A Laplace Eq. in Cylindrical coordinates

[chimay]

Hi,
I need to solve Laplace equation:$\nabla ^2 \Phi(x,r)=0$ in cylindrical domain $0<r<r_0$, $0<x<L$ and $0<\phi<2\pi$. The boundary conditions are the following ones:
$\left\{ \begin{aligned} &C_{di}\Phi(x,r_0)=\epsilon \frac{\partial \Phi(x,r)}{\partial r}\rvert_{r=r_0} \\ &\Phi(0,r)=f_1(r) \\ &\Phi(L,r)=f_2(r) \end{aligned} \right.$
$\epsilon$ being the dielectric constant of the medium and $C_{di}$ a constant capacitance; $f_1$ and $f_2$ are two known function.
Forcing no $\phi$ dependence, and solving the equation by separation of variable:
$\Phi(x,r)=(Ae^{\lambda x/r_0}+Be^{-\lambda x/r_0})J_0(\lambda r/r_0)$
$\lambda$ being the separation constant and $J_0$ the first type Bessel function of 0-order.
By applying the first boundary condition:
$\lambda \frac{J_1(\lambda)}{J_0(\lambda)}=C_r$
$C_r=r_0 C_{di}/\epsilon$
which can be solved to compute all the values of $\lambda$.
By applying the remaining boundary conditions, I obtain the following set of equations:
$\left\{ \begin{aligned} &\sum_{m=1}^{\infty} (A_m+B_m)J_0(\lambda_m r/r_0)=f_1(r) \\ &\sum_{m=1}^{\infty} (A_m e^{\lambda_m L/r_0}+B_m e^{-\lambda_m L/r_0})J_0(\lambda_m r/r_0)=f_2(r) \\ \end{aligned} \right.$
which allow me to compute the values of all the coefficients, by exploiting the orthogonality between differently scaled Bessel functions:
$\left\{ \begin{aligned} &A_i=\frac{1}{(2\sinh{\lambda_i L/r_0})(\frac{1}{2}{J_1(\lambda_i)}^2)}(I_2-I_1 e^{-\lambda_i L/r_0}) \\ &B_i=\frac{1}{(2\sinh{\lambda_i L/r_0})(\frac{1}{2}{J_1(\lambda_i)}^2)}(I_1e^{\lambda_i L/r_0}-I_1) \\ \end{aligned} \right.$
$I=\int_0^1 \frac{r}{r_0} f(r) J_0(\lambda_i r/r_0) d(\frac{r}{r_0})$

Now the point that is driving me crazy is the following: how do you expect $A_i$ and $B_i$ to change as a function of the order $i$? I expect them to decrease in modulus; indeed, this is the case for $A_i$, but $B_i$ presents a strange behaviour, oscillating in sign and increasing in modulus. Can you see any mistake? If you want, I can provide you with all the mathematical passages that I omitted for brevity, the trends of $A_i$ and $B_i$ or show you whatever plot you may need. Any suggestion is really appreciated here.

Thank you all.

 

[Charles Link]

I spotted one algebraic mistake I believe: On $B_i$, the last $-I_1$ should be $-I_2$. I didn't check every other detail, but the rest of it looks like it might be correct. $\\$ Editing: A little further reading on the Bessel functions shows that for the functions to be orthogonal, I believe it requires $J_o(\lambda_m)=0$. I'm not sure that your functions meet this condition. $\\$ Additional editing: From what I googled on the subject, you may need to also use $J_n$ of all orders in your series rather than limiting it to $J_o$.

 

[chimay]

The algebraic mistake s a transcription error; in my implementation $I_2$ correctly takes the place of $I_1$.

I am not very strong on Bessel equation and I completely forgot the condition you pointed me out!
Can you clarify your second comment? Do you mean:
$\left\{ \begin{aligned} &\sum_{m=1}^{\infty} (A_m+B_m)J_m(\lambda_m r/r_0)=f_1(r) \\ &\sum_{m=1}^{\infty} (A_m e^{\lambda_m L/r_0}+B_m e^{-\lambda_m L/r_0})J_m(\lambda_m r/r_0)=f_2(r) \\ \end{aligned} \right.$
?

Thank you a lot, your help is really really appreciated.

 

[Orodruin]

Editing: A little further reading on the Bessel functions shows that for the functions to be orthogonal, I believe it requires $J_o(\lambda_m)=0$. I'm not sure that your functions meet this condition.

This is not correct. Any homogeneous boundary condition, be it of Dirichlet, Neumann, or Robin type, will do. Each of the choices results in a Sturm-Liouville problem.

Additional editing: From what I googled on the subject, you may need to also use $J_n$ of all orders in your series rather than limiting it to $J_o$.

This would be true if the problem was not rotationally symmetric. As it stands, the rotational symmetry directly implies that only the $n = 0$ terms are of relevance.

 

[chimay]

I found this: http://www.hit.ac.il/staff/benzionS/Differential.Equations/Orthogonality_of_Bessel_functions.htm

In general, is the RHS of Eq. 96 the condition to be satisfied, isn't it?

 

[Orodruin]

I found this: http://www.hit.ac.il/staff/benzionS/Differential.Equations/Orthogonality_of_Bessel_functions.htm

In general, is the RHS of Eq. 96 the condition to be satisfied, isn't it?

Yes, this is equivalent to what I said. For a Dirichlet or Neumann condition, both terms are zero. For a Robin condition, the derivative is proportional to the function value itself, leading to the two terms cancelling.

 

[Orodruin]

I did not check your arithmetics, but let me point out that it is significantly simpler to use $C \cosh(\lambda x/r_0) + D \sinh(\lambda x/r_0)$ for the $x$-dependent part, since the boundary condition at $x = 0$ will give you one of the constants directly. You should find that both $C$ and $D$ are decrease sufficiently fast with $m$ as long as $I_1$ and $I_2$ do (which they should if $f_1$ and $f_2$ are in the correct Hilbert space).

 

[chimay]

I do not get how I can compute both constants if I cannot exploit orthogonality. Could you help me with this too?
Thank you very much.

 

[Orodruin]

Also note that it is not necessarily a problem that the coefficients increase in modulus. The norm of the basis functions $J_0(\lambda_m r/r_0)$ is not constant.

I do not get how I can compute both constants if I cannot exploit orthogonality. Could you help me with this too?
Thank you very much.

What do you mean? I just told you you can exploit orthogonality - it is a Sturm-Liouville problem!

 

[chimay]

As I mentioned before, I am very weak on this topic. Is this correct?

In general
$\int_0^1 x J_0(\lambda_m x)J_0(\lambda_i x)=\frac{{J^{'}_0(\lambda_m)}J_0(\lambda_i)-J_0(\lambda_m){J^{'}_0(\lambda_i)}}{{\lambda_i}^2-{\lambda_m}^2}$
In my case, since
$\lambda \frac{J_1(\lambda)}{J_0(\lambda)}=C_r$
I can write
$J^{'}_0(\lambda_m)=-J_1(\lambda_m)=-J_0(\lambda_m) \frac{C_r}{\lambda_m}$
and
$J^{'}_0(\lambda_i)=-J_1(\lambda_i)=-J_0(\lambda_i) \frac{C_r}{\lambda_m}$
and so
$\int_0^1 x J_0(\lambda_m x)J_0(\lambda_i x)=\frac{{J^{'}_0(\lambda_m)}J_0(\lambda_i)-J_0(\lambda_m){J^{'}_0(\lambda_i)}}{{\lambda_i}^2-{\lambda_m}^2}=0$
So I can exploit orthogonality even if $\lambda$ are not zeros of $J_0$.
If this is true, my calculations are correct, and I cannot explain the strange behaviour of the constants I depicted before...

 

[chimay]

Also note that it is not necessarily a problem that the coefficients increase in modulus. The norm of the basis functions $J_0(\lambda_m r/r_0)$ is not constant.

I tried to sum numerically the first 16 values of $A_i+B_i$ and I am very far from fitting the boundary condition in $x=0$ and $x=L$.
Also, $\sum_i^n B_i$ seems to increase with $n$.

 

[Orodruin]

I tried to sum numerically the first 16 values of $A_i+B_i$ and I am very far from fitting the boundary condition in $x=0$ and $x=L$.
Also, $\sum_i^n B_i$ seems to increase with $n$.

As I said, your basis functions do not have constant norm so there is no a priori reason to expect the coefficients to decrease. Without knowing more of the functions $f_i$, I don't know what more I can say about this.

As I mentioned before, I am very weak on this topic. Is this correct?

