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

Title: Problem 1
Post 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
Title: Re: Problem 1
Post by: Victor Ivrii on October 14, 2012, 03:19:38 PM
You need to find all points $(x,y)$ where solution is determined by initial conditions
Title: Re: Problem 1
Post by: Dana Kayes on October 14, 2012, 03:24:26 PM
Thanks!
Title: Re: Problem 1
Post by: Aida Razi on October 14, 2012, 10:06:18 PM
Professor Ivrii,

can we post solutions to this problem set?
Title: Re: Problem 1
Post by: Victor Ivrii on October 14, 2012, 10:17:15 PM
Professor Ivrii,

can we post solutions to this problem set?

Yes--it was mentioned.
Title: Re: Problem 1
Post by: Aida Razi on October 15, 2012, 01:33:21 AM
Solution is attached,
Title: Re: Problem 1
Post by: Victor Ivrii on October 15, 2012, 03:54:23 AM
(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^2-t^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: $\phi_{1,3}(\xi)= \pm \sqrt{\xi-1}$ and
\begin{equation*}
u(x,t)= \frac{x^2+1}{t^2+1}-1 =\left\{\begin{aligned}
&\sqrt{\frac{x^2-t^2}{t^2+1}}\qquad &&x>|t|,\\
-&\sqrt{\frac{x^2-t^2}{t^2+1}}\qquad &&x<-|t|
\end{aligned}\right.
\end{equation*}

Quote
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  >:(