Confused about the nature of Laplace vs Poisson equation in BVP

[yucheng]

Homework Statement

Two coaxial, nonconducting cylinders have surface charge densities $\sigma_a(\phi)$ on $\sigma_b(\phi)$ on the inner and outer cylinders, giving rise to potentials $V_a(\phi)$ and $V_b(\phi)$ on the two surfaces. Find the potential (a) for $r<a$ (b) $r>a$ and (c) for $r \in (a,b)$

(Vanderlinde, Example 5.3)

Relevant Equations

N/A

Hi!

The problem clearly states that there is a surface charge density, which somehow gives rise to a potential.

The author has solved the Laplace equation in cylindrical coordinates and applied the equation to the problem.

So $\nabla^2 V(r,\phi) = 0$, and $V(a,\phi) = V_a(\phi)$ (where the potential takes on this value at the boundary). However, because of the surface charge at $r=a$, doesn't the potential also satisfy Poisson's equation? Does this mean that given, $V_a(\phi)$, we can also find $\sigma_a$?

I cooked up an explanation to this:

$$\nabla^2 V(r,\phi) = \frac{\rho_a}{\epsilon_0} = \frac{ \sigma_a \delta(r-a)}{\epsilon_0}$$

But this is infinite at $r=a$ what's wrong? Is it because $V(r,\phi)$ is not valid at the boundary? I know it solves Laplace equation hence not valid where there are charges (that is why in the problem statement, the author only solves for $r<a$). What does $V(a,\phi) = V_a(\phi)$ mean then?

Thanks in advance!

P.S. looks like someone asked this before but never got a reply...

https://www.physicsforums.com/threads/electromagnetics-parallel-plates-poisson-laplace.134235/

 

[BvU]

Homework Statement: N/A

Not correct. You post a fraction of the problem statement

Two coaxial, nonconducting cylinders have surface charge densities $\sigma_a(\phi)$ on $\sigma_b(\phi)$ on the inner and outer cylinders, giving rise to potentials $V_a(\phi)$ and $V_b(\phi)$ on the two surfaces. Find the potential (a) for $r<a$.

But you don't reveal what $a$ is, and introduce more confusion by writing 'potential (a)' .

Where are the $\sigma$ ? What is the complete problem statement ?
Is there a drawing we don't know about ?

$\$

 

[yucheng]

Homework Statement: N/A

Not correct. You post a fraction of the problem statement

But you don't reveal what $a$ is, and introduce more confusion by writing 'potential (a)' .$\$

Sorry I completed it. I thought omitting a few parts will not cause any misunderstanding. Sorry again!

 

[BvU]

Homework Statement:: Two coaxial, nonconducting cylinders have surface charge densities $\sigma_a(\phi)$ on $\sigma_b(\phi)$ on the inner and outer cylinders, giving rise to potentials $V_a(\phi)$ and $V_b(\phi)$ on the two surfaces. Find the potential (a) for $r<a$ (b) $r>a$ and (c) for $r \in (a,b)$

(Vanderlinde)

Thank you. So much better ! Especially with the drawing we know which is inner and which is outer..

The 2006 thread is solved with Gauss (see e.g. here) . That way a divergence is circumvented.

In your exercise, you can do the same. Basically you have two zero-thickness cylinders of charge, and if I am not mistaken (it's 2:30 AM here), you can simply write the equation and solve. The infinities are discontinuities in the first derivative, i.e. in $E$.

(cf spherical example)

$\$

 

[yucheng]

Thank you. So much better ! Especially with the drawing we know which is inner and which is outer..

The 2006 thread is solved with Gauss (see e.g. here) . That way a divergence is circumvented.

In your exercise, you can do the same. Basically you have two zero-thickness cylinders of charge, and if I am not mistaken (it's 2:30 AM here), you can simply write the equation and solve. The infinities are discontinuities in the first derivative, i.e. in $E$.

(cf spherical example)

I think you should get a rest, for health's sake...

Actually I understand how to solve it; I am just confused with some theory and the infinite discontinuities, how to interpret and reconcile them.

P.S. did I now mention it's an example? The author actually solved it.

 

[vela]

I cooked up an explanation to this:

$$\nabla^2 V(r,\phi) = \frac{\rho_a}{\epsilon_0} = \frac{ \sigma_a \delta(r-a)}{\epsilon_0}$$

But this is infinite at $r=a$ what's wrong? Is it because $V(r,\phi)$ is not valid at the boundary? I know it solves Laplace equation hence not valid where there are charges (that is why in the problem statement, the author only solves for $r<a$). What does $V(a,\phi) = V_a(\phi)$ mean then?

Recall that $\vec E = -\nabla V$. If $V$ were discontinuous, $\vec E$ would be infinite, which isn't physically acceptable. So we need $V$ to be continuous, i.e., $V(r=a) = \lim_{r\to a+} V(r)$.

As @BvU noted, the equation you wrote down isn't saying $V$ is discontinuous; it's saying the divergence of $\vec E$ is discontinuous, which is what you'd expect as you cross a sheet of charge.

