Web Bonus Problem 1

[Victor Ivrii]

There are several subproblems and one can submit solution for any of them

Solve (for $t>0$)
\begin{align}
&u_t + u u_x=0,\label{eq-1}\\
&u|_{t=0}=f(x)\label{eq-2}
\end{align}
with one of the following initial data
\begin{align}
f(x)=&\left\{\begin{aligned}
-1& && x<-a,\\
x/a& && -a\le x \le a,\\
1& && x>a;
\end{aligned}\right.
\label{eq-3}\\
f(x)=&\left\{\begin{aligned}
1& && x<-a,\\
-x/a& && -a\le x \le a,\\
-1& && x>a;
\end{aligned}\right.
\label{eq-4}\\
f(x)=&\left\{\begin{aligned}
-1& && x<0,\\
1& && x> 0.\\
\end{aligned}\right.
\label{eq-5}
\end{align}
Here $a>0$ is a parameter. Plotting solutions for different $t>0$ would be appreciated

[Jessica Chen]

I interpret this question in the following:

This is a quasilinear equation with coefficient u.
As we solved in HA1, the general solution for $u$ is $ u = f(x_0) = f(x-ut)$, now consider the first set of initial data.
Case1. If $x_0 < -a$, then we have $u = -1$ and also $x_0 = x+t < -a$.
Case2. If  $x_0 > a$, then we have $u = 1$ and also $x_0 = x-t > a$.
Case3. If $-a\le x \le a$, then we have $u = \frac{x}{a}$.
\begin{equation}
x_0 = x - \frac{x}{a}t\rightarrow u = \frac{x}{a+t}
\end{equation}
Then the solution looks like
\begin{align}
u(x,t)=&\left\{\begin{aligned}
-1& && x<-a-t,\\
\frac{x}{a+t}& && -a-t\le x \le a+t,\\
1& && x>a+t;
\end{aligned}\right.
\\
\end{align}

[Victor Ivrii]

a. Jessica considered case: initial data (\ref{eq-3}) and for everybody convenience I just draw $u(x,t)$ for fixed $t$ (WBP1-1.png); as $t$ increases threshold points $-a-4$ and $a+t$ run to the left and right respectively, so the plot just stretches horizontally and solution is defined for all $t>0$. So we are done here!

b. initial data (\ref{eq-4})—any takers? I just remark that it is pretty similar to what happens in a as $t<0$

c. initial data (\ref{eq-5})—any takers?

[Jessica Chen]

The general solution for $u$ is $ u = f(x_0) = f(x-ut)$, now consider the first set of initial data.
When t = 0,
Case1. If $x < -a$, then we have $u = 1$
Case2. If  $x > a$, then we have $u = -1$
Case3. If $-a\le x \le a$, then we have $u =- \frac{x}{a}$.
\begin{equation}
u = -\frac{x-ut}{a}\rightarrow u = \frac{x}{t-a}
\end{equation}
Then the solution looks like
\begin{align}
u(x,t)=&\left\{\begin{aligned}
1& && x<-a+t,\\
\frac{x}{t-a}& && -a+t\le x \le a-t,\\
-1& && x>a-t;
\end{aligned}\right.
\\
\end{align}

Thus as t increase, -a+t and a-t runs toward each other, the the solution graph is squeezing. The solution is define for $0<t<a$.

[Victor Ivrii]

Thus as $t$ increases, $-a+t$ and $a-t$ runs toward each other, the the solution graph is squeezing. The solution is define for all $t>0$.

While the first sentence above is correct, the conclusion in the second sentence is wrong. Continuous solution exists for $t<a$ only (figure WBP1-2.png), as $t=a$ we get a step (figure WBP1-3.png), and for $t>a$ we get a Z-shaped "graph" (figure WBP1-4.png)--and obviously it cannot be a graph of any function on interval $(a-t,t-a)$!

c. Is obtained from a by setting $a=0$.

Remark. To consider b for $t\ge a$ we rewrite equation as $u_t+(\frac{1}{2}u^2)_x=0$ which is considered in the class of discontinuous functions and understood in the weak (or distributional) sense. Then  (figure WBP1-3.png) stays for $t>a$. We will detail it if time permits in the very end of the class.