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Given a pde $$u_y+e^uu_x=0$$ $x \in \mathbb{R}$, $t>0$ with the initial conditions $$u(x,0)=f(x)=\begin{cases} 2 & x<0 \\ 1 & x>0 \end{cases}$$ Solve the pde.

My attempt so far, I find the characteristic curves for $(s,0,f(s))$ $$\frac{dy}{dt}=1$$ which gives $$y=t$$ $$\frac{du}{dt}=0$$ which gives $$u=f(s)$$ and $$\frac{dx}{dt}=e^u \implies $$ $$x=e^{f(s)}t+s$$ So $$u(x,y)=f(x-e^uy)=\begin{cases} 2 & x< e^2 t \\ 1 & x>et \end{cases}$$

Now I know that to find the time the shock forms I must find $t$ such that $$(e^u,1) \cdot (0,-e^{f(s)}f'(s)-1)=0 \implies $$ which gives $$t=\frac{1}{e^{f(s)} f'(s)}$$

But the problem is that $f(s)$ is constant on each interval so the derivative vanishes, therefore the shock time becones infinite, this is what I don't understand, how can I find the time the shock forms?

EditPiAf
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  • In the wording of your question they are three variables $x,y,t$, but no $t$ in the PDE itself. It is written that $u=u(x,y)$ and that $u(x,0)$ is the initial condition. This isn't correct since $y=0$ is not an initial condition which should be at $t=0$. Would you mind clarify and rewrite correctly the question. – JJacquelin Dec 30 '20 at 08:57

1 Answers1

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The initial data is discontinuous. Therefore, this is a Riemann problem for the PDE $$ u_y + (e^u)_x = 0 $$ rewritten in conservation form, whose flux function $u\mapsto e^u$ is convex. In the present case, the base characteristics intersect, and the solution is a single shock wave with speed $$ s = \frac{e^2 - e^1}{2 - 1} = e\, (e-1) \approx 4.67 $$ deduced from the Rankine-Hugoniot condition. The shock forms immediately, at $y=0^+$.

EditPiAf
  • 21,328
  • But what is the breaking time? How do I find it? –  Dec 29 '20 at 20:30
  • My question was to find the time $t$ the shock forms, in this cases the time $t=\frac{1}{e^{f(s)}f'(s)}$ is not defined for $f'(s)=0$ so how can I find $t$? –  Dec 29 '20 at 20:41
  • Should I expect that in any Riemann problem with piecewise constant initial conditions where a shock forms, that the shock forms at $t=0$? –  Dec 29 '20 at 20:47
  • @lost-mid For Riemann problems, solutions are either shocks (forming at $t=0$) or smooth rarefaction waves (or combinations of those). Hence yes, shock formation cannot happen elsewhere for Riemann problems. – EditPiAf Dec 29 '20 at 20:49