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

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- Nov 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 - Nov 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:

$\displaystyle y^2\frac{\partial^2 p}{\partial z^2}-(z^a+2xy)\frac{\partial p}{\partial z}+x^2u=0$

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

$\displaystyle ay''-(b+x)y'+cy=0;\quad y=f(x)$

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):

$\displaystyle p(x;\{a,b,c,k_1,k_2\})$

Now, in terms of the PDE, the arbitrary constants are functions of the other two variables. Call them $\displaystyle f(x,y), g(x,y)$. Now, using the solution you get for y above (p), now make the substitutions $\displaystyle a=y^2,\; b=2xy,\; c=x^2,\; k_1=f(x,y),\; k_2=g(x,y)$. 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. :) - Nov 21st 2008, 07:12 AMshawsend
Alright, I'd like to know why this wouldn't work: I start with

$\displaystyle y^2\frac{\partial^2 p}{\partial z^2}-(z+2xy)\frac{\partial p}{\partial z}+x^2p=0$

but first solve:

$\displaystyle a y''-(t+b)y'+cy=0$

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

$\displaystyle y(t)=\sum_{n=0}^{\infty}k_n t^n$

where $\displaystyle k_n=\frac{\frac{b}{a}(n-1) k_{n-1}-\frac{1}{a}k_{n-2}(c-n+2)}{n(n-1)}$

with $\displaystyle k_0, k_1$ arbitrary. I now make the back-substitutions from the PDE and obtain a power series solution in z:

$\displaystyle p(x,y,z)=\sum_{n=0}^{\infty}f_n(x,y)z^n$

where:

$\displaystyle f_n(x,y)=\frac{\frac{2x}{y}(n-1)f_{n-1}(x,y)-\frac{1}{y^2}(x^2-1)f_{n-2}(x,y)(3-n)}{y^2 n(n-1)}$

with $\displaystyle f_0(x,y)$ and $\displaystyle f_1(x,y)$ arbitrary functions of x and y.