Find the shock curve and a weak solution for a PDE

Question

I have the following problem that asks to find the shock curve for the following IVP $$u_t + (u^2)_x = 0, \quad u(x,0) = \frac{1}{2}e^{-x},$$ to then obtain a weak solution from this curve using Rankine Hugoniot condition. I'm pretty lost on how to raise the problem. In Evans there is an example where it is easy to get u^- and u^+, so we can solve for the shock curve using an ODE. In this case, does anyone have an idea how to raise the problem?

Answer 1

Don't know how to get shock curve analytically, but i computed your PDE numerically with the additional boundary condition: $u(0,t)=\frac{1}{2}$

From the simulation it's to be seen that the shock curve is $x=t$. I know that my answer is mathematically not satisfying, sorry. Perhaps it helps you anyway.

Answer 2

$$u_t+2u\,u_x=0$$ The Charpit-Lagrange characteristic ODEs are : $$\frac{dt}{1}=\frac{dx}{2u}=\frac{du}{0}$$ A first characteristic equation comes from necessarily $du=0$ : $$u=c_1$$ A second characteristic equation comes from solving $\frac{dt}{1}=\frac{dx}{2c_1}$ $$x-2c_1t=c_2$$ The general solution of the PDE expressed on the form of implicit equation $c_1=F(c_2)$ is : $$\boxed{u=F(x-2u\,t)}$$ $F$ is an arbitrary function (to be determined according to the specified condition). $$u(x,0)=\frac12 e^{-x}=F(x)$$ So the function $F$ is determined. We put it into the above general solution where the argument is $(x-2u\,t)$ : $$u=\frac12 e^{-(x-2u\,t)}$$ This is the solution ( on implicit form) of the PDE which satisfies the condition.

In order to obtain the explicit form of the solution we have to solve for $u$ the implicit equation. This is not possible with the elementary functions. A special function is necessary, the LambertW function : https://mathworld.wolfram.com/LambertW-Function.html . The result is : $$\boxed{u(x,t)=-\frac{1}{2t}W\big(-e^{-x}t\big)}$$ For $\quad t\to 0\quad$ then $\quad W\big(- e^{-x}t\big)\to - e^{-x}t\quad$ which shows that the condition is satisfied.

Considering the properties of the Lambert $W(X)$ for negative argument $\quad X=-e^{-x}t\quad$ real solution exists in the range $-e^{-1}<X<0$. Moreover the function is multivalued. So they are two real solutions separated by $-e^{-1}=X=-e^{-x}t$ $$t=e^{x-1}$$ The problem is fully solved on the pure mathematical viewpoint. Since the Rankine Hugoniot condition is a more physical notion the context of the problem should be known in order to discuss the validity of the above solution.