Laplace equation- variable domain

[chimay]

Hi,
I need to solve Laplace equation $\nabla ^2 \Phi(z,r)=0$ in cylindrical coordinates in the domain $r_1<r<r_2$, $0<z<L$.
The boundary conditions are:
$\left\{ \begin{aligned} &\Phi(0,r)=V_B \\ &\Phi(L,r)=V_P \\ & -{C^{'}}_{ox} \Phi(x,r_2)=C_0 \frac{\partial \Phi(x,r)}{\partial r}\rvert_{r=r_2} \\ &\frac{\partial \Phi(x,r)}{\partial r}\rvert_{r=r_1}=0 \\ \end{aligned} \right.$
By separation of variables I obtain:
$\Phi_(z,r)=(A e^{-\lambda z} + B e^{+\lambda z})(C J_0(\lambda r) + D Y_0(\lambda r))$
$J_0$ and $Y_0$ being zero order first type and second type Bessel functions.
The general solution is:
$\Phi_{tot}= \sum_m (A_m e^{-\lambda_m z} + B_m e^{+\lambda_m z})(C_m J_0(\lambda_m r) + D_m Y_0(\lambda_m r))$
and all the constants che can be calculated exploiting the boundary conditions.

Now assume $L$ is not fixed, but it can vary in a certain range $0<L_1<L<L_2$. What I am thinking about is: is it possible to compute $L$ such that \frac{\partial \Phi_{tot}}{\partial z} \rvert_{z=L} =0?

Thank you!

 

[Dr Transport]

yes, you have the solution written, just take the derivative and find the solution

 

[chimay]

Form the boundary conditions I can write
<br /> \Phi_m=(\exp(\lambda_m z)+B_m \exp(-\lambda_m z) )C_m F_m(r)<br />
where $F_m$ is a function independent from $L$ and
<br /> \begin{cases}<br /> B_m=\frac{(K_{Bm}/K_{Pm}) \exp(\lambda_m L)-1}{1- (K_{Bm}/K_{Pm}) \exp(-\lambda_m L)} \\<br /> C_m=\frac{K_{Pm}-K_{Bm} \exp(-\lambda_m L)}{2 K_{1m} \sinh{\lambda_m L}}<br /> \end{cases}<br />
$K_{Pm}, K_{Bm}$ and $K_{1m}$ being constants.
By computing $\frac{\partial \Phi_{tot}}{\partial z} \rvert_{z=L} =0$ I get
<br /> \sum_m \lambda_m (\exp{\lambda_m L} - B_m \exp{-\lambda_m L}) C_m F_m(r) = 0<br />
How can I solve this equation? Note that $L$ must not be $m-$dependent.

Thank you for your interest.