### Author Topic: HA1 problem 6  (Read 670 times)

#### Victor Ivrii

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##### HA1 problem 6
« on: January 20, 2015, 06:55:26 AM »
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

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##### Re: HA1 problem 6
« Reply #1 on: January 22, 2015, 10:23:35 PM »
HA1-6

#### Jessica Chen

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##### Re: HA1 problem 6
« Reply #2 on: January 22, 2015, 10:35:35 PM »
a. This is a quasilinear equation.

\frac{d t}{1} = \frac{d x}{u} = \frac{d u}{0};

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

u(x, 0) = x

Therefore we have $u = f(x_0) = f(x-ut)$ along characteristics, so

u(x,y) = f(x-ut) = x-ut\\
u(x, y) = \frac{x}{1+t}

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.
« Last Edit: January 23, 2015, 09:03:53 AM by Victor Ivrii »

#### Ping Wei

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##### Re: HA1 problem 6
« Reply #3 on: January 23, 2015, 10:24:52 AM »
bï¼‰ 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.