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6 Convergence to an entropy solution

In the last section, we saw the existence of a convergent sequence of approximate. In this section, we show that the limit is a weak solution and satisfies the entropy condition in this section.

To prove the limit is a weak solution, it is sufficient that the integral

$\begin{eqnarray*} I & = & \int\hspace{-6pt}\int _{[0,T]\times[0,1]}(u^\Delta \p... ...+0,x)\phi(0,x)dx \\ & & - \int_0^1 u^\Delta (T+0,x)\phi(T,x)dx \end{eqnarray*}$

tends to zero as $\Delta x$ tends to zero for any $\phi\in C^1_0([0,T]\times[0,1])$ such that $\phi(t,0)=\phi(t,1) \hspace{1em}(0<t<T)$ , because the data $(\bar{u}^0_1,\bar{u}^0_3,\ldots,\bar{u}^0_{2L-1})$ is uniformly bounded and have a subsequence which converges to a function $\bar{u}(x)$ of $L^\infty(0,1)$ .

By the similar calculation in §5, it follows

$\begin{eqnarray*} I & = & \int\hspace{-6pt}\int _{[0,T]\times[0,1]} g\phi dtdx ... ...int_0^1 \phi(u^{2n}_{0,+}-u^{2n}_{+})dx \\ & = & I_1 + I_2 \\ \end{eqnarray*}$

since $\sigma[u^\Delta ]-[f^\Delta ]=0$ for shocks from the Rankine-Hugoniot relation and $\phi_0^{2n-1}=\phi_L^{2n-1}$ , where

$\begin{eqnarray*} I_1 & = & \sum_{n=1}^N \left\{ \sum_{j=1}^{L-1}\int_{E_{2j}^... ... dtdx + \sum_{n=1}^N \int_0^1 \phi(u^{2n}_{0,+}-u^{2n}_{+})dx. \end{eqnarray*}$

tends to zero because

$\begin{eqnarray*} \vert I_1\vert & \leq & \Vert\phi\Vert _{C^1}\Delta x \left(... ...leq & C(\Vert\phi\Vert _{C^1},M,T,G_0,\Lambda_2)\sqrt{\Delta x}. \end{eqnarray*}$

For

, we obtain

$\begin{eqnarray*} I_2 & = & \int\hspace{-6pt}\int _{[0,T]\times[0,1]} g\phi dtd... ...{E_{2j-1}^{2n}} g(t,x)\{\phi(t,x)-\tilde{\phi}^{2n}_{2j-1}\} dx, \end{eqnarray*}$

where

$\begin{displaymath} \tilde{\phi}^n_j = \frac{1}{m(E^n_j)}\int_{E^n_j}\phi(n\Delta t,x)dx. \end{displaymath}$

The function $\phi-\tilde{\phi}^{2n}_{2j-1}$ tends to zero because

$\begin{eqnarray*} \vert\phi-\tilde{\phi}^{2n}_{2j-1}\vert & \leq & \frac{1}{2... ...lta t) \\ & \leq & \Vert\phi\Vert _{C^1}\Delta x(2+1/\Lambda). \end{eqnarray*}$

Therefore we obtain

$\begin{displaymath} \vert I\vert\leq C(\Vert\phi\Vert _{C^1},M,T,G_0,\Lambda_2)\sqrt{\Delta t}. \end{displaymath}$

Lastly, we show that the limit satisfies the entropy condition (13). Let be a smooth entropy pair and be a convex. The weak form of the entropy condition is the inequality

$\begin{displaymath} \int\hspace{-6pt}\int _{t>0} (U\phi_t + F\phi_x + U'g\phi) dtdx\geq 0 \end{displaymath}$

for any $\phi\in C^1((0,\infty)\times{\mbox{\sl R}})$ . It is sufficient for the proof to show that

$\begin{displaymath} \bar{I} = \int_0^T dt\int_0^1(U(u^\Delta )\psi_t + F(u^\Delta )\psi_x + U'(u^\Delta )g\psi_x)dx \geq o(1) \end{displaymath}$

for $\psi\in C^1([0,T]\times[0,1])$ such that

$\begin{displaymath} \psi \geq 0,\hspace{1em}\psi(0,x)=\psi(T,x) \hspace{1em}(x\i... ...1]),\hspace{1em} \psi(t,0)=\psi(t,1) \hspace{1em}(t\in [0,T]). \end{displaymath}$

