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This is an exercise in Evans, Partial Differential Equations(1st edition), page 164, problem 13

Assume $F(0) = 0, u$ is a continuous integral solution of the conservation law: $$ \left\{ \begin{array}{rl} u_t + F(u)_x = 0 &\mbox{ if $\mathbb{R} \times (0,\infty)$} \\ u=g &\mbox{ if $\mathbb{R} \times \left\{t=0\right\} $} \end{array} \right. $$ and $u$ has compact support in $\mathbb{R} \times [0,\infty]$. Prove for all $t>0$ $$ \int_{-\infty}^{\infty} u(\cdot,t)\,dx = \int_{-\infty}^{\infty}g \,dx. $$

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Notice that in the equality you want to prove: $$ \int_{-\infty}^{\infty} u(x,\color{red}{t})\,dx = \int_{-\infty}^{\infty}g\, dx, $$ the left term is a function of the temporal variable while the right term is a real constant assuming that $u(x,0) = g\in L^1(\mathbb{R})$ and is compactly supported as well. If for the moment we assume $u$ is smooth, take the time derivative on the left: $$ \frac{d}{d\color{red}{t}}\int_{-\infty}^{\infty} u(x,t)\,dx = \int_{-\infty}^{\infty} \frac{\partial}{\partial \color{red}{t}}u(x,t)\,dx,\tag{1} $$ where we performed the differentiation under the integral sign for $u$ is smooth. The right side of above is: $$ \int_{-\infty}^{\infty} \frac{\partial}{\partial t}u(x,t)\,dx = -\int_{-\infty}^{\infty} \frac{\partial}{\partial x} F\big(u(x,t)\big)\,dx =- F\big(u(x,t)\big)\Big\vert^{\infty}_{-\infty}=0, $$ because of $u$ is compactly supported and $F(0) = 0$. Hence $$ \int_{-\infty}^{\infty} u(x,t)\,dx = \int_{-\infty}^{\infty}u(x,0)\, dx = \int_{-\infty}^{\infty}g\, dx. $$ This gives you an idea of why this is called conservation law.

Back to the integral solution $u$ of the conservation law (the definition is somewhere earlier in that conservation law chapter of Evans): $$ \int^{\infty}_0 \int^{\infty}_{-\infty} \Big(u v_t + F(u) v_x\Big)\,dxdt + \int^{\infty}_{-\infty} gv\,dx \big|_{t=0}= 0,\tag{2} $$ for $v\in C^{\infty}_c(\mathbb{R}\times [0,\infty))$. Now $u$ only lies in $L^{\infty}\cap C_c$ ($u$ may not be differentiable anymore, thinking all those blows up in time, and shock waves in space!), the trick in (1) is not applicable anymore, here the way to prove this is to choose proper test function $v$.

Think $u(x,\tau)$ for any $\tau>0$, and consider the problem when the time starts at $\tau$: $$ \left\{ \begin{aligned} u_t + F(u)_x &= 0 &\text{ in } \mathbb{R} \times (\tau,\infty), \\ u&=u(x,\tau) &\text{ at } \mathbb{R} \times \left\{t=\tau\right\} , \end{aligned} \right.\tag{$\star$} $$ The weak solution to $(\star)$ coincides with the original IVP if we assume the solution is unique (but unfortunately this is not true for the integral solution, that's why we solve for an entropy solution by artificially adding a diffusion perturbation, which is the so-called vanishing viscosity method). Then we can see $u$ satisfies: $$ \int^{\infty}_{\tau} \int^{\infty}_{-\infty} \Big(u v_t + F(u) v_x\Big)\,dxdt + \int^{\infty}_{-\infty} uv\,dx \big|_{t=\tau}= 0,\tag{3} $$ for $v\in C^{\infty}_c(\mathbb{R}\times [\tau,\infty))$ which can be easily extended to the whole time domain smoothly. The difference between (2) and (3) is $$ \int^{\tau}_{0} \int^{\infty}_{-\infty} \Big(u \color{blue}{v_t} + F(u) \color{blue}{v_x}\Big)\,dxdt + \int^{\infty}_{-\infty} gv\,dx \big|_{t=0} - \int^{\infty}_{-\infty} uv\,dx \big|_{t=\tau}= 0.\tag{4} $$ Now you can either argue by (A) the test function makes (2) and (3) together must vanish on $[0,\tau)$, hence the blue terms are gone, or (B) choosing $v = 1$ on a set containing the support of $u$ from $t=0$ to $t=\tau$, and the support of $F(u)$ (notice $F\big(u(x\to \infty,t)\big) = F(0) = 0$, the compactly supportedness of $u$ implies that $F(u)$ is compactly supported), then blue terms are gone as well.

Hence we have by (4): for any $\tau>0$ $$\int^{\infty}_{-\infty} gv \,dx \big|_{t=0} =\int^{\infty}_{-\infty} uv\,dx \big|_{t=\tau}.$$ By the choice of the test function $v$ above in (B), we have $$\int^{\infty}_{-\infty} g(x) \,dx =\int^{\infty}_{-\infty} u(x,\tau)\,dx.$$

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Hint: Consider $\int_\infty^\infty \int_0^t u_s ds dx $.

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What is $u_s$ in your hint and could you maybe elaborate on the idea? – user76189 May 6 '13 at 14:26
@user76189 $u_s$ is the same as $u_t$, you need a different name to avoid name clash. – Ma Ming May 7 '13 at 9:16
@MaMing Integral solution to the conservation laws is not generally differentiable, and you can't do this. – Shuhao Cao Jul 20 '13 at 20:20
@ShuhaoCao I am too naïve, good to know your solution. – Ma Ming Jul 22 '13 at 21:00
@MaMing No you are not, for smooth solution, your method works! Sorry for being rude for the previous comment, cheers. – Shuhao Cao Jul 31 '13 at 1:42

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