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laplace in polar coords

 
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Jul23-11, 06:32 PM   #1
 

laplace in polar coords


1. The problem statement, all variables and given/known data
solve the polar coord equation


2. Relevant equations
Urr+(1/r)Ur+(1/r2)U[itex]\vartheta[/itex][itex]\vartheta[/itex]=0; 1[itex]\leq[/itex]r[itex]\leq[/itex]2, 0[itex]\leq[/itex][itex]\vartheta[/itex][itex]\leq[/itex][itex]\pi[/itex]/4
U(1,[itex]\vartheta[/itex])=0 U(2,[itex]\vartheta[/itex])=[itex]\vartheta[/itex]
U[itex]\vartheta[/itex](r,0)=0 U([itex]\vartheta[/itex],[itex]\pi[/itex]/4)=0
and has the form
U(r,[itex]\vartheta[/itex])=Uo(r,[itex]\vartheta[/itex])+[itex]\sum[/itex]cnRn(r)[itex]\vartheta[/itex]n([itex]\vartheta[/itex])

3. The attempt at a solution
so I used variation of parameters and the homogeneous boundaries with [itex]\vartheta[/itex] to get [itex]\vartheta[/itex]n([itex]\vartheta[/itex])=cos(4n[itex]\vartheta[/itex]) then taking r2R''+rR'-cR=0 and using the Euler equation to get r(r-1)+r-c which is r=[itex]\pm[/itex]c, c=16n2
so that will give me Rn(r)=Cnr4n+dnr-4n and knowing condition U(1,[itex]\vartheta[/itex])=0 bn=0 so whats left is Cnr4n I got stuck here how do I solve for Cn do I try subbing in conditions to solve or is there some equation im missing
and then using the
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laplace transform, pde, polar coordinates
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