- #1
chimay
- 80
- 6
Hi,
I need to solve Laplace equation:##\nabla ^2 \Phi(x,r)=0## in cylindrical domain ##r_1<r<r_2##, ##0<z<+\infty##. The boundary conditions are:
##
\left\{
\begin{aligned}
&\Phi(0,r)=V_B \\
& -{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 \\
&\lim_{z\rightarrow \infty} \Phi(z,r)=0
\end{aligned}
\right.
##
By separation of variables, I can write the solution as:
##
\Phi_(z,r)=e^{-\lambda z}(A J_0(\lambda r)+BY_0(\lambda r))
##
By applying the first boundary condition I get:
##
\sum_{m=1}^{\infty} A_m J_0(\lambda_m r)+B_m Y_0(\lambda_m r)=V_B
##
Now I can compute a first relation between ##A_m## and ##B_m## by exploiting orthogonality between different scaled Bessel functions (##i \ne m##):
##
\left\{
\begin{aligned}
& \int_{r_1}^ {r_2} r J_0(\lambda_i r)J_0(\lambda_m r) dr =0 \\
& \int_{r_1}^ {r_2} r J_0(\lambda_i r)Y_0(\lambda_m r) dr =0 \\
& \int_{r_1}^ {r_2} r Y_0(\lambda_i r)Y_0(\lambda_m r) dr =0
\end{aligned}
\right.
##
by virtue of the second and the third boundary conditions, as it is explained here (http://www.hit.ac.il/staff/benzionS/Differential.Equations/Orthogonality_of_Bessel_functions.htm)
and it was pointed out also here
(https://www.physicsforums.com/threads/laplace-eq-in-cylindrical-coordinates.915420/)
My question is the following:
Assume the second boundary condition was ## {C^{'}}_{ox} (V_G - \Phi(x,r_2))=C_0 \frac{\partial \Phi(x,r)}{\partial r}\rvert_{r=r_2} ##, ##V_G=constant##. It seems to me that in this case the orthogonality conditions are not valid anymore, because of the presence of the constant term ##V_G##.
1) Is this true ?
2) How can I proceed in this case?
Thank you all
Edit: I am having some doubts also about the integrals concerning ##Y_0##. Are they the same as the ones that involve ##J_0##?
I need to solve Laplace equation:##\nabla ^2 \Phi(x,r)=0## in cylindrical domain ##r_1<r<r_2##, ##0<z<+\infty##. The boundary conditions are:
##
\left\{
\begin{aligned}
&\Phi(0,r)=V_B \\
& -{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 \\
&\lim_{z\rightarrow \infty} \Phi(z,r)=0
\end{aligned}
\right.
##
By separation of variables, I can write the solution as:
##
\Phi_(z,r)=e^{-\lambda z}(A J_0(\lambda r)+BY_0(\lambda r))
##
By applying the first boundary condition I get:
##
\sum_{m=1}^{\infty} A_m J_0(\lambda_m r)+B_m Y_0(\lambda_m r)=V_B
##
Now I can compute a first relation between ##A_m## and ##B_m## by exploiting orthogonality between different scaled Bessel functions (##i \ne m##):
##
\left\{
\begin{aligned}
& \int_{r_1}^ {r_2} r J_0(\lambda_i r)J_0(\lambda_m r) dr =0 \\
& \int_{r_1}^ {r_2} r J_0(\lambda_i r)Y_0(\lambda_m r) dr =0 \\
& \int_{r_1}^ {r_2} r Y_0(\lambda_i r)Y_0(\lambda_m r) dr =0
\end{aligned}
\right.
##
by virtue of the second and the third boundary conditions, as it is explained here (http://www.hit.ac.il/staff/benzionS/Differential.Equations/Orthogonality_of_Bessel_functions.htm)
and it was pointed out also here
(https://www.physicsforums.com/threads/laplace-eq-in-cylindrical-coordinates.915420/)
My question is the following:
Assume the second boundary condition was ## {C^{'}}_{ox} (V_G - \Phi(x,r_2))=C_0 \frac{\partial \Phi(x,r)}{\partial r}\rvert_{r=r_2} ##, ##V_G=constant##. It seems to me that in this case the orthogonality conditions are not valid anymore, because of the presence of the constant term ##V_G##.
1) Is this true ?
2) How can I proceed in this case?
Thank you all
Edit: I am having some doubts also about the integrals concerning ##Y_0##. Are they the same as the ones that involve ##J_0##?
Last edited: