HA1 problem 6

[Victor Ivrii]

Solutions to be posted as a "Reply" only after January 22, 21:00

a. Solve IVP
\begin{align}
&u_t+uu_x=0,
\label{eq-HA1.10}\\
&u|_{t=0}=x
\label{eq-HA1.11}
\end{align}
b. Describe domain in (x,t) where this solution is properly defined. Consider separately $t>0$ and $t<0$.

[Biao Zhang]

HA1-6

[Jessica Chen]

a. This is a quasilinear equation.
\begin{equation}
\frac{d t}{1} = \frac{d x}{u} = \frac{d u}{0};
\end{equation}
Then we get $ut-x= C$ for some arbitrary constant C.
Consider the initial condition $u|_{t=0}=x$, take an initial point $(0, x_0)$ such that
\begin{equation}
u(x, 0) = x
\end{equation}
 Therefore we have $u = f(x_0) = f(x-ut)$ along characteristics, so
\begin{equation}
u(x,y) = f(x-ut) = x-ut\\
u(x, y) = \frac{x}{1+t}
\end{equation}


b.
When $t>-1$, the solution is clearly hold.
However when $t<-1$ the solution breaks because the characteristics lines cannot be interpreted if they are intersected or undefined.

[Ping Wei]

b） when t < 0 t not equal 0 
      when t > 0 all t are fine

[Victor Ivrii]

This is a quasilinear equation as coefficients depend on $u$. Then we cannot solve for $x,t$ separately; we are looking for 3-dimensional curves defined by $\frac{dt}{1}=\frac{dx}{u}=\frac{du}{0}$ but their $(x,t)$ projections intersect and as soon as it happens we cannot assign a value to $u$.

In this example it happens as $t=-1$: all integral lines intersect in the same points but it is rather an exception (but even less dramatic intersections break things; see Web bonus problems). So, solution as we understand it exists for $t>-1$ only.

Note. However there is a workout via discontinuous solutions which we discuss if time permits in the end of the class. This particular equation is a toy-model for gas dynamics and jumps in solutions are interpreted as shock waves. This is much more advanced topic.

Note. Web Bonus Problem 1 is related.