Can someone help me to solve this equation or even classify it..

(z^a)*p[x,y,z]=(x^2)*p[x,y,z]-(2*x*y)*p'[x,y,z]+(y^2)*p''[x,y,z]

where p'[x,y,z],p''[x,y,z] means partial derivative with respect to z

and 0<a<1

Thanks

Printable View

- November 19th 2008, 12:17 PMtursecond order partial differential equation
Can someone help me to solve this equation or even classify it..

(z^a)*p[x,y,z]=(x^2)*p[x,y,z]-(2*x*y)*p'[x,y,z]+(y^2)*p''[x,y,z]

where p'[x,y,z],p''[x,y,z] means partial derivative with respect to z

and 0<a<1

Thanks - November 21st 2008, 03:52 AMshawsend
Here's my suspicion but I'm not entirely sure: Don't let the partials intimidate you. It's still an ordinary DE in the variable z:

Alright let's down-grade it first: drop the a for now and consider just the ordinary-looking DE:

For constants a, b, and c. See the similarity? Ok, we can probably solve this and end up with a solution containing two arbitrary constants along with the a,b, and c (I'll call the solution p now):

Now, in terms of the PDE, the arbitrary constants are functions of the other two variables. Call them . Now, using the solution you get for y above (p), now make the substitutions . I bet a dollar this solution will then solve the PDE for a=1.

Yea, I know, it's not the one with variable a.. . . a journey begins with a first step. :) - November 21st 2008, 07:12 AMshawsend
Alright, I'd like to know why this wouldn't work: I start with

but first solve:

via power series. My initial results show the solution to be:

where

with arbitrary. I now make the back-substitutions from the PDE and obtain a power series solution in z:

where:

with and arbitrary functions of x and y.