3
$\begingroup$

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).

$\endgroup$
1
  • 1
    $\begingroup$ The particular case $u_l = 0$, $u_r = 1$ is tackled here. $\endgroup$
    – EditPiAf
    Apr 18 '19 at 13:35
4
$\begingroup$

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 $s=\frac{u_l+u_r}{2}$.

$\endgroup$
3
  • $\begingroup$ Many thanks. Please, could you explain why and how did you use the limit $x/s$ at third line? Thank you very much. $\endgroup$ Apr 17 '19 at 15:10
  • 1
    $\begingroup$ In the second line, first term line I have $0 < t<\infty$, $-\infty < x < st$. After the change of the order of the integration I have $-\infty < x < \infty$, but I still have other conditions for $t$ that must be satisfied: the condition $x<st$ that can be rewritten as $t>x/s$ (assuming $s>0$), and condition $t>0$. I can write them both as $t>{\rm max}\{0,x/s\}$ and I get $$ \int_0^\infty dt \int_{-\infty}^{st} dx = \int_{-\infty}^\infty dx \int_{{\rm max}\{0,x/s\}}^\infty dt = \int_{-\infty}^0 dx \int_0^\infty dt + \int_0^\infty dx \int_{x/s}^\infty dt$$ $\endgroup$ Apr 17 '19 at 15:17
  • $\begingroup$ Great... I didn't know that I can do this. Many thanks $\endgroup$ Apr 17 '19 at 15:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.