In general
$\int_0^1 x J_0(\lambda_m x)J_0(\lambda_i x)=\frac{{J^{'}_0(\lambda_m)}J_0(\lambda_i)-J_0(\lambda_m){J^{'}_0(\lambda_i)}}{{\lambda_i}^2-{\lambda_m}^2}$
In my case, since
$\lambda \frac{J_1(\lambda)}{J_0(\lambda)}=C_r$
I can write
$J^{'}_0(\lambda_m)=-J_1(\lambda_m)=-J_0(\lambda_m) \frac{C_r}{\lambda_m}$
and
$J^{'}_0(\lambda_i)=-J_1(\lambda_i)=-J_0(\lambda_i) \frac{C_r}{\lambda_m}$
and so
$\int_0^1 x J_0(\lambda_m x)J_0(\lambda_i x)=\frac{{J^{'}_0(\lambda_m)}J_0(\lambda_i)-J_0(\lambda_m){J^{'}_0(\lambda_i)}}{{\lambda_i}^2-{\lambda_m}^2}=0$
So I can exploit orthogonality even if $\lambda$ are not zeros of $J_0$.

Yes, this is correct. The orthogonality follows from the $\lambda$ being such that the Robin boundary condition is satisfied.

 

[chimay]

These are the remaining functions:

$f_1=\Phi_1-\Phi_0(r)$ and $f_2=\Phi_2-\Phi_0(r)$,
$\Phi_0(r)=V-k_a \log(\frac{k_b}{{(4k_b-r^2)}^2}k_c)$
where $V$ and $k_{a,b,c}$ are some constants.

Can you guess any $A_i$ and $B_i$ behaviour?
I am starting to think there is some hidden mistake in my implementation.

Thank you a lot!

 

[Orodruin]

What are the possible values of $k_b$?

 

[chimay]

$k_b$ is the solution of another equation.
Do you mean numerical values ?

 

[Orodruin]

I am just trying to figure out if that function is square integrable or not. Without looking at it in detail I am not sure I can help you. A suggestion would be to try to see what happens with your expressions in some cases (different fs) where you know the solutions to check your implementation.

 

[chimay]

I will take another look tomorrow in the morning. Anyway, my approach should be correct, and the equations for $A_i$ and $B_i$ in the first post should be valid...right?

 

[chimay]

Hi,
I may have found one tricky point. Can someone tell me the result of $\int_0^1 x J_0(\lambda_m x)J_0(\lambda_i x)$ when $\lambda_m=\lambda_i$ and $\lambda_i$ is not a zero of the Bessel function?
Edit:
I have finally found the mistake. I used the following relation:
$\int_0^1 x J_0^2(\lambda x) dx=\frac{1}{2}J_1^2(\lambda)$
but when $\lambda$ is not a zero of $J_0$, like in my case, this holds:
$\int_0^1 x J_0^2(\lambda x) dx=\frac{1}{2} (J_0^2(\lambda)+J_1^2(\lambda))$.
By using the correct relation I find $B_i$ to descrease...

 

[chimay]

Hi,
since you helped me so much in my previous doubt, I would like to make another question about Bessel functions.
Assume that my boudary condition is expressed like this:
$\sum_{m=1}^{\infty} A_m J_0(\lambda_m r/r_0) + B_m Y_0(\lambda_m r/r_0)=C$
$C$ being a certain constant and $Y$ second type Bessel function.
How can I proceed to obtain a relation between $A_m$ and $B_m$ now, following a similar path as before?

Thank you

 

[Orodruin]

The same way as before. First you need to find the appropriate boundary conditions so that you can find what combination of Bessel functions form a complete basis on the domain of interest.

 

[berkeman]

Hi,
since you helped me so much in my previous doubt, I would like to make another question about Bessel functions.
Assume that my boudary condition is expressed like this:
$\sum_{m=1}^{\infty} A_m J_0(\lambda_m r/r_0) + B_m Y_0(\lambda_m r/r_0)=C$
$C$ being a certain constant and $Y$ second type Bessel function.
How can I proceed to obtain a relation between $A_m$ and $B_m$ now, following a similar path as before?

Thank you

Hi @chimay -- First, please create a new thread with your new question. It is very confusing for you to ask a new question in your existing thread.

Second, advanced schoolwork questions are only allowed in the technical forums if you show a lot of effort on your part to try to answer it, as you did at the top of this existing thread. Asking for help on an advanced problem in the technical PF forums and showing no effort is not allowed.

This thread is locked. Please re-post as a new thread, and show your best efforts again to start to solve the problem. That will make this advanced schoolwork question eligible for help in the PF technical forums.

Thank you.