(PDF file: paper9.pdf)

4 Decay estimates

In this section, we obtain the estimate for the approximation $u^\Delta (t,x)$ constructed in the last section by the similar way to Tadmor ([13]) for fractional step Lax-Friedrichs difference approximation.

The step values $u^n_j=u(n\Delta t,j\Delta x)$ are able to be described as forms of the Lax-Friedrichs difference scheme

$$\begin{array}{l} \left\{\begin{array}{lll} u^{2n+1}_j & =... ...nd{array}\right.\\ \hspace{2em}(n=0,1,2,\ldots), \end{array}$$ (18)

where $f^n_j=f(u^n_j)$ and

$$g^n_j = \frac{1}{4\Delta t\Delta x}\int_{(n-1)\Delta t}^{(n+1)\Delta t} dt \int_{(j-1)\Delta x}^{(j+1)\Delta x} g(t,x) dx.$$

Let $v^n_j$ be the $x$-backward difference of $u^n_j$

$$v^n_j=\frac{u^n_j-u^n_{j-2}}{2\Delta x}.$$

Then,

$$\begin{eqnarray*} %% v^{2n+1}_j & = & \frac{u^{2n+1}_j - u^{2n+1}_{j-2}}{2\Dx} ... ...1}) -\Delta t\delta\frac{(v^{2n}_{j+1})^2+(v^{2n}_{j-1})^2}{2}. \end{eqnarray*}$$

Similarly,

$$\begin{eqnarray*} v^{2n+2}_j & \leq & \frac{v^{2n+1}_{j+1}+v^{2n+1}_{j-1}}{2} ... ...frac{(v^{2n+1}_{j+1})^2+(v^{2n+1}_{j-1})^2}{2} - 2 G_1\right\}, \end{eqnarray*}$$

where $G_1$ is the value in (15). Let $N^n=\max_{j}v^n_j$. $N^n$ is non-negative because the summation

$$\sum_{j\in J_n,0\leq j< 2L}v^n_j$$

equals zero from the space periodicity of $u^n_j$. A function

$$h(x) = \frac{A}{2}x - \frac{\Delta t\delta}{2}x^2 \hspace{2em}(A>0)$$

increases for $x\leq A/(2\Delta t\delta)$. Hence, if

$$N^{2n}\leq \frac{1}{2\Delta t\delta}(1- \frac{\Delta t}{\Delta x}\Lambda)$$

then

$$N^{2n+1}=\max_{j}v^{2n+1}_j \leq N^{2n} - \Delta t\delta(N^{2n})^2.$$

Similarly,

$$N^{2n+1}\leq \frac{1}{2\Delta t\delta}(1- \frac{\Delta t}{\Delta x}\Lambda)$$

yields

$$N^{2n+2}=\max_{j}v^{2n+1}_j \leq N^{2n} - \Delta t\{\delta(N^{2n})^2-2G_1\}.$$

The simple estimate for $N^n$

$$N^n = \max_j v^n_j = \max_j \frac{u^n_j-u^n_{j-2}}{2\Delta x} \leq \frac{M+TG_0}{\Delta x}$$

shows that if

$$2\delta(M+TG_0)+\Lambda \leq \frac{\Delta x}{\Delta t},$$ (19)

then

$$\frac{N^{2n+2}-N^{2n}}{2\Delta t} \leq G_1 - \frac{\delta}{2}(N^{2n})^2$$ (20)

for $n \leq N-1$ since

$$\begin{eqnarray*} N^{2n+1} & \leq & N^{2n} - \Delta t\delta (N^{2n})^2 \leq N^{... ...\\ & \leq & N^{2n} - \Delta t\delta (N^{2n})^2 + 2\Delta tG_1. \end{eqnarray*}$$

