Linear, constant coefficient problems

3.1 Linear, constant coefficient problems

We consider an evolution equation on $n$ -dimensional space of the following form:

Here, $m u = u (t,x ) ∈ ℂ$ is the state vector, and $ut$ its partial derivative with respect to $t$ . Next, the $A ν$ ’s denote complex, $m × m$ matrices where $ν = (ν1,ν2,...,νn )$ denotes a multi-index with components $νj ∈ {0, 1,2,3,...}$ and $|ν| := ν1 + ...+ νn$ . Finally, $D ν$ denotes the partial derivative operator

|ν| D := ----∂-------- ν ∂xν11⋅ ⋅ ⋅ ∂xνnn

of order $|ν |$ , where $D := I 0$ . Here are a few representative examples:

Example 1. The advection equation $u (t,x ) = λu (t,x) t x$ with speed $λ ∈ ℝ$ in the negative $x$ direction.

Example 2. The heat equation $ut(t,x ) = Δu (t,x )$ , where

-∂2- -∂2- -∂2- Δ := ∂x2 + ∂x2 + ...+ ∂x2 1 2 n

denotes the Laplace operator.

Example 3. The Schrödinger equation $u (t,x) = iΔu (t,x) t$ .

Example 4. The wave equation $Utt = ΔU$ , which can be cast into the form of Eq. (3.1 ),

We can find solutions of Eq. (3.1 ) by Fourier transformation in space,

Applied to Eq. (3.1

) this yields the system of linear ordinary differential equations

for each wave vector $n k ∈ ℝ$ where $P (ik )$ , called the symbol of the differential operator $P (∂∕∂x )$ , is defined as

The solution of Eq. (3.4

) is given by

where $ˆu(0,k)$ is determined by the initial data for $u$ at $t = 0$ . Therefore, the formal solution of the Cauchy problem

with given initial data $f$ for $u$ at $t = 0$ is

where $ˆf(k) = ---1--∫ e− ik⋅xf (x)dnx (2π)n∕2$ .

3.1.1 Well-posedness

At this point we have to ask ourselves if expression (3.9 ) makes sense. In fact, we do not expect the integral to converge in general. Even if $ˆf$ is smooth and decays rapidly to zero as $|k| → ∞$ we could still have problems if $|eP(ik)t|$ diverges as $|k| → ∞$ . One simple, but very restrictive, possibility to control this problem is to limit ourselves to initial data $f$ in the class $ω 𝒮$ of functions, which are the Fourier transform of a $C∞$ -function with compact support, i.e., $f ∈ 𝒮 ω$ , where

A function in this space is real analytic and decays faster than any polynomial as $|x| → ∞$ .¹ If $ω f ∈ 𝒮$ the integral in Eq. (3.9

) is well-defined and we obtain a solution of the Cauchy problem (3.7

, 3.8

), which, for each $t ≥ 0$ lies in this space. However, this possibility suffers from several unwanted features:

The space of admissible initial data is very restrictive. Indeed, since $ω f ∈ 𝒮$ is necessarily analytic it is not possible to consider nontrivial data with, say, compact support, and study the propagation of the support for such data.
For fixed $t > 0$ , the solution may grow without bound when perturbations with arbitrarily small amplitude but higher and higher frequency components are considered. Such an effect is illustrated in Example 6 below.
The function space $𝒮ω$ does not seem to be useful as a solution space when considering linear variable coefficient or quasilinear problems, since, for such problems, the different $k$ modes do not decouple from each other. Hence, mode coupling can lead to components with arbitrarily high frequencies.²

For these reasons, it is desirable to consider initial data of a more general class than $𝒮ω$ . For this, we need to control the growth of $eP(ik)t$ . This is captured in the following

Definition 1. The Cauchy problem (3.7 , 3.8 ) is called well posed if there are constants $K ≥ 1$ and $α ∈ ℝ$ such that

The importance of this definition relies on the property that for each fixed time $t > 0$ the norm $|eP(ik)t|$ of the propagator is bounded by the constant $C (t) := Ke αt$ , which is independent of the wave vector $k$ . The definition does not state anything about the growth of the solution with time other that this growth is bounded by an exponential. In this sense, unless one can choose $α ≤ 0$ or $α > 0$ arbitrarily small, well-posedness is not a statement about the stability in time, but rather about stability with respect to mode fluctuations.

