Toronto Math Forum
APM3462015S => APM346Home Assignments => Web Bonus Problems => Topic started by: Victor Ivrii on January 08, 2015, 03:43:12 AM

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{eq1}\\
&u_{t=0}=f(x)\label{eq2}
\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{eq3}\\
f(x)=&\left\{\begin{aligned}
1& && x<a,\\
x/a& && a\le x \le a,\\
1& && x>a;
\end{aligned}\right.
\label{eq4}\\
f(x)=&\left\{\begin{aligned}
1& && x<0,\\
1& && x> 0.\\
\end{aligned}\right.
\label{eq5}
\end{align}
Here $a>0$ is a parameter. Plotting solutions for different $t>0$ would be appreciated

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(xut)$, 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 $x_0 = xt > a$.
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<at,\\
\frac{x}{a+t}& && at\le x \le a+t,\\
1& && x>a+t;
\end{aligned}\right.
\\
\end{align}

a. Jessica considered case: initial data (\ref{eq3}) and for everybody convenience I just draw $u(x,t)$ for fixed $t$ (WBP11.png); as $t$ increases threshold points $a4$ 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{eq4})â€”any takers? I just remark that it is pretty similar to what happens in a as $t<0$
c. initial data (\ref{eq5})â€”any takers?

The general solution for $u$ is $ u = f(x_0) = f(xut)$, 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{xut}{a}\rightarrow u = \frac{x}{ta}
\end{equation}
Then the solution looks like
\begin{align}
u(x,t)=&\left\{\begin{aligned}
1& && x<a+t,\\
\frac{x}{ta}& && a+t\le x \le at,\\
1& && x>at;
\end{aligned}\right.
\\
\end{align}
Thus as t increase, a+t and at runs toward each other, the the solution graph is squeezing. The solution is define for $0<t<a$.

Thus as $t$ increases, $a+t$ and $at$ 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 WBP12.png), as $t=a$ we get a step (figure WBP13.png), and for $t>a$ we get a Zshaped "graph" (figure WBP14.png)and obviously it cannot be a graph of any function on interval $(at,ta)$!
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 WBP13.png) stays for $t>a$. We will detail it if time permits in the very end of the class.