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 Logarythmic Nov1-06 03:56 PM

Laplace equation

Does

$$\nabla ^2 u(r,\theta) = 0$$

with the boundary conditions

$$u(1,\theta) = u(2,\theta) = \sin^2 \theta$$

have any solutions?

This was a problem on my exam but someone must have written the conditions wrong, or am I stupid?

 TMFKAN64 Nov1-06 04:08 PM

Why not? It's a well-defined, well-behaved boundary condition over an enclosed space (the region between two concentric spheres).

The general solution in spherical coordinates is sum (Ai r^i + Bi / r^(i+1))Pi(cos theta). Plug in r=1 and r=2, and rewrite the sin^2 theta as 1 - cos^2 theta... then you only need the first few Legendre polynomials and you can solve for the coefficients.

 Logarythmic Nov1-06 04:53 PM

But the problem is to be done in plane polar coordinates...?

 TMFKAN64 Nov1-06 05:26 PM

Ah, right, sorry, I'm used to looking at these things in spherical coordinates... oh well, same wine, different bottle.

The general solutions look like A0 ln r + B0 + sum (Ai r^i + Bi / r^i)(Ci cos (i theta) + Di sin (i theta)). Use the double angle formula to get your cos^2 term.

 Logarythmic Nov2-06 02:25 AM

How am I ever gonna get $$u(2,\theta)$$, wich contains an $$ln 2$$, to equal $$\sin^2 \theta$$? And how can you possibly write $$\sin^2 \theta$$ as a linear combination of sin and cos?

 arildno Nov2-06 12:42 PM

$$\sin^{2}(\theta)=\frac{1}{2}-\frac{\cos(2\theta)}{2}$$

 TMFKAN64 Nov2-06 01:08 PM

Quote:
 Quote by Logarythmic How am I ever gonna get $$u(2,\theta)$$, wich contains an $$ln 2$$, to equal $$\sin^2 \theta$$?