The solution $y(t)$ of ordinary differential equation

$$\left\{\begin{array}{l} \displaystyle y'=G_1 - \frac{\delta}{2}y^2 [3ex] y(0)=N^0\end{array}\right.$$

tends to the value

$$\alpha=\sqrt{2G_1/\delta}$$ (21)

as $t$ tends to infinity. If $N^0>\alpha$ then $y(t)$ is monotone decreasing convex function, and if $N^0<\alpha$ then $y(t)$ is monotone increasing concave one. In the latter case, the tangent line of $y(t)$ started $t=t_0$ across the line $y=\alpha$ at

$$t=t_0 + \frac{2}{\delta(y(t_0)+\alpha)}$$

and the time is not smaller than $t_0 + 1/(\delta\alpha)$. Hence, if $N^{2n}\leq\alpha$ then $N^{2m}\leq\alpha$ for $n\leq m\leq N$ provided that

$$2\Delta t\leq \frac{1}{\delta\alpha}$$ (22)

from the inequality (20). In the case $N^0>\alpha$, it is easy to show that

$$N^{2m} \leq y(2m\Delta t) \leq \alpha+\frac{2\alpha}{\mbox{\large\it e}^{2m\alpha\delta\Delta t}-1} \hspace{2em}(m\leq N).$$ (23)

We consider the estimate for $u^{2N}_j$ using above estimates for $v^n_j$. For the summation of $u^{2N}_j$

$$\begin{eqnarray*} \lefteqn{2\Delta x\sum_{j=1}^L u^{2N}_{2j-1}} \\ & = & 2\De... ...t_0^1 u_0(x)dx + \int_0^T dt\int_0^1 g(t,x)dx \\ & = & \bar{u} \end{eqnarray*}$$

from (3).

Next simple lemma is used to the estimate for $u^{2N}_j$.

LEMMA 3   Let $p_j$ be real values which satisfy

$$p_1+p_2+\cdots+p_K = 0,$$

and let

$$q_j = \left\{\begin{array}{ll} p_j-p_{j-1} & (j=2,3,\ldots,K),\\ p_1-p_K & (j=1). \end{array}\right.$$

Then

$$\max_j \vert p_j\vert\leq \sum_{j=1}^K \vert q_j\vert \leq 2K\max_j q_j.$$

The proof is omitted.

Let $K=L$, $p_j=u^{2N}_{2j-1}-\bar{u}$ in Lemma 3, then $q_j=2v^{2N}_{2j-1}\Delta x$ and

$$\vert u^{2N}_j-\bar{u}\vert \leq 2L \max_j 2\Delta xv^{2N}_j... ...\alpha + \frac{4\alpha}{\mbox{\large\it e}^{\alpha\delta T}-1}$$

from (23).

PROPOSITION 4   Under assumptions (19), (22)

$$\vert u^{2N}_j-\bar{u}\vert\leq 2\alpha + \frac{4\alpha}{\mbox{\large\it e}^{\alpha\delta T}-1}.$$ (24)

Proposition 4 give the time-periodic solution of the difference approximation.

Let $A$ be sufficiently large number such that

$$A > 2\alpha + \frac{4\alpha}{\mbox{\large\it e}^{\alpha\delta T}-1},$$

and let $M=\vert\bar{u}\vert+A$. We take mesh lengths $\Delta x$ and $\Delta t$ to satisfy (19), (22) and

$$\frac{\Delta x}{\Delta t} \leq \Lambda_2\equiv 2\{2\delta(M+TG_0)+\Lambda\}$$ (25)

for the estimation of the entropy (§5). Let $D_L$ be a set

$$\begin{eqnarray*} \lefteqn{ D_L=D_L(\bar{u},A) }\\ & = & \{(u_1,u_2,\ldots,u_... ...u_j = \bar{u}, \hspace{1em}\max \vert u_j-\bar{u}\vert\leq A\}. \end{eqnarray*}$$

If we take $(u^0_1,u^0_3,\ldots,u^0_{2L-1})$ from $D_L$, set $u^0_j$ periodically for any $j\in J_0$, then we can construct the approximation $u^\Delta (t,x)$ by the way in §3. Obviously $D_L$ is a closed convex set and the range of the mapping

$$D_L \ni (u^0_1,u^0_3,\ldots u^0_{2L-1}) \mapsto (u^{2N}_1,u^{2N}_3,\ldots u^{2N}_{2L-1})$$

is contained in $D_L$ from Proposition 4. Hence, we obtain the fixed point

$$(\bar{u}^0_1,\bar{u}^0_3,\ldots,\bar{u}^0_{2L-1})$$ (26)

of the mapping from Brouwer's fixed point theorem in $D_L$. Of course, the fixed point depends on $\Delta x$.