Let us illustrate the meaning of Definition 1 with a few examples:

Example 5. The heat equation $ut(t,x ) = Δu (t,x )$ .
Fourier transformation converts this equation into $ˆu (t,k) = − |k |2ˆu(t,k) t$ . Hence, the symbol is $2 P (ik) = − |k|$ and $P(ik)t − |k|2t |e | = e ≤ 1$ . The problem is well posed.

Example 6. The backwards heat equation $ut(t,x) = − Δu (t,x)$ .
In this case the symbol is $P(ik) = + |k|2$ , and $|eP(ik)t| = e|k|2t$ . In contrast to the previous case, $eP(ik)t$ exhibits exponential frequency-dependent growth for each fixed $t > 0$ and the problem is not well posed. Notice that small initial perturbations with large $|k |$ are amplified by a factor that becomes larger and larger as $|k|$ increases. Therefore, after an arbitrarily small time, the solution is contaminated by high-frequency modes.

Example 7. The Schrödinger equation $ut(t,x) = iΔu (t,x)$ .
In this case we have $2 P(ik) = i|k |$ and $P (ik)t |e | = 1$ . The problem is well posed. Furthermore, the evolution is unitary, and we can evolve forward and backwards in time. When compared to the previous example, it is the factor $i$ in front of the Laplace operator that saves the situation and allows the evolution backwards in time.

Example 8. The one-dimensional wave equation written in first-order form,

The symbol is $P(ik) = ikA$ . Since the matrix $A$ is symmetric and has eigenvalues $±1$ , there exists an orthogonal transformation $U$ such that

Therefore, $|eP(ik)t| = 1$ , and the problem is well posed.

Example 9. Perturb the previous problem by a lower-order term,

The symbol is $P(ik) = ikA + λI$ , and $|eP(ik)t| = eλt$ . The problem is well posed, even though the solution grows exponentially in time if $λ > 0$ .

More generally one can show (see Theorem 2.1.2 in [259 ]):

Lemma 1. The Cauchy problem for the first-order equation $ut = Aux + B$ with complex $m × m$ matrices $A$ and $B$ is well posed if and only if $A$ is diagonalizable and has only real eigenvalues.

By considering the eigenvalues of the symbol $P (ik )$ we obtain the following simple necessary condition for well-posedness:

Lemma 2 (Petrovskii condition). Suppose the Cauchy problem (3.7 , 3.8 ) is well posed. Then, there is a constant $α ∈ ℝ$ such that

for all eigenvalues $λ$ of $P (ik)$ .

Proof. Suppose $λ$ is an eigenvalue of $P(ik)$ with corresponding eigenvector $v$ , $P (ik )v = λv$ . Then, if the problem is well posed,

for all $t ≥ 0$ , which implies that $eRe (λ)t ≤ Ke αt$ for all $t ≥ 0$ , and hence $Re (λ) ≤ α$ . □

Although the Petrovskii condition is a very simple necessary condition, we stress that it is not sufficient in general. Counterexamples are first-order systems, which are weakly, but not strongly, hyperbolic; see Example 10 below.

3.1.2 Extension of solutions

Now that we have defined and illustrated the notion of well-posedness, let us see how it can be used to solve the Cauchy problem (3.7 , 3.8 ) for initial data more general than in $𝒮ω$ . Suppose first that $f ∈ 𝒮 ω$ , as before. Then, if the problem is well posed, Parseval’s identities imply that the solution (3.9 ) must satisfy

Therefore, the $𝒮ω$ -solution satisfies the following estimate

for all $ω f ∈ 𝒮$ . This estimate is important because it allows us to extend the solution to the much larger space $2 n L (ℝ )$ . This extension is defined in the following way: let $2 n f ∈ L (ℝ )$ . Since $𝒮 ω$ is dense in $L2 (ℝn )$ there exists a sequence ${fj}$ in $𝒮 ω$ such that $∥fj − f∥ → 0$ . Therefore, if the problem is well posed, it follows from the estimate (3.18

