How to test the weak solution to hyperbolic conservation law?

Solution 1:

To show that $u$ is a weak solution of this initial-value problem, we show that $$ \int_{0}^{\infty}\int_{-\infty}^{\infty} \left[ \phi_t u + \phi_x f(u)\right] \mathrm{d}x\, \mathrm{d}t = -\int_{-\infty}^{\infty} \phi(x,0)\, u (x,0)\,\mathrm{d}x $$ is satisfied for all $\phi$ in $C_0^1(\mathbb{R}\times \mathbb{R}^+)$. Let us prove this identity in the case $$ u(x,t) = u_1(x,t) = \left\lbrace\begin{aligned}&0 &&\text{if } x < t/2\, ,\\ &1 &&\text{if } x > t/2\, ,\end{aligned}\right. $$ with the flux function of Burgers' equation $f(u) = \frac{1}{2}u^2$. To do so, we split the integral in two parts, and switch the integrals according to the Fubini theorem: \begin{aligned} \int_{0}^{\infty}\!\int_{-\infty}^{\infty} \left[ \phi_t u + \phi_x f(u)\right] \mathrm{d}x\, \mathrm{d}t &= \int_{-\infty}^{\infty}\int_{0}^{\infty} \phi_t u\, \mathrm{d}t\, \mathrm{d}x + \int_{0}^{\infty}\!\int_{-\infty}^{\infty} \phi_x f(u)\, \mathrm{d}x\, \mathrm{d}t \\ &= \int_{0}^{\infty}\int_{0}^{2x} \phi_t \, \mathrm{d}t\, \mathrm{d}x + \frac{1}{2}\int_{0}^{\infty}\!\int_{t/2}^{\infty} \phi_x \, \mathrm{d}x\, \mathrm{d}t \\ &= \int_{0}^{\infty}\left[\phi(x,2x) - \phi(x,0)\right]\mathrm{d}x - \frac{1}{2}\int_{0}^{\infty} \phi(t/2,t)\, \mathrm{d}t \\ &= -\int_{0}^{\infty}\phi(x,0)\,\mathrm{d}x \\ &= -\int_{-\infty}^{\infty}\phi(x,0)\,u(x,0)\,\mathrm{d}x \, . \end{aligned} In the case where $u(x,t) = u_2(x,t)$, the proof is similar.

This is a particular case of exercise 3.4 p 29 of the book [1].

[1] R.J. LeVeque: Numerical Methods for Conservation Laws, 2nd ed., Birkhäuser, 1992.