3

I am considering the following two problems: $$ u_t + \left( \frac{u^2}{2} \right)_x = 0 $$ and $$ \left( \frac{u^2}{2} \right)_t + \left( \frac{u^3}{3} \right)_x = 0 $$

I know that these Equations have the same continuous solutions.

I am now considering the Riemann problem with $u = 2$ at $x<0$ and $u = 1$ at $x>0$. I am trying to find the discontinuous solutions and explain why they are different.

For the first problem, I used the Rankine-Hugoniot relation to find the shock speed $s=\frac{3}{2}.$ This gives (I think) the weak solution of $u(x,t) = 2$ for $x \lt st$ and $u(x,t) = 1$ for $x \gt st$.

However, I am stuck on the second problem. I tried to rewrite it as $ w_t + (f(w))_x = 0$ where $w=\frac{u^2}{2}$ and $f(w) = \frac{ 2^{3/2} w^{3/2} }{3}$. Then the initial condition for $w$ is $w(x,0) = 2$ for $x<0$ and $w(x,0) = 1/2$ for $x>0.$ Applying the Rankine-Hugoniot relation to this $f(w)$ with $w^-=2$ and $w^+= \frac{1}{2}$ I obtain a shock speed of $s=\frac{14}{9}.$ Is this is so, the solution I obtain is $ w(x,t) = 2$ for $x<\frac{14t}{9}$ and $w(x,t) = 1/2$ for $ x> \frac{14t}{9}.$ Plugging in $w=\frac{u^2}{2}$ and solving for $u$ yields the same solution as for the first problem but with a different shock speed.

Furthermore, I am not sure of how to explain why their discontinuous solutions are different?

Math Undergrad Student
  • 123
  • For the Riemann problem the discontinuity is "in space" for the initial condition at $t=0$, so it is not clear to me why you are mentioning $t=0^+$ and $t=0^-$. See e.g. https://math.stackexchange.com/q/2827384/532409 – Quillo Dec 08 '24 at 17:05
  • @LorenzoPompili if I continue with that approach, I will get a different answer. However, I am still not sure of the explanation for why the problems have the same continuous solution and different discontinuous solutions. Any thoughts? – Math Undergrad Student Dec 08 '24 at 17:08
  • 1
    @Quillo I think who edited the post miswrote that part – Lorenzo Pompili Dec 08 '24 at 17:09
  • @LorenzoPompili I just added my attempt! – Math Undergrad Student Dec 08 '24 at 17:39
  • @LorenzoPompili yes sorry I will fix that now! Is the answer then just that the have the solutions have a discontinuous solution with a similar structure but that moves at a different speed? Why? – Math Undergrad Student Dec 08 '24 at 17:47
  • 1
    About your question. To show that the two PDEs have the same continuous solutions, you essentially need that $(u^2/2)x=u(u^3/3)_x$, and similarly $(u^2/2)_t=uu_t$. I think the main point is that these identities only hold and make sense for functions that are regular enough, but they break down if the function is discontinuous. In case you are wondering, yes, the two equations make sense for a large class of functions that are not continuous, which includes the shock-type solutions (this needs the language of distributions: https://en.wikipedia.org/wiki/Distribution(mathematics) ) – Lorenzo Pompili Dec 08 '24 at 17:54
  • @LorenzoPompili could you explain why we require $(u^2/2)_x = u(u^2/3)_x$ and $(u^2/2)_t = uu_t$ for the continuous solutions to be the same? – Math Undergrad Student Dec 08 '24 at 18:01
  • (* I meant $(u^3/3)_x=u(u^2/2)_x$). We need these identities because by substitution, the second PDE becomes $u(u_t+(u^2/2)_x)=0$. So if a function solves the first PDE, it also solves the second PDE (the term between parentheses is zero). The other implication is a little more messy to prove rigorously, but morally, if $u$ solves the second PDE, then either $u=0$ (and this solves the first PDE), or the term in parentheses is zero, and so $u$ solves the first PDE. – Lorenzo Pompili Dec 08 '24 at 18:06
  • I think it may become more clear by writing these out as integral conservation laws. The R-H conditions guarantee unique solutions by satisfying these integral conservation laws across discontinuities, but the integral equations are not the same, hence the solutions should not necessarily be the same. The only reason they are the same in the case of strong solutions is because of the chain rule as Lorenzo pointed out – whpowell96 Dec 08 '24 at 20:17

0 Answers0