solution to the initial value problem for the diffusion equation in the region in the region $ [ s , + \infty ) \times \mathbf R $. The restriction of a bounded function which is twice differentiable in the spatial variable to the initial line $ \{ s \} \times \mathbf R $ is a twice continuously differentiable bounded function so we can't expect a classical solution to satisfy the initial conditions \[\tag{2.7.1} u ( t , x ) = f ( x )\] unless $ f $ is bounded and twice continuously differentiable. We do not yet know though whether there is a solution under these conditions. I claim that in fact \[\tag{2.7.2} u ( t , x ) = \int _ { - \infty } ^ { + \infty } \exp ( - \pi z ^ 2 ) f \left ( x + z \sqrt { 4 \pi k ( t - s ) } \right ) \, d z\] is a solution. First of all, note that \[\tag{2.7.3} u ( s , x ) = \int _ { - \infty } ^ { + \infty } \exp ( - \pi z ^ 2 ) f ( x ) \, d z = f ( x ) \int _ { - \infty } ^ { + \infty } \exp ( - \pi z ^ 2 ) \, d z = f ( x ) ,\] so it at least satisfies the initial conditions. Also, the change of variable \[\tag{2.7.4} y = x + \sqrt { 4 \pi k ( t - s ) } , \qquad z = \frac { y - x } { \sqrt { 4 \pi ( t - s ) } } ,\] permissible when $ s > t $, converts the integral into \[\tag{2.7.5} u ( t , x ) = \int _ { - \infty } ^ { + \infty } K ( t - s , x - y ) f ( y ) \, d y ,\] where \[\tag{2.7.6} K ( t , x ) = \frac 1 { \sqrt { 4 \pi k t } } \exp \left ( \frac { - x ^ 2 } { \sqrt { 4 k t } } \right ) .\] We met this $ K $ before when we considered Gallilean invariant solutions. The function appearing in the integral is just a translation of this function, and so is also a solution of the diffusion equation, at least for $ t > 0 $, which is the only place it's defined. Assuming we can differentiate inside the integral we therefore have \[\tag{2.7.7} \frac { \partial u } { \partial t } ( t , x ) - k \frac { \partial ^ 2 u } { \partial x ^ 2 } ( t , x ) = \int _ { - \infty } ^ { + \infty } \left ( \frac { \partial K } { \partial t } ( t - s , x - y ) - k \frac { \partial ^ 2 K } { \partial x ^ 2 } ( t - s , x - y ) \right ) f ( y ) \, d y = 0 .\] It is not hard to show that differentiation under the integral sign is indeed justified. First of all, we have \[\tag{2.7.8} \frac { \partial ^ { i + j } K } { \partial t ^ i \partial x ^ j } ( t , x ) = ( - 1 ) ^ j 2 ^ { - ( 2 i + j + 2 ) / 2 } \pi ^ { - 1 / 2 } k ^ { - ( j + 1 ) / 2 } t ^ { - ( 2 i + j + 1 ) / 2 } H _ { 2 i + j } \left ( \frac { x - y } { \sqrt { 2 k t } } \right ) \exp \left ( - \frac { x ^ 2 } { 4 k t } \right ) .\] The $ H _ n $ are the Hermite polynomials. There are two different conventions for the meaning of those. The one I'm using here is the one which makes the leading coefficients 1 rather than $ 2 ^ n $. The only property of them we need here though is that they are polynomials and the only properties of the constants appearing in the equation are that they are constant, so if you are used to the other convention you can ignore this.
The same method we used for the wave equation, the method of reflection, can be used to treat boundary value problems for the diffusion equation, provided the boundary conditions are of Dirichlet or Neumann type. Suppose, for example, that we are given data $ f $ which are continuous on the closed interval $ [ a , b ] $ and are looking for a solution to the initial value problem in the region $ [ s , + \infty ) \times [ a , b ] $ satisfying a Dirichlet condition at the left endpoint and a Neumann condition at the right endpoint, \[\tag{2.8.1} u ( t , a ) = 0 , \quad \frac { \partial u } { \partial x } ( t , b ) = 0 ,\] just as we did for the wave equation. We'll need $ f ( a ) = 0 $ in order to have a chance of solving this equation. As long as we're looking for a solution which is merely continuous for $ t \ge s $ and not one which is continously differentiable in $ t $ and twice continuously differentiable in $ x $ there we don't need to impose the conditions $ f ' ( b ) = 0 $ or $ f '' ( a ) = 0 $ which we imposed for the wave equation, and indeed it wouldn't make sense to impose them since we're merely assuming that $ f $ is continuous.
We can extend $ f $ to all of $ \mathbf R $ in the same way as we did for the wave equation, namely \[\tag{2.8.2} f ( x ) = \begin{cases} f ( a + ( b - a ) r ( x ) ) & \mbox{ if ${ l ( x ) = 0 }$,} \\ f ( b - ( b - a ) r ( x ) ) & \mbox{ if ${ l ( x ) = 1 }$,} \\ - f ( a + ( b - a ) r ( x ) ) & \mbox{ if ${ l ( x ) = 2 }$,} \\ - f ( b - ( b - a ) r ( x ) ) & \mbox{ if ${ l ( x ) = 3 }$,} \\ \end{cases}\] where \[\tag{2.8.3} \frac { x - a } { b - a } = 4 m ( x ) + l ( x ) + r ( x )\] and $ r ( x ) \in [ 0 , 1 ) $, and $ m ( x ) $ and $ l ( x ) $ both integers with $ 0 \le l ( x ) < 4 $. We then define \[\tag{2.8.4} u ( t , x ) = \int _ { - \infty } ^ { + \infty } \frac 1 { \sqrt { 4 \pi k ( t - s ) } } \exp \left ( - \frac { ( y - x ) ^ 2 } { 4 k ( t - s ) } \right ) f ( y ) \, d y ,\] where the $ f $ in the integrand is this extended $ f $. To prevent subsequent equations from getting very messy we will write this in terms of the fundamental solution: \[\tag{2.8.5} u ( t , x ) = \int _ { - \infty } ^ { + \infty } K ( t - s , x - y ) f ( y ) \, d y ,\] The argument that this is a solution to the initial value problem with the given boundary conditions, and is the only solution, is essentially the same as for the wave equation.
It's also possible to write the solution in terms of the original, unextended, $ f $ as follows. We split first split the integral into pieces: \[\tag{2.8.6} u ( t , x ) = \sum _ { l = 0 } ^ 3 \sum _ { m = - \infty } ^ { + \infty } \int _ { a + ( 4 m + l ) ( b - a ) } ^ { b + ( 4 m + l ) ( b - a ) } K ( t - s , x - y ) f ( y ) \, d y ,\] or, after a linear change of variable, \[\tag{2.8.7} u ( t , x ) = \sum _ { l = 0 } ^ 3 \sum _ { m = - \infty } ^ { + \infty } \int _ { a } ^ { b } K ( t - s , x - y - ( 4 m + 1 ) ( b - a ) ) f ( y + ( 4 m + l ) ( b - a ) ) \, d y .\] Now \[\tag{2.8.8} f ( y + ( 4 m + l ) ( b - a ) ) = \begin{cases} f ( y ) & \mbox { if ${ l = 0 }$,} \\ f ( a + b - y ) & \mbox { if ${ l = 1 }$,} \\ - f ( y ) & \mbox { if ${ l = 2 }$,} \\ - f ( a + b - y ) & \mbox { if ${ l = 3 }$.} \end{cases}\] We make a further change of variable in the odd cases, replacing $ y $ by $ a + b - y $, obtaining \[\tag{2.8.9} \begin{split} u ( t , x ) & = \sum _ { m = - \infty } ^ { + \infty } K ( t - s , x - y - 4 m ( b - a ) ) f ( y ) \, d y \\ & \quad{} + \sum _ { m = - \infty } ^ { + \infty } K ( t - s , x + y - 2 a - ( 4 m + 2 ) ( b - a ) - a - b ) f ( y ) \, d y \\ & \quad {} - \sum _ { m = - \infty } ^ { + \infty } K ( t - s , x - y - ( 4 m + 2 ) ( b - a ) ) f ( y ) \, d y \\ & \quad {} - \sum _ { m = - \infty } ^ { + \infty } K ( t - s , x + y - 2 a - ( 4 m + 4 ) ( b - a ) - a - b ) f ( y ) \, d y . \end{split}\] Here the $ f $'s in the integrands all refer to the original, unextended, $ f $.
The diffusion equation has a conservation law which applies when either there is no boundary or the boundary conditions are all Neumann conditions. First we consider the case without boundary. Suppose that \[\tag{2.9.1} \int _ { - \infty } ^ { + \infty } | f ( y ) | \, d y = 0 .\] Then \[\tag{2.9.2} \begin{split} \int _ { - \infty } ^ { + \infty } u ( t , x ) \, d x & = \int _ { - \infty } ^ { + \infty } \int _ { - \infty } ^ { + \infty } K ( t - s , x - y ) f ( y ) \, d y \, d x \\ & = \int _ { - \infty } ^ { + \infty } \int _ { - \infty } ^ { + \infty } K ( t - s , x - y ) f ( y ) \, d x \, d y \\ & = \int _ { - \infty } ^ { + \infty } f ( y ) \int _ { - \infty } ^ { + \infty } K ( t - s , x - y ) \, d x \, d y \\ & = \int _ { - \infty } ^ { + \infty } f ( y ) \, d y . \end{split}\] The interchange of the two integrals is justified by Fubini's theorem. So the quantity \[\tag{2.9.3} \int _ { - \infty } ^ { + \infty } u ( t , x ) \, d x\] is independent of $ t $.
Next we consider the case of a finite interval $ [ a , b ] $ with Neumann conditions at both endpoints. We apply Green's theorem with \[\tag{2.9.4} p = - u , \quad q = - k \frac { \partial u } { \partial x } .\] For the region we take the rectangle $ R = [ t _ 1 , t _ 2 ] \times [ a , b ] $, where $ t _ 2 > t _ 1 > s $. Green's theorem gives \[\tag{2.9.5} \sum _ { j = 1 } ^ 4 \int _ { C _ j } ( p ( t , x ) \, d x + q ( t , x ) \, d t ) = \int _ R \frac { \partial q } { \partial x } - \frac { \partial p } { \partial t } \, d A .\] The curves $ C _ 1 $, $ C _ 2 $, $ C _ 3 $ and $ C _ 4 $ will be the line segments from $ ( t _ 1 , a ) $ to $ ( t _ 1 , b ) $, $ ( t _ 1 , b ) $ to $ ( t _ 2 , b ) $, $ ( t _ 2 , b ) $ to $ ( t _ 2 , a ) $, and $ ( t _ 2 , a ) $ to $ ( t _ 1 , a ) $, respectively. The integrals along $ C _ 2 $ and $ C _ 4 $ are zero because of the Neumann condition. The right hand side will be zero for solutions of the diffusion equation. What we are left with is \[\tag{2.9.6} \int _ a ^ b u ( t _ 2 , x ) \, d x - \int _ a ^ b u ( t _ 1 , x ) \, d x = 0 .\] So \[\tag{2.9.7} \int _ a ^ b u ( t , x ) \, d x\] is independent of $ t $. This argument doesn't work for $ t = s $ because Green's theorem requires $ p $ and $ q $ to be continuously differentiable in all of $ R $ but we can take the limit as $ t _ 1 $ tends to $ s $ from above since the hypotheses of our theorem on continuity of integrals are satisfied.
In the original application of the diffusion equation to heat conduction Dirichlet boundary conditions correspond to conducting boundaries and Neumann boundary conditions correspond to insulating boundary conditions. The integral of $ u $ corresponds to the total energy. The theorem we've just proved is then conservation of energy for a thermally isolated system. We can't expect the theorem to apply with Dirichlet boundary conditions because energy can enter or leave the system through the conducting boundary.
Another important physical quantity in the original physical application is entropy, given by the integral \[\tag{2.9.8} - \int u ( t , x ) \log u ( t , x ) \, d x .\] Of course this only makes sense if $ u $ is positive, which it always is in the study of heat conduction. We've already seen that if the initial data are positive then the solution will remain positive forever. Unlike energy, we don't expect entropy to be conserved. The second law of thermodynamics says that it should be increasing, or at least non-decreasing. Again, we expect this only for isolated systems, so it should hold either when there is no boundary or when all boundary conditions are of Neumann type.
This time we'll treat the case of a finite interval first. Taking $ R $ as before we set \[\tag{2.9.9} p = u \log u , \quad q = k \frac { \partial u } { \partial x } \log u .\] Then \[\tag{2.9.10} \frac { \partial q } { \partial x } - \frac { \partial p } { \partial t } = \frac { k u } { \left ( \partial u / \partial x \right ) ^ 2 } - \left ( \frac { \partial u } { \partial t } - k \frac { \partial ^ 2 u } { \partial x ^ 2 } \right ) \log u .\] The second term on the right hand side is zero for solutions of the diffusion equation. As with our proof of energy conservation, the integrals over $ C _ 2 $ and $ C _ 4 $ vanish because of the Neumann boundary condition are we are left with \[\tag{2.9.11} \int _ a ^ b u ( t _ 2 , x ) \, d x - \int _ a ^ b u ( t _ 1 , x ) \, d x = \int _ R \frac { k u } { \left ( \partial u / \partial x \right ) ^ 2 } \, d A .\] Since $ u $ is positive the integrand on the right hand side is positive everywhere and so the integral is positive. It follows that \[\tag{2.9.12} \int _ a ^ b u ( t _ 2 , x ) \, d x > \int _ a ^ b u ( t _ 1 , x ) \, d x .\] As before, the use of Green's theorem presupposes $ t _ 2 > t _ 1 > s $ but we can use continuity to get the same result for $ t _ 2 > t _ 1 \ge s $.
