- #1
vemarli
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Homework Statement
Two concentric cylinders with radii a & b (b>a) with an infinitely long grounded strip along the z-direction are given potentials [tex]\phi_1[/tex] and [tex]\phi_2[/tex].
Find [tex]\Phi(r,\phi)[/tex] for a<r<b
Boundary conditions:
[tex]\Phi(r,2n\pi)=0[/tex]
[tex]\Phi(a,\phi)=\phi_1[/tex]
[tex]\Phi(b,\phi)=\phi_2
[/tex]
Homework Equations
The laplace equation in polar coordinates (as the solution is independent of z)
[tex]
\Delta^2\Phi=0[/tex]
Ansatz:
[tex]\Phi=R(r)Q(\phi)[/tex],
compute
[tex]\frac{r^2}{R(r)Q(\phi)}\Delta^2(R(r)Q(\phi))=0[/tex]
[tex]\frac{1}{r}\frac{d}{dr}(r\frac{dR}{dr})+\frac{1}{Q(\phi)}\frac{dQ(\phi}{d\phi^2}=0[/tex]
Assume that the first term equals n^2, which means that the second one equals -n^2
-> Solutions: (n=0):
[tex]R(r)=a_0 +b_0ln(r);[/tex]
[tex]Q(\phi)=A_0+B_0\phi[/tex]
and (n>1):
[tex]R(r)=a_nr^n +b_nr^{-n};[/tex]
[tex]Q(\phi)=A_ncos(n\phi)+B_nsin(n\phi) [/tex]
[tex]\Phi(r,\phi)=(a_0 +b_0ln(r))(A_0+B_0\phi)+\sum_{n=1}^\infty(a_nr^n +b_nr^{-n})(A_ncos(n\phi)+B_nsin(n\phi))[/tex]
The Attempt at a Solution
[/B]
From the boundary conditions we see that the solution needs to be periodic in [tex]\phi=2n\pi; n=0,1,...[/tex] which leads to
[tex]A_n=A_0=B_0=0[/tex]
This gives the solution:
[tex]\Phi(r,\phi)=\sum_{n=1}^\infty(a_nr^n +b_nr^{-n})sin(n\phi)[/tex]
Where I have now put B_n coefficients into a_n and b_n.
Now I need to satisfy
[tex]\Phi(a,\phi)=\phi_1[/tex]
[tex]\Phi(b,\phi)=\phi_2[/tex]
[tex]\Phi(a,\phi)=\sum_{n=1}^\infty(a_na^n +b_na^{-n})sin(n\phi)=\phi_1[/tex]
[tex]\Phi(a,\phi)=\sum_{n=1}^\infty(a_nb^n +b_nb^{-n})sin(n\phi)=\phi_2[/tex]
The problem with this is that I would say that the solutions need to be symmetric for phi = [0,pi] and phi = [0,-pi] i.e. The solution should be mirrored in these two domains Phi(r,phi)=Phi(r,-phi), and thus it would be sufficient to solve the boundary conditions only in one of these domains. But only having sinus functions would lead to a discontinuity at phi=plusminus pi... which makes me question the legitimity of this solution.
Furthermore I don't know how to satisfy the boundary conditions on r=a,b; I guess that if I assume phi_1>phi_2 then a_n=0, likewise, phi_1<phi_2 gives b_n=0 and phi_1=phi_2 gives R(r) = constant.
But, to satisfy the finite constant potentials at r=a,b means that we can not have sin(n\phi) solutions i.e. n=0 Which also means that we can't satisfy the fact that the strip is grounded (Phi=0).
So all in all I would assume that the ansatz is wrong... can anyone give a helping hand??