\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2007(2007), No. 08, pp. 1--10.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2007 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2007/08\hfil A regularity criterion] {A regularity criterion for the angular velocity component in axisymmetric Navier-Stokes equations} \author[O. Kreml, M. Pokorn\'y\hfil EJDE-2007/08\hfilneg] {Ond\v rej Kreml, Milan Pokorn\'y} % in alphabetical order \address{Ond\v{r}ej Kreml \newline Mathematical Institute of Charles University, Sokolovsk\'a 83, 186 75 Praha 8, Czech Republic} \email{kreml@karlin.mff.cuni.cz} \address{Milan Pokorn\'y \newline Mathematical Institute of Charles University, Sokolovsk\'a 83, 186 75 Praha 8, Czech Republic} \email{pokorny@karlin.mff.cuni.cz} \thanks{Submitted October 5, 2006. Published January 2, 2007.} \thanks{O. Kreml was supported by project LC06052 from the Jind\v{r}ich Ne\v{c}as Center for \hfill\break\indent Mathematical Modeling. \hfill\break\indent M. Pokorn\'y was supported by projects MSM 0021620839 from MSMT, 201/06/0352 \hfill\break\indent from the Czech Science Foundation, and LC06052 from Jind\v{r}ich Ne\v{c}as Center for \hfill\break\indent Mathematical Modeling.} \subjclass[2000]{35Q30, 76D05} \keywords{Axisymmetric flow; Navier-Stokes equations; \hfill\break\indent regularity of systems of PDE's} \begin{abstract} We study the non-stationary Navier-Stokes equations in the entire three-dimensional space under the assumption that the data are axisymmetric. We extend the regularity criterion for axisymmetric weak solutions given in \cite{pok}. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \section{Introduction} Consider the Navier-Stokes equations in the entire three-dimensional space; i.e., the system of PDE's \begin{equation} \label{1} \begin{gathered} \frac{\partial \textbf{v}}{\partial t} + \textbf{v}\cdot\nabla \textbf{v} - \nu\Delta \textbf{v} + \nabla p = \textbf{0} \quad \text{in } (0,T) \times \mathbb{R}^3\\ \mathop{\rm div} \textbf{v} = 0 \quad \text{in } (0,T) \times \mathbb{R}^3 \\ \textbf{v}(0,\textbf{x}) = \textbf{v}_0(\textbf{x}) \quad \text{in } \mathbb{R}^3\, , \end{gathered} \end{equation} where $\textbf{v}: (0,T)\times \mathbb{R}^3=\Omega_T \mapsto \mathbb{R}^3$ is the velocity field, $p:\Omega_T \mapsto \mathbb{R}$ is the pressure, $0 0$ is constant viscosity coefficient and $\textbf{v}_0$ is the initial velocity. To avoid technical difficulties, we take the forcing term on the right-hand side equal to zero. However, it is not difficult to formulate conditions on $\textbf{f}$ under which the statement of Theorem 1 remains true. We leave this relatively easy exercise to the kind reader. The question of smoothness and uniqueness of weak solutions to \eqref{1} is one of the most challenging problems in the theory of PDE's. The solution is known to be unique (in the class of all weak solutions satisfying the energy inequality) if it belongs to the class $L^{t,s}(\Omega_T)$ with $\frac{2}{t} + \frac{3}{s} \leq 1$, $t\in [2,+\infty]$, $s\in[ 3,+\infty]$ (see \cite {KoSo,Pr}). Moreover, if the weak solution belongs to $L^{t,s}(\Omega_T)$ with $\frac{2}{t} + \frac{3}{s} \leq 1$, $t\in[ 2,+\infty]$, $s\in[3,+\infty]$ and the input data are "smooth enough" then the solution is smooth. (See \cite{Se} for $s>3$, \cite{EsSeSv} for $s=3$.) In the case of the planar flow the weak solution is known to be unique and smooth as the data of the problem allow (see \cite {Le}, \cite{La1}). Thus a natural question, namely what can be said about the axisymmetric flow, appears. The first results in this direction were obtained in the late sixties for $v_\varphi = 0$ (see \cite{lad,Uch}) and later also in \cite{leo}. The case $v_\varphi \neq 0$, including the $z$-axis, was for the first time considered in the paper \cite{neu} where for $v_r \in L^{t,s}(\Omega_T)$ with $\frac{2}{t}+\frac{3}{s}\leq 1$, $t\in [2,+\infty]$, $s\in( 3,+\infty]$ the smoothness and thus also the uniqueness in the class of weak solutions satisfying the energy inequality was obtained. In the same paper the authors prove a regularity criterion for the angular velocity component. This criterion was improved in \cite{pok}. The author shows the smoothness and the uniqueness in the class of weak solutions satisfying the energy inequality for $v_\varphi \in L^{t,s}(\Omega_T)$ with $t\in ( 2,+\infty]$, $s\in(4, +\infty]$, $\frac{2}{t}+\frac{3}{s} < 1$. See also \cite{ChLe}, where the authors give several other smoothness criteria for the vorticity components. Another approach to this problem, based on the smallness of the swirl, can be found in \cite{Zaj}. Note that, except for the $L^{\infty,3}(\Omega_T)$ case, the criterion for $v_r$ is optimal from the scaling argument (see \cite{pok2} for discussion of this issue). On the other hand, for $v_\varphi$, we would like to have rather equality than strict inequality $\frac{2}{t} + \frac{3}{s} < 1$. Moreover, the restriction $s > 4$ seems to be artificial. In this paper, we will give a partial answer to the latter problem. Note that, unfortunately, we do not get $s$ close to $3$ and moreover, the criterion is not optimal from the viewpoint of the scaling. However, our main result reads as follows. \begin{theorem} \label{veta} Let $\mathbf{v}$ be a weak solution to problem \eqref{1} satisfying the energy inequality with $\mathbf{v}_0 \in W^{2,2}(\mathbb{R}^3)$ so that $\nabla \mathbf{v}_0 \in L^1(\mathbb{R}^3)$ and $(v_0)_\varphi r \in L^\infty(\mathbb{R}^3)$. Let $\mathbf{v}_0$ be axisymmetric. Suppose further that the angular component $v_\varphi$ of $\mathbf{v}$ belongs to $L^{t,s}(\Omega_T)$ for some $t \in \big(\frac{8s}{7s - 24}, \infty\big]$, $s \in (\frac{24}{7}, 4]$, $\frac{2}{t} + \frac{3}{s} < \frac{7}{4} - \frac{3}{s}$. Then $(\mathbf{v},p)$, where $p$ is the corresponding pressure, is the axisymmetric strong solution to problem \eqref{1} which is unique in the class of all weak solutions satisfying the energy inequality. \end{theorem} Note that the case $s > 4$ is successfully solved in \cite{pok}. Theorem \ref{veta} extends the result from \cite{pok} for $s \in (\frac{24}{7}, 4]$. Under an axisymmetric solution we understand a pair $(\mathbf{v}, p)$ such that in cylindrical coordinates $(r,\varphi,z)$, $r \in [0,\infty)$, $\varphi \in [0,2\pi)$ and $z \in \mathbb{R}$, $v_r$, $v_\varphi$ and $v_z$, considered in cylindrical coordinates, are independent of $\varphi$, and $p$, written in cylindrical coordinates, is also independent of $\varphi$. \section {Preliminaries} Denote by $(v_r, v_\varphi, v_z)$ the cylindrical coordinates of the