(PDF file: paper9.pdf)

2 Preliminaries

A function $u(t,x)$ is called the weak solution of the time-periodic problem (1) and

$$\left\{\begin{array}{ll} u(t,0)=u(t,1) & (0<t<T),\\ u(0,x)=u(T,x) & (0<x<1) \end{array}\right.$$ (7)

with period $T$ if the function $u(t,x)$ is bounded measurable in the region $(0,T)\times(0,1)$, and there exists a bounded measurable function $\bar{u}(x)$ such that space-periodic extensions of $u(t,x)$, $g(t,x)$ and $\bar{u}(x)$ satisfy

$$\displaystyle \int\hspace{-6pt}\int _{0<t<T} (u\phi_t + f(u... ...+ \int_{{\mbox{\sl R}}} \bar{u}(x)\{\phi(0,x)-\phi(T,x)\}dx = 0$$ (8)

for any $\phi(t,x) \in C^1_0([0,T]\times{\mbox{\sl R}}_x)$.

Note that the definition has other equivalent forms. One is the following:

$$\displaystyle \int\hspace{-6pt}\int _{[0,T]\times[0,1]} (u\... ...\phi) dxdt + \int_0^1 \bar{u}(x)\{\phi(0,x)-\phi(T,x)\}dx = 0$$ (9)

for any $\phi(t,x) \in C^1_0([0,T]\times[0,1])$, $\phi(t,0)=\phi(t,1) \hspace{1em}(0<t<T)$. Another is expressed in terms of the space-periodic extension of $\bar{u}$ and for the space and time periodic extension of $u(t,x)$. That is, it is required to satisfy

$$\displaystyle \int\hspace{-6pt}\int _{t>0} (u\phi_t + f(u)\... ...\phi) dxdt + \int_{{\mbox{\sl R}}} \bar{u}(x)\phi(0,x)dx = 0,$$ (10)

for any $\phi(t,x) \in C^1_0([0,\infty)\times{\mbox{\sl R}})$. The conclusion is that $u(t,x)$ satisfies (9) if and only if the followings valid

1. $u$ is solution of (1) in weak sense
   
   $$u_t+f(u)_x=g \hspace{2em}\mbox{distribution sense in } (0,T)\times(0,1).$$
   
   

2. $u(t,x)$ converges in weak sense to $u(x)$ as $t$ tends to zero and as $t$ tends to $T$
   
   $$\frac{1}{\varepsilon }\int_0^\varepsilon u(t,x)dt, \hspace{... ...\hspace{1em} \mbox{as} \hspace{1em}\varepsilon \downarrow 0$$
   
   
   in $L^\infty(0,1)$ weak$\ast$.
3. $f(u(t,\varepsilon ))$ and $f(u(t,1-\varepsilon ))$ converge in weak sense to the same value
   
   $$\frac{1}{\varepsilon }\int_0^\varepsilon f(u(t,x))dx, \hspa... ... \hspace{1em}\mbox{as} \hspace{1em}\varepsilon \downarrow 0$$
   
   
   in $L^\infty(0,T)$ weak$\ast$ for some function $\bar{f}(t)$ of $L^\infty(0,T)$.

It seems that the function $u$ which satisfies above conditions is a solution of the problem (1) satisfying

$$\left\{\begin{array}{ll} f(u(t,0))=f(u(t,1)) & (0<t<T),\\ u(0,x)=u(T,x) & (0<x<1)\\ \end{array}\right.$$ (11)

instead of (7). Certainly, both definitions are equivalent for the weak solution. However, these include different means for the entropy condition. The boundary condition of the original problem (7) seems to say that the space-periodic extension satisfies the equation (1), but the problem (11) does not seem to require it. Hence in the case that boundary values of the weak solution $u(t,0)$ and $u(t,1)$ are different, these should satisfy the entropy condition, that is,

$$f'(u(t,1))>0>f'(u(t,0))$$

for (7), but should not for (11). It remains an open problem whether the boundary condition (11) is well-posed.

A smooth function pair of $u$

$$(U(u),F(u))$$

is an entropy pair for the scalar conservation law

$$u_t+f(u)_x=0$$ (12)

if

$$U(u(t,x))_t + F(u(x,t))_x=0$$

for a smooth solution of (12). This is equivalent that $U$ and $F$ satisfy

$$F'(u)=U'(u)f'(u).$$

The function $U$ called entropy and $F$ called entropy flux. A weak solution $u$ satisfies the entropy condition if

$$U(u(t,x))_t + F(u(x,t))_x\leq U'(u(t,x))g(t,x) \hspace{1em}\mbox{in } (0,T)\times{\mbox{\sl R}}$$ (13)

for any smooth entropy pair with convex entropy $U$.

We suppose that the function $f(u)$ is smooth,

$$f''(u)\geq \delta >0 \hspace{1em}(u\in{\mbox{\sl R}}),\hspace{2em}\int_0^1 dx\int_0^T g(t,x)dt = 0,$$ (14)

and $g(t,x)$ is a time-periodic function with period $T$. The last relation for $g(t,x)$ is need for the existence of a time periodic solution.

We also assume that the space-extension of $g(t,x)$ satisfies

$$g(t,x)-g(t,y) \leq G_1(x-y) \hspace{2em}(x>y)$$ (15)

for any $x,y$ and $t$, where $G_1$ is a constant. The condition is necessary by a technical reason.

It is well-known that the solution of the Riemann problem for the scalar conservation law

$$\left\{\begin{array}{l} u_t+f(u)_x=0 \hspace{1em}(t>0, x\i... ...x<0),\\ u_r & (x>0)\\ \end{array}\right. \end{array}\right.$$ (16)

is each of the two typical waves. In the case $u_l<u_r$ the solution is the rarefaction wave

$$u(t,x) = \left\{\begin{array}{ll} u_l & (x/t < f'(u_l)),\\... ...) < x/t < f'(u_r)),\\ u_r & (x/t > f'(u_r))\end{array}\right.$$

and in the case $u_l>u_r$ the solution is the shock wave

$$u(t,x) = \left\{\begin{array}{ll} u_l & (x/t < s),\\ u_r & (x/t > s), \end{array}\right.$$

where the shock speed $s$ is determined by the Rankine-Hugoniot relation

$$f(u_r)-f(u_l)=s(u_r-u_l)$$

and Lax's entropy condition

$$f'(u_l)>s>f'(u_r)$$

(cf. [7], [12]).

We note that the average on the line $t=\Delta t$

$$\frac{1}{2\Delta x}\int_{-\Delta x}^{\Delta x}u(\Delta t,x)dx$$

for the solution of the Riemann problem is equal to the Lax-Friedrichs difference approximation

$$\frac{u_r+u_l}{2} - \frac{\Delta t}{2\Delta x}\{f(u_r)-f(u_l)\}$$

provided the Courant-Friedrichs-Lewy (CFL) condition

$$\frac{\Delta x}{\Delta t} > \max\{\vert f'(u_l)\vert,\vert f'(u_r)\vert\}.$$