### Author Topic: Problem 1  (Read 7301 times)

#### Dana Kayes

• Jr. Member
•  • Posts: 6
• Karma: 0 ##### Problem 1
« 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

#### Victor Ivrii ##### Re: Problem 1
« Reply #1 on: October 14, 2012, 03:19:38 PM »
You need to find all points $(x,y)$ where solution is determined by initial conditions

#### Dana Kayes

• Jr. Member
•  • Posts: 6
• Karma: 0 ##### Re: Problem 1
« Reply #2 on: October 14, 2012, 03:24:26 PM »
Thanks!

#### Aida Razi

• Sr. Member
•    • Posts: 62
• Karma: 15 ##### Re: Problem 1
« Reply #3 on: October 14, 2012, 10:06:18 PM »
Professor Ivrii,

can we post solutions to this problem set?

#### Victor Ivrii ##### Re: Problem 1
« Reply #4 on: October 14, 2012, 10:17:15 PM »
Professor Ivrii,

can we post solutions to this problem set?

Yes--it was mentioned.

#### Aida Razi

• Sr. Member
•    • Posts: 62
• Karma: 15 ##### Re: Problem 1
« Reply #5 on: October 15, 2012, 01:33:21 AM »
Solution is attached,

#### Victor Ivrii ##### Re: Problem 1
« Reply #6 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) 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}
Well, I meant, I am not going to look at your colourful scans anymore 