# Separation of Variables, Solutions to Laplace's Eqtns with Boundary Conditions

1. Jun 9, 2012

### mateomy

Not really a specific problem, but just a general question:

Does anyone have any good references (preferably online) for solving E&M problems with this method? I'm using Griffith's Electrodynamics book for my class and I'm trying to get ready for a final. This is the only part I'm having problems with. I don't feel comfortable solving these on my own.

Thanks.

2. Jun 10, 2012

### gabbagabbahey

I learned undergrad Electrodynamics from Griffiths, and found that working through as many of the problems at the end of each unit and chapter as I had time for was more than sufficient to get a firm grasp at solving these types of problems.

If you are having difficulties, I would suggest that you try a few of the related problems and post your attempts along with where you get stuck. After working through enough of the problems with some help, I'm sure you will find it much easier to tackle more of them on your own.

3. Jun 10, 2012

### mateomy

I'll post an example right now...

4. Jun 10, 2012

### mateomy

This is from page 141 of the 3rd edition....(Example 3.8)

An uncharged metal sphere of radius R is placed in an otherwise uniform E field= $E_0\,\hat{z}$. (etc etc)...Find the potential in the region outside of the sphere.

It is stated that the sphere is an equipotential "we may as well set it to zero". The by symmetry the entire xy plane is at potential zero. This time however, V does not go to zero at large z. In fact, far from the sphere the field is $E_0\,\hat{z}$, and hence V -> -$E_0 z +C$.

The boundary conditions are

i) V=0 when r=R
ii) V-> -$E_0\,r\,cos\theta$ for r >> R

First condition (using the general solution formula for spherical coords)

$$A_l R^l + \frac{B_l}{R^{l+1}}=0$$
$$B_l= -A_l R^{2l+1}$$
so,
$$\sum_{l=0}^{\infty} A_l(r^l - \frac{R^{2l+1}}{r^{l+1}}) P_l (cos\theta).$$
For r>>R, the second term in parenthesis is negligible, and therefore condition (ii) requires that
$$\sum_{l=0}^{\infty} A_l r^l P_l (cos\theta)=\,E_0 r cos\theta.$$
This is what I don't understand (and what I dont understand for most all of these problems)...Griffith goes on to state, "Evidently, only one term is present: l=1. In fact, since $P_l (cos\theta)=cos\theta$, we can read off immediately
$$A_1 = -E_0,\,all\,other\,A_l's\,zero.$$
Finally concluding that...
$$V(r,\theta)=-E_0(r-\frac{R^3}{r^2})cos\theta$$

What does he mean one term is present? Because there's no power greater than 1? And if it were larger than 1, would I have to solve each succeeding solution with the next higher Legendre polynomial? Ugh, I feel completely lost...

5. Jun 10, 2012

### mateomy

As a side note I think it's important to note that I still haven't taken diff EQ (formally). I know very little bits I've taught myself along the way to survive this far, I'll be taking it in the summer session. I was worried about signing up for the E&M course due to that limitation but I was assured by the head of the department that I shouldn't have an issue and for the most part he's been right, I'm heading into this final with an A in the class and I would like to keep it. So any assistance with that in mind would be much appreciated.

6. Jun 10, 2012

### mateomy

Here's another one that I just tried...

Find the potential outside a charged metal sphere (charge Q, radius R) placed in an otherwise uniform electric field $E_0$. Explain clearly where you are setting the zero of potential.

So just looking at the general solution for a spherically based problem I have the standard
$$A_l R^l +\frac{B_l}{R^{l+1}}=0$$
Can I make my boundary conditions as: $V\,\rightarrow\,E_0\,as\,R\rightarrow\,\infty$ and $V=0\,at\,R$? My problem here is I'm not sure where I should make my potential equal to zero. I was thinking about doing it at the spherical surface, but I'm not sure if that works.

I want to go forward from here but I ultimately get lost on how to deal with those l's.

7. Jun 10, 2012

### mateomy

Thinking further into it, I realized that my potential will go to zero far from the sphere but since I'm located within an E-field I can't just choose any point far away. After drawing a little diagram and studying another diagram from the example I initially posted I realized I could only have my potential go to zero on the equatorial plane...far away. And because the sphere is charged I will have a potential at the surface namely Q/R (plus constants). I'm still pushing around on different web-sites and books to make the determination on how to pick the various Legendre polynomials (i.e. how many to use).

