How to solve a 2nd order pde with constant a?

[maggie56]

Homework Statement

I have a pde,
16d²u/dxdy + du/dx + du/dy + au = 0 where a is constant.

Homework Equations

The Attempt at a Solution

I have tried to solve this pde using the substitutions x=e^t and y=e^s so t=ln(x) and s=ln(y) then finding
Du/dx= 1/x du/dt and du/dy= 1/y du/ds
For d²u/dxdy i am unsure if my answer is correct,
1/xy d²u/dsdt - 1/x du/dt - 1/y du/ds

When i substitute these into the pde i get 16/xy d²u/dsdt + au = 0
I could integrate this with respect to s and t but don't think that helps me.

Am i using the correct method here or is there a method that is better suited to my equation

Thank you for any help

 

[obafgkmrns]

Try setting u(x,y) = s(x)t(y) and separating the equation.

 

[maggie56]

Sorry, not sure i follow, what do you suggest i could set them as?

 

[obafgkmrns]

If you substitute u(x,y) = s(x)t(y) into your PDE, you'll obtain

16 (ds/dx)(dt/dy) + (ds/dx) t + s (dt/dy) + a s t = 0

Rearrange & factor:

(dt/dy)[16 (ds/dx) + s] = -t [(ds/dx) + a s]

Separate the variables:

(dt/dy) / t = -[(ds/dx) + s] / [16 (ds/dx ) + a s]

The left side is a function of x only & the right side is a function of y only. The only way that can be so is if both sides are equal and constant. So

(dt/dy) / t = C

-[(ds/dx) + s] / [16 (ds/dx + a s] = C

That leaves you with two linear ODEs. C is an eigenvalue that will be determined by the PDE's boundary conditions.

For more info, check out

http://mathworld.wolfram.com/SeparationofVariables.html

or

http://en.wikipedia.org/wiki/Separation_of_variables