The argument for the case without boundaries is more subtle. Let \[\tag{2.9.13} \varphi ( z ) = z \log z\] and \[\tag{2.9.14} w ( t , x , y ) = \varphi ( f ( y ) ) - \varphi ( u ( t , x ) ) - \varphi ' ( u ( t , x ) ) ( f ( y ) - u ( t , x ) ) .\] Multiplying by $ K ( t - s , x - y ) $ and integrating with respect to both $ x $ and $ y $ we get \[\tag{2.9.15} \begin{split} & \int _ { - \infty } ^ { + \infty } \int _ { - \infty } ^ { + \infty } K ( t - s , x - y ) w ( t , x , y ) \, d x \, d y \\ & \quad{} = \int _ { - \infty } ^ { + \infty } \int _ { - \infty } ^ { + \infty } K ( t - s , x - y ) \varphi ( f ( y ) ) \, d x \, d y \\ & \qquad {} - \int _ { - \infty } ^ { + \infty } \int _ { - \infty } ^ { + \infty } K ( t - s , x - y ) \varphi ( u ( t , x ) ) \, d y \, d x \\ & \qquad {} - \int _ { - \infty } ^ { + \infty } \int _ { - \infty } ^ { + \infty } K ( t - s , x - y ) \varphi ' ( u ( t , x ) ) f ( y ) \, d y \, d x \\ & \qquad {} + \int _ { - \infty } ^ { + \infty } \int _ { - \infty } ^ { + \infty } K ( t - s , x - y ) \varphi ' ( u ( t , x ) ) u ( t , x ) ) \, d y \, d x . \end{split}\] Here we've used Fubini's theorem to change the order of integration in some, but not all cases. Performing the inner integration in each of the integrals on the right we have \[\tag{2.9.16} \begin{split} \int _ { - \infty } ^ { + \infty } \int _ { - \infty } ^ { + \infty } K ( t - s , x - y ) w ( t , x , y ) \, d x \, d y & = \int _ { - \infty } ^ { + \infty } \varphi ( f ( y ) ) \, d y \\ & \quad {} - \int _ { - \infty } ^ { + \infty } \varphi ( u ( t , x ) ) \, d x \\ & \quad {} - \int _ { - \infty } ^ { + \infty } \varphi ' ( u ( t , x ) ) u ( t , x ) \, d x \\ & \quad {} + \int _ { - \infty } ^ { + \infty } \varphi ' ( u ( t , x ) ) u ( t , x ) ) \, d x . \end{split}\] The last two cancel so we are left with \[\tag{2.9.17} \begin{split} \int _ { - \infty } ^ { + \infty } \int _ { - \infty } ^ { + \infty } K ( t - s , x - y ) w ( t , x , y ) \, d x \, d y & = \int _ { - \infty } ^ { + \infty } \varphi ( f ( y ) ) \, d y \\ & \quad {} - \int _ { - \infty } ^ { + \infty } \varphi ( u ( t , x ) ) \, d x . \end{split}\] The fundamental solution is positive everywhere and $ w ( t , x , y ) $ is non-negative everywhere as a result of the convexity of $ \varphi $ so the integrand on the left hand side is non-negative and therefore so is its integral. It follows that \[\tag{2.9.18} \int _ { - \infty } ^ { + \infty } \varphi ( f ( y ) ) \, d y \le \int _ { - \infty } ^ { + \infty } \varphi ( u ( t , x ) ) \, d x\] or, equivalently, \[\tag{2.9.19} \int _ { - \infty } ^ { + \infty } \varphi ( u ( s , y ) ) \, d y \le \int _ { - \infty } ^ { + \infty } \varphi ( u ( t , x ) ) \, d x .\] In other words, the entropy at later times is always greater than or equal to the initial entropy. Of course if we have $ t _ 2 > t _ 1 \ge s $ then we can just view the solution restricted to $ [ t _ 1 , + \infty ) $ as the solution to an initial value problem with data prescribed at time $ t _ 1 $, so we find that the entropy at time $ t _ 2 $ is greater than or equal the entropy at time $ t _ 1 $, so entropy is non-decreasing.
We can sharpen the result above by noting that $ \varphi $ is strictly convex, so if $ w ( t , x , y ) $ is positive except when $ u ( t , x ) = f ( y ) $ and both $ w $ and $ K $ are continuous so if $ K w $ is positive anywhere then it's positive on an open set and so the integral is positive. It follows that entropy is strictly decreasing unless $ u ( t , x ) = f ( y ) $ for all $ t $, $ x $ and $ y $, which happens only if $ u $ is constant.
The argument we used for entropy applies to prove monotonicity of other interesting integrals. If $ \varphi $ is convex and differentiable then \[\tag{2.9.20} \int \varphi ( u ( t , x ) ) \, d x\] is a decreasing function of $ t $, and strictly decreasing if $ \varphi $ is strictly convex and $ u $ is not constant. This applies, for example to \[\tag{2.9.21} \int u ( t , x ) ^ 2 \, d x\] or, more generally, to \[\tag{2.9.22} \int | u ( t , x ) | ^ p \, d x\] for $ p > 1 $. A slight variant of the argument applies when $ \varphi $ is convex but not necessarily differentiable, and so includes the case $ p = 1 $ above.
We didn't do a lot of explicit examples for the initial value problem for the diffusion equation. One reason is that there are very few choices of initial data $ f $ for which the integral can be evaluated in closed form. If we don't care about getting a closed form solution, and in applications we generally don't, one option would be to evaluate the integral numerically. In practice what people do is somewhat different. Assuming $ u $ is continuously differentiable in time and twice continuously differentiable in space we have the limits \[\tag{2.10.1} \lim _ { \Delta t \to 0 } \frac { u ( t + \Delta x , x ) - u ( t , x ) } { \Delta x } = \frac { \partial u } { \partial t } ( t , x )\] and \[\tag{2.10.2} \lim _ { \Delta x \to 0 } \frac { u ( t , x + \Delta x ) - 2 u ( t , x ) + u ( t , x + \Delta x ) } { \Delta x ^ 2 } = \frac { \partial ^ 2 u } { \partial x ^ 2 } ( t , x ) .\] Indeed the first of these is just the definition of the partial derivative. For the second we have \[\tag{2.10.3} u ( t , x + \Delta x ) - u ( t , x ) = \Delta x \int _ 0 ^ 1 \frac { \partial u } { \partial x } ( t , x + r \Delta x ) \, d p\] and \[\tag{2.10.4} u ( t , x - \Delta x ) - u ( t , x ) = - \Delta x \int _ 0 ^ 1 \frac { \partial u } { \partial x } ( t , x - r \Delta x ) \, d p\] by the fundamental theorem of calculus, so \[\tag{2.10.5} \begin{split} & u ( t , x + \Delta x ) - 2 u ( t , x ) + u ( t , x - \Delta x ) \\ & \qquad = \Delta x \int _ 0 ^ 1 \left [ \frac { \partial u } { \partial x } ( t , x + r \Delta x ) - \frac { \partial u } { \partial x } ( t , x - r \Delta x ) \right ] d p \end{split}\] and, using the fundamental theorem of calculus again, \[\tag{2.10.6} u ( t , x + \Delta x ) - 2 u ( t , x ) + u ( t , x - \Delta x ) = \frac { \Delta x ^ 2 } 2 \int _ 0 ^ 1 \int _ { - 1 } ^ 1 \frac { \partial ^ 2 u } { \partial x ^ 2 } ( t , x + q r \Delta x ) \, d q \, d r .\] Dividing by $ \Delta x ^ 2 $ and taking the limit as $ \Delta x $ goes to $ 0 $, which is permissible since we have a uniform limit of continuous functions on a product of integrals, we get \[\tag{2.10.7} \lim _ { \Delta x \to 0 } \frac { u ( t , x + \Delta x ) - 2 u ( t , x ) + u ( t , x + \Delta x ) } { \Delta x ^ 2 } = \frac { \partial ^ 2 u } { \partial x ^ 2 } ( t , x )\] as expected.
It follows from the equations above that for classical solutions $ u $ to the diffusion equation we have \[\tag{2.10.8} \lim _ { \Delta t \to 0 , \Delta x \to 0 } \frac { u ( t + \Delta t , x ) - u ( t , x ) } { \Delta t } - k \frac { u ( t , x + \Delta x ) - 2 u ( t , x ) + u ( t , x - \Delta x ) } { \Delta x ^ 2 } = 0 .\] In some sense then the equation \[\tag{2.10.9} u ( t + \Delta t , x ) = p u ( t , x + \Delta x ) + ( 1 - 2 p ) u ( t , x ) + p u ( t , x - \Delta x )\] should hold approximately. for classical solutions to the diffusion equation, where \[\tag{2.10.10} p = k \frac { \Delta t } { \Delta x ^ 2 } .\] We now make a leap of faith and conjecture that if \[\tag{2.10.11} u ( t + \Delta t , x ) = p u ( t , x + \Delta x ) + ( 1 - 2 p ) u ( t , x ) + p u ( t , x - \Delta x )\] holds exactly then $ u $ is approximately equal to a classical solution. This equation allows us to determine the values of $ u $ at time $ t + \Delta t $ from the values of $ u $ at the time $ t $. In particular, given the initial values at time $ s $ we can find the values at time $ s + j \Delta t $ for all positive integers $ j $. So we hope to solve the diffusion equation numerically but choosing appropriate values of $ \Delta t $ and $ \Delta x $, finding $ p $, then using the initial conditions to get $ u ( s , j \Delta x ) = f ( j \Delta x ) $, and then finally applying the equation \[\tag{2.10.12} u ( t + \Delta t , x ) = p u ( t , x + \Delta x ) + ( 1 - 2 p ) u ( t , x ) + p u ( t , x - \Delta x )\] repeatedly to get $ u ( s + i \Delta t , j \Delta x ) $ for all positive $ i $, thus giving us the values of $ u $ on a grid.
The procedure described above is an example of what's called a finite difference method. As indicated, it relies on a leap of faith, although a rather plausible one. Before trying to prove that this leap is justified we should examine some numerical evidence. For this we need to pick a problem for which we know the exact answer. It's easy to check that \[\tag{2.10.13} u ( t , x ) = \exp ( - \pi ^ 2 t ) \cos ( \pi x )\] satisfies the diffusion equation with $ k = 1 $. By the uniqueness theorem proved earlier it is the only bounded solution. It has \[\tag{2.10.14} u ( 1 , 0 ) = \exp ( - \pi ^ 2 ) = 0.000051723.\] We can compute the finite difference approximation with $ \Delta t = 1 / M $ and $ \Delta x = 1 / N $ for various values of $ M $ and $ N $. Our expectation, based on the limit relations which motivated the method, should be that increasing $ M $ and $ N $ should reduce the error. The following table give percentage errors for a variety of choices of $ M $ and $ N $. \[\tag{2.10.15} \begin{array}{r|r|r} M & N & \mathrm{\%\,error} \\ \hline 250 & 10 & 10.9421 \\ 375 & 10 & 4.7931 \\ 500 & 10 & 1.6050 \\ 625 & 10 & 0.3499 \\ 750 & 10 & 1.6604 \\ 1000 & 10 & 3.3223 \\ 1500 & 10 & 5.0040 \\ 2000 & 10 & 5.8523 \\ 2000 & 15 & 1.1840 \\ 2000 & 18 & 0.0720 \\ 2000 & 20 & 0.4047 \\ 2000 & 25 & 1.1323 \\ 2000 & 30 & 1.5254 \\ 2000 & 32 & \mathrm{??????} \end{array}\] The first few lines conform to our expectations. Increasing $ M $ for fixed $ N $ seems to decrease the error. After a certain point though the error starts to increase again. The next lines show the effect of increasing $ M $ with $ N $ fixed. Initially this reduces the error, but it then starts to increase again. The error in the last line is omitted from the table. It would have been $ 229566039412762781639954661376\,\%$!
The main point of this exercise is to show that numerical solution of partial differential equations is a subtle problem and that seemingly reasonable ideas can give disastrously wrong answers. If that's all you retain from this section then it has served its main purpose, but we can investigate this a bit further.
Consider a random walk where at each time interval $ \Delta t $ particles can move left or right by a distance $ \Delta x $, each with probability $ p $, or can stay where they are. Probabilities are non-negative and probabilities of an exhaustive set of mutually exclusive probabilities add up to $ 1 $ so we can only consider $ 0 \le p \le 1 / 2 $. If $ u ( t , x ) $ is the expected number of particles at the position $ x $ at time $ t $ then by the usual rules for computing expectations using conditional probabilities we will have the equation \[\tag{2.10.16} u ( t + \Delta t , x ) = p u ( t , x + \Delta x ) + ( 1 - 2 p ) u ( t , x ) + p u ( t , x - \Delta x ) ,\] which should look familiar. So our finite difference approximation for the diffusion equation models a discrete diffusion process. In fact the diffusion equation usually arises as the continuum limit of discrete diffusion processes, so in most cases we're really just modelling one discrete diffusion process with another.
The point of view from the preceding paragraph gives some clues into how to prove that the finite difference scheme converges to a genuine solution in those cases where it does. We want to prove that the limit of sum of independent random variables approaches the fundamental solution we found earlier, which is Gaussian in form. This looks like a job for the central limit theorem. In fact, if we didn't know the explicit solution formula in advance we could derive it from the central limit theorem.
What about the catastrophic failure we saw earlier though? Recall that \[\tag{2.10.17} p = k \frac { \Delta t } { \Delta x ^ 2 } .\] Our probabilistic interpretation of the finite difference scheme only makes sense if $ p $ lies in the interval $ ( 0 , 1 / 2 ] $. Let's look at the example we considered before, but now with the $ p $ values indicated. \[\tag{2.10.18} \begin{array}{r|r|r|r} M & N & p & \mathrm{\%\,error} \\ \hline 250 & 10 & 0.4000 & 10.9421 \\ 375 & 10 & 0.2667 & 4.7931 \\ 500 & 10 & 0.2000 & 1.6050 \\ 625 & 10 & 0.1600 & 0.3499 \\ 750 & 10 & 0.1333 & 1.6604 \\ 1000 & 10 & 0.1000 & 3.3223 \\ 1500 & 10 & 0.0667 & 5.0040 \\ 2000 & 10 & 0.5000 & 5.8523 \\ 2000 & 15 & 0.1125 & 1.1840 \\ 2000 & 18 & 0.1620 & 0.0720 \\ 2000 & 20 & 0.2000 & 0.4047 \\ 2000 & 25 & 0.3125 & 1.1323 \\ 2000 & 30 & 0.4500 & 1.5254 \\ 2000 & 32 & 0.5120 & 229566039412762781639954661376 \end{array}\] All of the $ p $ values except the last one are below the critical threshold of $ 1 / 2 $ except for the last one, which is barely above it. Even going just barely above it leads to disaster though.
A somewhat less obvious property of the table above is that the approximations, which generally have a somewhat disappointing error percentage, seem to be anomolously good for $ p \approx 1 / 6 $. To see if this is an accident we can try $ M = 2400 $, $ N = 20 $, which has $ p = 1 / 6 $. These $ M $ and $ N $ are in the same general range as the ones in the second half of the table, but give a much better error of $ 0.0011\,\% $. This is not an accident. The value $ p = 1 / 6 $ is special. This fact admits an analytic explanation and a probabilistic explanation but we won't go into either of them here.
We won't discuss numerical solution of any other partial differential equations in these notes. As you might have guessed from the example above the numerical solution of partial differential equations is a complicated subject, too complicated for an introductory module like this one. What we saw for the diffusion equation is typical of other equations though, except for the existence of a nice probabilistic interpretation, which is special to the diffusion equation. In particular solution methods for partial differential equations generally involve discretising all of the independent variables and the characteristic sizes of those discretisations generally have to satisfy one or more conditions to ensure numerical stability. Crossing that critical threshold, even slightly, leads to catastrophic numerical instability.
The Black-Scholes equation, (0.1.8), we mentioned in the introduction. As a reminder, it was \[\tag{2.11.1} \frac { \partial v } { \partial \tau } + \frac 1 2 \sigma ^ 2 s ^ 2 \frac { \partial ^ 2 v } { \partial s ^ 2 } + r s \frac { \partial v } { \partial s } - r v = 0 .\] It describes the evolution of the value of a derivative, although neither value nor derivative mean what they usual do in mathematics. Value means what you would expect in a financial context: the price at which an asset can be bought or sold. Derivative means an asset whose value depends on the value of some other asset. Usually this means an option on a stock, i.e. a contract giving one the right to buy or sell a stock at a given price on a given date. In the equation above $ v $ is the value of the derived asset, $ s $ is the price of the underlying asset, $ r $ is the rate of return on a risk free asset, $ \sigma $ is the volatility of the price of the underlying asset, and $ \tau $ is time.
The change of variable For now let's assume the derivative is an option on a stock which allows us to buy or sell it at a given price $ K $, usually called the strike price, at time $ T $, usually called the expiry date of the option. If make the changes of variable \[\tag{2.11.2} t = T - \tau , \quad u = v \exp ( r t ) , \quad x = \log ( s / K ) + \left ( r - \frac 1 2 \sigma ^ 2 \right ) t , \sigma = \sqrt { 2 k }\] in the Black-Scholes equation we get the diffusion equation. of $ t $ and $ s $. Note that $ t $ is the time remaining until expiry, so $ t = 0 $ corresponds to expiry and positive values of $ t $ correspond to times before the option expires, which are the times at which we'd like to compute its value. The value at expiry is a known function of $ s $ and therefore of $ s $. The specific function depends on whether our option is an option to sell, usually called a put, or an option to buy, usually called a call. So the problem of computing the value of the option at earlier times is an initial value problem for the diffusion equation, although it's really a final value problem for the Black-Scholes equation due to the time reversal in our change of variables.