) that the corresponding solutions $uj$ defined by Eq. (3.9

) form a Cauchy-sequence in $L2(ℝn )$ , and we can define

where the limit exists in the $L2(ℝn )$ sense. The linear map $U(t) : L2(ℝn ) → L2 (ℝn)$ satisfies the following properties:

$U (0 ) = I$ is the identity map.
$U (t + s) = U (t)U (s)$ for all $t,s ≥ 0$ .
For $ω f ∈ 𝒮$ , $u (t,.) = U (t)f$ is the unique solution to the Cauchy problem (3.7 , 3.8 ).
$αt ∥U (t)f ∥ ≤ Ke ∥f∥$ for all $2 n f ∈ L (ℝ )$ and all $t ≥ 0$ .

The family ${U (t) : t ≥ 0}$ is called a semi-group on $2 n L (ℝ )$ . In general, $U (t)$ cannot be extended to negative $t$ as the example of the backwards heat equation, Example 6, shows.

For $f ∈ L2(ℝn )$ the function $u (t,x ) := U (t)f (x)$ is called a weak solution of the Cauchy problem (3.7 , 3.8 ). It can also be constructed in an abstract way by using the Fourier–Plancharel operator $ℱ : L2(ℝn ) → L2(ℝn )$ . If the problem is well posed, then for each $f ∈ L2 (ℝn )$ and $t ≥ 0$ the map $P(ik)t k ↦→ e ℱ(f )(k )$ defines an $2 n L (ℝ )$ -function, and, hence, we can define

According to Duhamel’s principle, the semi-group $U (t)$ can also be used to construct weak solutions of the inhomogeneous problem,

where $2 n F : [0,∞ ) → L (ℝ )$ , $t ↦→ F(t,⋅)$ is continuous:

For a discussion on semi-groups in a more general context see Section 3.4.

3.1.3 Algebraic characterization

In order to extend the solution concept to initial data more general than analytic, we have introduced the concept of well-posedness in Definition 1. However, given a symbol $P(ik)$ , it is not always a simple task to determine whether or not constants $K ≥ 0$ and $α ∈ ℝ$ exist such that $P(ik)t αt |e | ≤ Ke$ for all $t ≥ 0$ and $k ∈ ℝn$ . Fortunately, the matrix theorem by Kreiss [257 ] provides necessary and sufficient conditions on the symbol $P(ik)$ for well-posedness.

Theorem 1. Let $P(ik)$ , $n k ∈ ℝ$ , be the symbol of a constant coefficient linear problem, see Eq. (3.5 ), and let $α ∈ ℝ$ . Then, the following conditions are equivalent:

There exists a constant $K ≥ 0$ such that

for all $t ≥ 0$ and $k ∈ ℝn$ .
There exists a constant $M > 0$ and a family $H (k)$ of $m × m$ Hermitian matrices such that

for all $k ∈ ℝn$ .

A generalization and complete proof of this theorem can be found in [259 ]. However, let us show here the implication (ii) $⇒$ (i) since it illustrates the concept of energy estimates, which will be used quite often throughout this review (see Section 3.2.3 below for a more general discussion of these estimates). Hence, let $H (k)$ be a family of $m × m$ Hermitian matrices satisfying the condition (3.25 ). Let $k ∈ ℝn$ and $v0 ∈ ℂm$ be fixed, and define $v (t) := eP(ik)tv0$ for $t ≥ 0$ . Then we have the following estimate for the “energy” density $∗ v(t) H (k )v(t)$ ,

d --v(t)∗H (k )v (t) = [P (ik)v(t)]∗H (k )v(t) + v(t)∗H (k)P(ik)v(t) dt ∗ ∗ = v(t) [P (ik) H (k) + H (k)P (ik )]v(t) ≤ 2αv (t)∗H (k )v (t),

which implies the differential inequality

Integrating, we find

which implies the inequality (3.24

) with $K = M$ .

