I'm taking Non-Linear PDEs course this semester. I'm stuck in this HW problem.
Solve the IVP:
$ u_t+uu_x =0, \ \ x \in \mathbb{R}, \ t\ge 0 $
$$
u(x,0) = \left\lbrace
\begin{aligned}
& 1 &&\text{for}\; x < -1\\
& -x &&\text{for}\; {-1}\leqslant x \leqslant 0 \\
& 1 &&\text{for}\; x > 0\, .
\end{aligned}
\right.
$$
This is my attempt.
$\frac{du}{dt}=0$ along $\frac{dx}{dt} = u$
$\implies u = \xi$ along $x = tu_0(\xi)+\xi$
For $\xi < -1$, $u = 1$ along the characteristic $x = t+\xi$
For $\xi \in (-1,0)$, $u = -\xi$ along $x = -\xi t+\xi$ \implies $u = \frac{x}{t-1}$
For $\xi >0$, $u = 1$ along the characteristic $x = t+\xi$
Plotting these characteristics (on desmos), we can observe that there is a void. The horizontal axis is $x$, and the vertical axis is $t$.
Hence, a fan of characteristics will emerge from $(x,t) =(0,0)$. In that region $u =\frac{x}{t}$ as shown below.
For $t>1$, the fan of characteristics emerging from origin intersects the characteristics $x= -\xi t+\xi$ emerging from $-1 <x<0$. Hence, a shock would emerge. From Rankine-Hugoniot Conditions, shock speed can be calculated as:
$$\frac{ds}{dt} = \frac{u^{+}+u^{-}}{2} = \frac{\frac{s}{t}+\frac{s}{t-1}}{2}$$
On solving this ODE with the initial condition as $s(t = 1+) = 0$, we get: $s = C \sqrt{t(t-1)}$ where $C$ is an arbitrary constant. Now, here is the problem,
- How to get $C$?
- On plotting the shock path, another void seems to be created which doesn't seem quite right!
Can you please clarify whether my approach is correct? Any help would be greatly appreciated.
PS: Apologies for poor formatting/images.
Best Answer
JJacquelin's answer resolves the solution for $x<0$ and $0<t<1$. Notice that there is a rarefaction fan of the form $u = x/t$ starting at $(0,0)$, so this should also contain a condition where $u(x,t) = x/t$ for $0\leq x \leq t$ for $0\leq t\leq1$. From this and JJaacquelin's answer, we can determine that the solution at $t=1$ is given by $$ u(x,1) = \begin{cases} 1 & x<0 \\ x & 0<x<1 \\ 1 & x>1 \end{cases}. $$ We now solve this initial value problem starting at $t=1$. We can see that a shock will occur at $x=0$, so our solution for $t>1$ will look like $$ u(x,t) = \begin{cases} 1 & x<s(t) \\ x/t & s(t) < x < t \\ 1 & x > t \end{cases}, $$ where $x = s(t)$ is the shock curve. To find the shock curve, we apply the Rankine-Hugoniot conditions, which say $$ s'(t) = \frac{f(u_+) - f(u_-)}{u_+ - u_-} = \frac{u_+ + u_-}{2} = \frac{1}{2} + \frac{s}{2t}, \, s(1) = 0. $$ Notice here that we substituted $u_+ = x/t$ and $x = s(t)$ to obtain this ODE. Solving obtains $s(t) = t - \sqrt{t}$. We can combine this with the $0<t<1$ solution to obtain a final solution.