8. Jun 10, 2012

### gabbagabbahey

It's only slightly more complicated than that. Considering that $P_3(\cos\theta)$ contains not only a $\cos^3\theta$ term but also a $\cos\theta$ term
and higher order Legendre Polynomials likewise contain more than ione power of $\cos (\theta)$, it is natural to wonder if some linear combination of higher order Legendre polynomials $P_\ell(\cos\theta)$ will also produce a net result of $$V(r,\theta)=E_0r cos\theta$$. However, orthogonality of the Legendre Polynomials guarantees this is not the case (they are linearly independent). As a result, the highest power of $\cos\theta$ that occurs in a linear combination of Legendre Polynomials, will always be the highest order Legendre Polynomial with non-zero coefficient present.

9. Jun 10, 2012

### mateomy

I think I understand. So, if that condition to be met stated that I needed say, $E_0rcos^2 \theta$ I could simply choose which Legendre had the highest power matching my boundary condition and by their linear independence and orthogonality state finally that the solution would be
$$V(r,\theta)=-\frac{1}{2}E_0(r-\frac{R^3}{r^2})(3cos\theta -1)$$

*I know this doesn't necessarily work into the solution but I'm just trying to get a feel for applying these polys.

10. Jun 14, 2012

### gabbagabbahey

Sort of.... say for example you know that $V(\mathbf{r})\approx \alpha(r^2\cos^2\theta +5r\cos\theta-\frac{1}{3}r^2)$, for $r\ll 1$, where $\alpha$ is just some constant, and you are asked to say which Legendre polynomials are present (have non-zero coefficients), and what the values of $A_{\ell}$ are for those terms.

You would probably start by noticing that the highest power of $\cos\theta$ present is 2, and recognizing that all higher order terms are therefore zero, so you have

$$\begin{array}{ll} V(\mathbf{r})\approx \alpha(r^2\cos^2\theta +5r\cos\theta-\frac{1}{3}r^2) &= \sum_{\ell = 0}^{\infty}A_{\ell}r^{\ell}P_{\ell}(\cos\theta) \\ &= \sum_{\ell = 0}^{2}A_{\ell}r^{\ell}P_{\ell}(\cos\theta) \\ &= A_{2}r^{2}P_{2}(\cos\theta) + A_{1}r^{1}P_{1}(\cos\theta) +A_{0}r^{0}P_{0}(\cos\theta)\end{array}$$

Substituting in for the Legendre polynomials and collecting terms in powers of $\cos(\theta)$ gives you

$$\alpha(r^2\cos^2\theta +5r\cos\theta-\frac{1}{3}r^2) = \frac{3}{2}A_2r^2\cos^2\theta + A_1r\cos\theta + A_0 - \frac{1}{2}A_2r^2$$

Comparing the $\cos^2\theta$ terms gives you $A_2 = \frac{2}{3}\alpha$. Then comparing the $\cos\theta$ terms gives you $A_1 = 5\alpha$. Finally, comparing the constant terms gives you $A_0 = 0$.

11. Jun 14, 2012

### gabbagabbahey

Pretend for a second, that there is no sphere. What would be the potential if all you have is a uniform electric field, say $\mathbf{E}=E_0\hat{z}$? Now pretend, that there is no uniform electric field, what would you expect the potential of a charged metal sphere to look like at great distances?

12. Jun 21, 2012

### mateomy

Sorry I didn't see any of this. Finals finished. Needless to say I was raked by a question using the ideas above. Still interested in figuring it out though.

If there were no electric field I would expect the potential to go to zero. In a uniform electric field $E_0 z$ I would expect the potential to go to "just" the value of the electric field at great distances.

Last edited: Jun 21, 2012
13. Jun 21, 2012

### gabbagabbahey

At infinity yes. But what about slightly closer than that (but still far away)?

You mean you would expect it to go to $E_0 z$ since $\mathbf{E}=-\mathbf{\nabla}(E_0 z) = E_o\mathbf{\hat{z}}$, right?