Hi,

Could anyone check the attached and see where i have gone wrong on this problem. Asked to find parametric form of pde and hence relationship between u and x,y. The general solution does not match the solution for initial conditions

Thanks

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- Dec 22nd 2010, 01:49 PMbugatti79Parametric form of Quasi Linear PDE
Hi,

Could anyone check the attached and see where i have gone wrong on this problem. Asked to find parametric form of pde and hence relationship between u and x,y. The general solution does not match the solution for initial conditions

Thanks - Dec 22nd 2010, 02:38 PMJester
I believe you changed sign going from

$\displaystyle - \dfrac{1}{x} = t + A \; \text{to}\; x = \dfrac{1}{-t - A}$

Edit. No mistake - my bad. - Dec 22nd 2010, 04:53 PMAckbeet
But isn't it true that if

$\displaystyle -\dfrac{1}{x}=t+A,$ then

$\displaystyle \dfrac{1}{x}=-t-A,$ and hence

$\displaystyle x=\dfrac{1}{-t-A}?$ So, I think that's all right.

I think the mistake might be in applying the initial conditions. Are you told that when $\displaystyle u=0,$ it's also true that $\displaystyle t=0?$ I don't see that in the original problem statement. So, you can only say that

$\displaystyle s=\dfrac{1}{-t-A}.$

You've parametrized the equation in terms of $\displaystyle t$ but there's no guarantee that that parametrization has its origin just anywhere you want it, right? I would go through and re-apply the initial conditions after your integrations, and just leave $\displaystyle t$ in there and see what happens. - Dec 23rd 2010, 01:51 AMbugatti79
Danny and Ackbeet,

The problem statement is all that was given however examples in the lectures notes are based on the idea that we parametrise the initial conditions so that they correspond to t=0. So I am assuming this for the above problem, ie

A characteristic passes through $\displaystyle \Gamma$ at point $\displaystyle P (x=s, y=1-s, u=0) at t=0$

This is the only way we have been taught. I spotted an error where s should be equal to $\displaystyle \frac{1}{t+\frac{1}{x}}$ Ie, the minus is changed to a plus. This still doesnt solve the problem.... - Dec 23rd 2010, 02:45 AMAckbeet
So your final equation there is

$\displaystyle \displaystyle y=\frac{u^{2}}{2}+1-\left[\frac{1}{u+\frac{1}{x}}\right],$ incorporating the sign correction.

Evaluating at $\displaystyle u=0, x=s$ yields

$\displaystyle \displaystyle y=1-\left[\frac{1}{\frac{1}{s}}\right]=1-s,$

as required. Which initial condition is not being satisfied here? - Dec 23rd 2010, 03:23 AMbugatti79
mmmm...that makes perfect sense what you did but im going the other way, ie I want to see if u goes to 0 when i sub x=s and y=1-s into general solution.

Afer some manipulation I get

$\displaystyle -s=\frac{u{^2}}{2}-\frac{s}{us+1}$ I cant see how this gets u=0. Is this to do with the general solution being defined implicitly?

Thanks - Dec 23rd 2010, 04:57 AMAckbeet
Your equation is correct. You can keep going, you know. Get a common denominator on the RHS, multiply through by that common denominator, get all the $\displaystyle u$'s over to one side, etc. I get the following:

$\displaystyle u^{3}s+u^{2}+2us^{2}=0,$ which implies either that $\displaystyle u=0,$ or that

$\displaystyle u=\dfrac{-1\pm\sqrt{1-8s^{3}}}{2s}.$

Does $\displaystyle u$ have to be real for all $\displaystyle s?$ If so, then you can definitely rule out both of these solutions, and you're forced to conclude that $\displaystyle u=0.$ Otherwise, I'm not sure you can. - Dec 23rd 2010, 06:04 AMbugatti79
- Dec 23rd 2010, 06:11 AMAckbeet
Hmm. Well then, there's no

*a priori*reason to rule out the nonzero solutions. Incidentally, for those nonzero solutions, you'll get real $\displaystyle u$ if and only if $\displaystyle s \le 1/2$. I don't know how much more can be said on the matter.