Prove that shock wave is weak solution of Burgers' equation (Riemann problem) In math modeling studies, I need to prove that

$$u(x,t)=\begin{cases}u_l\qquad x<st\\ u_r\qquad x>st\end{cases}$$
  where $$s=(u_l+u_r)/2$$

is a weak solution for the Riemann problem of Burgers' equation $u_t+uu_x=0$ with the Riemann data

$$u(x,0)=\begin{cases}u_l\qquad x<0\\
u_r\qquad x>0\end{cases}$$

Integrating with a test function $\phi\in C^1_0$, I got
$$
\int_0^\infty\int_{-\infty}^{\infty}\left[u\phi_t+\frac{u^2}{2}\phi_x\right]dx \ dt=\dfrac{u^2_l-u_r^2}{2}\int_0^\infty \phi(st,t)dt-\int_{-\infty}^{\infty} \phi(x,0)u(x,0)dx.
$$
How can I cancel $\displaystyle \dfrac{u^2_l-u_r^2}{2}\int_0^\infty \phi(st,t)dt$?
Many thanks for a help.

This is Exercise 3.4 p. 29 of the book Numerical Methods for Conservation Laws by R.J. LeVeque (Birkäuser, 1992).
 A: I believe that you've missed a certain term in the calculations, one that would cancel the one you have a problem with.
My calculations, assuming $s>0$:
\begin{align} & \int_0^\infty dt \int_{-\infty}^\infty dx (u \phi_t + \frac12 u^2 \phi_x) = \\
&= \int_0^\infty dt \int_{-\infty}^{st} dx \big( u_l \phi_t + \frac12 u_l^2 \phi_x \big) + \int_0^\infty dt \int_{st}^\infty dx \big( u_r \phi_t + \frac12 u_r^2 \phi_x \big) = \\
&=u_l \Big(\int_{-\infty}^0 dx \int_{0}^\infty dt \,\phi_t + \int_0^\infty dx \int_{x/s}^\infty dt \,\phi_t\Big) + \frac12 u_l^2 \int_0^\infty dt \int_{-\infty}^{st} dx \,\phi_x + \\
&\qquad + u_r \int_0^\infty dx \int_0^{x/s} dt \,\phi_t + \frac12 u_r^2 \int_0^\infty dt \int_{st}^\infty dx \,\phi_x = \\
&= -u_l \int_{-\infty}^0 dx \,\phi(x,0) - u_l \int_0^\infty dx\, \phi (x,x/s) + \frac12 u_l^2 \int_0^\infty dt \,\phi(st,t) + \\
&\qquad -u_r \int_0^\infty dx \,\phi(x,0) + u_r \int_0^\infty dx \,\phi(x,x/s) - \frac12 u_r^2 \int_0^\infty dt \,\phi(st,t) = \\
&= -\int_{-\infty}^\infty dx\, u(x,0)\phi(x,0) + (u_r-u_l) \int_0^\infty dx \,\phi(x,x/s) + \frac{u_l^2-u_r^2}{2} \int_0^\infty dt \,\phi(st,t)\end{align}
After a change of variables the second term cancels the third, since $s=\frac{u_l+u_r}{2}$.
