Laplace problem-analytically

In which we show that the method of separation of variables gives an analytic solution to Laplace's equation for azimuthal ($\phi$) symmetry.

Let's relate this to:

Imagine: We start with a uniform electric field, $\myv E=E_0\uv z$ (out to infinity). We insert a conducting sphere in the field (centered at the origin) which is grounded ($V=0$). Now what's the electric field?

  • Conducting sphere $\Rightarrow$ Its surface is an equipotential.
  • Azimuthal symmetry: The solution must be independent of the $\phi$ coordinate. $V(r,\theta)$.
  • Laplace's equation, $\grad^2V=0$, holds outside of the sphere.
  • Inside: $\myv E=0$.
  • Just outside the sphere, the electrical field should be perpendicular to the surface: When $r=R=$ radius of the sphere we should only have a radial ($r$) component to the field, that is $\left. \myv E(\theta,r)\right|^{r=R}=E(\theta,R)\uv r$.

This meets the conditions of the second uniqueness theorem so there should be a unique solution.

All we have to do is solve the Laplace equation, and match boundary conditions to find it!

In spherical coordinates, assuming azimuthal symmetry (that is, $\frac{\del V}{\del \phi}=0$): $$\grad^2 V = \frac{1}{r^2}\frac{\del}{\del r}\left(r^2 \frac{\del V}{\del r}\right)+\frac{1}{r^2 \sin \theta}\frac{\del}{\del \theta}\left(\sin \theta \frac{\del V}{\del \theta}\right)=0$$.

Separation of variables

The method of separation of variables consists in guessing that the solutions to a partial differential equation consist of the products of functions, each of which is a function of just one coordinate: $$V(\theta,r) = \Theta(\theta)R(r)$$.

Substituting this in to Laplace's equation gives... $$\begineq\grad^2 V = 0 &= \frac{1}{r^2}\frac{\del}{\del r}\left(r^2 \frac{\del \Theta(\theta)R(r)}{\del r}\right)+\frac{1}{r^2 \sin \theta}\frac{\del}{\del \theta}\left(\sin \theta \frac{\del \Theta(\theta)R(r)}{\del \theta}\right)\\ &=\frac{\Theta(\theta)}{r^2}\frac{d}{d r}\left(r^2 \frac{dR(r)}{d r}\right)+\frac{R(r)}{r^2 \sin \theta}\frac{d}{d \theta}\left(\sin \theta \frac{d \Theta(\theta)}{d \theta}\right)\endeq$$

Multiplying through by $r^2/(\Theta(\theta)R(r))$:

$$ \frac{1}{R(r)}\frac{d}{d r}\left(r^2 \frac{dR(r)}{d r}\right)= -\frac{1}{\Theta(\theta) \sin \theta}\frac{d}{d \theta}\left(\sin \theta \frac{d \Theta(\theta)}{d \theta}\right).$$

The only way for this to hold, since $\theta$ and $r$ are independent variables, is if the left chunk (only a function of $r$) and the right chunk (only a function of $\theta$) are each equal to a constant--the same constant, called the separation constant $k$. So this one equation is really two equations.

$\Theta(\theta)$ equation

$$ -\frac{1}{\Theta(\theta) \sin \theta}\frac{d}{d \theta}\left(\sin \theta \frac{d \Theta(\theta)}{d \theta}\right)=k=l(l+1).$$

It turns out that solutions are only possible for certain values of the separation constant: $k=l(l+1)$ where $l>=0$ is an integer. Re-arranging our equation for $\Theta(\theta)$ we get: $$\frac{1}{\sin \theta}\frac{d}{d \theta}\left(\sin \theta \frac{d \Theta(\theta)}{d \theta}\right) + l(l+1)\Theta(\theta) = 0.$$

The solutions that work for physical situations are the Legendre polynomials $P_l(x)$:

$l$$P_l(\cos \theta)$
0$1$
1$\cos \theta$
2$\frac{1}{2}(3 \cos^2 \theta -1)$
3$\frac{1}{2}(5\cos^3 \theta - 3\cos \theta)$


$R(r)$ equation

$$\frac{d}{dr}\left(r^2\frac{dR}{dr}\right) = l(l+1) R.$$ Since this is a 2nd order differential equation in one variable $\Rightarrow$ There are two independent solutions for any particular value of $l$.

