Laplace's Equation in Cylindrical Coordinates (Potential)

[V0ODO0CH1LD]

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:
[tex] \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 [/tex]
Boundary conditions (specific to this problem):
[tex]\begin{eqnarray}
u(r,\phi,0)=0\text{ for }0≤r≤a \\
u(a,\phi,z)=0\text{ for }0≤z≤L \\
u(r,\phi,L)=u_0\text{ for }0≤r≤a
\end{eqnarray}[/tex]

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
[tex] \frac{1}{rR(r)}\frac{d}{dr}(rR'(r))+\frac{1}{r^2\Phi(\phi)}\Phi''(\phi)+\frac{1}{Z(z)}Z''(z)=0. [/tex]
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
[tex] \frac{r}{R(r)}\frac{d}{dr}(rR'(r))-m^2+(rk)^2=r\frac{d}{dr}(rR'(r))+[(rk)^2-m^2]R(r)=0, [/tex]
where we can substitute ##y=rk## to get
[tex] y^2R''(y)+yR'(y)+[y^2-m^2]R(r)=0, [/tex]
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?

 

[vela]

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:
[tex] \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 [/tex]
Boundary conditions (specific to this problem):
[tex]\begin{eqnarray}
u(r,\phi,0)=0\text{ for }0≤r≤a \\
u(a,\phi,z)=0\text{ for }0≤z≤L \\
u(r,\phi,L)=0\text{ for }0≤r≤a
\end{eqnarray}[/tex]

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
[tex] \frac{1}{rR(r)}\frac{d}{dr}(rR'(r))+\frac{1}{r^2\Phi(\phi)}\Phi''(\phi)+\frac{1}{Z(z)}Z''(z)=0. [/tex]
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
[tex] \frac{r}{R(r)}\frac{d}{dr}(rR'(r))-m^2+(rk)^2=r\frac{d}{dr}(rR'(r))+[(rk)^2-m^2]R(r)=0, [/tex]
where we can substitute ##y=rk## to get
[tex] y^2R''(y)+yR'(y)+[y^2-m^2]R(r)=0, [/tex]
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##?

 

[V0ODO0CH1LD]

The last one should be equal to ##u_0##, not 0.

Thanks! Already fixed the typo.

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?

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?

 

[vela]

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.

 

[V0ODO0CH1LD]

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?

 

[vela]

What are the solutions for ##R## in terms of the constants ##m## and ##k##?

 

[V0ODO0CH1LD]

I still don't see what the next step is.. How do I keep solving the problem from here?

What are the solutions for ##R## in terms of the constants ##m## and ##k##?

I'm not sure.. Are you talking about
[tex] R(r)=\sum_{n=0}^\infty{}b_n(rk)^{n+m}, [/tex]
the solution for the Bessel equation? Which would mean that
[tex] \sum_{n=0}^\infty{}b_n(ak)^{n+m}=0. [/tex]
Is that what I am supposed to do? Where do I go from here?

 

[vela]

I don't understand where you got that expression for R from.

 

[V0ODO0CH1LD]

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.

 

[vela]

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.