3.1.4 First-order systems

Many systems in physics, like Maxwell’s equations, the Dirac equation, and certain formulation of Einstein’s equations are described by first-order partial-differential equations (PDEs). In fact, even systems, which are given by a higher-order PDE, can be reduced to first order at the cost of introducing new variables, and possibly also new constraints. Therefore, let us specialize the above results to a first-order linear problem of the form

where $A1,...,An, B$ are complex $m × m$ matrices. We split $P (ik) = P (ik ) + B 0$ into its principal symbol, $∑n P0 (ik ) = i kjAj j=1$ , and the lower-order term $B$ . The principal part is the one that dominates for large $|k|$ and hence the one that turns out to be important for well-posedness. Notice that $P0(ik)$ depends linearly on $k$ . With these observations in mind we note:

A necessary condition for the problem to be well posed is that for each $k ∈ ℝn$ with $|k| = 1$ the symbol $P0 (ik )$ is diagonalizable and has only purely imaginary eigenvalues. To see this, we require the inequality

for all $t ≥ 0$ and $k ∈ ℝn$ , $k ⁄= 0$ , replace $t$ by $t∕|k|$ , and take the limit $|k | → ∞$ , which yields $|eP0(ik′)t| ≤ K$ for all $k′ ∈ ℝn$ with $|k′| = 1$ . Therefore, there must exist for each such $k′$ a complex $m × m$ matrix $′ S (k )$ such that $′− 1 ′ ′ ′ S (k) P0 (ik )S (k ) = iΛ(k )$ , where $′ Λ (k )$ is a diagonal real matrix (cf. Lemma 1).
In this case the family of Hermitian $m × m$ matrices $H (k′) := (S (k′)− 1)∗S (k′)− 1$ satisfies

for all $k ′ ∈ ℝn$ with $|k ′| = 1$ .
However, in order to obtain the energy estimate, one also needs the condition $− 1 ′ M I ≤ H (k ) ≤ M I$ , that is, $′ H (k )$ must be uniformly bounded and positive. This follows automatically if $′ H (k)$ depends continuously on $k′$ , since $k ′$ varies over the $(n − 1)$ -dimensional unit sphere, which is compact.³ In turn, it follows that $H (k ′)$ depends continuously on $k′$ if $S (k′)$ does. However, although this may hold in many situations, continuous dependence of $S(k′)$ on $k′$ cannot always be established; see Example 12 for a counterexample.

These observations motivate the following three notions of hyperbolicity, each of them being a stronger condition than the previous one:

Definition 2. The first-order system (3.28 ) is called

weakly hyperbolic, if all the eigenvalues of its principal symbol $P0(ik)$ are purely imaginary.
strongly hyperbolic, if there exists a constant $M > 0$ and a family of Hermitian $m × m$ matrices $H (k)$ , $k ∈ Sn− 1$ , satisfying

for all $k ∈ Sn−1$ , where $Sn−1 := {k ∈ ℝn : |k| = 1}$ denotes the unit sphere.
symmetric hyperbolic, if there exists a Hermitian, positive definite $m × m$ matrix $H$ (which is independent of $k$ ) such that

for all $k ∈ Sn−1$ .

The matrix theorem implies the following statements:

Strongly and symmetric hyperbolic systems give rise to a well-posed Cauchy problem. According to Theorem 1, their principal symbol satisfies

and this property is stable with respect to lower-order perturbations,

The last inequality can be proven by applying Duhamel’s formula (3.23 ) to the function $ˆu(t) := eP (ik)t ˆf$ , which satisfies $ˆu(t) = P (ik)ˆu(t) + F(t) t 0$ with $F (t) = B ˆu(t)$ . The solution formula (3.23 ) then gives $ˆ ∫ t |ˆu(t)| ≤ K (|f| + |B| 0 |ˆu(s)|ds)$ , which yields $K|B|t ˆ |ˆu(t)| ≤ Ke |f|$ upon integration.
As we have anticipated above, a necessary condition for well-posedness is the existence of a complex $m × m$ matrix $S(k)$ for each $k ∈ Sn−1$ on the unit sphere, which brings the principal symbol $P0(ik)$ into diagonal, purely imaginary form. If, furthermore, $S (k)$ can be chosen such that $|S(k)|$ and $|S(k )− 1|$ are uniformly bounded for all $k ∈ Sn −1$ , then $H (k) := (S(k)−1)∗S(k )− 1$ satisfies the conditions (3.31 ) for strong hyperbolicity. If the system is well posed, Theorem 2.4.1 in [259 ] shows that it is always possible to construct a symmetrizer $H (k)$ satisfying the conditions (3.31 ) in this manner, and hence, strong hyperbolicity is also a necessary condition for well-posedness. The symmetrizer construction $H (k) := (S (k)−1)∗S(k)−1$ is useful for applications, since $S (k)$ is easily constructed from the eigenvectors and $− 1 S(k)$ from the eigenfields of the principal symbol; see Example 15.
Weakly hyperbolic systems are not well posed in general because $|eP0(ik)t|$ might exhibit polynomial growth in $|k|t$ . Although one might consider such polynomial growth as acceptable, such systems are unstable with respect to lower-order perturbations. As the next example shows, it is possible that $|eP(ik)t|$ grows exponentially in $|k |$ if the system is weakly hyperbolic.