While the Black-Scholes equation applies to the value of any option the example described above, an option which can be exercised only at expiry, is only one particular type of option, usually called a European option. The more common type of option, even in Europe, is what's called an American option, where the option can be exercised at any time. The effect of this is to convert our pure initial value problem into a boundary value problem, but this boundary value problem is of a very different type from the ones we've considered previously. The boundary conditions, in terms of the original variables, are \[\tag{2.11.3} v ( \tau , g ( \tau ) ) = \max ( 0 , v ( T , g ( \tau ) ) )\] and \[\tag{2.11.4} \frac { \partial v } { \partial x } ( \tau , g ( \tau ) ) = 1 .\] What is $ g ( \tau ) $? It is the price at which one should choose to exercise the option early at time $ \tau $. This is not a given function. Rather, finding this function is part of solving the problem. So, unlike the boundary value problems we've considered previously, the location of the boundary is not known in advance but rather has to be solved for. In some sense the lack of information about the location of the boundary is compensated for by the fact that we have two boundary conditions to be satisfied on the boundary rather than one.
Boundaries whose location is not known in advance are called free boundaries. They don't just arise in financial mathematics and indeed didn't first arise there. The classical example of a free boundary problem is fluid flow with a fluid which has a boundary, for example a bubble within the fluid region or the top surface of a water wave. The peculiarity of free boundary problems is that even when the equation is linear, as the Black-Scholes equation and the equations for irrotational incompressible fluid flow are, the methods needed to study them look much more like those of the theory of nonlinear differential equations.
We've already seen Burgers' equation \[\tag{3.0.1} \frac { \partial u } { \partial t } + u \frac { \partial u } { \partial x } = 0\] Burgers' equation is a vastly simplified model of the evolution of the free boundary for fluid flow without viscosity.
There is a general theory which applies to first order scalar differential equations, of which this is one, but here we'll just do everything by hand in this special case.
Suppose $ u $ is a continuously differentiable solution to this equation in a neighbourhood of the point $ ( t _ 0 , x _ 0 ) $ and set \[\tag{3.1.1} p ( t ) = u ( t , x _ 0 + v t - v t _ 0 ) - v\] where $ v = u ( t _ 0 , x _ 0 ) $. Then $ p ( t _ 0 ) = 0 $ and the chain rule gives \[\tag{3.1.2} p ' ( t ) = \frac { \partial u } { \partial t } ( t , x _ 0 + v t - v t _ 0 ) + v \frac { \partial u } { \partial x } ( t , x _ 0 + v t - v t _ 0 )\] or, using the fact that $ u $ satisfies the differential equation, \[\tag{3.1.3} p ' ( t ) = - p ( t ) \frac { \partial u } { \partial x } ( t , x _ 0 + v t - v t _ 0 ) .\] Defining \[\tag{3.1.4} q ( t ) = p ( t ) \exp \left ( \int _ { t _ 0 } ^ t \frac { \partial u } { \partial x } ( t , x _ 0 + v s - v t _ 0 ) \, d s \right )\] we find $ q ( t _ 0 ) = 0 $ and using the product rule and the fundamental theorem of calculus we find that \[\tag{3.1.5} q ' ( t ) = 0\] so $ q $ is zero everywhere. It follows that $ p $ is also zero everywhere and \[\tag{3.1.6} u ( t , x _ 0 + v t - v t _ 0 ) = v .\] In other words, $ u $ is constant on the line $ x - v t = x _ 0 - v t _ 0 $. So to solve the initial value problem \[\tag{3.1.7} u ( t _ 0 , x ) = f ( x )\] it suffices to solve eliminate $ x _ 0 $ from the system of equations \[\tag{3.1.8} x - u t = x _ 0 - u t _ 0 , \quad u = f ( x _ 0 ) .\]
As a simple example consider linear initial conditions \[\tag{3.1.9} u ( t _ 0 , x ) = c x .\] Eliminating $ x _ 0 $ from \[\tag{3.1.10} x - u t = x _ 0 - u t _ 0 , \quad u = c x _ 0 .\] gives \[\tag{3.1.11} u ( t , x ) = \frac { c x } { 1 + c ( t - t _ 0 ) } .\] That this satisfies the differential equation and initial conditions is easy to check directly. The behaviour depends on the sign of $ c $. If $ c $ nonnegative then the solution exists for all nonnegative values of $ t $ while if $ c $ is negative the solution exists up until $ t = t _ 0 - 1 / c $ but there is no continuously differentiable solution afterwards. Unlike the wave or diffusion equations we can therefore not expect global solutions for Burgers' equation, even for very nice initial data.
If you know something about the existence and uniqueness theorems for ordinary differential equations this should not surprise you. The existence results for linear ordinary differential equations give global existence, while the ones for nonlinear ordinary differential equations only give existence in a finite time interval, whose length depends on the choice of initial data. Since the wave and diffusion equations are linear while Burgers' is nonlinear it isn't particularly unexpected that we get global existence for the first two and only existence in a finite interval for the last one. In fact the situation is worse than that though. Consider the initial conditions \[\tag{3.2.1} u ( 0 , x ) = \cos \left ( \pi x ^ 2 \right ) .\] The solution, for as long as it exists, should be equal to $ ( - 1 ) ^ k $ on the lines \[\tag{3.2.2} x = \sqrt { k } + ( - 1 ) ^ k t ,\] where $ k $ is a nonnegative integer. Considering the cases $ k = 2 j $ and $ k = 2 j + 1 $, where $ j $ is a positive integer, we see that at the point \[\tag{3.2.3} ( t , x ) = \left ( \frac { \sqrt { 2 j + 1 } - \sqrt { 2 j } } 2 , \frac { \sqrt { 2 j + 1 } + \sqrt { 2 j } } 2 \right )\] $ u ( t , x ) $ should be equal to both $ + 1 $ and $ - 1 $, so the solution cannot extend as far forward in time as \[\tag{3.2.4} t = \frac { \sqrt { 2 j + 1 } - \sqrt { 2 j } } 2 .\] Similarly, considering the cases $ k = 2 j - 1 $ and $ k = 2 j $ we see that at the point \[\tag{3.2.5} ( t , x ) = \left ( - \frac { \sqrt { 2 j } - \sqrt { 2 j - 1 } } 2 , \frac { \sqrt { 2 j - 1 } + \sqrt { 2 j } } 2 \right )\] $ u ( t , x ) $ should again be equal to both $ + 1 $ and $ - 1 $, so the solution cannot extend as far backward in time as \[\tag{3.2.6} t = - \frac { \sqrt { 2 j } - \sqrt { 2 j - 1 } } 2 .\] But these remarks apply to all integers $ j $, and both $ \sqrt { 2 j + 1 } - \sqrt { 2 j } $ and $ \sqrt { 2 j } - \sqrt { 2 j - 1 } $ tend to zero as $ j $ tends to infinity, so there is no time interval of positive length on which we have a continuously differentiable solution to this initial value problem, even though the initial data is bounded and infinitely differentiable!
There is much more to be said about Burgers' equation. It was originally introduce to model fluid flow and, in particular, shock formation. There is a natural way to extend solutions beyond the singularities we've seen above, although not as a continuously differentiable, or even continuous function. This is true also for the more complicated, but also more physically relevant, Euler equations. That is a topic for a more advanced text though.
The Laplace equation \[\tag{4.1.1} \frac { \partial ^ 2 u } { \partial x ^ 2 } + \frac { \partial ^ 2 u } { \partial y ^ 2 } = 0\] is linear and so has a scaling symmetry in the dependent variable. It has constant coefficients and so has translational symmetry in both variables. \[\tag{4.1.2} ( T _ { \xi , \eta } u ) ( x , y ) = u ( x - \xi , y - \eta ) .\] Somewhat more interestingly, it has reflectional symmetry in either the $ x $ or $ y $ variable, a scaling symmetry in the variables $ x $ and $ y $ \[\tag{4.1.3} ( S _ \mu u ) ( x , y ) = u ( x / \mu , y / \mu )\] and a rotational symmetry \[\tag{4.1.4} ( R _ \theta u ) ( x , y ) = u ( x \cos \theta + y \sin \theta , - x \sin \theta + y \cos \theta )\] This is far from a complete list of symmetries though. We also have an inversion symmetry \[\tag{4.1.5} ( J u ) ( x , y ) = u \left ( \frac x { x ^ 2 + y ^ 2 } , \frac y { x ^ 2 + y ^ 2 } \right ) .\] This one is more complicated to check than the previous ones.
Instead of immediately taking derivatives, which gets rather messy, it's better at this point to ask more generally under what conditions on the functions $ p $ and $ q $ it is the case that \[\tag{4.1.6} \tilde u ( x , y ) = u ( p ( x , y ) , q ( x , y ) )\] is a symmetry of the Laplace equation, and then checking whether the particular $ p $ and $ q $ which appear in the definition of $ J $ satisfy those conditions. Taking one derivative using the chain rule, \[\tag{4.1.7} \frac { \partial \tilde u } { \partial x } ( x , y ) = \frac { \partial p } { \partial x } ( x , y ) \frac { \partial u } { \partial x } ( p ( x , y ) , q ( x , y ) ) + \frac { \partial q } { \partial x } ( x , y ) \frac { \partial u } { \partial y } ( p ( x , y ) , q ( x , y ) )\] Taking another derivative, \[\tag{4.1.8} \begin{split} \frac { \partial ^ 2 \tilde u } { \partial x ^ 2 } ( x , y ) & = \left ( \frac { \partial p } { \partial x } ( x , y ) \right ) ^ 2 \frac { \partial ^ 2 u } { \partial x ^ 2 } ( p ( x , y ) , q ( x , y ) ) \\ & + 2 \frac { \partial p } { \partial x } ( x , y ) \frac { \partial q } { \partial x } ( x , y ) \frac { \partial ^ 2 u } { \partial x \partial y } ( p ( x , y ) , q ( x , y ) ) \\ & + \left ( \frac { \partial q } { \partial x } ( x , y ) \right ) ^ 2 \frac { \partial ^ 2 u } { \partial y ^ 2 } ( p ( x , y ) , q ( x , y ) ) \\ & + \frac { \partial p ^ 2 } { \partial x ^ 2 } ( x , y ) \frac { \partial u } { \partial x } ( p ( x , y ) , q ( x , y ) ) \\ & + \frac { \partial q ^ 2 } { \partial x ^ 2 } ( x , y ) \frac { \partial u } { \partial y } ( p ( x , y ) , q ( x , y ) ) . \end{split}\] Taking $ y $ derivatives gives \[\tag{4.1.9} \begin{split} \frac { \partial ^ 2 \tilde u } { \partial y ^ 2 } ( x , y ) & = \left ( \frac { \partial p } { \partial y } ( x , y ) \right ) ^ 2 \frac { \partial ^ 2 u } { \partial x ^ 2 } ( p ( x , y ) , q ( x , y ) ) \\ & + 2 \frac { \partial p } { \partial y } ( x , y ) \frac { \partial q } { \partial y } ( x , y ) \frac { \partial ^ 2 u } { \partial x \partial y } ( p ( x , y ) , q ( x , y ) ) \\ & + \left ( \frac { \partial q } { \partial y } ( x , y ) \right ) ^ 2 \frac { \partial ^ 2 u } { \partial y ^ 2 } ( p ( x , y ) , q ( x , y ) ) \\ & + \frac { \partial p ^ 2 } { \partial y ^ 2 } ( x , y ) \frac { \partial u } { \partial x } ( p ( x , y ) , q ( x , y ) ) \\ & + \frac { \partial q ^ 2 } { \partial y ^ 2 } ( x , y ) \frac { \partial u } { \partial y } ( p ( x , y ) , q ( x , y ) ) . \end{split}\] Adding these equations gives \[\tag{4.1.10} \begin{split} \frac { \partial ^ 2 \tilde u } { \partial x ^ 2 } ( x , y ) + \frac { \partial ^ 2 \tilde u } { \partial y ^ 2 } ( x , y ) & = \left [ \left ( \frac { \partial p } { \partial x } ( x , y ) \right ) ^ 2 + \left ( \frac { \partial p } { \partial y } ( x , y ) \right ) ^ 2 \right ] \frac { \partial ^ 2 u } { \partial x ^ 2 } ( p ( x , y ) , q ( x , y ) ) \\ & + 2 \left [ \frac { \partial p } { \partial x } ( x , y ) \frac { \partial q } { \partial x } ( x , y ) + \frac { \partial p } { \partial y } ( x , y ) \frac { \partial q } { \partial y } ( x , y ) \right ] \frac { \partial ^ 2 u } { \partial x \partial y } ( p ( x , y ) , q ( x , y ) ) \\ & + \left [ \left ( \frac { \partial q } { \partial x } ( x , y ) \right ) ^ 2 + \left ( \frac { \partial q } { \partial y } ( x , y ) \right ) ^ 2 \right ] \frac { \partial ^ 2 u } { \partial y ^ 2 } ( p ( x , y ) , q ( x , y ) ) \\ & + \left [ \frac { \partial p ^ 2 } { \partial x ^ 2 } ( x , y ) + \frac { \partial p ^ 2 } { \partial y ^ 2 } ( x , y ) \right ] \frac { \partial u } { \partial x } ( p ( x , y ) , q ( x , y ) ) \\ & + \left [ \frac { \partial q ^ 2 } { \partial x ^ 2 } ( x , y ) + \frac { \partial q ^ 2 } { \partial y ^ 2 } ( x , y ) \right ] \frac { \partial u } { \partial y } ( p ( x , y ) , q ( x , y ) ) . \end{split}\] This looks quite complicated but it's easy to check that if either the pair of equations \[\tag{4.1.11} \frac { \partial p } { \partial x } + \frac { \partial q } { \partial y } = 0 , \quad \frac { \partial p } { \partial y } - \frac { \partial q } { \partial x } = 0\] or the pair of equations \[\tag{4.1.12} \frac { \partial p } { \partial x } - \frac { \partial q } { \partial y } = 0 , \quad \frac { \partial p } { \partial y } + \frac { \partial q } { \partial x } = 0\] holds then the coefficients of the first derivatives and the mixed second partial derivative are all zero and the other coefficients are equal, so \[\tag{4.1.13} \frac { \partial ^ 2 \tilde u } { \partial x ^ 2 } ( x , y ) + \frac { \partial ^ 2 \tilde u } { \partial y ^ 2 } ( x , y )\] is equal to \[\tag{4.1.14} \frac 1 2 \left [ \left ( \frac { \partial p } { \partial x } ( x , y ) \right ) ^ 2 + \left ( \frac { \partial p } { \partial y } ( x , y ) \right ) ^ 2 + \left ( \frac { \partial q } { \partial x } ( x , y ) \right ) ^ 2 + \left ( \frac { \partial q } { \partial y } ( x , y ) \right ) ^ 2 \right ]\] times \[\tag{4.1.15} \frac { \partial ^ 2 u } { \partial x ^ 2 } ( p ( x , y ) , q ( x , y ) ) + \frac { \partial ^ 2 u } { \partial y ^ 2 } ( p ( x , y ) , q ( x , y) ) .\] As long as the the derivative of the mapping defined by $ p $ and $ q $ is non-singular, which we need in order for this to be a symmetry, the sum of squares of partial derivatives is positive and so \[\tag{4.1.16} \frac { \partial ^ 2 \tilde u } { \partial x ^ 2 } ( x , y ) + \frac { \partial ^ 2 \tilde u } { \partial y ^ 2 } ( x , y ) = 0\] if and only if \[\tag{4.1.17} \frac { \partial ^ 2 u } { \partial x ^ 2 } ( p ( x , y ) , q ( x , y ) ) + \frac { \partial ^ 2 u } { \partial y ^ 2 } ( p ( x , y ) , q ( x , y)) = 0 .\] So the transformation is a symmetry provided that its derivative is non-singular and $ p $ and $ q $ satisfy one of the pairs of equations above. These equations are of independent interest. The first pair are the equations for two dimensional incompressible irrotational fluid flow while the second pair are the Cauchy-Riemann equations, which are important in complex analysis. In both cases the usual names for the variables I've been calling $ p $ and $ q $ are $ u $ and $ v $ but we already have a $ u $ so it was necessary to relabel them here.
