Laplace's Equation in Cylindrical Coordinates (Potential)

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Homework Help Overview

The discussion revolves around solving Laplace's equation in cylindrical coordinates for a hollow cylinder with specified boundary conditions. The problem involves finding the potential function ##u(r,\phi,z)## given that the base and sides are at null potential while the lid is at a potential ##u_0##.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the separation of variables method, substituting the potential function into Laplace's equation, and deriving forms for the functions in terms of Bessel's equation. Questions arise regarding the application of boundary conditions and the implications of the conditions on the constants involved.

Discussion Status

Participants are actively exploring the implications of boundary conditions on the potential function and the constants derived from the separation of variables. Some have offered clarifications on the nature of the boundary conditions, while others express uncertainty about the next steps in the solution process.

Contextual Notes

There is a noted discrepancy in the boundary conditions as one participant corrects the potential at the top lid from 0 to ##u_0##. Additionally, the requirement for the potential to be single-valued around the cylinder introduces constraints on the allowed values of certain parameters.

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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&#039;(r))+\frac{1}{r^2\Phi(\phi)}\Phi&#039;&#039;(\phi)+\frac{1}{Z(z)}Z&#039;&#039;(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&#039;(r))-m^2+(rk)^2=r\frac{d}{dr}(rR&#039;(r))+[(rk)^2-m^2]R(r)=0,
where we can substitute ##y=rk## to get
y^2R&#039;&#039;(y)+yR&#039;(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?
 
Last edited:
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V0ODO0CH1LD said:

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)=0\text{ for }0≤r≤a<br /> \end{eqnarray}
The last one should be equal to ##u_0##, not 0.

You should also have ##u(r,\phi,z) = u(r,\phi+2\pi,z)## because you want u to be single-valued. This imposes a constraint on the allowed values of ##m##.

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&#039;(r))+\frac{1}{r^2\Phi(\phi)}\Phi&#039;&#039;(\phi)+\frac{1}{Z(z)}Z&#039;&#039;(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&#039;(r))-m^2+(rk)^2=r\frac{d}{dr}(rR&#039;(r))+[(rk)^2-m^2]R(r)=0,
where we can substitute ##y=rk## to get
y^2R&#039;&#039;(y)+yR&#039;(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?
Take the condition that the base is held at 0. That requires Z(0)=0 since it has to hold for all values of ##r## and ##\phi##. What does this tell you about A and B?

Take the condition that the side is held at 0. The requires that R(a)=0. What can you infer from this about the allowed values of ##k##?
 
vela said:
The last one should be equal to ##u_0##, not 0.
Thanks! Already fixed the typo.
vela said:
You should also have ##u(r,\phi,z) = u(r,\phi+2\pi,z)## because you want u to be single-valued. This imposes a constraint on the allowed values of ##m##.
Could you expand on that?
vela said:
Take the condition that the base is held at 0. That requires Z(0)=0 since it has to hold for all values of ##r## and ##\phi##. What does this tell you about A and B?

Take the condition that the side is held at 0. The requires that R(a)=0. What can you infer from this about the allowed values of ##k##?
The condition ##Z(0)=0## implies that ##A=-B##, right?. Do I use ##R(a)=0## before ##Z(L)=u_0##? And does that mean that I would have to solve Bessel's equation to completely solve this problem?
 
V0ODO0CH1LD said:
Could you expand on that?
If you increase ##\phi## by ##2\pi##, you've gone around the cylinder once and end up back in the same spot. You can't have two different values of ##u## for the same point.

The condition ##Z(0)=0## implies that ##A=-B##, right?
Yes, so ##Z(z) = A(e^{kr}-e^{-kr}) = A'\sinh kr##.

Do I use ##R(a)=0## before ##Z(L)=u_0##?
You can't say that ##Z(L)=u_0##. You can only say ##u(r,\phi,L)=u_0##.

And does that mean that I would have to solve Bessel's equation to completely solve this problem?
I'd expect you can just say that each R is a Bessel function without having to derive that result over again.
 
vela said:
If you increase ##\phi## by ##2\pi##, you've gone around the cylinder once and end up back in the same spot. You can't have two different values of ##u## for the same point.Yes, so ##Z(z) = A(e^{kr}-e^{-kr}) = A'\sinh kr##.You can't say that ##Z(L)=u_0##. You can only say ##u(r,\phi,L)=u_0##.I'd expect you can just say that each R is a Bessel function without having to derive that result over again.
I don't see how I am supposed to use ##R(a)=0## to move forward.. What am I supposed to be doing at this point?
 
What are the solutions for ##R## in terms of the constants ##m## and ##k##?
 
I still don't see what the next step is.. How do I keep solving the problem from here?
vela said:
What are the solutions for ##R## in terms of the constants ##m## and ##k##?
I'm not sure.. Are you talking about
R(r)=\sum_{n=0}^\infty{}b_n(rk)^{n+m},
the solution for the Bessel equation? Which would mean that
\sum_{n=0}^\infty{}b_n(ak)^{n+m}=0.
Is that what I am supposed to do? Where do I go from here?
 
I don't understand where you got that expression for R from.
 
vela said:
I don't understand where you got that expression for R from.
I used the Frobenius method on ##R##, which gives the answer as a power series of that form. But I honestly don't know what the next step is, after I found ##Z(z)=A(e^{kz}-e^{-kz})## I don't know what to do.
 
  • #10
At this point, I think you need to go over a similar example in your textbook and get an understanding of the logic behind each step.
 

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