Example 10. Consider the weakly hyperbolic system [259 ]

with $a ∈ ℝ$ a parameter. The principal symbol is $( ) P0(ik) = ik 1 1 0 1$ and

Using the tools described in Section 2 we find for the norm

┌ ------------------------------ ││ ∘ (---------)2----- |eP0(ik)t| = ∘ 1 + k2t2 + 1 + k2t2 − 1, (3.37 ) 2 2

which is approximately equal to $|k|t$ for large $|k|t$ . Hence, the solutions to Eq. (3.35

) contain modes, which grow linearly in $|k |t$ for large $|k|t$ when $a = 0$ , i.e., when there are no lower-order terms.

However, when $a ⁄= 0$ , the eigenvalues of $P (ik )$ are

which, for large $k$ has real part $Re (λ ) ≈ ± ∘ |a∥k-|∕2 ±$ . The eigenvalue with positive real part gives rise to solutions, which, for fixed $t$ , grow exponentially in $|k|$ .

Example 11. For the system [353 ],

the principal symbol, $( ) k1 + k2 k1 P0(ik) = i 0 2(k1 + k2)$ , is diagonalizable for all vectors $1 k = (k1,k2) ∈ S$ except for those with $k + k = 0 1 2$ . In particular, $P (ik) 0$ is diagonalizable for $k = (1,0)$ and $k = (0,1 )$ . This shows that in general, it is not sufficient to check that the $n$ matrices $1 A$ , $2 A$ ,…, $n A$ alone are diagonalizable and have real eigenvalues; one has to consider all possible linear combinations $n ∑ Ajk j=1 j$ with $k ∈ Sn −1$ .

Example 12. Next, we present a system for which the eigenvectors of the principal symbol cannot be chosen to be continuous functions of $k$ :

The principal symbol $( ) P0(ik) = i k1 k2 k2 − k1$ has eigenvalues $∘ -------- λ ±(k) = ±i k2 + k2 1 2$ and for $(k1,k2) ⁄= (0,0)$ the corresponding eigenprojectors are

When $(k1,k2) → (0,0)$ the two eigenvalues fall together, and $A (k)$ converges to the zero matrix. However, it is not possible to continuously extend $P (k ,k ) ± 1 2$ to $(k ,k ) = (0,0) 1 2$ . For example,

for positive $h > 0$ . Therefore, any choice for the matrix $S(k)$ , which diagonalizes $A(k)$ , must be discontinuous at $k = (0,0,±1 )$ since the columns of $S(k)$ are the eigenvectors of $A (k)$ .

Of course, $A (k)$ is symmetric and so $S (k )$ can be chosen to be unitary, which yields the trivial symmetrizer $H (k) = I$ . Therefore, the system is symmetric hyperbolic and yields a well-posed Cauchy problem; however, this example shows that it is not always possible to choose $S(k )$ as a continuous function of $k$ .

Example 13. Consider the Klein–Gordon equation

in two spatial dimensions, where $m ∈ ℝ$ is a parameter, which is proportional to the mass of the field $Φ$ . Introducing the variables $u = (Φ,Φt, Φx,Φy )$ we obtain the first-order system

( ) ( ) ( ) 0 0 0 0 0 0 0 0 0 1 0 0 | 0 0 1 0 | | 0 0 0 1| | − m2 0 0 0| ut = |( |) ux + |( |) uy + |( |) u. (3.44 ) 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0

The matrix coefficients in front of $ux$ and $uy$ are symmetric; hence the system is symmetric hyperbolic with trivial symmetrizer $2 H = diag(m ,1,1,1)$ .⁴ The corresponding Cauchy problem is well posed. However, a problem with this first-order system is that it is only equivalent to the original, second-order equation (3.43

) if the constraints $(u1)x = u3$ and $(u1)y = u4$ are satisfied.

An alternative symmetric hyperbolic first-order reduction of the Klein–Gordon equation, which does not require the introduction of constraints, is the Dirac equation in two spatial dimensions,

This system implies the Klein–Gordon equation (3.43

) for either of the two components of $v$ .

Yet another way of reducing second-order equations to first-order ones without introducing constraints will be discussed in Section 3.1.5.

Example 14. In terms of the electric and magnetic fields $u = (E, B )$ , Maxwell’s evolution equations,

constitute a symmetric hyperbolic system. Here, $J$ is the current density and $∇$ and $∧$ denote the nabla operator and the vector product, respectively. The principal symbol is

and a symmetrizer is given by the physical energy density,

in other words, $−1 H = 2 I$ is trivial. The constraints $∇ ⋅ E = ρ$ and $∇ ⋅ B = 0$ propagate as a consequence of Eqs. (3.46

, 3.47

), provided that the continuity equation holds: $(∇ ⋅ E − ρ)t = − ∇ ⋅ J − ρt = 0$ , $(∇ ⋅ B)t = 0$ .

Example 15. There are many alternative ways to write Maxwell’s equations. The following system [353 , 287 ] was originally motivated by an analogy with certain parametrized first-order hyperbolic formulations of the Einstein equations, and provides an example of a system that can be symmetric, strongly, weakly or not hyperbolic at all, depending on the parameter values. Using the Einstein summation convention, the evolution system in vacuum has the form

∂ E = ∂j(W − W ) − α(∂ W j − ∂jW ), (3.50 ) t i ij ji i j ij ∂ W = − ∂ E − β-δ ∂kE , (3.51 ) t ij i j 2 ij k

where $E i$ and $W = ∂ A ij i j$ , $i = 1,2,3$ , represent the Cartesian components of the electric field and the gradient of the magnetic potential $Aj$ , respectively, and where the real parameters $α$ and $β$ determine the dynamics of the constraint hypersurface defined by $k ∂ Ek = 0$ and $∂kWij − ∂iWkj = 0$ .

In order to analyze under which conditions on $α$ and $β$ the system (3.50 , 3.51 ) is strongly hyperbolic we consider the corresponding symbol,

Decomposing $Ei$ and $Wij$ into components parallel and orthogonal to $ki$ ,

where in terms of the projector $γij := δij − kikj$ orthogonal to $k$ we have defined $E¯ := klEl$ , $E¯i := γijEj$ and $¯W := kikjWij$ , $W¯i := γikWkjkj$ , $¯Vj := kiWik γkj$ , $¯U := γijWij$ , and $W¯ij := (γikγjl − 2 −1γijγkl)Wkl$ ,⁵ we can write the eigenvalue problem $P (ik)u = iλu 0$ as

λE¯= − α¯U , λ¯U = − β ¯E, ( β) λW¯ = − 1 + -- E¯, 2 λE¯i = (1 + α) ¯Wi − V¯i, λ¯V = − ¯E , i i λW¯i = 0, λW¯ = 0. ij

It follows that $P0(ik)$ is diagonalizable with purely complex eigenvalues if and only if $αβ > 0$ . However, in order to show that in this case the system is strongly hyperbolic one still needs to construct a bounded symmetrizer $H (k)$ . For this, we set $μ := √ α-β$ and diagonalize $P (ik) = iS(k)Λ (k)S(k)− 1 0$ with $Λ (k) = diag(μ,− μ,0,1,− 1,0, 0)$ and

( ¯ μ ¯ ) E − βμU || E¯+( β ¯U ) || || βW¯ − 1 + β U¯ || S(k )− 1u = | E¯ − V¯ + (1 + 2α)W¯ | . (3.54 ) || ¯i ¯i ¯i|| | Ei + Vi − (1 + α)Wi | ( ¯Wi ) W¯ij

