Question on Dirichlet problem in cylinder with radial symmetry.

[yungman]

This is not homework.

I have problem deriving the solution for cylinder with radial symmetry given:

\nabla^2U(\rho,z)=R''+\frac{1}{\rho}R'+\frac{Z''}{Z}=0

Which give \rho^2 R''+ \rho R' -k\rho^2 R=0 \hbox { and } Z''+kZ=0

With given boundary conditions U(\rho,0) = U(\rho,h) =0 \hbox { for } \rho<a \hbox{ and } U(a,z) = f(z)

The book claimed k=-\lambda^2 = -ve can only produce trivial solution of R. I cannot verify this.

k=-ve give \rho^2 R''+ \rho R' +\lambda^2 R=0

This is parametric Bessel's equation of order zero which give R(\rho)=c_1 J_0(\lambda_n\rho) + c_2 Y_0(\lambda_n\rho)

R(0) is bounded \Rightarrow c_2 = 0 \Rightarrow R(\rho)= c_1 J_0(\lambda_n\rho)

That's where I get stuck. I cannot rule this out R(\rho)= c_1 J_0(\lambda_n\rho) as a solution with the given boundary condition.

The book claimed only k=+ve would give solution. Please give me some suggestion.

 

[LCKurtz]

It is the Z boundary value problem which restricts the values of k. Your Z bvp is:

Z'' + k Z = 0
Z(0) = Z(h) = 0

If you solve this you will find that only k = μ² > 0 gives non-trivial solutions. So these are the values that must be used in the other equation too.

 

[yungman]

It is the Z boundary value problem which restricts the values of k. Your Z bvp is:

Z'' + k Z = 0
Z(0) = Z(h) = 0

If you solve this you will find that only k = μ² > 0 gives non-trivial solutions. So these are the values that must be used in the other equation too.

I am embarrassed! Thank you very much.

One more verification. Z''+kZ=0 \Rightarrow Z=d_1cos(\sqrt{k} z) + d_2 sin (\sqrt{k} z)

for k=0 \Rightarrow Z=d_1.\;\;U(\rho,0)=0 \Rightarrow d_1=0 \
I don't know why I cannot edit the d_2 portion.


Therefore k=0 give trivial solution also. Am I correct?

Thanks a million
Alan

 

[LCKurtz]

When k = 0 the solution to the equation does not have a {sine,cosine} form. It is just:

Z'' = 0.

Apply the boundary conditions to the general solution of that and see what happens.

 

[yungman]

Thanks