for some function u(x,y)

Rewrite the PDE and IC's in terms of by expressing

and in terms of and and solve.

My attempt: I believe the first equation is of the form

therefore but they are not in terms of and

Have I gone one step too far?

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- February 11th 2011, 09:06 AMbugatti79Solve Linear PDE via a change of variables
for some function u(x,y)

Rewrite the PDE and IC's in terms of by expressing

and in terms of and and solve.

My attempt: I believe the first equation is of the form

therefore but they are not in terms of and

Have I gone one step too far? - February 11th 2011, 09:11 AMAckbeet
Could be way off here, but couldn't you just use this:

- February 11th 2011, 09:21 AMbugatti79
- February 11th 2011, 09:43 AMAckbeet
Right. And, of course, the

are all easy to compute. - February 11th 2011, 09:57 AMbugatti79
- February 11th 2011, 10:03 AMAckbeet
- February 11th 2011, 11:34 AMbugatti79
Wow, that looks good I must say...never saw this implemented in pde's before. Transposing of the matrix involves making the first row the first column and the second row the scond column. so if im right then

. Rewriting the pde becomes

=1

The IC being for

Is this correct...Can this be solved using the method of characteristics? ill try.. - February 11th 2011, 11:45 AMAckbeet
From

you should do regular matrix multiplication to arrive at

Where does that take you? - February 11th 2011, 11:53 AMbugatti79
- February 11th 2011, 11:56 AMAckbeet
Assuming you've plugged everything in correctly, doesn't that simplify down quite a bit? You've got a cancellation, and if I'm not mistaken, the identity might be useful.

- February 11th 2011, 01:48 PMbugatti79
- February 11th 2011, 01:50 PMAckbeet
No, I think when you integrate you're going to have to have

right? - February 11th 2011, 02:01 PMbugatti79
- February 11th 2011, 02:05 PMAckbeet
It's because the integration you did was sort of "partial integration w.r.t. ", not a total integration like you'd have in an ODE setting. You remember doing exact ODE's? When you integrated w.r.t. , you'd get an arbitrary function of . In your case, period. It

*is*a function of two variables. If you doubt the extra function , then just do a partial differentiation and see if you get back to the original DE. You do, right?

What is the IC, exactly? - February 11th 2011, 02:15 PMbugatti79