Let $l=3$. Find two different functions $R(r)$ that solve this equation. [Hint: They are both powers of $r$.]





In general: $$R_l(r) = r^l \ \hbox{and}\ R_l(r)=r^{-(l+1)}.$$

Now, if $V_1$ and $V_2$ are each separately solutions to Laplace's equation $\grad^2 V=0$, then $A_1V_1+B_2V_2$ is also a solution. So the most general solution to Laplace's equation in terms of these harmonic functions is...

$$V(r,\theta) = \sum_{l=0}^{\infty} \left[A_l r^l + B_l r^{-l-1}\right] P_l(\cos \theta).$$

Legendre polynomials as unit vectors

Writing the unit vectors like this: $$\uv e_1\equiv \uv x; \ \ \uv e_2\equiv \uv y; \ \ \uv e_3\equiv \uv z$$ then here are some familiar properties of 3-d vectors (which involve the dot product):

  1. Orthogonality: $$ \uv e_i\cdot \uv e_j=\delta_{ij}\equiv \begin{cases} 0 & \text{ if }i\neq j \\ 1 & \text{ if }i=j \end{cases} $$
  2. Expressing a vector in terms of its components: $$\myv v=v_1\uv e_1+v_2\uv e_2+v_3\uv e_3=\sum_{i=1}^3 v_i\uv e_i.$$ where the components are found by dotting the unit vectors with the vector: $$v_i=\myv v\cdot\uv e_i.$$
  3. Completeness: Every vector in 3-dimensional space has a unique decomposition in 3 components. There is no vector than cannot be expressed this way!

The Legendre polynomials have analogous properties over the "vector space" of functions of $\theta$ defined on the interval $[0,\pi]$:

Orthogonality

Consider the "space" of bounded, continuous functions which are defined on the interval $\theta \in [0,\pi]$ which is an appropriate interval for functions of the spherical polar coordinate $\theta$. Let's define this as a "dot product": $$f(\theta)\cdot g(\theta)\equiv \int_0^{\pi}f(\theta)g(\theta)\sin\theta d\theta.$$

The Legendre polynomials, $P_0(\theta)=1$, $P_1(\theta)=\cos\theta$, $P_2(\theta)=\frac12(3\cos^\theta-1)$, ... are orthogonal in this sense: $$P_n(x) \cdot P_m(x) = \begin{cases} 0 & \ \hbox{ if }\ n \neq m\\ \frac{2}{2n+1} & \ \hbox{ if }\ n=m \end{cases} $$

For example... $$\begineq P_0(\theta)\cdot P_2(\theta) &= \int_0^\pi1\cdot\frac{1}{2}(3\cos^2\theta-1) \sin \theta\, d \theta\\ &=-\int_{0}^{\pi}\frac{1}{2}(3\cos^2\theta-1)\, d (\cos\theta) \\ &\equiv \int_{{\color{purple} \cos 0}}^{{\color{purple}\cos\pi}}{\color{purple}-}\frac{1}{2}(3x^2-1)\, d x\\ &= \int_{{\color{purple}-1}}^{{\color{purple}+1}}{\color{purple}+}\frac{1}{2}(3x^2-1)\, d x\\ &=\frac{1}{2}\left[(3x^3/3-x)\right]_{x=-1}^{1} \\ &= \frac{1}{2}\left[(1-1)-(-1+1)\right]=0\endeq$$

Components/completeness

It turns out that *any* bounded function of $\theta$ on the interval $\theta \in [0,\pi]$ can be expressed as a unique sum: $$f(\theta)=\sum_{i=0}^{\infty}a_l\,P_l(\theta)$$ where the components $a_l$ are given by $$a_l=\frac{2l+1}{2}f(\theta)\cdot P_l(\theta) =\frac{2l+1}{2}\int_0^\pi f(\theta) P_l(\theta)\sin\theta\, d\theta.$$

Express the function $f(\theta)=\cos(3\theta)$ as a sum of Legendre polynomials, $$f(\theta)=\sum_{l=0}^\infty a_lP_l(\cos\theta)$$

Using the formula above with $P_0=1$... $$\begineq a_0 &= \frac{2*(0)+1}{2}\int_0^\pi \cos{3\theta}*1*\sin\theta\,d\theta\\ &=\frac12[\text{Mathematica...}]=\frac12*0=0\endeq$$

..By the way, there is a symmetry argument that could have told you this by sketching out $\sin \theta$ and $\cos(3\theta)$. It turns out all the $a_l$'s with $l$ even are 0.

