
#1
Nov106, 03:56 PM

P: 286

Does
[tex]\nabla ^2 u(r,\theta) = 0 [/tex] with the boundary conditions [tex]u(1,\theta) = u(2,\theta) = \sin^2 \theta [/tex] have any solutions? This was a problem on my exam but someone must have written the conditions wrong, or am I stupid? 



#2
Nov106, 04:08 PM

P: 1,079

Why not? It's a welldefined, wellbehaved 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. 



#3
Nov106, 04:53 PM

P: 286

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




#4
Nov106, 05:26 PM

P: 1,079

Laplace equation
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. 



#5
Nov206, 02:25 AM

P: 286

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




#6
Nov206, 12:42 PM

Sci Advisor
HW Helper
PF Gold
P: 12,016

[tex]\sin^{2}(\theta)=\frac{1}{2}\frac{\cos(2\theta)}{2}[/tex]




#7
Nov206, 01:08 PM

P: 1,079

Basically, you are free to choose any combination of the coefficients you need to match the boundary conditions. There are some integrals you can do to solve this analytically, but in most cases that you are likely to see in homework or on a test, you can just pick out the coefficients by inspection. 



#8
Nov206, 01:23 PM

P: 286

Damn it. I'm not worth those 4 points. ;)



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