It is in fact possible to show that the transformation \[\tag{4.1.18} \tilde u ( x , y ) = u ( p ( x , y ) , q ( x , y ) )\] is a symmetry if and only if the conditions above are satisfied. The non-singularity of the derivative is needed since symmetries must be invertible functions. To see that one of the pairs of first order order equations for $ p $ and $ q $ must hold we note that the coefficients of the pure second order equations must be equal and the coefficient of the mixed partial derivative must vanish, i.e. \[\tag{4.1.19} \left ( \frac { \partial p } { \partial x } \right ) ^ 2 + \left ( \frac { \partial p } { \partial y } \right ) ^ 2 - \left ( \frac { \partial q } { \partial x } \right ) ^ 2 - \left ( \frac { \partial q } { \partial y } \right ) ^ 2 = 0\] and \[\tag{4.1.20} 2 \left [ \frac { \partial p } { \partial x } \frac { \partial q } { \partial x } + \frac { \partial p } { \partial y } \frac { \partial q } { \partial y } \right ] = 0 .\] Squaring and adding we obtain, after a little bit of algebra, the equation \[\tag{4.1.21} \left [ \left ( \frac { \partial p } { \partial x } + \frac { \partial q } { \partial y } \right ) ^ 2 + \left ( \frac { \partial p } { \partial y } - \frac { \partial q } { \partial x } \right ) ^ 2 \right ] + \left [ \left ( \frac { \partial p } { \partial x } - \frac { \partial q } { \partial y } \right ) ^ 2 + \left ( \frac { \partial p } { \partial y } + \frac { \partial q } { \partial x } \right ) ^ 2 \right ] = 0 .\] For this to be zero we need one of the factors to be zero, which means that each of the two squared terms in that factor must be zero. This leads us to the two pairs of equations given earlier.
Solving either of the pairs of first order equations for $ p $ and $ q $ is no easier than solving the Laplace equation, but verifying that given $ p $ and $ q $ satisfy one or the other is just a matter of computing their first partial derivatives. For example, with \[\tag{4.1.22} p ( x , y ) = \frac { x } { x ^ 2 + y ^ 2 }\] and \[\tag{4.1.23} q ( x , y ) = \frac { y } { x ^ 2 + y ^ 2 } ,\] as in the definition of $ J $, we have \[\tag{4.1.24} \frac { \partial p } { \partial x } ( x , y ) = - \frac { x ^ 2 - y ^ 2 } { \left ( x ^ 2 + y ^ 2 \right ) ^ 2 } ,\] \[\tag{4.1.25} \frac { \partial p } { \partial y } ( x , y ) = - \frac { 2 y } { \left ( x ^ 2 + y ^ 2 \right ) ^ 2 } ,\] \[\tag{4.1.26} \frac { \partial q } { \partial x } ( x , y ) = - \frac { 2 y } { \left ( x ^ 2 + y ^ 2 \right ) ^ 2 } ,\] and \[\tag{4.1.27} \frac { \partial p } { \partial x } ( x , y ) = \frac { x ^ 2 - y ^ 2 } { \left ( x ^ 2 + y ^ 2 \right ) ^ 2 } ,\] and so the pair \[\tag{4.1.28} \frac { \partial p } { \partial x } + \frac { \partial q } { \partial y } = 0 , \quad \frac { \partial p } { \partial y } - \frac { \partial q } { \partial x } = 0\] is satisfied and $ J $ is therefore a symmetry of the Laplace equation.
We can find more symmetries by composing the ones we've just seen. For example, we can write any reflection through a line in the plane as a composition of a translation, a rotation, a reflection about the $ x $ or $ y $ axis, another rotation, and another translation. Since these are all symmetries it follows without further computation that all such reflections are symmetries of the Laplace equation.
For a more complicated example, set \[\tag{4.1.29} G = T _ { 0 , 1 } J T _ { 0 , - 1 / 2 } J S _ 2 .\] We then have that \[\tag{4.1.30} \begin{split} ( G u ) ( x , y ) & = ( T _ { 0 , 1 } J T _ { 0 , - 1 / 2 } J S _ 2 u ) ( x , y ) \\ & = ( J T _ { 0 , - 1 / 2 } J S _ 2 u ) ( x , y - 1 ) \\ & = ( T _ { 0 , - 1 / 2 } J S _ 2 u ) \left ( \frac x { x ^ 2 + ( y - 1 ) ^ 2 } , \frac { y - 1 } { x ^ 2 + ( y - 1 ) ^ 2 } \right ) \\ & = ( J S _ 2 u ) \left ( \frac x { x ^ 2 + ( y - 1 ) ^ 2 } , \frac 1 2 \frac { x ^ 2 + y ^ 2 - 1 } { x ^ 2 + ( y - 1 ) ^ 2 } \right ) \\ & = ( S _ 2 u ) \left ( 4 \frac x { x ^ 2 + ( y + 1 ) ^ 2 } , 2 \frac { x ^ 2 + y ^ 2 - 1 } { x ^ 2 + ( y + 1 ) ^ 2 } \right ) \\ & = u \left ( 2 \frac x { x ^ 2 + ( y + 1 ) ^ 2 } , \frac { x ^ 2 + y ^ 2 - 1 } { x ^ 2 + ( y + 1 ) ^ 2 } \right ) . \end{split}\] Since we already know that $ T _ { 0 , 1 } $, $ J $, $ T _ { 0 , - 1 / 2 } $, and $ S _ 2 $ are symmetries we can conclude that $ G $ is a symmetry without directly checking in the way we did for $ J $. What transformations can we write as a product of the ones we already have? This is a question we'll return to later.
Now that we have some symmetries we can ask which solutions are invariant under particular groups of symmetries. For most groups the answers are uninteresting. For example the only solutions invariant under translations are the constant solutions. In some cases we get more interesting examples though. A function is invariant under the rotations if and only if it is of the form \[\tag{4.1.31} u ( t , x ) = \varphi ( \sqrt { x ^ 2 + y ^ 2 } )\] for some function $ \varphi $. Substituting this into the Laplace equation we see that $ \varphi $ must satisfy the equation \[\tag{4.1.32} r \varphi '' ( r ) + \varphi ' ( r ) = 0 .\] Letting $ s = \varphi ' ( r ) $ we have, at least formally, $r \frac { d s } { d r } + s = 0$ so \[\tag{4.1.33} \frac { d s } s + \frac { d r } r = 0\] and so \[\tag{4.1.34} \log s + \log r = \log ( s r )\] is constant and therefore so is $ s r $. In other words, \[\tag{4.1.35} s = \frac { c _ 1 } r\] for some constant $ c _ 1 $. Then \[\tag{4.1.36} \varphi ' ( r ) = s = \frac { c _ 1 } r\] so, integrating again, \[\tag{4.1.37} \varphi ( r ) = c _ 1 \log r + c _ 2 .\] The preceding argument can be made rigorous but there's no real need to do so since we can check directly that this is a solution to the the ordinary differential equation for $ \varphi $ and the uniqueness of solutions to the initial value problem for ordinary differential equations then shows that these are the only solutions. So our rotationally invariant solutions are \[\tag{4.1.38} u ( x , y ) = c _ 1 \log ( \sqrt { x ^ 2 + y ^ 2 } ) + c _ 2 .\]
Once we have solution we can use our symmetries to get more solutions. For example, once we know that \[\tag{4.1.39} u ( x , y ) = \log ( x ^ 2 + y ^ 2 )\] is a solution we see that \[\tag{4.1.40} ( T _ { \xi , \eta } u ) ( x , y ) = \log ( x ^ 2 + y ^ 2 - 2 \xi x - 2 \eta y + \xi ^ 2 + \eta ^ 2 )\] is a solution. This wouldn't be terribly difficult to check directly, but it would be considerably more painful to check directly that \[\tag{4.1.41} ( G u ) ( x , y ) = \log \left ( \frac { x ^ 2 + ( y + 1 ) ^ 2 } { x ^ 2 + ( y - 1 ) ^ 2 } \right )\] is a solution.
Suppose we have a solution $ u $ of the Laplace equation in the open disc of radius $ \rho $ about the origin. By rotational invariance then, as discussed above, \[\tag{4.2.1} ( R _ \theta u ) ( x , y ) = u ( x \cos \theta + y \sin \theta , - x \sin \theta + y \cos \theta )\] is also a solution in this disc. We can average this over all values of $ \theta $. \[\tag{4.2.2} \begin{split} \tilde u ( x , y ) & = \frac 1 { 2 \pi } \int _ { - \pi } ^ { \pi } R _ \theta ( x , y ) \, d \theta \\ & = \frac 1 { 2 \pi } \int _ { - \pi } ^ { \pi } u ( x \cos \theta + y \sin \theta , - x \sin \theta + y \cos \theta ) \, d \theta . \end{split}\] This average will also be a solution of the wave equation, as we can see by differentiating under the integral sign, and will be rotationally invariant. Taking $ ( x , y ) = ( 0 , 0 ) $ in the equation above we see that \[\tag{4.2.3} \tilde u ( 0 , 0 ) = u ( 0 , 0 ) .\] We've already described the rotationally invariant solutions of the Laplace equation and the only ones which are defined at the origin are the constant ones, so in fact \[\tag{4.2.4} \tilde u ( x , y ) = u ( 0 , 0 )\] throughout the disc.
For any $ r < \rho $ we have, by taking $ ( x , y ) = ( r , 0 ) $, \[\tag{4.2.5} u ( 0 , 0 ) = \frac 1 { 2 \pi } \int _ { - \pi } ^ { \pi } u ( r \cos \theta , r \sin \theta ) \, d \theta .\] If we suppose further that $ u $ is continuous on the closed disc of radius $ \rho $ then the right hand side is a continuous function of $ r $ on the closed interval $ [ 0 , r ] $ by our theorem on continuity of integrals, and so we have, by taking limits as $ r $ approaches $ \rho $ from below, \[\tag{4.2.6} u ( 0 , 0 ) = \frac 1 { 2 \pi } \int _ { - \pi } ^ { \pi } u ( r \cos \theta , r \sin \theta ) \, d \theta\] for $ r = \rho $ as well, and hence for all $ r \le \rho $.
The choice of the origin as the centre of our disc simplified the calculations above, but we know that spatial translations are symmetries of the Laplace equation so the following theorem then follows from this special case:
The following is called the weak maximum principle.
There is also a strong maximum principle, which has an additional hypothesis, the connectedness of $ \Omega $, but a stronger conclusion, that the maximum of $ u $ occurs only on the boundary unless $ u $ is constant.
This is proved using the mean value property. Suppose, as above, that $ \Omega $ is a bounded connected open subset of $ \mathbf R ^ 2 $. Then its closure $ \bar \Omega $ is also bounded. Suppose further that $ u $ is a continuous function on $ \bar \Omega $ which is twice continuously differentiable in $ \Omega $ and satisfies the Laplace equation there. Let \[\tag{4.3.5} V = \{ ( x , y ) \in \Omega : u ( x , y ) = \max _ { \bar \Omega } u \} .\] Note that $ \bar \Omega $ is a closed bounded subset of $ \mathbf R ^ 2 $, so the maximum refered to above exists. This $ V $ is closed, as a subset of $ \Omega $, because $ u $ is continuous.
$ V $ also an open subset of $ \Omega $ though. We can see this as follows. Suppose $ ( x , y ) \in V $. Then $ ( x , y ) \in \Omega $ and, since $ \Omega $ is open there is some positive $ \rho $ such that the disc of radius $ \rho $ about $ ( x , y ) $ is contained in $ \Omega $. For any $ r < \rho $ then $ u ( x , y ) $ is equal to the average value of $ u $ on the circle of radius $ r $ about $ ( x , y ) $. But $ u \in V $ so $ u ( x , y ) = \max _ { \bar \Omega } u $ and this circle is contained in $ \bar \Omega $ so $ u $ is less than or equal to $ u ( x , y ) $ everywhere on this circle. In other words, the value of $ u $ on the circle is everywhere less than or equal to its average value on the circle. Since $ u $ is continuous this can only happen if the value is equal to its average everywhere. The whole circle of radius $ r $ then also belongs to $ V $. This holds for all $ r < \rho $ so the open disc of radius $ \rho $ about $ ( x , y ) $ is a subset of $ V $. We've now established that for every point of $ V $ there is an open disc centred at $ ( x , y ) $ which is also in $ V $, and so $ V $ is open.
We've now seen that the connected set $ \Omega $ has a subset $ V $ which is, when considered as a subset of $ \Omega $, both open and closed. By the definition of connectedness this means that $ V $ is either empty or is all of $ \Omega $. Either possibility can occur. If $ V $ is empty then the maximum of $ u $ over $ \bar \Omega $ does not occur in $ \Omega $ and so must occur only on its boundary, $ \partial \Omega $. If $ V $ is all of $ \Omega $ then $ u $ is constant on $ \Omega $ and so, by continuity, also on $ \bar \Omega $.
Stating the maximum principles as maximum principles is a matter of convention. There are also minimum principles, obtained by replacing the word maximum by minimum everywhere in the statements above. The proofs are the same, except that in addition to replacing the word maximum by minimum everywhere we need to reverse the sign of all inequalities. Alternatively, since multiplication by non-zero constants is a symmetry of the equation we can just apply a maximum principle to $ - u $ to obtain the corresponding minimum principle.
Two consequences of the minimum principles are a non-negativity theorem and a positivity theorem. With hypotheses as in the statement of the strong maximum principle, if $ u $ is non-negative on $ \partial \Omega $ then it is non-negative in $ \Omega $ as well. If it is non-negative everywhere in $ \partial \Omega $ and is positive somewhere then it is positive everywhere in $ \Omega $.
The open upper half-plane $ \Omega = \{ ( x , y ) \in \mathbf R ^ 2 \colon y > 0 \} $ is unbounded so neither version of the maximum principle applies to it and indeed it is possible for a continuous function $ u $ on $ \bar \Omega $ which satisfies the Laplace equation in $ \Omega $ to take values greater than $ \max _ { \partial \Omega } u $ in $ \Omega $. $ u ( x , y ) = y $ is an example. We saw a similar situation when we looked at the diffusion equation and were able to get a maximum priniciple in an unbounded region by imposing an additional hypothesis: that $ u $ is bounded. The same is true here.
Suppose $ r > 0 $ and consider the function $ v $ defined by \[\tag{4.3.7} v ( x , y ) = \frac \pi 4 - \arctan \frac { r ^ 2 - 2 r y - x ^ 2 - y ^ 2 } { r ^ 2 + 2 r y - x ^ 2 - y ^ 2 }\] in the open disc of radius $ r \sqrt 2 $ about the point $ ( 0 , r ) $. The denominator is non-zero in this disc so $ v $ is a continuous function. It takes the value $ 0 $ where this disc intersects the $ x $ axis and the value $ 1 $ where it intersects the circle of radius $ r $ about the point $ ( 0 , 0 ) $ and takes values between $ 0 $ and $ 1 $ in the half between them. It is also, as a tedious but straightforward calculation shows, a solution of the Laplace equation.