vector field $\mathbf{v}$ and by $(\omega_r, \omega_\varphi, \omega_z)$ the cylindrical coordinates of $\mathop{\rm curl} \mathbf{v}$, i.e. $\omega_r = -\frac{\partial v_\varphi}{\partial z}$, $\omega_\varphi = \frac{\partial v_r}{\partial z} - \frac{\partial v_z}{\partial r}$ and $\omega_z = \frac{1}{r} \frac{\partial }{\partial r}(rv_\varphi)$ for $\mathbf{v}$ an axisymmetric field. Moreover, let $\mathbf{w} = (w_{1}, w_{2}, w_{3})$ denote the cartesian coordinates of $\mathop{\rm curl} \mathbf{v}$. We will use the standard notation for the Lebesgue spaces $L^p(\mathbb{R}^3)$ endowed with the standard norm $\|\cdot\|_{p,\mathbb{R}^3}$ and the Sobolev spaces $W^{k,p}(\mathbb{R}^3)$ endowed with the standard norm $\|\cdot\|_{k,p, \mathbb{R}^3}$. By $L^{t,s}(\Omega_T)$, $\Omega_T = (0,T)\times \mathbb{R}^3$ we understand the anisotropic Lebesgue space $L^t(0,T;L^s(\mathbb{R}^3))$. If no confusion can arise we skip writing $\mathbb{R}^3$ and $\Omega_T$, respectively. Vector-valued functions are printed boldfaced. Nonetheless, we do not distinguish between $L^q(\mathbb{R}^3)^3$ and $L^q(\mathbb{R}^3)$. In order to keep a simple notation, all generic constants will be denoted by $C$; thus $C$ can have different values from term to term, even in the same formula. By $D\mathbf{v}$ we mean the gradient of $\mathbf{v}$ expressed in the cartesian coordinates, while $\nabla v_r$ denotes the derivatives with respect to $r$ and $z$ only ($\mathbf{v}$ is axisymmetric). Similarly for $v_\varphi$ and $v_z$. We will use the following inequalities. For the proofs of Lemmas \ref{lem1}--\ref{lem2a} see \cite{neu}. \begin{lemma} \label{lem1} Let $\mathbf{v}$ be a sufficiently smooth vector field. Then there exists a constant $C(p)>0$, independent of $\mathbf{v}$, such that for $10$ and $(\mathbf{v}, p)$, a weak solution\footnote{It means that $\mathbf{v}$ is the weak solution and $p$ is the corresponding pressure which can be easily computed and, up to an additive constant, is uniquely determined.} to system \eqref{1}, which is a strong solution on the time interval $(0,t_0)$ such that $\mathbf{v} \in L^2(0,t_0; W^{3,2})\cap L^\infty(0,t_0;W^{2,2})$ with $\frac{\partial \mathbf{v}}{\partial t}$ and $\nabla p \in L^2(0,t_0;W^{1,2})$. Moreover, if $\mathbf{v}_0$ is axisymmetric then also the strong solution is axisymmetric. \end{lemma} Now let $\mathbf{v}_0$ be as in Lemma \ref{lem9} (axisymmetric). We define: \[ t^* = \sup \big\{t>0: \mbox{there exists an axisymmetric strong solution to \eqref{1} on }(0,t)\big\} \] It follows from Lemma \ref{lem9} that $t^*>0$. Now let $\mathbf{v}$ be a weak solution to the Navier-Stokes system as in Theorem \ref{veta}. Due to the uniqueness property (thus the energy inequality is required!), it coincides with the strong solution from Lemma \ref{lem9} on any compact subinterval of $[0,t^*)$. There are two possibilities. Either $t^*=\infty$ and we have the global-in-time regular solution, or $t^*<\infty$. We will exclude the latter by showing that $v_r$ satisfies on $(0,t^*)$ the assumptions of Lemma \ref{lem4}. To this aim we will essentially use both the information about the better regularity of one velocity component and the fact that the solution is axisymmetric. Now, let $0<\bar{t}0$ and $\psi(r)$ a cut-off function equal to zero