For convex entropy

we can see

$\begin{displaymath} \sigma[U]-[F] \geq 0 \end{displaymath}$

(cf. [3]). Hence,

$\begin{eqnarray*} \bar{I} & = & \sum_{n=1}^N \int_0^1 [U\psi]_{(2n-1)\Delta t+0... ...}_1 + \bar{L}_2 + \bar{L}_3 +\int\hspace{-6pt}\int U'g\psi dtdx \end{eqnarray*}$

since $u^\Delta (T+0,x)=u^\Delta (+0,x)$ , where

$\begin{eqnarray*} \bar{L}_1 & = & \sum_{n=1}^N\left\{\sum_{n=1}^L \psi^{2n-1}_{... ...2j-1}}(\psi-\psi^{2n}_{2j-1}) (U^{2n}_{-} - U^{2n}_{0,+})dx.\\ \end{eqnarray*}$

By $U''\geq 0$ and $\psi^{2n-1}_0=\psi^{2n-1}_{2L}$ , $\bar{L}_1\geq 0$ . It follows by the similar way in §5 that

$\begin{eqnarray*} \vert\bar{L}_3\vert & \leq & \Vert\psi\Vert _{C^1}\Delta x\ma... ... & C(\Vert\psi\Vert _{C^1},U',M,T,G_0,\Lambda_2)\sqrt{\Delta x}. \end{eqnarray*}$

Hence we have

$\begin{displaymath} \bar{I} \geq \int\hspace{-6pt}\int _{[0,T]\times[0,1]}U'g\psi dtdx + \bar{L}_2 +O(\sqrt{\Delta x}).\end{displaymath}$

(29)

From Taylor's expansion theorem

$\begin{eqnarray*} \bar{L}_2 & = & \sum_{n=1}^N \int_0^1 \psi(2n\Delta t,x)\{U'(... ...-1}U'(u^{2n}_{2j-1}) \int_{E^{2n}_{2j-1}} \psi(2n\Delta t,x)dx. \end{eqnarray*}$

First two terms of the right hand side of (29) become

$\begin{eqnarray*} & & \int\hspace{-6pt}\int _{[0,T]\times[0,1]}U'g\psi dtdx + \... ...\int_0^1\{U'-U'(u^{2n}_{0,+})\} \psi^{2n}_{+}gdx + O(\Delta x), \end{eqnarray*}$

where $\psi^{2n}_{+}$ is a step value function defined by

$\begin{displaymath} \psi^{2n}_{+}(x)=\psi^{2n}_{2j-1} \hspace{2em}(x\in E^{2n}_{2j-1}). \end{displaymath}$

Remember the following lemma proved in [8].

LEMMA 6 Let

be the solution of the Riemann problem (16), $\bar{u}$ be the average of $u(\Delta t,x)$

$\begin{displaymath} u(t,x)=R(t,x;u_l,u_r), \hspace{2em} \bar{u}=\frac{1}{2\Delta x}\int_{-\Delta x}^{\Delta x}u(\Delta t,x)dx \end{displaymath}$

and the maximum wave speed $\Lambda=\max\{\vert f'(u_l)\vert,\vert f'(u_r)\vert\}$ satisfy $\Lambda < \Delta x/\Delta t$ . Then

$\begin{displaymath} \int_{-\Delta x}^{\Delta x}\vert\bar{u}-u(t,x)\vert dx \leq... ...u}-u(\Delta t,x)\vert dx \hspace{2em}(0\leq t\leq \Delta t), \end{displaymath}$

where

is a positive constant

$\begin{displaymath} C=(1-\Lambda \Delta t/\Delta x)^{-1}. \end{displaymath}$

For the approximation $u^\Delta (t,x)$ the constant is

$\begin{displaymath} \left(1-\frac{\Lambda}{2\delta(M+TG_0)+\Lambda}\right)^{-1} =1+\frac{\Lambda}{2\delta(M+TG_0)} \end{displaymath}$

from (19). By Lemma 6 we estimate the rest term

$\begin{eqnarray*} & & \left\vert\sum_{n=1}^N \int_{(2n-2)\Delta t}^{2n\Delta t}... ... u^{2n}_{-}-u^{2n}_{0,+}\vert)dx \\ & \leq & C\sqrt{\Delta x}. \end{eqnarray*}$

Therefore, we obtain $\bar{I}\geq -C\sqrt{\Delta x}$ .

Next: Bibliography Up: Time-periodic solutions Previous: 5 Compactness of entropy

Shigeharu TAKENO
15 January 2002