Suppose now that \[\tag{4.3.8} \sup _ { \bar \Omega } u > \sup _ { \partial \Omega } u .\] and let \[\tag{4.3.9} w = u - \left ( \sup _ { \bar \Omega } u - \sup _ { \partial \Omega } u \right ) v .\] Then $ w $ is less than or equal to $ \sup _ { \partial \Omega } u $ on the part of the real axis between $ ( - r , 0 ) $ and $ ( r , 0 ) $ and also on the part of the circle of radius $ r $ about $ ( 0 , 0 ) $ between them.
We would like to apply the weak maximum principle to $ w $ in the region \[\tag{4.3.10} \Omega ' = \{ ( x , y ) \in \mathbf R ^ 2 \colon y > 0 , x ^ 2 + y ^ 2 < r ^ 2 \}\] to conclude that $ w $ is less than or equal to $ \sup _ { \partial \Omega } u $ in the closure of this set, i.e. the upper half of the closed disc of radius $ r $ about $ ( 0 , 0 ) $, but unfortunately $ v $, and hence $ w $, is undefined at the points $ ( - r , 0 ) $ and $ ( r , 0 ) $ on its boundary. $ u $ is continuous at those points though so for every positive $ \epsilon $ there is a positive $ \delta $ such that if $ ( x , y ) $ is contained in the disc of radius $ \delta $ about $ ( r , 0 ) $ then $ u ( x , y ) $ is within $ \epsilon $ of $ u ( r , 0 ) $ and hence less than or equal $ \sup _ { \partial \Omega } u + \epsilon $, and similarly if it's contained in the disc of radius $ r $ about $ ( - r , 0 ) $. If in addition $ y \ge 0 $ then $ v ( x , y ) \ge 0 $ so $ w ( x , y ) \le u ( x , y ) $ and so $ w ( x , y ) $ is also less than or equal to $ \sup _ { \partial \Omega } u + \epsilon $. So if instead of the region above we consider the region \[\tag{4.3.11} \Omega '' \{ ( x , y ) \in \mathbf R ^ 2 \colon y > 0 , x ^ 2 + y ^ 2 < r ^ 2 , ( x + r ) ^ 2 + y ^ 2 < \rho ^ 2 , ( x - r ) ^ 2 + y ^ 2 < \rho ^ 2 \}\] for some positive $ \rho $ less than $ \delta $ then this region does satisfy the hypotheses of the weak maximum principle for the function $ w $, which is less than or equal to $ \sup _ { \partial \Omega } u + \epsilon $ on its boundary, so \[\tag{4.3.12} w ( x , y ) \le \sup _ { \partial \Omega } u + \epsilon \] at all points in the region $ \Omega '' $. For every $ ( x , y ) \in \Omega ' $ though we can choose $ \rho $ sufficiently small that $ ( x , y ) \in \Omega ' $ so in fact the above inequality holds for all $ ( x , y ) \in \Omega ' $. Since it holds for all positive $ \epsilon $ we also have the stronger inequality \[\tag{4.3.13} w ( x , y ) \le \sup _ { \partial \Omega } u \] there. In other words, \[\tag{4.3.14} u ( x , y ) \le \sup _ { \partial \Omega } u + \left ( \sup _ { \bar \Omega } u - \sup _ { \partial \Omega } u \right ) v ( x , y ) .\] This is valid for any $ ( x , y ) \in \Omega $ and any $ r > \sqrt { x ^ 2 + y ^ 2 } $ and so for all sufficiently large $ r $. We may therefore take the limit as $ r $ tends to infinity, which gives \[\tag{4.3.15} u ( x , y ) \le \sup _ { \partial \Omega } u .\] This was proved for $ ( x , y ) \in \Omega $ but of course also holds for $ ( x , y ) \in \partial \Omega $ and hence for all $ ( x , y ) \in \bar \Omega $. So \[ \sup _ { \bar \Omega } u \le \sup _ { \partial \Omega } u . \] This inequality was proved on the assumption that \[ \sup _ { \bar \Omega } u > \sup _ { \partial \Omega } u \] though, and is incompatible with it, so that assumption must be false, which means that \[ \sup _ { \bar \Omega } u \le \sup _ { \partial \Omega } u \] holds without any assumption. The reverse inequality follows immediately from $ \partial \Omega \subseteq \bar \Omega $ so in fact \[ \sup _ { \bar \Omega } u = \sup _ { \partial \Omega } u , \] as claimed.
The Dirichlet problem for the Laplace equation on $ \Omega $ with boundary data $ f $, a continuous function on $ \partial \Omega$, is to find a continuous function $ u $ on $ \bar \Omega $ which is twice continuously differentiable in $ \Omega $ and satisfies the Laplace equation there, and such that the restriction of $ u $ to $ \partial \Omega $ is equal to $ f $.
As formulated above, the Dirichlet problem may fail to have any solutions, even if we assume $ \Omega $ is a bounded open subset of $ \mathbf R ^ 2 $. With those assumptions though we can at least show that it has at most one solution.
This follows almost immediately from the weak maximum principle and minimum principle. Suppose $ v $ and $ w $ solve the Dirichlet problem on the same region $ \Omega $ with the same boundary data $ f $. Let \[\tag{4.4.1} u = v - w . \] Then $ u $ satisfies the hypotheses of the maximum principle and minimum principle and so takes its maximum and minimum on $ \partial u $. But $ u $ is zero there, so its maximum and minimum are both zero and so $ u $ must be identically zero. In other words, $ u = v $.
The above proof works without change for the upper half-plane, provided we use the corresponding version of the maximum principle and add the requirement that $ u $ is bounded.
What about other unbounded regions $ \Omega $. It's clear that no such theorem can hold for $ \Omega = \mathbf R ^ 2 $. In that case $ \partial \Omega $ is empty so the boundary condition would hold vacuously and the theorem would say that there's at most one bounded solution to the Laplace equation in $ \mathbf R ^ 2 $, but every constant is a solution. In fact those are the only bounded solutions in this case and for every proper subset of $ \mathbf R ^ 2 $ the assumption that solutions are bounded is sufficient for uniqueness, but we won't prove this.
The connectedness assumption only really matters for the strong maximum principle. It's not needed for the weak maximum principle or the uniqueness theorem. On the other hand there's not much interesting to be said about disconnected regions. We can always treat them by considering each connected component separately. This is why we looked only at intervals for the wave and diffusion equations. Solving those equations in a disjoint union of intervals just requires solving it in each interval separately.
The Dirichlet problem for the upper half plane can only have a bounded solution when the initial data are bounded, since the restriction of a bounded function can't be unbounded. For bounded data it does indeed have a bounded solution as the following theorem shows.
The proof is to write down an equation for the solution and check that it works: \[\tag{4.5.1} u ( x , y ) = \frac 1 \pi \int _ { - \infty } ^ { + \infty } \frac { f ( x + t y ) } { 1 + t ^ 2 } \, d t .\] Strictly speaking $ f $ should be a function on the $ x $ axis in $ \mathbf R ^ 2 $ rather than on the real line so we should write $ f ( z , 0 ) $ in place of $ f ( z ) $ but no one ever does this. Everyone tacitly identifies the $ x $ axis with the real line in the obvious way. The integrand satisfies the bounds \[\tag{4.5.2} \left | \frac { f ( x + t y ) } { 1 + t ^ 2 } \right | \le \frac { \sup | f | } { 1 + t ^ 2 } ,\] the right hand side of which is independent of $ x $ and $ y $ and is an integrable function on the interval of integration, which is all of $ \mathbf R $ in this case, so by an earlier theorem $ u $ is a continuous function. From the same inequality we get the bound \[\tag{4.5.3} | u ( x , y ) | \le \frac { \sup | f | } \pi \int _ { - \infty } ^ { + \infty } \frac { d t } { 1 + t ^ 2 } = \sup | f | ,\] so $ u $ is bounded. Also, \[\tag{4.5.4} u ( x , 0 ) = \frac 1 \pi \frac { f ( x ) } { 1 + t ^ 2 } \, d t = f ( x )\] so $ u $ satisfies the boundary condition. The only remaining thing to check is that $ u $ is twice continuously differentiable in the open upper half-plane and solves the Laplace equation there.
The equation which defines $ u $ was convenient for everything we've done so far but it's ill suited to proving properties of derivatives of $ u $ because if we formally differentiate under the integral sign we will get derivatives of $ f $ appearing and we have no reason to believe those derivatives exist. Instead we need to make a change of variable in the integral. Set \[\tag{4.5.5} z = x + t y .\] Then \[\tag{4.5.6} u ( x , y ) = \frac 1 \pi \int _ { - \infty } ^ { + \infty } \frac { y f ( z ) } { ( x - z ) ^ 2 + y ^ 2 } \, d z .\] Note that this change of variable is only valid for $ y > 0 $ so we can't expect the equation above to hold for $ y = 0 $ and indeed it doesn't. If it did we would have \[\tag{4.5.7} u ( x , 0 ) = 0\] but we've already seen that \[\tag{4.5.8} u ( x , 0 ) = f ( x ) .\]
We can see by induction that the derivatives of the integrand are \[\tag{4.5.9} \begin{split} & \frac { \partial ^ { j + k } } { \partial x ^ j \partial y ^ k } \frac { y f ( z ) } { ( x - z ) ^ 2 + y ^ 2 } \\ & \quad = \sum ( - 1 ) ^ { j + \frac { k + l - 1 } 2 } \frac { ( j + k ) ! ( j + k + 1 ) ! } { ( j + k - l + 1 ) ! l ! } \frac { ( x - z ) ^ { j + k - 1 + 1 } y ^ l f ( z ) } { ( ( x - z ) ^ 2 + y ^ 2 ) ^ { j + k + 1 } } , \end{split}\] where the sum is over all $ l $ between $ 0 $ and $ j + k + 1 $ such that $ k + l $ is odd, so that the factorials and the power of $ - 1 $ are well defined. The precise details aren't important but each summand is bounded by a constant times \[\tag{4.5.10} \frac { \sup | f | } { ( ( x - z ) ^ 2 + y ^ 2 ) ^ { \frac { j + k + 1 } 2 } } ,\] which is integrable as a function of $ z $ uniformly for $ ( x , y ) $ in a any closed rectangle in the open upper half-plane. So differentiation under the integral sign is justfied. It follows that $ u $ is not just twice continuously differentiable in the open upper half-plane but in fact infinitely differentiable. Also we get \[\tag{4.5.11} \frac { \partial ^ 2 u } { \partial x ^ 2 } = \int _ { - \infty } ^ { + \infty } \frac { ( 6 ( x - z ) ^ 2 y - 2 y ^ 3 ) f ( z ) } { ( ( x - z ) ^ 2 + y ^ 2 ) ^ 3 } \, d z\] and \[\tag{4.5.12} \frac { \partial ^ 2 u } { \partial y ^ 2 } = \int _ { - \infty } ^ { + \infty } \frac { ( - 6 ( x - z ) ^ 2 y + 2 y ^ 3 ) f ( z ) } { ( ( x - z ) ^ 2 + y ^ 2 ) ^ 3 } \, d z\] so $ u $ satisfies the Laplace equation there.
The equation \[\tag{4.5.13} u ( x , y ) = \frac 1 \pi \int _ { - \infty } ^ { + \infty } \frac { y f ( z ) } { ( x - z ) ^ 2 + y ^ 2 } \, d z ,\] which was used to prove the differentiability properties of $ u $ and to prove that it satisfies the hypotheses of the Laplace equation, has a name. It is called the Poisson formula, or more precisely the Poisson formula for the upper half-plane, to distinguish it from the Poisson formula for the unit disc, which we will meet later. The equation \[\tag{4.5.14} u ( x , y ) = \frac 1 \pi \int _ { - \infty } ^ { + \infty } \frac { f ( x + t y ) } { 1 + t ^ 2 } \, d t ,\] which was used to show the continuity of $ u $ on the closed upper half-plane and to show that it satisfies the boundary conditions, does not have a name. This is somewhat unfortunate since the second of these two equations is more fundamental than the first, both because it holds in the full domain of definition of $ u $ rather than just at those points where $ y > 0 $ and, relatedly, because it can be used to derive the other equation, but not vice versa. We can get from either equation to the other via a change of variable, but this change of variable is only valid for $ y > 0 $ and even if it were valid when $ y = 0 $ we couldn't use it to justify deriving the unnamed equation from the Poisson formula there because the Poisson formula isn't correct there. By contrast, we can use the change of variable to derive Poisson formula for $ y > 0 $, which is the only region in which it is correct.
We encountered something similar with the solution formula for the diffusion equation, but there at least the solution formula had a $ \sqrt t $ in the denominator to warn us not to apply it at $ t = 0 $. The denominator in the Poisson formula is well defined on the line $ y = 0 $, except at the single point $ z = x $ and it's a well known property of integration that changing the integrand at a single point doesn't affect the integrand, so there's nothing in the form of the equation to warn us not to apply it.
As a final observation, the continuity of $ y $ at the boundary together with the validity of Poisson formula for $ y > 0 $ means it must be true that \[\tag{4.5.15} \lim _ { y \to 0 ^ + } \frac 1 \pi \int _ { - \infty } ^ { + \infty } \frac { y f ( z ) } { ( x - z ) ^ 2 + y ^ 2 } \, d z = f ( x ) .\] This is notable because blindly taking the limit inside the integral would give the wrong answer. It does not, of course, satisfy the hypotheses of any of our theorems for exchanging limits and integrals so there is no contradiction here, just a warning. It's important to check the hypotheses of theorems not just because that's what mathematicians do but but also because not doing so will lead to wrong answers in real examples.
We can check, using our criterion from earlier, that \[\tag{4.6.1} \tilde u ( x , y ) = u \left ( \frac { 2 y } { ( x + 1 ) ^ 2 + y ^ 2 } , \frac { 1 - x ^ 2 - y ^ 2 } { ( x + 1 ) ^ 2 + y ^ 2 } \right )\] is a symmetry of the Laplace equation. This is equivalent to \[\tag{4.6.2} u ( x , y ) = \tilde u \left ( \frac { 1 - x ^ 2 - y ^ 2 } { x ^ 2 + ( y + 1 ) ^ 2 } , \frac { 2 x } { x ^ 2 + ( y + 1 ) ^ 2 } \right ) .\] If $ u $ satisfies the Laplace equation in the open upper half-plane then $ \tilde u $ satisfies the Laplace equation in the open unit disc and vice versa. If $ u $ is bounded and continuous on the closed upper half-plane then $ \tilde u $ is bounded and continuous on the closed unit disc with the point $ ( - 1 , 0 ) $ removed. Given a function $ \tilde f $ on the unit circle we can define a corresponding function $ f $ on the real line by \[\tag{4.6.3} f ( x , y ) = \tilde f \left ( \frac { 1 - x ^ 2 - y ^ 2 } { x ^ 2 + ( y + 1 ) ^ 2 } , \frac { 2 x } { x ^ 2 + ( y + 1 ) ^ 2 } \right ) .\] Of course $ y = 0 $ at all these points and as discussed earlier it's usual to identify the $ x $ axis with the real line so we can write this more simply as \[\tag{4.6.4} f ( x ) = \tilde f \left ( \frac { 1 - x ^ 2 } { 1 + x ^ 2 } , \frac { 2 x } { 1 + x ^ 2 } \right ) .\] If $ \tilde f $ was bounded and continuous then so is $ f $. The converse to this statement isn't true because $ \tilde f $ can be discontinuous at $ ( - 1 , 0 ) $ even if $ f $ was continuous everywhere, but we won't need the converse. In any case, from a bounded continuous $ f $ we can use the Poisson formula to get a bounded continuous $ u $ on the closed upper half-plane satisfying the Laplace equation in the open upper half-plane: \[\tag{4.6.5} u ( x , y ) = \frac 1 \pi \int _ { - \infty } ^ { + \infty } \frac { f ( x + t y ) } { 1 + t ^ 2 } \, d t .\] This $ u $ will satisfy the boundary condition \[\tag{4.6.6} u ( x , 0 ) = f ( x ) .\] Strictly speaking what we are using here is not the Poisson formula but the other, unnamed, formula which is related to it, as discussed earlier, to get an expression for $ u $ which is correct in the closed half-plane, not just the open half-plane. Using this $ u $ we define $ \tilde u $ as above, obtaining a function $ \tilde u $ which is continuous on the open unit disc, continuous and bounded on the closed unit disc minus the point $ ( - 1 , 0 ) $, and whose restriction to the unit circle minus that point is equal to $ \tilde f $.
