# Toronto Math Forum

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

Title: TT1 = Problem 1
Post 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{eq-1}
\end{equation}

• (a) Find the characteristic curves and sketch them in the $(x,t)$ plane.
• (b) Write the general solution.
• (c) Solve  equation (\ref{eq-1})  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{eq-2}
\end{equation}
is fully determined by the initial condition $u(x,0)=g(x)$ ($x>0$)?
Title: Re: TT1 = Problem 1
Post by: Djirar on October 16, 2012, 08:25:30 PM
My solution. Please check there might be mistakes. Sorry for quality of scan and handwriting.
Title: Re: TT1 = Problem 1
Post by: Aida Razi on October 16, 2012, 08:25:40 PM
Title: Re: TT1 = Problem 1
Post by: Aida Razi on October 16, 2012, 08:27:12 PM
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  :)
Title: Re: TT1 = Problem 1
Post by: Djirar on October 16, 2012, 08:32:04 PM
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 x-axis and the horizontal is y-axis. I hope I didn't forget this on the test  :(
Title: Re: TT1 = Problem 1
Post by: Zarak Mahmud on October 16, 2012, 08:41:46 PM
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 x-axis and the horizontal is y-axis. I hope I didn't forget this on the test  :(

Do you mean the $t$ axis?
Title: Re: TT1 = Problem 1
Post by: Qitan Cui on October 16, 2012, 08:42:02 PM
I have a different solution for bonus
Title: Re: TT1 = Problem 1
Post by: Djirar on October 16, 2012, 08:46:51 PM
I think Qitan Cui is correct. In my haste I must have messed up the integration of the bonus part.
Title: Re: TT1 = Problem 1
Post by: Jinchao Lin on October 16, 2012, 08:52:58 PM
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 $\{(t,x) | tx>-1 \}. Title: Re: TT1 = Problem 1 Post by: Djirar on October 16, 2012, 08:54:19 PM My solution. Please check there might be mistakes. Sorry for quality of print and hand writing. Djirar, I posted my solution 10 seconds after you :) I was scribbling my solutions as fast as I could :) Edit: I forgot to write down the axes to part a. The vertical is x-axis and the horizontal is y-axis. I hope I didn't forget this on the test :( Do you mean the$t$axis? Yes$t$not$y$. Thank you. Title: Re: TT1 = Problem 1 Post by: Ian Kivlichan on October 16, 2012, 09:24:06 PM Subqueston (d):$ \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?
Title: Re: TT1 = Problem 1
Post by: Victor Ivrii on October 16, 2012, 09:43:12 PM
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 $\log |x|-t$ 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.