Toronto Math Forum
APM3462012 => APM346 Math => Home Assignment X => Topic started by: Dana Kayes on October 14, 2012, 03:13:49 PM

For part C, 'Where the solution is fully determined by the initial condition u(x, 0) = g(x).', is supposed to be this a question, 'Where is the solution fully determined ...' or is it asking us to do something?
Thanks

You need to find all points $(x,y)$ where solution is determined by initial conditions

Thanks!

Professor Ivrii,
can we post solutions to this problem set?

Professor Ivrii,
can we post solutions to this problem set?
Yesit was mentioned.

Solution is attached,

(a), (c), (d) solved correctly. On attached picture (built with online plotter http://math.rice.edu/~dfield/dfpp.html (http://math.rice.edu/~dfield/dfpp.html)) where notations are a bit different ($(x,y)$ instead of $(t,x)$) one can see characteristics and initial line $t=0$  bold). The characteristics cross it no more than once (good, no contradiction) and initial data define solution on the characteristic which cross initial line (so $x>t$ which consists of two sectors)
(b), (e) contain glitches:
(b) AR believes that solution has the form $\phi \bigl( \frac{x^2+1}{t^2+1}\bigr)$ which would mean that the solution must be an even function with respect to $t$ and with respect to $x$ which is not necessary the case. In reality $\frac{x^2+1}{t^2+1}$ marks not the characteristic but the whole curve consisting of two disjoint components and on these components $u$ by no means is the same. Therefore absolutely correct answer is
In $k$th sector ($\{x>t\}$, $\{x<t\}$ as $k=1,3$ and $\{t>x\}$, $\{t<x\}$ as $k=2,4$) general solution is
$u(x,t)=\phi _k \bigl( \frac{x^2+1}{t^2+1}\bigr)$ with $\phi_{1,3}(\xi)$ defined for $\xi>1$ and $\phi_{2,4}(\xi)$ defined for $0<\xi<1$.
(one can give other equivalent descriptions).
So, again, initial data define solutions in sectors $1,3$.
In (d) initial function$x^2$ is even with respect to $x$ and therefore the above remark has no effect and the answer is $\phi (\xi)=\xi 1$ and
\begin{equation*}
u(x,t)= \frac{x^2+1}{t^2+1}1 =\frac{x^2t^2}{t^2+1}.
\end{equation*}
In (e) initial function$x$ is odd with respect to $x$ and therefore the above remark works in full: $\phi_{1,3}(\xi)= \pm \sqrt{\xi1}$ and
\begin{equation*}
u(x,t)= \frac{x^2+1}{t^2+1}1 =\left\{\begin{aligned}
&\sqrt{\frac{x^2t^2}{t^2+1}}\qquad &&x>t,\\
&\sqrt{\frac{x^2t^2}{t^2+1}}\qquad &&x<t
\end{aligned}\right.
\end{equation*}
PS Quo usque tandem abutere, Catilina, patientia nostra?
To what length will you abuse our patience, Catiline?
Well, I meant, I am not going to look at your colourful scans anymore >:(