We now have equations relating $ \tilde u $ to $ u $, $ u $ to $ f $, and $ f $ to $ \tilde f $, which we can combine to get an equation relating $ \tilde u $ to $ \tilde f $. \[\tag{4.6.7} \begin{split} \tilde u ( x , y ) & = u \left ( \frac { 2 y } { ( x + 1 ) ^ 2 + y ^ 2 } , \frac { 1 - x ^ 2 - y ^ 2 } { ( x + 1 ) ^ 2 + y ^ 2 } \right ) \\ & = \frac 1 \pi \int _ { - \infty } ^ { + \infty } \frac { f \left ( \frac { 2 y + t - t x ^ 2 - t y ^ 2 } { ( x + 1 ) ^ 2 + y ^ 2 } \right ) } { 1 + t ^ 2 } \, d t \\ & = \frac 1 \pi \pi \int _ { - \infty } ^ { + \infty } \frac { \tilde f ( \xi ( t , x , y ) , \eta ( t , x , y ) ) } { 1 + t ^ 2 } \, d t , \end{split}\] where \[\tag{4.6.8} \begin{split} \xi ( t , x , y ) & = \frac { ( 2 y + t - t x ^ 2 - t y ^ 2 ) ^ 2 - ( ( x + 1 ) ^ 2 + y ^ 2 ) ^ 2 } { ( 2 y + t - t x ^ 2 - t y ^ 2 ) ^ 2 } , \\ \eta ( t , x , y ) & = \frac { 2 ( 2 y + t - t x ^ 2 - t y ^ 2 ) ^ 2 ( ( x + 1 ) ^ 2 + y ^ 2 ) ^ 2 } { ( 2 y + t - t x ^ 2 - t y ^ 2 ) ^ 2 } . \end{split}\]
This is ugly, but mainly because of an unsuitable choice of variable of integration. We have \[\tag{4.6.9} \xi ( t , x , y ) ^ 2 + \eta ( t , x , y ) ^ 2 = 1\] because our symmetry takes the upper-half plane to the unit disc. We can therefore write \[\tag{4.6.10} \xi ( t , x , y ) = \cos \theta , \quad \eta ( t , x , y ) = \sin \theta \] for some $ \theta $. We use this $ \theta $ as our new variable of integration, which introduces a Jacobian factor, which we can calculate as follows. \[\tag{4.6.11} \frac { d \theta } { d t } = 2 \frac { ( x - \cos \theta ) ^ 2 + ( y - \sin \theta ) ^ 2 } { ( 1 + t ^ 2 ) ( 1 - x ^ 2 - y ^ 2 ) }\] so \[\tag{4.6.12} \frac { d t } { 1 + t ^ 2 } = \frac 1 2 \frac { 1 - x ^ 2 - y ^ 2 } { ( x - \cos \theta ) ^ 2 + ( y - \sin \theta ^ 2 ) } \, d \theta\] and \[\tag{4.6.13} \tilde u ( x , y ) = \frac 1 { 2 \pi } \int _ { - \pi } ^ { + \pi } \frac { ( 1 - x ^ 2 - y ^ 2 ) \tilde f ( \cos \theta , \sin \theta ) } { ( x - \cos \theta ) ^ 2 + ( y - \sin \theta ^ 2 ) } \, d \theta .\]As usual, we have to be careful about where our change of variable is valid. The factor of $ 1 - x ^ 2 - y ^ 2 $ in the denominator alerts us to the fact that this change of variable is not permitted if $ ( x , y ) $ lies on the unit circle, but it does work everywhere in the open unit disc. We can still evaluate $ \tilde u $ on the unit circle though, since we already have the equation \[\tag{4.6.14} \tilde u ( x , y ) = \tilde f ( x , y )\] there, except possibly at the point $ ( - 1 , 0 ) $.
What we now have is a bounded continuous function $ \tilde u $ on the closed unit disc minus the point $ ( - 1 , 0 ) $ which is equal to $ \tilde f $ on the unit circle, except possibly at the same point, and which satisfies the Laplace equation in the open unit disc. The point $ ( - 1 , 0 ) $ is special only because of the particular choice of symmetry though. Since we're mapping the disc to the upper half-plane, and hence the unit circle to the $ x $-axis, we will always miss some point, but we can choose which point that will be. We could, for example, have chosen \[\tag{4.6.15} \tilde u ( x , y ) = u \left ( \frac { 2 y } { ( x - 1 ) ^ 2 + y ^ 2 } , \frac { 1 - x ^ 2 - y ^ 2 } { ( x - 1 ) ^ 2 + y ^ 2 } \right )\] instead, which would have given us the exact same equation for $ \tilde u $, but now with the point $ ( 1 , 0 ) $ being the potential point of discontinuity. But that means that our $ \tilde u $ does in fact extend continuously to the whole closed unit disc.
We have now proved the following existence theorem.
As with the Poisson formula for the upper half-plane, none of these formulae should be used on the boundary, where they would give $ u ( x , y ) = 0 $ if applied. For the upper half-plane we had an alternative equation which could be applied everywhere. Do we have one here? Almost. The ugly $ t $ integral which we rejected earlier is valid except at the point $ ( - 1 , 0 ) $ and we can find a similar equation valid except at an other choice of boundary point. More interestingly, there is an alternate formula valid at all points except for $ ( 0 , 0 ) $, \[\tag{4.6.20} u ( x , y ) = \frac 1 { 2 \pi } \int f ( p ( \psi ) , q ( \psi ) ) \, d \psi ,\] where $ ( p ( \psi ) , q ( \psi ) ) $ is the unique point on the unit circle with a circle passing through the point $ ( x , y ) $, making an angle $ \psi $ with the ray from $ ( 0 , 0 ) $, and perpendicular to the unit circle. There is such a point for each angle $ \psi $ and each $ ( x , y ) $ in the open unit disc except for $ ( 0 , 0 ) $. It's possible to write down the function $ p $ and $ q $ explicitly, which unfortunately spoils the simplicity of the equation above, and these equations continue to make sense even for $ ( x , y ) $ on the unit circle and the equation for $ u $ above will continue to hold there, unlike the Poisson formula.
The idea behind the derivation of the Poisson formula for the unit disc was that once we know how to solve the Dirichlet problem in a region we can solve the Dirichlet problem in the image of that region under any symmetry. Now that we know how to solve the Dirichlet problem in the unit disc we can solve it in any other disc because any two discs are related by a translation and scaling. We can also solve the Dirichlet problem in any half-plane because any half-plane is related to the upper half-plane by a translation and rotation.
Everything we have done so far in this chapter can be generalised to the Laplace equation in higher dimensions. In almost all cases the generalisation is a simple matter of adding more independent variables the the equations. There are a few places where the generalisation is a bit less straightforward. The non-constant radial solution in higher dimensions involves powers of the radial coordinate rather than logarithms, for example, and the action of the the higher dimensional Lorentz group is slightly more complicated, since we need to multiply the function $ u $ by a certain factor in addition to transforming its arguments. Also, averaging over rotations in dimensions greater than two is harder to describe explicitly. Even though in these cases the generalisation to higher dimensions requires more work, it does exist. What we'll discuss in this brief section though is specific to two dimensions and does not have a good analogue in higher dimensions.
In multivariable calculus one learns that a continuously differentiable vector field $ \mathbf w = ( w _ 1 , \ldots , w _ n ) $ on a simply connected domain is the gradient of a twice continuously differentiable function $ v $, i.e. $ \mathbf w = \mathbf \nabla v $, if and only if it satisfies the equations \[\tag{4.7.1} \frac { \partial w _ j } { \partial w _ k } = \frac { \partial w _ k } { \partial w _ j }\] for all $ 1 \le j < k \le n $. The fact that this is necessary, even without the simple connectedness assumption, follows from equality of mixed second partial derivatives. In particular, in two dimensions we have that \[\tag{4.7.2} w _ 1 = \frac { \partial v } { \partial x } , \quad w _ 2 = \frac { \partial v } { \partial y }\] if and only if \[\tag{4.7.3} \frac { \partial w _ 1 } { \partial y } = \frac { \partial w _ 2 } { \partial x } .\] Here, as usual, we've written $ x $ for $ x _ 1 $ and $ y $ for $ x _ 2 $. Now if $ u $ is a solution of the Laplace equation in a simply connected region and $ w _ 1 $ and $ w _ 2 $ are defined by \[\tag{4.7.4} w _ 1 = - \frac { \partial u } { \partial y } , \quad w _ 2 = \frac { \partial u } { \partial x }\] then \[\tag{4.7.5} \frac { \partial w _ 1 } { \partial y } = \frac { \partial w _ 2 } { \partial x }\] so there must be a twice continuously differentiable $ v $ such that \[\tag{4.7.6} w _ 1 = \frac { \partial v } { \partial x } , \quad w _ 2 = \frac { \partial v } { \partial y } ,\] i.e. such that \[\tag{4.7.7} \frac { \partial u } { \partial x } - \frac { \partial v } { \partial y } = 0 , \quad \frac { \partial u } { \partial y } + \frac { \partial v } { \partial x } = 0 .\] These are the Cauchy-Riemann equations, which we met once before as a sufficient condition for a transformation of the plane to give rise to a symmetry of the Laplace equation. Taking derivatives we get \[\tag{4.7.8} \frac { \partial ^ 2 u } { \partial y \partial x } - \frac { \partial ^ 2 v } { \partial y ^ 2 } = 0 , \quad \frac { \partial ^ 2 u } { \partial x \partial y } + \frac { \partial ^ 2 v } { \partial x ^ 2 } = 0\] so subtracting the first equation from the second and using the equality of mixed partial derivatives we see that $ v $ also satisfies the Laplace equation \[\tag{4.7.9} \frac { \partial ^ 2 v } { \partial x ^ 2 } + \frac { \partial ^ 2 v } { \partial y ^ 2 } = 0 .\] This $ v $ is usually called the harmonic conjugate of $ u $, although it would make more sense to call it a harmonic conjugate rather than the harmonic conjugate, since adding any constant to $ v $ would also give a function satisfying the same conditions. This is similar to the situation with indefinite integrals, where people usually say the indefinite integral even though it would be more correct to say an indefinite integral.
Suppose that $ u $ is the solution to the Dirichlet problem in the upper half plane with boundary data $ f $, so that \[\tag{4.7.10} u ( x , y ) = \frac 1 \pi \int _ { \infty } ^ { + \infty } \frac { y f ( z ) } { ( x - z ) ^ 2 + y ^ 2 } \, d z\] for $ ( x , y ) $ in the open upper half-plane. We've already seen that differentiation under the integral sign is justified in the upper half-plane, so we have \[\tag{4.7.11} \frac { \partial u } { \partial x } ( x , y ) = - \frac 1 \pi \int _ { - \infty } ^ { + \infty } \frac { 2 ( x - z ) y f ( z ) } { \left [ ( x - z ) ^ 2 + y ^ 2 \right ] ^ 2 } \, d z\] and \[\tag{4.7.12} \frac { \partial u } { \partial y } ( x , y ) = \frac 1 \pi \int _ { - \infty } ^ { + \infty } \frac { \left [ ( x - z ) ^ 2 - y ^ 2 \right ] f ( z ) } { \left [ ( x - z ) ^ 2 + y ^ 2 \right ] ^ 2 } \, d z\] In this case we can write down a harmonic conjugate explicitly: \[\tag{4.7.13} v ( x , y ) = \frac 1 \pi \int _ { - \pi } ^ \pi \frac { ( x - z ) f ( z ) } { ( x - z ) ^ 2 + y ^ 2 } \, d z .\] Differentiation under the integral sign is again justified and gives \[\tag{4.7.14} \frac { \partial v } { \partial x } ( x , y ) = - \frac 1 \pi \int _ { - \infty } ^ { + \infty } \frac { \left [ ( x - z ) ^ 2 - y ^ 2 \right ] f ( z ) } { \left [ ( x - z ) ^ 2 + y ^ 2 \right ] ^ 2 } \, d z\] and \[\tag{4.7.15} \frac { \partial v } { \partial y } ( x , y ) = - \frac 1 \pi \int _ { - \infty } ^ { + \infty } \frac { 2 ( x - z ) y f ( z ) } { \left [ ( x - z ) ^ 2 + y ^ 2 \right ] ^ 2 } \, d z\] so $ \partial u / \partial x - \partial v / \partial y = 0 $ and $ \partial u \partial y + \partial v / \partial x = 0 $, as required.
If we make the same change of variable $ t = ( z - x ) / y $ that gave us the alternate formula \[\tag{4.7.16} u ( x , y ) = \frac 1 \pi \int _ { - \infty } ^ { + \infty } \frac { f ( x + t y ) } { 1 + t ^ 2 } \, d t\] then we get \[\tag{4.7.17} v ( x , y ) = \frac 1 \pi \int _ { - \infty } ^ { + \infty } \frac { t f ( x + t y ) } { 1 + t ^ 2 } \, d t .\] Here we can't simply take limits as $ y $ goes to $ 0 $ because the function $ t / ( 1 + t ^ 2 ) $, unlike $ 1 / ( 1 + t ^ 2 ) $, is not absolutely integrable. Indeed the limit \[\tag{4.7.18} g ( x ) = \lim _ { y \to 0 ^ + } \frac 1 \pi \int _ { - \infty } ^ { + \infty } \frac { t f ( x + t y ) } { 1 + t ^ 2 } \, d t\] need not exist, even if $ f $ is bounded and continuous. If it does exist it is called the Hilbert transform of $ f $.
A classical solution of the Laplace equation is one which is twice continuously differentiable. Our solutions of the Dirichlet problem in the upper half-plane are classical solutions in the open upper half-plane but not necessarily in the closed upper half-plane. Now the restriction of a twice continuously differentiable solution is twice continuously differentiable so there's no hope of finding a solution with boundary data $ f $ unless $ f $ is twice continuously differentiable. We might hope that this necessary condition is also sufficient, but unfortunately this is not true, and we can now see why.
Suppose $ u $ is a bounded classical solution of the Laplace equation in the closed upper half-plane with boundary data $ f $, i.e. $ u ( x , 0 ) = f ( x ) $. As discussed above, $ f $ must be twice continuously differentiable. We will assume the $ f '' $ is not just continuous but bounded. It's possible to dispense with this assumption, but it would make the calculations which follow much trickier. Under this additional assumption differentiation under the integral in \[\tag{4.8.1} u ( x , y ) = \frac 1 \pi \int _ { - \infty } ^ { + \infty } \frac { f ( x + t y ) } { 1 + t ^ 2 } \, d t\] is justified and gives \[\tag{4.8.2} \frac { \partial ^ 2 u } { \partial x \partial y } ( x , y ) = \frac 1 \pi \int _ { - \infty } ^ { + \infty } \frac { t f '' ( x + t y ) } { 1 + t ^ 2 } \, d t .\] Since we've assumed $ u $ is a classical solution in the closed upper half-plane the limit as $ y $ goes to $ 0 $ must exist and be continuous. In other words, the Hilbert transform of $ f '' $ must exist and be continuous. So now we have an additional necessary condition for the existence of classical solutions to the Dirichlet problem: In addition to requiring $ f '' $ to be continuous we need it to have a continuous Hilbert transform. We found this condition under the additional assumption that $ f '' $ is bounded but, as mentioned earlier, that condition is just used to simplify the proof and is not actually needed.
