Toronto Math Forum
APM3462012 => APM346 Math => Term Test 1 => Topic started by: Victor Ivrii on October 16, 2012, 06:26:37 PM

Consider the first order equation:
\begin{equation}
u_t + x u_x = 0.
\label{eq1}
\end{equation}
 (a) Find the characteristic curves and sketch them in the $(x,t)$ plane.
 (b) Write the general solution.
 (c) Solve equation (\ref{eq1}) with the initial condition $u(x,0)= \cos(2x)$.
Explain why the solution is fully determined by the initial condition.
 (d) bonus Describe domain in which solution of
\begin{equation}
u_t + x^2 u_x = 0, \qquad x>0
\label{eq2}
\end{equation}
is fully determined by the initial condition $u(x,0)=g(x)$ ($x>0$)?

My solution. Please check there might be mistakes. Sorry for quality of scan and handwriting.

Check the attachment please,

My solution. Please check there might be mistakes. Sorry for quality of print and hand writing.
Djirar, I posted my solution 10 seconds after you :)

My solution. Please check there might be mistakes. Sorry for quality of print and hand writing.
Djirar, I posted my solution 10 seconds after you :)
I was scribbling my solutions as fast as I could :)
Edit: I forgot to write down the axes to part a. The vertical is xaxis and the horizontal is yaxis. I hope I didn't forget this on the test :(

My solution. Please check there might be mistakes. Sorry for quality of print and hand writing.
Djirar, I posted my solution 10 seconds after you :)
I was scribbling my solutions as fast as I could :)
Edit: I forgot to write down the axes to part a. The vertical is xaxis and the horizontal is yaxis. I hope I didn't forget this on the test :(
Do you mean the $t$ axis?

I have a different solution for bonus

I think Qitan Cui is correct. In my haste I must have messed up the integration of the bonus part.

Subqueston (d):
$ \frac{dt}{1} = \frac{dx}{x^2} $
$ t = x^{1}+c $
so the general solution is $ u(t,x)=f(t+x^{1})$.
$u(0,x)=f(x^{1})=g(x)$
$f(y)=g(y^{1})$
Since $y=x^{1}$, so when $x>0$ we have $y>0$ as well.
$u(t,x)=f(t+x^{1})$
We need $t+x^{1}>0$
Since $x>0$,
therefore $tx+1>0$
so the domain be defined is $\{(t,x)  tx>1 \}.

My solution. Please check there might be mistakes. Sorry for quality of print and hand writing.
Djirar, I posted my solution 10 seconds after you :)
I was scribbling my solutions as fast as I could :)
Edit: I forgot to write down the axes to part a. The vertical is xaxis and the horizontal is yaxis. I hope I didn't forget this on the test :(
Do you mean the $t$ axis?
Yes $t$ not $y$. Thank you.

Subqueston (d):
$ \frac{dt}{1} = \frac{dx}{x^2} $
$ t = x^{1}+c $
so the general solution is $ u(t,x)=f(t+x^{1})$.
$u(0,x)=f(x^{1})=g(x)$
$f(y)=g(y^{1})$
Since $y=x^{1}$, so when $x>0$ we have $y>0$ as well.
$u(t,x)=f(t+x^{1})$
We need $t+x^{1}>0$
Since $x>0$,
therefore $tx+1>0$
so the domain be defined is $\{(t,x)  tx>1 \}.
I think Jinchao has the most correct solution.
Qitan, is it possible to only have the one discontinuity in your solution  won't your characteristic curves be "blocked" by the discontinuity at tx=1, and not able to go any further?

Djirar solved (a), (c) correctly and made a mistake in (b), Aida solved (a)(c) correctly. If we consider $x$ which could be negative and positive $\log xt$ is not good enough as it is the same on two disjoint curves, but $xe^{t}$ works. If initial function was not even then Djirar would not be able honestly to satisfy it from the "general" solution.
Jinchao (BTW, change your name with proper capitalization and as on BlackBoard) is correct and Ian' remark explain fallacy of Qui solutions.
In two attached pictures one can see vector fields and curves for (1) and (2) respectively; one can see that certain characteristics never intersect initial curve. Condition $x>0$ was given only for a sake of simplicity.