Then, the quadratic form associated with the symmetrizer is

u ∗H (k )u = u∗(S(k)− 1)∗S (k)−1u | ( ) |2 ¯ 2 α- ¯ 2 || ¯ β- ¯|| ¯i ¯ = 2|E | + 2β |U | + |βW − 1 + 2 U | + 2E Ei [ i i][ ] i ij + 2 V¯ − (1 + α )W ¯ ¯Vi − (1 + α ) ¯Wi + W¯ W¯i + ¯W ¯Wij,

and $H (k)$ is smooth in $k ∈ S2$ . Therefore, the system is indeed strongly hyperbolic for $α β > 0$ .

In order to analyze under which conditions the system is symmetric hyperbolic we notice that because of rotational and parity invariance the most general $k$ -independent symmetrizer must have the form

with strictly positive constants $a$ , $b$ , $c$ and $d$ , where $Wˆij := W (ij) − δijW ∕3$ denotes the symmetric, trace-free part of $Wij$ and $W := W jj$ its trace. Then,

[ 2α ] u∗HP0 (ik )u = ia (Ei )∗ (α + 2)kjW [ij] + αkjWˆij − --kiW ( 3 ) [ij] ∗ ˆ ij ∗ 3β- ∗ i + ib(W ) Eikj − ic(W ) Eikj − id 1 + 2 W k Ei.

For $H$ to be a symmetrizer, the expression on the right-hand side must be purely imaginary. This is the case if and only if $a(α + 2) = b$ , $− a α = c$ and $2aα∕3 = d (1 + 3 β∕2)$ . Since $a$ , $b$ , $c$ and $d$ are positive, these equalities can be satisfied if and only if $− 2 < α < 0$ and $β < − 2 ∕3$ . Therefore, if either $α$ and $β$ are both positive or $α$ and $β$ are both negative and $α ≤ − 2$ or $β ≥ − 2∕3$ , then the system (3.50

, 3.51

) is strongly but not symmetric hyperbolic.

3.1.5 Second-order systems

An important class of systems in physics are wave problems. In the linear, constant coefficient case, they are described by an equation of the form

where $m v = v(t,x) ∈ ℂ$ is the state vector, and $ij ji j j A = A ,B ,C ,D, E$ denote complex $m × m$ matrices. In order to apply the theory described so far, we reduce this equation to a system that is first order in time. This is achieved by introducing the new variable $∑n w := vt − Bj ∂∂xjv j=1$ .⁶ With this redefinition one obtains a system of the form (3.1

) with $u = (v,w )T$ and

( ) ∑n 0 I P(∂ ∕∂x) = Bj -∂--+ ( ∑n jk j k ∂2 n∑ j j ∂ ) . (3.57 ) ∂xj (A + B B )∂xj∂xk + (C + DB )∂xj + E D j=1 j,k=1 j=1

Now we could apply the matrix theorem, Theorem 1, to the corresponding symbol $P (ik )$ and analyze under which conditions on the matrix coefficients $Aij,Bj, Cj,D, E$ the Cauchy problem is well posed. However, since our problem originates from a second-order equation, it is convenient to rewrite the symbol in a slightly different way: instead of taking the Fourier transform of $v$ and $w$ directly, we multiply $vˆ$ by $|k|$ and write the symbol in terms of the variable $ˆU := (|k|ˆv, ˆw )T$ . Then, the $L2$ -norm of $Uˆ$ controls, through Parseval’s identity, the $L2$ -norms of the first partial derivatives of $v$ , as is the case for the usual energies for second-order systems. In terms of $Uˆ$ the system reads

in Fourier space, where

n ( 0 |k|I ) ∑ jˆ ( n∑ ∑n ) Q (ik ) = i|k| B kj + − |k | (Ajk + BjBk )ˆkjˆkk + i (Cj + DBj )ˆkj + 1|k|E D (3.59 ) j=1 j,k=1 j=1

with $ˆ kj := kj∕|k|$ . As for first-order systems, we can split $Q (ik)$ into its principal part,