Are our necessary conditions sufficient? In other words, given an $ f $ which is twice continuously differentiable and bounded and whose second derivative has a continuous Hilbert transform, is there a bounded classical solution of the Laplace equation in the closed upper half-plane with $ f $ as boundary values? The answer is yes, although we won't prove it.
Suppose $ G $ is a non-singular symmetric matrix. We will choose a particular $ G $ soon, but some of what follows doesn't depend on the particular choice of $ G $ and this will be more apparent if we delay choosing one. We say that a matrix $ A $ is orthogonal with respect to $ G $ if \[\tag{4.9.1} A ^ T G A = G .\] If $ G = I $ then this is the usual notion of orthogonality, but in general it is different. The identity matrix is clearly orthogonal with respect to $ G $ for any $ G $. Also, if $ A $ and $ B $ are orthogonal matrices with respect to $ G $ then \[\tag{4.9.2} ( A B ) ^ T G ( A B ) = B ^ T A ^ T G A B = B ^ T ( A ^ T G A ) B = B ^ T G B = G\] so $ A B $ is also an orthogonal matrix with respect to $ G $. Also, \[\tag{4.9.3} ( G ^ { - 1 } A ^ T G ) A = G ^ { - 1 } ( A ^ T G A ) = G ^ { - 1 } G = I ,\] so $ G ^ { - 1 } A ^ T G $ is a left inverse to $ A $. Now any left inverse to a square matrix is also a right inverse and so $ A $ is invertible, with \[\tag{4.9.4} A ^ { - 1 } = G ^ { - 1 } A ^ T G .\] Another way to see that $ A $ is invertible is to take determinants in the equation $ A ^ T G A = G $ and use the fact that determinants are multiplicative and the fact that the determinant of a transpose is the same as the determinant of the original matrix. This gives $ \det ( A ) ^ 2 \det ( G ) = \det ( G ) $ or, since $ G $ is non-singular and hence has non-zero determinant, \[\tag{4.9.5} \det ( A ) ^ 2 = 1 ,\] so $ \det ( A ) = 1 $ or $ \det ( A ) = - 1 $. Now that we know $ A ^ { - 1 } $ exists we can calculate \[\tag{4.9.6} A ^ { - T } G A ^ { - 1 } = A ^ { - T } ( A ^ T G A ) A ^ { - 1 } = \left ( A ^ { - T } A ^ T \right ) G \left ( A A ^ { - 1 } \right ) = I G I = G .\] So $ A ^ { - 1 } $ is an orthogonal matrix with respect to $ G $. In other words, the orthogonal matrices with respect to $ G $ form a group. Those with determinant $ 1 $ form a subgroup, called the special orthogonal matrices with respect to $ G $.
I assumed $ G $ was symmetric, but if you look closely you'll see that I never actually used that assumption. Unless you're working over a field with characteristic $ 2 $ though there's nothing to be gained by allowing $ G $ not to be symmetric since $ A $ will be orthogonal with respect to $ G $ if and only if its orthogonal with respect to the symmetrised $ \frac 1 2 G + \frac 1 2 G ^ T $. In our case the field is the field of real numbers, which is of characteristic $ 0 $, and from now on we will work with a particular choice of $ G $, $G = \left [ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \end{matrix} \right ] ,$ which is clearly symmetric. One consequence of symmetry is that if we define \[\tag{4.9.7} \langle \mathbf v , \mathbf w \rangle = \mathbf v ^ T G \mathbf w\] then \[\tag{4.9.8} \langle \mathbf v , \mathbf w \rangle = \mathbf v ^ T G \mathbf w = ( \mathbf v ^ T G \mathbf w ) ^ T = \mathbf w ^ T G ^ T \mathbf v = \mathbf w ^ T G \mathbf v = \langle \mathbf w , \mathbf v \rangle .\] The second equation holds because $ 1 \times 1 $ matrices are always symmetric.
Some examples of matrices orthogonal with respect to this $ G $ are \[\tag{4.9.9} \left [ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & \cos \theta & - \sin \theta & 0 \\ 0 & \sin \theta & \cos \theta & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right ] , \quad \left [ \begin{matrix} \cosh \log \mu & 0 & 0 & - \sinh \log \mu \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ - \sinh \log \mu & 0 & 0 & \cosh \log \mu \end{matrix} \right ] , \quad \left [ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & -1 \end{matrix} \right ] ,\] \[\tag{4.9.10} \left [ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \end{matrix} \right ] , \quad \left [ \begin{matrix} 1 + ( \xi ^ 2 + \eta ^ 2 ) / 2 & \xi & \eta & ( \xi ^ 2 + \eta ^ 2 ) / 2 \\ \xi & \xi & 0 & \xi \\ \eta & 0 & \eta & \eta \\ - ( \xi ^ 2 + \eta ^ 2 ) / 2 & - \xi & - \eta & 1 - ( \xi ^ 2 + \eta ^ 2 ) / 2 \end{matrix} \right ] .\]
One further special property of the $ G $ we have chosen is that $ G ^ 2 = I $, so if $ A $ is orthogonal with respect to $ G $ then \[\tag{4.9.11} \begin{split} A G A ^ T & = A G ^ { - 1 } A ^ T = A \left ( A ^ T G A \right ) ^ { - 1 } A ^ T = A \left ( A ^ { - 1 } G ^ { - 1 } A ^ { - T } \right ) A ^ T \\ & = \left ( A A ^ { - 1 } \right ) G ^ { - 1 } \left ( A ^ { - T } A ^ T \right ) = I G ^ { - 1 } I = G ^ { - 1 } = G , \end{split}\] so $ A ^ T $ is also orthogonal with respect to $ G $. This would not have been true for general choices of $ G $.
For future reference we note that the equation $ A ^ { - 1 } = G ^ { - 1 } A ^ T G $ in this case is \[\tag{4.9.12} \left [ \begin{matrix} a _ { 0 0 } & a _ { 0 1 } & a _ { 0 2 } & a _ { 0 3 } \\ a _ { 1 0 } & a _ { 1 1 } & a _ { 1 2 } & a _ { 1 3 } \\ a _ { 2 0 } & a _ { 2 1 } & a _ { 2 2 } & a _ { 2 3 } \\ a _ { 3 0 } & a _ { 3 1 } & a _ { 3 2 } & a _ { 3 3 } \end{matrix} \right ] ^ { - 1 } = \left [ \begin{matrix} a _ { 0 0 } & - a _ { 1 0 } & - a _ { 2 0 } & - a _ { 3 0 } \\ - a _ { 0 1 } & a _ { 1 1 } & a _ { 2 1 } & a _ { 3 1 } \\ - a _ { 0 2 } & a _ { 1 2 } & a _ { 2 2 } & a _ { 3 2 } \\ - a _ { 0 3 } & a _ { 1 3 } & a _ { 2 3 } & a _ { 3 3 } \end{matrix} \right ] .\] Note that we number indices from $ 0 $ rather than following the usual linear algebra convention of numbering them from $ 1 $. We do this because the group of matrices orthogonal with respect to this particular $ G $, known as the Lorentz group, is particularly important in special relativity, where the modern convention is to number from $ 0 $. For our purposes, which are to describe symmetries of the Laplace equation, the choice doesn't matter but some choice has to be made.
The equation $ A ^ T G A = G $ is an equation for $ 4 \times 4 $ matrices and so gives an equation for each entry. The one for the upper left entry is \[\tag{4.9.13} a _ { 0 0 } ^ 2 - a _ { 0 1 } ^ 2 - a _ { 0 2 } ^ 2 - a _ { 0 3 } ^ 2 = 1 ,\] from which it follows that $ a _ { 0 0 } ^ 2 \ge 1 $ and hence $ a _ { 0 0 } \ge 1 $ or $ a _ { 0 0 } \le - 1 $. In the first case the matrix is said to be orthochronous. Note that since the transpose of a Lorentz matrix is a Lorentz matrix we also have the equation \[\tag{4.9.14} a _ { 0 0 } ^ 2 - a _ { 1 0 } ^ 2 - a _ { 2 0 } ^ 2 - a _ { 3 0 } ^ 2 = 1 .\] Suppose that $ A $ and $ B $ are isochronous Lorentz matrices. Then the entry in the upper left corner of the product $ A B $ is \[\tag{4.9.15} a _ { 0 0 } b _ { 0 0 } - a _ { 0 1 } b _ { 1 0 } - a _ { 0 2 } b _ { 2 0 } - a _ { 0 3 } b _ { 3 0 }\] and, by Cauchy-Schwarz, \[\tag{4.9.16} \begin{split} a _ { 0 1 } b _ { 1 0 } + a _ { 0 2 } b _ { 2 0 } + a _ { 0 3 } b _ { 3 0 } & \le \sqrt { a _ { 0 1 } ^ 2 + a _ { 0 2 } ^ 2 + a _ { 0 3 } ^ 2 } \sqrt { b _ { 1 0 } ^ 2 + b _ { 2 0 } ^ 2 + b _ { 3 0 } ^ 2 } \\ & = \sqrt { a _ { 0 0 } ^ 2 - 1 } \sqrt { b _ { 0 0 } ^ 2 - 1 } \end{split}\] and therefore \[\tag{4.9.17} a _ { 0 0 } b _ { 0 0 } - a _ { 0 1 } b _ { 1 0 } - a _ { 0 2 } b _ { 2 0 } - a _ { 0 3 } b _ { 3 0 } \ge a _ { 0 0 } b _ { 0 0 } - \sqrt { a _ { 0 0 } ^ 2 - 1 } \sqrt { b _ { 0 0 } ^ 2 - 1 } > 0 .\] It follows that $ A B $ is orthochronous. It's also clear from our equation for the inverse matrix that $ A ^ { - 1 } $ is orthochronous whenever $ A $ is, so the orthochronous Lorentz matrices form a subgroup of the Lorentz matrices and the orthochronous special Lorentz matrices form a subgroup of the special Lorentz matrices, which is known as the restricted Lorentz group.
We describe non-zero vectors $ \mathbf w $ in $ \mathbf R ^ 4 $ as time-like, light-like or space-like according to whether $ \langle \mathbf w , \mathbf w \rangle $ is positive, zero, or negative. For the time-like and light-like vectors we have \[\tag{4.9.18} w _ 0 ^ 2 \ge w _ 1 ^ 2 + w _ 2 ^ 2 + w _ 3 ^ 2 \ge 0 .\] Now if $ \mathbf w $ is time-like the first inequality is strict while if it is lightlike the first inequality is an equation, but in that case we can use the fact that $ \mathbf w $ is non-zero to see that that second inequality is strict. Here it's important that we only applied the classification of vectors into time-like, light-like and space-like to non-zero vectors. In any case, for a time-like or light-like vector we have $ w _ 0 ^ 2 $ and hence $ w _ 0 > 0 $ or $ w _ 0 < 0 $. In the first case we call $ \mathbf w $ future directed while in the second case we call it past directed. Multiplication by Lorentz matrices takes time-like vectors to time-like, light-like to light-like, and space-like to space-like, as we see from the special case $ \mathbf v = \mathbf w $ of the identity \[\tag{4.9.19} \begin{split} \langle A \mathbf v , A \mathbf w \rangle & = ( A \mathbf v ) ^ T G ( A \mathbf w ) = \left ( \mathbf v ^ T A ^ T \right ) G ( A \mathbf w ) \\ & = \mathbf v ^ T \left ( A ^ T G A \right ) \mathbf w = \mathbf v ^ T G \mathbf w = \langle \mathbf v , \mathbf w \rangle . \end{split}\] Multiplication by isochronous Lorentz matrices also takes future directed time-like and light-like vectors to future directed time-like and space-like vectors and take past directed ones to past directed ones.
We call the set of positive multiples of a non-zero vector a ray. All of the vectors in a ray have the same character, time-like, light-like or space-like, and if they are time-like or space-like then they are either all future directed or past directed. We can therefore use the same terms to refer to rays as we did for non-zero vectors.
Suppose $ \mathbf v $ and $ \mathbf w $ are future directed time-like or light-like vectors, so that \[\tag{4.9.20} v _ 0 \ge \sqrt { v _ 1 ^ 2 + v _ 2 ^ 2 + v _ 3 ^ 2 } , \quad w _ 0 \ge \sqrt { w _ 1 ^ 2 + w _ 2 ^ 2 + w _ 3 ^ 2 } ,\] with inequalities which are strict in the timelike case and weak in lightlike case. By the Cauchy-Schwarz inequality we have \[\tag{4.9.21} v _ 1 w _ 1 + v _ 2 w _ 2 + v _ 3 w _ 3 \le \sqrt { v _ 1 ^ 2 + v _ 2 ^ 2 + v _ 3 ^ 2 } \sqrt { w _ 1 ^ 2 + w _ 2 ^ 2 + w _ 3 ^ 2 }\] with an inequality which is strict unless one of $ ( v _ 1 , v _ 2 , v _ 3 ) $ and $ ( w _ 1 , w _ 2 , w _ 3 ) $ is a non-negative multiple of the other. It follows that \[\tag{4.9.22} \langle \mathbf v , \mathbf w \rangle = \mathbf v ^ T G \mathbf w = v _ 0 w _ 0 - v _ 1 w _ 1 - v _ 2 w _ 2 - v _ 3 w _ 3 \ge 0\] This inequality is strict unless $ \mathbf v $ and $ \mathbf w $ are both light-like and one of $ ( v _ 1 , v _ 2 , v _ 3 ) $ and $ ( w _ 1 , w _ 2 , w _ 3 ) $ is a non-negative multiple of the other. But then $ v _ 0 = \sqrt { v _ 1 , v _ 2 , v _ 3 } $ and $ w _ 0 = \sqrt { w _ 1 , w _ 2 , w _ 3 } $ are related in the same proportion and hence $ \mathbf v $ and $ \mathbf w $ belong to the same light-like ray.
