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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).

ViktorStein
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Quiet_waters
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1 Answers1

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

Adam Latosiński
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  • Many thanks. Please, could you explain why and how did you use the limit $x/s$ at third line? Thank you very much. – Quiet_waters Apr 17 '19 at 15:10
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    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 $$ \int_0^\infty dt \int_{-\infty}^{st} dx = \int_{-\infty}^\infty dx \int_{{\rm max}{0,x/s}}^\infty dt = \int_{-\infty}^0 dx \int_0^\infty dt + \int_0^\infty dx \int_{x/s}^\infty dt$$ – Adam Latosiński Apr 17 '19 at 15:17
  • Great... I didn't know that I can do this. Many thanks – Quiet_waters Apr 17 '19 at 15:23