( ) ∑n j n 0 I Q0 (ik) := i|k| B ˆkj + |k|( − ∑ (Ajk + BjBk )ˆk ˆk 0 ) , (3.60 ) j=1 j,k=1 j k

which dominates for $|k | → ∞$ , and the remaining, lower-order terms. Because of the homogeneity of $Q0 (ik)$ in $k$ we can restrict ourselves to values of $k ∈ Sn−1$ on the unit sphere, like for first-order systems. Then, it follows as a consequence of the matrix theorem that the problem is well posed if and only if there exists a symmetrizer $H (k)$ and a constant $M > 0$ satisfying

for all such $k$ . Necessary and sufficient conditions under which such a symmetrizer exists have been given in [261

] for the particular case in which the mixed–second-order derivative term in Eq. (3.56

) vanishes; that is, when $Bj = 0$ . This result can be generalized in a straightforward manner to the case where the matrices $Bj = βjI$ are proportional to the identity:

Theorem 2. Suppose $Bj = βjI$ , $j = 1, 2,...,n$ . (Note that this condition is trivially satisfied if $m = 1$ .) Then, the Cauchy problem for Eq. (3.56 ) is well posed if and only if the symbol

has the following properties: there exist constants $M > 0$ and $δ > 0$ and a family $h(k )$ of Hermitian $m × m$ matrices such that

for all $n−1 k ∈ S$ .

Proof. Since for $Bj = βjI$ the advection term $n i|k| ∑ Bjˆk j=1 j$ commutes with any Hermitian matrix $H (k)$ , it is sufficient to prove the theorem for $Bj = 0$ , in which case the principal symbol reduces to

We write the symmetrizer $H (k )$ in the following block form,

where $H11(k)$ , $H22(k )$ and $H12(k)$ are complex $m × m$ matrices, the first two being Hermitian. Then,

Now, suppose $h (k )$ satisfies the conditions (3.63

). Then, choosing $H12 (k) := 0$ , $H22 (k) := h (k)$ and $H (k) := h (k)R(k ) 11$ we find that $H (k)Q (ik) + Q (ik )∗H (k) = 0 0 0$ . Furthermore, $M −1I ≤ H (k) ≤ M I 22$ and $δI ≤ H11(k ) = h (k)R(k ) ≤ M CI$ where

is finite because $R (k)u$ is continuous in $k$ and $u$ . Therefore, $H (k )$ is a symmetrizer for $Q0(ik)$ , and the problem is well posed.

Conversely, suppose that the problem is well posed with symmetrizer $H (k)$ . Then, the vanishing of $∗ H (k)Q0 (ik) + Q0 (ik) H (k)$ yields the conditions $∗ H11(k) = H22 (k)R(k ) = R (k) H22 (k)$ and the conditions (3.63 ) are satisfied for $h(k) := H22(k)$ . □

Remark: The conditions (3.63 ) imply that $R (k)$ is symmetric and positive with respect to the scalar product defined by $h (k )$ . Hence it is diagonalizable, and all its eigenvalues are positive. A practical way of finding $h(k)$ is to construct $T(k)$ , which diagonalizes $R(k)$ , $− 1 T (k ) R(k)T (k) = P (k)$ with $P(k )$ diagonal and positive. Then, $−1 ∗ −1 h(k ) := (T (k ) )T (k)$ is the candidate for satisfying the conditions (3.63 ).

Let us give some examples and applications:

Example 16. The Klein–Gordon equation $vtt = Δv − m2v$ on flat spacetime. In this case, $ij ij A = δ$ and $j B = 0$ , and $2 R (k) = |k |$ trivially satisfies the conditions of Theorem 2.

Example 17. In anticipation of the following Section 3.2, where linear problems with variable coefficients are treated, let us generalize the previous example on a curved spacetime $(M, g)$ . We assume that $(M, g )$ is globally hyperbolic such that it can be foliated by space-like hypersurfaces $Σt$ . In the ADM decomposition, the metric in adapted coordinates assumes the form

with $α > 0$ the lapse, $i β$ the shift vector, which is tangent to $Σt$ , and $i j γijdx ⊗ dx$ the induced three-metric on the spacelike hypersurfaces $Σt$ . The inverse of the metric is given by