Consider three distinct future directed light-like rays, say the rays through the vectors $ \mathbf u $, $ \mathbf v $ and $ \mathbf w $. Since they're future directed and distinct the numbers $ \langle \mathbf u , \mathbf v \rangle $, $ \langle \mathbf v , \mathbf w \rangle $ and $ \langle \mathbf w , \mathbf u \rangle $ are all positive. It therefore makes sense to define \[\tag{4.9.23} \mathbf p = \mathbf u \sqrt { \frac { \langle \mathbf v , \mathbf w \rangle } { \langle \mathbf w , \mathbf u \rangle \langle \mathbf u , \mathbf v } } , \quad \mathbf q = \mathbf v \sqrt { \frac { \langle \mathbf w , \mathbf u \rangle } { \langle \mathbf u , \mathbf v \rangle \langle \mathbf v , \mathbf w } } , \quad \mathbf r = \mathbf w \sqrt { \frac { \langle \mathbf u , \mathbf v \rangle } { \langle \mathbf v , \mathbf w \rangle \langle \mathbf w , \mathbf u } } .\] Note that I've written scalar multiplication with the scalar on the right rather than the usual convention of writing it on the left. Unlike the usual convention, this one is consistent with identifying scalars with $ 1 \times 1 $ matrices and it has the advantage of making a number of facts which would otherwise require special consideration, like the fact that scalar multiplication commutes with multiplication by a matrix, into instances of the associative law for matrix multiplication, so I will write scalar multiples from the right from now on. In any case, we have \[\tag{4.9.24} \langle \mathbf p , \mathbf q \rangle = \left \langle \mathbf u \sqrt { \frac { \langle \mathbf v , \mathbf w \rangle } { \langle \mathbf w , \mathbf u \rangle \langle \mathbf u , \mathbf v } } , \mathbf v \sqrt { \frac { \langle \mathbf w , \mathbf u \rangle } { \langle \mathbf u , \mathbf v \rangle \langle \mathbf v , \mathbf w } } \right \rangle = \frac { \langle \mathbf u , \mathbf v \rangle } { \langle \mathbf u , \mathbf v \rangle } = 1\] and, similarly, \[\tag{4.9.25} \langle \mathbf q , \mathbf r \rangle = \langle \mathbf r , \mathbf p \rangle = 1 .\] The vectors $ \mathbf p $, $ \mathbf q $ and $ \mathbf r $ are all light-like so we also have \[\tag{4.9.26} \langle \mathbf p , \mathbf p \rangle = \langle \mathbf q , \mathbf q \rangle = \langle \mathbf r , \mathbf r \rangle = 0 .\] The rays through $ \mathbf u $, $ \mathbf v $ and $ \mathbf w $ are the same as the rays through $ \mathbf p $, $ \mathbf q $ and $ \mathbf r $, but it's more convenient to use the second of these representations. Suppose \[\tag{4.9.27} \mathbf s = \mathbf p + \mathbf q + \mathbf r , \quad \mathbf t = \mathbf p a + \mathbf q b + \mathbf r c\] for some scalars $ a $, $ b $ and $ c $. Then we can compute \[\tag{4.9.28} \langle \mathbf s , \mathbf t \rangle = 2 , \quad \langle \mathbf s , \mathbf t \rangle = 2 a + 2 b + 2 c , \quad \langle \mathbf t , \mathbf t \rangle = 2 a b + 2 b c + 2 c a .\] So \[\tag{4.9.29} a ^ 2 + b ^ 2 + c ^ 2 = \frac { \langle \mathbf s , \mathbf t \rangle ^ 2 - 2 \langle \mathbf s , \mathbf s \rangle \langle \mathbf t , \mathbf t \rangle } 4 .\] The right hand side is zero whenever $ \mathbf t $ is zero, which means the left hand side must also be zero and therefore $ a $, $ b $ and $ c $ must all be zero. In other words, $ \mathbf p $, $ \mathbf q $ and $ \mathbf r $ are linearly independent. Let $ \mathbf n $ be vector linearly independent from $ \mathbf p $, $ \mathbf q $ and $ \mathbf r $ and set \[\tag{4.9.30} \mathbf o = \mathbf n 2 - \mathbf p \langle \mathbf p , \mathbf n \rangle - \mathbf q \langle \mathbf q , \mathbf n \rangle - \mathbf r \langle \mathbf r , \mathbf n \rangle .\] Then \[\tag{4.9.31} \langle \mathbf o , \langle \mathbf p \rangle = \langle \mathbf o , \langle \mathbf q \rangle = \langle \mathbf o , \langle \mathbf r \rangle = 0 .\] Now $ \mathbf o $ is also linearly independent from $ \mathbf p $, $ \mathbf q $ and $ \mathbf r $ so the determinant of the matrix \[\tag{4.9.32} B = \left [ \begin{matrix} o _ 0 & p _ 0 & q _ 0 & r _ 0 \\ o _ 1 & p _ 1 & q _ 1 & r _ 1 \\ o _ 2 & p _ 2 & q _ 2 & r _ 2 \\ o _ 3 & p _ 3 & q _ 3 & r _ 3 \end{matrix} \right ]\] is non-zero and we can, by multiplying $ \mathbf o $ by an appropriate non-zero factor, arrange for it to be $ 1 $, which we will assume from now on that we have done. Because of how determinants work, if you'd listed $ \mathbf u $, $ \mathbf v $ and $ \mathbf w $ in a different order then $ \mathbf p $, $ \mathbf q $ and $ \mathbf r $ would also appear in a different order and you'd get either the same $ \mathbf o $ or $ - \mathbf o $, depending on whether the permutation is even or odd. The construction of $ \mathbf o $, $ \mathbf p $, $ \mathbf q $ and $ \mathbf r $ means that \[\tag{4.9.33} B ^ T G B = \left [ \begin{matrix} \langle \mathbf o , \mathbf o \rangle & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{matrix} \right ] .\] Taking determinants on both sides and using the fact that $ \det ( B ^ T ) = \det ( B ) = 1 $ we find that \[\tag{4.9.34} \langle \mathbf o , \mathbf o \rangle = - \frac 1 2 .\] We can summarise the results of our investigations above as follows.
Suppose now that we have two such triples, and apply the lemma to both of them, obtaining vectors $ \mathbf o ' $, $ \mathbf p ' $, $ \mathbf q ' $, and $ \mathbf r ' $ for one of them and $ \mathbf o '' $, $ \mathbf p '' $, $ \mathbf q '' $, and $ \mathbf r '' $ for the other. As before we form matrices with those vectors as columns: \[\tag{4.9.37} B ' = \left [ \begin{matrix} o ' _ 0 & p ' _ 0 & q ' _ 0 & r ' _ 0 \\ o ' _ 1 & p ' _ 1 & q ' _ 1 & r ' _ 1 \\ o ' _ 2 & p ' _ 2 & q ' _ 2 & r ' _ 2 \\ o ' _ 3 & p ' _ 3 & q ' _ 3 & r ' _ 3 \end{matrix} \right ] , \quad B '' = \left [ \begin{matrix} o '' _ 0 & p '' _ 0 & q '' _ 0 & r '' _ 0 \\ o '' _ 1 & p '' _ 1 & q '' _ 1 & r '' _ 1 \\ o '' _ 2 & p '' _ 2 & q '' _ 2 & r '' _ 2 \\ o '' _ 3 & p '' _ 3 & q '' _ 3 & r '' _ 3 \end{matrix} \right ] .\] Then \[\tag{4.9.38} { B ' } ^ T G B ' = \left [ \begin{matrix} - \frac 1 2 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{matrix} \right ] = { B '' } ^ T G B '' .\] Defining $ A $ by \[\tag{4.9.39} A = B '' { B ' } ^ { - 1 }\] we see that \[\tag{4.9.40} \begin{split} A ^ T G A & = \left ( B '' { B ' } ^ { - 1 } \right ) ^ T G \left ( B '' { B ' } ^ { - 1 } \right ) = \left ( { B ' } ^ { - T } { B '' } ^ T \right ) G \left ( B '' { B ' } ^ { - 1 } \right ) \\ & = { B ' } ^ { - T } \left ( { B '' } ^ T G B '' \right ) { B ' } ^ { - 1 } = { B ' } ^ { - T } G { B ' } ^ { - 1 } = { B ' } ^ { - T } \left ( { B ' } ^ T G B ' \right ) { B ' } ^ { - 1 } \\ & = \left ( { B ' } ^ { - T } { B ' } ^ T \right ) G \left ( B ' { B ' } ^ { - 1 } \right ) = I G I = G \end{split}\] so $ A $ is a Lorentz matrix. It is in fact a member of the special Lorentz group since both $ B ' $ and $ B '' $ have determinant $ 1 $ and therefore so does $ A $. Also, by the definition of the inverse, \[\tag{4.9.41} A \left [ \begin{matrix} o ' _ 0 & p ' _ 0 & q ' _ 0 & r ' _ 0 \\ o ' _ 1 & p ' _ 1 & q ' _ 1 & r ' _ 1 \\ o ' _ 2 & p ' _ 2 & q ' _ 2 & r ' _ 2 \\ o ' _ 3 & p ' _ 3 & q ' _ 3 & r ' _ 3 \end{matrix} \right ] = B '' { B ' } ^ { - 1 } B = B '' = \left [ \begin{matrix} o '' _ 0 & p '' _ 0 & q '' _ 0 & r '' _ 0 \\ o '' _ 1 & p '' _ 1 & q '' _ 1 & r '' _ 1 \\ o '' _ 2 & p '' _ 2 & q '' _ 2 & r '' _ 2 \\ o '' _ 3 & p '' _ 3 & q '' _ 3 & r '' _ 3 \end{matrix} \right ] .\] Left multiplication works column by column, so this is equivalent to the four vector equations \[\tag{4.9.42} A \mathbf o ' = \mathbf o '' , \quad A \mathbf p ' = \mathbf p '' , \quad A \mathbf q ' = \mathbf q '' , \quad A \mathbf r ' = \mathbf r '' .\] Since $ A $ takes future directed light-like vectors to future directed light-like vectors it is orthochronous. Since we've already seen that it has determinant $ 1 $ it follows that it is a member of the restricted Lorentz group. We therefore have the following lemma.
In the previous section Lorentz matrices always acted on $ \mathbf R ^ 4 $ but left multiplication. In this section we make them act on $ \mathbf R ^ 2 $. To start with we define a function $ i \colon \mathbf R ^ 2 \to \mathbf R ^ 4 $ and a function $ p \colon \mathbf R ^ 4 \to \mathbf R ^ 2 $. \[\tag{4.10.1} i ( x , y ) = \left ( \frac { 1 + x ^ 2 + y ^ 2 } 2 , x , y , \frac { 1 - x ^ 2 - y ^ 2 } 2 \right )\] and \[\tag{4.10.2} p ( T , X , Y , Z ) = \left ( \frac X { T + Z } , \frac Y { T + Z } \right ) .\]
This $ p $ isn't really a function defined on $ \mathbf R ^ 4 $. Instead it's defined on the complement of the hyperplane $ T + Z = 0 $ in $ \mathbf R ^ 4 $. This is a situation commonly encountered for rational functions and there are two ways of dealing with it. One is to be very careful with domains of definition. In this approach we define the domain of a composition $ f \circ g $ by saying that $ x $ belongs to its domain if it belongs to the domain of $ g $ and $ g ( x ) $ belongs to the domain of $ f $. The other approach, the usual one in algebra, is to define all operations formally. To see the difference, consider the functions $ f ( x ) = - 1 / x $ and $ g = f $. Formally, \[\tag{4.10.3} ( f \circ g ) ( x ) = f ( g ( x ) ) = - \frac 1 { g ( x ) } = - \frac 1 { - 1 / x } = x\] so we consider $ f \circ g $ to be the identity function on all of $ \mathbf R $, even though according to the other convention it would be defined only on $ \mathbf R $ minus the point $ 0 $. It's not immediately obvious that this second approach can be made logically consistent, but it can be, and it's better suited to discussing groups of rational functions, which is what we'll be doing here.
It's easy to see that \[\tag{4.10.4} p ( i ( x , y ) ) = ( x , y )\] and \[\tag{4.10.5} i ( p ( T , X , Y , Z ) ) = ( T , X , Y , Z ) \frac 1 { T + Z } - ( 1 , 0 , 0 , - 1 ) \frac { T ^ 2 - X ^ 2 - Y ^ 2 - Z ^ 2 } { 2 ( T + Z ) ^ 2 } .\] Using the notation of the previous section \[\tag{4.10.6} ( i \circ p ) ( \mathbf w ) = \mathbf w \frac 1 { \langle \mathbf w , \mathbf e \rangle } - \mathbf e \frac { \langle \mathbf w , \mathbf w \rangle } { 2 \langle \mathbf e , \mathbf w \rangle ^ 2 } ,\] where $ \mathbf e = ( 1 , 0 , 0 , - 1 ) $. Also \[\tag{4.10.7} p ( \mathbf w \lambda ) = p ( \mathbf w )\] for any positive $ \lambda $. We can therefore think of $ p $ as a function from future directed light-like rays to $ \mathbf R ^ 2 $. Strictly speaking the $ T + Z $ in the denominator makes it a function on future directed light-like rays except the ray through the point $ \mathbf e $. Rather than removing a point from the domain though it's more convenient to add one to the codomain, adding an extra point at infinity as the image of the the ray through $ \mathbf e $.
For any orthochronous Lorentz matrix $ A $ we define a function $ \varphi _ A $ by \[\tag{4.10.8} \varphi _ A = p \circ \mu _ A \circ i ,\] where $ \mu _ A $ is left multiplication by $ A $ as a function on $ \mathbf R ^ 4 $. If we consider $ \varphi $ as a function from $ \mathbf R ^ 2 $ to itself then we may have to remove a point from the domain but we can instead think of it as a function from the extended plane, with an added point at infinity, to itself. If $ A $ and $ B $ are both orthochronous Lorentz matrices then \[\tag{4.10.9} \begin{split} ( \varphi _ A \circ \varphi _ B ) ( x , y ) & = p ( A i ( p ( B i ( x , y ) ) ) ) = p \left ( A B i ( x , y ) \lambda \right ) \\ & = p \left ( A B i ( x , y ) \right ) = \varphi _ { A B } ( x , y ) , \end{split}\] where \[\tag{4.10.10} \lambda = \frac 1 { \langle B i ( x , y ) , \mathbf e \rangle } .\] We've used here the fact that $ i ( x , y ) $ is light-like and therefore so is $ B i ( x , y ) $ in order to remove the second term which would otherwise appear in $ ( i \circ p ) ( B i ( x , y ) ) $. Since the equation above holds for all $ ( x , y ) $ we have \[\tag{4.10.11} \varphi _ A \circ \varphi _ B = \varphi _ { A B } .\] In other words, the correspondence between orthochronous Lorentz matrices and functions from the extended plane to itself converts matrix multiplication into function composition. In the standard terminology of algebra we therefore have an action of the orthochronous Lorentz group on the extended plane.
What is all of this doing in a chapter on the Laplace equation? It turns out that if $ A $ is an orthochronous Lorentz transformation then \[\tag{4.10.12} ( L _ A u ) ( x , y ) = u ( \varphi _ { A ^ { - 1 } } ( x , y ) )\] Is a symmetry of the Laplace equation. The inverse in this formula is not terribly important, since the inverse of any orthochronous Lorentz matrix is an orthochronous Lorentz matrix, but it does ensure that $ L _ A ( L _ B u ) $ is equal to $ L _ { A B } u $ rather than $ L _ { B A } u $.
Returning to our five examples of Lorentz matrices from the previous section, we get the following five values for $ \varphi _ A ( x , y ) $ \[\tag{4.10.13} \left ( \cos ( \theta ) x - \sin ( \theta ) y , \sin ( \theta ) x + \cos ( \theta ) y \right ) , \quad \left ( \mu x , \mu y \right ) ,\] \[\tag{4.10.14} \left ( \frac y { x ^ 2 + y ^ 2 } , \frac x { x ^ 2 + y ^ 2 } \right ) , \quad \left ( \frac { 2 y } { ( x + 1 ) ^ 2 + y ^ 2 } , \frac { 1 - x ^ 2 - y ^ 2 } { ( x + 1 ) ^ 2 + y ^2 } \right ) , \quad \left ( x + \xi , y + \eta \right )\] and the corresponding actions on functions have $ ( L _ A u ) ( x , y ) $ as follows \[\tag{4.10.15} u \left ( \cos ( \theta ) x + \sin ( \theta ) y , - \sin ( \theta ) x + \cos ( \theta ) y \right ) , \quad u \left ( \frac x \mu , \frac y \mu \right ) ,\] \[\tag{4.10.16} u \left ( \frac y { x ^ 2 + y ^ 2 } , \frac x { x ^ 2 + y ^ 2 } \right ) , \quad u \left ( \frac { 1 - x ^ 2 - y ^ 2 } { x ^ 2 + ( y + 1 ) ^ 2 } , \frac { 2 x } { x ^ 2 + ( y + 1 ) ^ 2 } \right ) , \quad u ( x - \xi , y - \eta ) .\]