A similar situation appears in quantum mechanics with delta function potentials.

https://quantummechanics.ucsd.edu/ph130a/130_notes/node141.html#derive:continuity

 

[yucheng]

The topic of discontinuities is interesting indeed.

Actually, what I had in mind was that:

Suppose we consider $r\leq a$; the potential takes the form $$V = D_0 + \sum r^m(A_m cos m\varphi + B_m sin m\varphi)$$
which clearly satisfies Laplace equations in the region $r<a$.
But since there are surface charges on the boundary, it should satisfy Poisson's equation at $r=a$? And it should be infinite...

Am I right?

However, if we differentiate naively,

$$\nabla^2 V \bigg\rvert_{r = a} = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial V}{\partial r} \right) + \frac{1}{r^2} \frac{\partial ^2 V}{\partial \varphi ^2} \bigg \rvert_{r=a}= 0 \neq \frac{\sigma_a \delta(r-a) }{\epsilon_0} \bigg\rvert_{r=a} = \infty$$

Hence satisfies Laplace equation at the boundary ?

 

[Orodruin]

If you are looking at the problem in the mentioned regions away from the surfaces, the charge density is zero and so you obtain Laplace equation. If you want to solve the problem in the entire space then yes, you are then solving Poisson’s equation with formally infinite charge densities at the surfaces (since the surface volume is zero and the charge non-zero). You can either solve Laplace equation in the regions separately and use the surface charges to find the matching conditions on the boundaries or you can solkve the Poisson equation in the full space.

 

[yucheng]

However, if we differentiate naively,

$$\nabla^2 V \bigg\rvert_{r = a} = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial V}{\partial r} \right) + \frac{1}{r^2} \frac{\partial ^2 V}{\partial \varphi ^2} \bigg \rvert_{r=a}= 0 \neq \frac{\sigma_a \delta(r-a) }{\epsilon_0} \bigg\rvert_{r=a} = \infty$$

Hence satisfies Laplace equation at the boundary ?

I think I understand why I got confused; a solution to the differential equation does not mean a solution to the physical problem.

If our problem was to solve for the region $r<a$ i.e. a domain excluding charges, then we can solve Laplace equation. We still need a boundary condition at $V(r=a)$ i.e. the region with charges! Which means that the potential we've found takes on the right values at the region without and with charges, but it's laplacian is wrong on the boundary! The reason: a mathematical solution to a boundary value problem (here Laplace, not Poisson) does not mean that it is a physical everywhere the solution is defined (not the boundary with charges), and actually solves the problem!

P.S. clearly, solving Laplace and Poisson equation will give us the same potential function (at least at the region with charges, and the boundaries), with a different laplacian.

Hence, the domain of validity for $V$ after solving Laplace is mathematically the region $r \leq a$ but physically $r < a$. Am I right?

If we are to include the boundary charges, we need to solve for Poisson's equation.

I suppose the key takeaway is that physics is all about modelling. Just because the mathematical model says something, does not mean it is consistent with physical interpretations, and that you've solved the right problem...

You can either solve Laplace equation in the regions separately and use the surface charges to find the matching conditions on the boundaries

I don't get how this is done... Maybe I'll just look for an example somewhere :)

 

[vanhees71]

What's confusing for me is that he gives both the surface-charge distributions and the boundary values of the potential.

A way to read the question is that he want's to say that the potentials are given along the two coaxial cylindrical shells and that there are only surface-charge densities on these shells and no other charges anywhere else. Then the problem is well defined, and you can solve it via a Fourier-series ansatz in $\phi$:
$$V(r,\varphi)=A_0 \ln(r/r_0) +B_0 + \sum_{m \in \mathbb{Z} \setminus \{0\}} (A_m r^m + B_m r^{-m}) \exp(\mathrm{i} m \varphi),$$
where $A_m,B_m \in \mathbb{C}$.

Now you can use the boundary conditions at $r=0$, $r=a$, $r=b$, and $r \rightarrow \infty$ to determine the Fourier coefficients $A_m$ and $B_m$ for the three regions $0<r<a$, $a<r<b$, and $r>b$. Also note that the potentials must be continuous everywhere with the electric-field components $E_r=-\partial_r V$ making jumps at $r=a$ and $r=b$, which determine the surface charges uniquely.

I don't know, whether the textbook you use, maybe uses the real cos and sin Fourier series, but that's of course at the end leading to the same result.

 

[yucheng]

Also note that the potentials must be continuous everywhere with the electric-field components Er=−∂rV making jumps at r=a and r=b, which determine the surface charges uniquely.

It took me quite a while to realize this...

Just curios, it's possible that the potential of the geometry is specified, but there are no surface charges; to show this, one should compute the difference between the directional derivative ("below" and "above" the surface) of the potential along the boundary, right?

 

[Orodruin]

It took me quite a while to realize this...

Just curios, it's possible that the potential of the geometry is specified, but there are no surface charges; to show this, one should compute the difference between the directional derivative ("below" and "above" the surface) of the potential along the boundary, right?

Sure, if you put the potentials on the surfaces such that they coincide with the potential of a solution to the Laplace equation in all of space, then there will be no surface charge required.