I believe you changed sign going from
Edit. No mistake - my bad.
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
But isn't it true that if
then
and hence
So, I think that's all right.
I think the mistake might be in applying the initial conditions. Are you told that when it's also true that I don't see that in the original problem statement. So, you can only say that
You've parametrized the equation in terms of 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 in there and see what happens.
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 at point
This is the only way we have been taught. I spotted an error where s should be equal to Ie, the minus is changed to a plus. This still doesnt solve the problem....
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
I cant see how this gets u=0. Is this to do with the general solution being defined implicitly?
Thanks
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 's over to one side, etc. I get the following:
which implies either that or that
Does have to be real for all If so, then you can definitely rule out both of these solutions, and you're forced to conclude that Otherwise, I'm not sure you can.