near $r=0$. Now we integrate the equality over $\mathbb{R}^3$ then pass first with $\psi(r)$ to the identity function and finally with $\delta$ to zero. Note that we cannot take directly $\delta = 0$ as some integrals cannot be controlled, cf. \cite{leo}. We get \begin{equation} \label{3.4} \frac {1}{q} \frac{d}{dt} \big\|\frac{\omega_\varphi}{r}\big\|_q^q + \frac{4(q-1)}{q^2} \nu \int \Big|\nabla \Big(\big|\frac{\omega_\varphi}{r} \big|^{q/2}\Big)\Big|^2 \leq \Big|\int \frac{2}{r} v_\varphi \frac{\partial v_\varphi}{\partial z} \frac{\omega_\varphi}{|\omega_\varphi|^{2-q}}\frac{1}{r^q}\Big|\, . \end{equation} To prove Theorem \ref{veta} we sum (\ref{3.3}) and (\ref{3.4}) and estimate all terms on the right-hand side with $q = \frac{5}{6}p$. First we will estimate the term $I_1 = \int \frac{v_r}{r} |\omega_\varphi|^p$, where we basically follow \cite{pok}. Using the H\"older inequality, the interpolations, the Sobolev embedding inequality, Lemma \ref{lem2} and the Young inequality we finally get \begin{equation*} I_1 \leq \delta\big\|\nabla\big(|\frac{\omega_\varphi}{r}|^{q/2}\big) \big\|_2^2 + C(\delta)\|\omega_\varphi\|_2^2\big(\|\omega_\varphi\|_p^p + \big\|\frac{\omega_\varphi}{r}\big\|_q^q\big) \end{equation*} with arbitrarily small positive $\delta$. The first term can be included into the left-hand side while the second term can be estimated later on, using the Gronwall inequality. Next we want to estimate the other term on the right-hand side of (\ref{3.3}), namely $I_2$. However, the term $I_2 = \int \frac{2}{r} v_\varphi\frac{\partial v_\varphi}{\partial z} \frac{\omega_\varphi}{|\omega_\varphi|^{2-p}}$ can be estimated same way as the term on the right-hand side of (\ref{3.4}), $I_3 = |\int \frac{2}{r} v_\varphi\frac{\partial v_\varphi}{\partial z}\frac{\omega_\varphi}{|\omega_\varphi|^{2-q}}\frac{1}{r^q}|$. This is due to the fact that main problems are near the $z$-axis and $I_2$ is of lower order than $I_3$. (Here we also use the fact that $v_\varphi r \in L^\infty(Q_T)$.) Choose $\varepsilon > 0$. \begin{equation*} I_3 \leq 2\int\big(\frac{|\omega_\varphi|^{q - 1}}{r^X}\big) \big(\frac{|v_\varphi|^\alpha}{r^Y}\big) \big|\frac{\partial v_\varphi}{\partial z}\big|\big(\frac{|v_\varphi|^{1 - \alpha}}{r^Z}\big), \end{equation*} where \begin{equation*} X = q + 1 - \frac{2}{q} - \varepsilon,\quad Y = \big(\frac{1 + \varepsilon}{2} + \frac{1}{q}\big)(2 - q),\quad Z = \frac{2}{q} + \varepsilon - Y,\quad \alpha = 2 - q. \end{equation*} Using the Young inequality we get \begin{equation*} I_3 \leq \delta\int\frac{\omega_\varphi^q}{r^\frac{qX}{q - 1}} + C(\delta)\Big( \int\frac{|v_\varphi|^{2q}}{r^{q(1 + \varepsilon) + 2}} + \int\big|\frac{\partial}{\partial z}\big|\frac{v_\varphi^q}{r^{\frac{q(1 + \varepsilon)}{2}}}\big|\big|^2\Big). \end{equation*} Since $\frac{qX}{q - 1} = q + 2 - \varepsilon\frac{q}{q-1}$, we can use Lemma \ref{lem3} and \begin{align*} I_3 &\leq \delta C(\varepsilon)\Big(\int\big|\nabla\big( |\frac{\omega_\varphi}{r}|^{q/2}\big)\big|^2 + \int |\frac{\omega_\varphi}{r}|^q\Big) \\ &+ C(\delta)\Big(\int\frac{|v_\varphi|^{2q}}{r^{q(1 + \varepsilon) + 2}} + \int|\frac{\partial}{\partial z}|\frac{v_\varphi^q}{r^{\frac{q(1 + \varepsilon)}{2}}} ||^2\Big). \end{align*} Taking sufficiently small $\delta$ we can include the first term to