V0ODO0CH1LD
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Homework Statement
A hollow cylinder with radius ##a## and height ##L## has its base and sides kept at a null potential and the lid on top kept at a potential ##u_0##. Find ##u(r,\phi,z)##.
Homework Equations
Laplace's equation in cylindrical coordinates:
\nabla^2u=\frac{1}{r}\frac{\partial}{\partial{}r}\left(r\frac{\partial{}u}{\partial{}r}\right)+\frac{1}{r^2}\frac{\partial^2u}{\partial\phi^2}+\frac{\partial^2u}{\partial{}z^2}=0
Boundary conditions (specific to this problem):
\begin{eqnarray}<br /> u(r,\phi,0)=0\text{ for }0≤r≤a \\<br /> u(a,\phi,z)=0\text{ for }0≤z≤L \\<br /> u(r,\phi,L)=u_0\text{ for }0≤r≤a<br /> \end{eqnarray}
The Attempt at a Solution
After substituting ##u=R(r)\Phi(\phi)Z(z)## and dividing the equation by ##R(r)\Phi(\phi)Z(z)## we get
\frac{1}{rR(r)}\frac{d}{dr}(rR'(r))+\frac{1}{r^2\Phi(\phi)}\Phi''(\phi)+\frac{1}{Z(z)}Z''(z)=0.
Since the ##z## term is isolated we can set it to a constant ##k^2## to get ##Z(z)=Ae^{kz}+Be^{−kz}##. Then we multiply through by ##r^2## and isolate the ##\phi## term and set it to a constant ##-m^2## to get ##\Phi(\phi)=C\cos(m\phi)+D\sin(m\phi)##. The original equation now looks like
\frac{r}{R(r)}\frac{d}{dr}(rR'(r))-m^2+(rk)^2=r\frac{d}{dr}(rR'(r))+[(rk)^2-m^2]R(r)=0,
where we can substitute ##y=rk## to get
y^2R''(y)+yR'(y)+[y^2-m^2]R(r)=0,
the Bessel equation.
How do I use the boundary conditions to move forward from here? What does the format of the answer I am looking for looks like?
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