Recent content by Chingon

Ok, so since ∂F/∂y = 0, that means it has a homogeneous solution of the form Y=Ʃcos(n∏y/(b/2)) correct? So that would cancel out the Y terms in the Fourier Expansion of f(x,y)?

Ok, so based on my boundary conditions, I think I have to expand f(x,y) as a double Fourier cosine expansion, so: f(x,y)= 1/2Ʃ(C2cos(n∏x/a) + C3cos(n∏y/b)) + ƩƩC4cos(m∏x/a)cos(n∏y/b) Also, I can let F(x,y)=u(x,y)=X(x)Y(y) Plugging that into the original Poisson equation I end up with...

Homework Statement Solve the equation: ∂2F/∂x2 + ∂2F/∂y2 = f(x,y) Boundary Conditions: F=Fo for x=0 F=0 for x=a ∂F/∂y=0 for y=0 and y=b Homework Equations How can I find Eigengunctions of F(x,y) for expansion along Y in terms of X? The Attempt at a Solution I can't imagine...

Hmm... I suppose I should just give that a shot. Thanks!

Homework Statement I need to expand 1/y(x)2 , where y(x)=x1/2Ʃ(-1)n/(n!)2 * (3x/4)n for n=0 to ∞ Homework Equations How does one arrive at the correct solution (-coefficients seem to vanish, only + remain)? The Attempt at a Solution I know that x1/2Ʃ(-1)n/(n!)2 * (3x/4)n expands...

Yes, but I was focused on the brute force of integration by parts method as opposed to using this theorem. I think I'm just going to go the easy route and use this method - hopefully I won't lose too many points!

So I guess I'm wondering how to handle the partial derivatives inside the integrand. Take the first term for example: ∫∂/∂x(AyBz)dxdydz Do the ∂x and dx cancel out?

No, we haven't studied those so I don't believe that's a path I can take. I have used that identity to reduce the problem to ∫Div(A×B)=0 This reduces to: ∫[∂xAy*Bz-∂xAz*By+∂yAz*By-∂yAx*BZ+∂ZAx*By-∂zAy*Bx]dV

Homework Statement ∫Bdot[∇×A]dV=∫Adot[∇×B]dV Prove this by integration by parts. A(r) and B(r) vanish at infinity. Homework Equations I'm getting stuck while trying to integrate by parts - I end up with partial derivatives and dV, which is dxdydz? The Attempt at a Solution I...