the left-hand side of (\ref{3.4}), the second term can be later estimated using the Gronwall inequality. We have to deal with the last two terms on the right-hand side. Summing up the estimates of $I_1$ and $I_3$, \begin{equation}\label{GGG} \begin{aligned} &\frac {d}{dt} \big(\|\omega_\varphi\|_p^p + \|\frac{\omega_\varphi}{r}\|_q^q\big) + \int\Big(\frac{|\omega_\varphi|^p}{r^2} + |\nabla|\omega_\varphi|^{p/2}|^2 + |\nabla(|\frac{\omega_\varphi}{r}|^{q/2})|^2\Big)\\ &\leq C \int|\frac{\omega_\varphi}{r}|^q +C\|\omega_\varphi\|_2^2\big(\|\omega_\varphi\|_p^p + \|\frac{\omega_\varphi}{r}\|_q^q\big)\\ &\quad + C\Big( \int \frac{|v_\varphi|^{2q}}{r^{q(1 + \varepsilon) + 2}} + \int |\nabla|\frac{v_\varphi^q}{r^{\frac{q(1 + \varepsilon)}{2}}}||^2\Big). \end{aligned} \end{equation} To estimate the last two terms, we test the equation for $v_\varphi$ by $|v_\varphi|^{2q - 2} v_\varphi/r^{q(1 + \varepsilon)}$. We get \begin{align*} &\frac{2}{q}\frac{d}{dt}\int\frac{|v_\varphi|^{2q}}{r^{q(1 + \varepsilon)}} + \frac{2q - 1}{q^2}\nu\int|\nabla|\frac{v_\varphi^q}{r^{\frac{q(1 + \varepsilon)}{2}}} ||^2 + \frac{(2q)^2 - q^2(1 + \varepsilon)^2}{(2q)^2}\nu \int\frac{|v_\varphi|^{2q}}{r^{q(1 + \varepsilon) + 2}} \\ &= (-1 - \frac{1 + \varepsilon}{2})\int\frac{|v_\varphi|^{2q}}{r^{q(1 + \varepsilon)}}\frac{v_r}{r}, \end{align*} i.e. together with (\ref{GGG}), \begin{equation}\label{HHH} \begin{aligned} &\frac{d}{dt}\Big(\|\omega_\varphi\|_p^p + \|\frac{\omega_\varphi}{r}\|_q^q + \int\frac{|v_\varphi|^{2q}}{r^{q(1 + \varepsilon)}}\Big) \\ &+ \int \Big(\frac{|\omega_\varphi|^p}{r^2} + |\nabla|\omega_\varphi|^{p/2}|^2 + |\nabla(|\frac{\omega_\varphi}{r}|^{q/2})|^2 + |\nabla|\frac{v_\varphi^q}{r^{\frac{q(1 + \varepsilon)}{2}}}||^2 + \frac{|v_\varphi|^{2q}}{r^{q(1 + \varepsilon) + 2}}\Big) \\ &\leq C \int|\frac{\omega_\varphi}{r}|^q + C \int \frac{|v_\varphi|^{2q}}{r^{q(1 + \varepsilon)}}\frac{|v_r|}{r} + C\|\omega_\varphi\|_2^2\big(\|\omega_\varphi\|_p^p + \|\frac{\omega_\varphi}{r}\|_q^q\big). \end{aligned} \end{equation} Denoting by $I_4$ the second integral on the right-hand side, we have \begin{equation*} I_4 = \int \big(\frac{|v_r|}{r^{1 + k}}\big) \big(\frac{|v_\varphi|^{2q}}{r^{q(1+\varepsilon)}}\big)^\alpha \big(\frac{|v_\varphi|^{2q}}{r^{q(1 + \varepsilon) + 2}}\big)^\beta |v_\varphi|^{\gamma}, \end{equation*} where $k \in [0,1]$, $\gamma = \frac{5+k}{3}$, $\beta = \frac{5}{12}(1-k) + \varepsilon\frac{5+k}{12}$ and $\alpha = \frac{5k+7}{12} - \frac{5+k}{5p} - \varepsilon\frac{5+k}{12}$. Recall that $q = \frac{5}{6}p$. Let $1 < a < \frac{2}{k}$. Hence \begin{equation}\label{I4} I_4 \leq \big\|\frac{v_r}{r^{1+k}}\big\|_a \big\|\frac{|v_\varphi|^{2q}}{r^{q(1+\varepsilon)}}\big\|_1^\alpha \big\|\frac{|v_\varphi|^{2q}}{r^{q(1+\varepsilon) + 2}}\big\|_1^\beta \big\|v_\varphi\big\|_{\frac{\gamma a}{a(1 - \alpha - \beta) - 1}}^\gamma. \end{equation} Using Lemma \ref{c1} we get \begin{equation*} \|\frac{v_r}{r^{1+k}}\|_a \leq C\|\frac{\omega_\varphi}{r^k}\|_a \end{equation*} and furthermore \begin{equation}\label{hol} \|\frac{\omega_\varphi}{r^k}\|_a \leq \|\omega_\varphi\|_{a}^{1-k}\|\frac{\omega_\varphi}{r}\|_{a}^k. \end{equation} Since $k \in (0,1]$, $1 0$ such that $\frac{\omega_\varphi}{r}$ is bounded in $L^{\infty,1+\eta}$ and in $L^{1 + \eta, 3(1+\eta)}$ for $0 < \eta \leq \eta_0$. Thus \begin{equation*} I_5 \leq C\|\frac{\omega_\varphi}{r}\|_{3(1+\eta)}\|v_r\| _{\frac{3(1+\eta)}{1 + 2\eta}}\|\omega_\varphi\|_{\frac{3}{2}}^{\frac{1}{2}} \leq C\|\frac{\omega_\varphi}{r}\|_{3(1+\eta)}\|\omega_\varphi\|_{\frac{3}{2}}^{\frac{3}{2} - \frac{2\eta}{1 + \eta}}\|v_r\|_2^{\frac{2\eta}{1+\eta}} \end{equation*} and we can estimate $I_5$ using the Gronwall inequality. Finally, to estimate $I_6$ we will use the fact that there exists $\eta_1 > 0$ such that for $0 < \eta \leq \eta_1$, $\frac{|v_\varphi|^{1+\eta}}{r^{\frac{1+\eta}{2}(1+\varepsilon)}} \in L^{\infty,2}$ and its gradient is bounded in $L^{2,2}$. In fact we will use the same information for $|\frac{v_\varphi}{r^{\frac{1}{2}}}|^{1+\eta}$, but this information is weaker since the main problems are near the axis (recall that due to Lemma \ref{lem2a} $v_\varphi r \in L^{\infty}(\Omega_T)$). \begin{align*} I_6 &= \big|\int\frac{2}{r}v_\varphi\frac{\partial v_\varphi}{\partial z}| \omega_\varphi|^{-1/2}\omega_\varphi\big| \\ &\leq C\big|\int |\omega_\varphi|^{\frac{1}{2}}(\frac{v_\varphi}{\sqrt{r}})^{1-\eta} \frac{\partial}{\partial z}(\frac{v_\varphi}{\sqrt{r}})^{1+\eta}\big| \\ &\leq C\|\omega_\varphi\|_{\frac{3}{2}}^{\frac{1}{2}}\| (\frac{v_\varphi}{\sqrt{r}})^{6(1-\eta)}\|_1^{\frac{1}{6}}\|\nabla| \frac{v_\varphi}{\sqrt{r}}|^{1+\eta}\|_2 \\ &= C\|\omega_\varphi\|_{\frac{3}{2}}^{\frac{1}{2}}\| (\frac{v_\varphi}{\sqrt{r}})^{1+\eta}\|_r^{\frac{r}{6}}\|\nabla| \frac{v_\varphi}{\sqrt{r}}|^{1+\eta}\|_2, \end{align*} where $ r = 6\frac{1-\eta}{1+\eta} < 6$. Thus $I_6$ can be again estimated by means of the Gronwall inequality. Since \begin{gather*} s = \frac{\gamma a}{a(1 - \alpha - \beta) - 1} = \frac{10ap\gamma}{6a\gamma - 10p},\\ t = \gamma B = \frac{10ap\gamma}{10ap(1-\beta) + 15p - 15a - 3ka}, \\ \gamma = \frac{5+k}{3}, \end{gather*} we compute \begin{equation*} \frac{2}{t} + \frac{6}{s} = \frac{7+5k}{2(5+k)} + \frac{9}{5p} - \frac{9}{a(5+k)} - \frac{\varepsilon}{2}. \end{equation*} Now, using that $a< \frac 2k$, $\varepsilon>0$, we get \begin{equation*} \frac 2t + \frac 6s < \frac{7-4k}{2(5+k)} + \frac{9}{5p}. \end{equation*} Recall that $\alpha = \frac{5k+7}{12} - \frac{5+k}{5p} - \varepsilon\frac{5+k}{12}$ needs to be greater than zero. This implies $p > \frac{12}{5}\frac{5+k}{5k+7}$. Using this we get \begin{equation}\label{BB} \frac{2}{t} + \frac{6}{s} < \frac{7-4k}{2(5+k)} + \frac{9}{5p} < \frac{7}{4}. \end{equation} Note that taking $a$ sufficiently close to $\frac 2k$ and $p$ close to $\frac{12}{5}\frac{5+k}{5k+7}$, we get $\frac{2}{t} + \frac{6}{s}$ arbitrarily close to $\frac 74$. Moreover we need $\frac{1}{t} \geq 0$. Thus \begin{equation*} \frac{1}{t} = \frac{7+5k}{4(5+k)} + \frac{9}{2a(5+k)} - \frac{9}{10p} - \frac{\varepsilon}{4} \geq 0 \end{equation*} and consequently $p > \frac{18}{35}\frac{5+k}{1+2k}$. The lowest $s$ we get taking $\frac{1}{t} = 0$ and $\alpha$ almost equal to zero. Therefore we take \begin{equation*} p = \frac{12}{5}\frac{5+k}{5k+7} = \frac{18}{35}\frac{5+k}{1+2k}, \end{equation*} which implies $k = \frac{7}{13}$, $p = \frac{48}{35}$ and $s = \frac{24}{7} + \varepsilon$ for arbitrarily small $\varepsilon$. Note that since $k \in (\frac {7}{13},1)$ and $p \in (\frac 65, \frac {48}{35})$, taking $a$ close to $\frac 2k$, we indeed have $p