One more, using the same formula, with $P_1=cos\theta$. $$\begineq a_1 &= \frac{2*(1)+1}{2}\int_0^\pi \cos{3\theta}*\cos\theta*\sin\theta\,d\theta\\ &=\frac32[\text{Mathematica...}]=\frac32*-\frac25=-\frac35\endeq$$

Calculating a few more, you should find that $a_2=0$ and $a_3 \gt 0$. And if you graph $a_1P_1(\cos\theta)+a_3P_3(\cos\theta)$ you should find it identical to the graph of $\cos(3\theta)$.

Find $a_3$ and graph $a_1P_1(\cos\theta)+a_3P_3(\cos\theta)$ and $\cos(3\theta)$ (slightly offset).

Back to our original problem

For a problem with azimuthal symmetry, we know the general solution in terms of the Legendre polynomials: $$V(r,\theta) = \sum_{l=0}^{\infty} \left[A_l r^l + B_l r^{-l-1}\right] P_l(\cos \theta),$$ where $P_0=1$, $P_1=\cos \theta$, $P_2=\frac{1}{2}(3 \cos^2 \theta -1)$, etc. That is to say... all we need to do is find the $2*\infty$ coefficients....

$\color{#f00}{ A_0=?\\A_1=?\\A_2=?\\ A_3=?\\A_4=?\\ A_5=?\\A_6=?\\ ...\\ }$
$\color{#00f}{ B_0=?\\ B_1=?\\B_2=?\\ B_3=?\\B_4=?\\ B_5=?\\B_6=?\\ ...\\ }$

The particular solution should match the boundary conditions of the problem. There are boundary conditions at infinity, and on the surface of the sphere at $r=R$.

Apply the boundary conditions

...to find the precise solution.

We know that the solution can be expressed as...

Step 1: What terms in $V(r,\theta)$ can be neglected in the limit $r\to\infty$?

$\lim_{r\to\infty}B_lr^{-l-1}=0$, so $$\begineq V(r\gg R,\theta) \to& \sum_{l=0}^{\infty} A_l r^l P_l(\cos \theta)\\ &=A_0+A_1r\cos\theta+A_2r^2\frac 12(3 \cos^2\theta-1)+... \endeq $$

Step 2: Apply the boundary condition $V(r\gg R,\theta)\to -E_0 r \cos\theta$. Most of the $A_l$ have to be zero, except for one. Which one (which $l$) and what is it's value?

Ah, let $A_1=-E_0$ and $A_{l\neq 1}=0$. With these coefficients, the general solution now reads $$V(r,\theta)=-E_0r\cos\theta + \sum_{l=0}^\infty B_l r^{-l-1}P_l(\cos \theta).$$ The field was initially uniform in the $\uv{z}$ direction before the introduction of the sphere. Far, far away from the sphere, it should still be uniform in the $\uv z$ direction.

One potential that does that... $$\lim_{\frac{r}{R}\to +\infty}V(x,y,z) = -E_0z $$ Or, in spherical polar coordinates, $z=r\cos\theta$ so $$V(r \gg R,\theta) \to -E_0 r\cos \theta.$$ This boundary condition says nothing about the $B_l$, since they are multiplied by $r^{-(l +1)}$ and $\lim_{r\to \infty} r^{-(l+1)} = 0.$ But the $A_l$ are multiplied by $r^l$ which do not disappear at large $r$.

There is only one term that involves the combination $r\cos\theta$, (the $l=1$ term) so we conclude that $A_1=-E_0$ and *all* the other $A_l$s must vanish.

$\color{#faa}{ A_0=0\\} \color{#f00}{A_1=-E_0}\\ \color{#faa}{A_2=0\\ A_3=0\\A_4=0\\ A_5=0\\A_6=0\\ ...\\ }$
$\color{#00f}{ B_0=?\\ B_1=?\\B_2=?\\ B_3=?\\B_4=?\\ B_5=?\\B_6=?\\ ...\\ }$

[Formally because of the linear independence of the $P_l(\cos \theta)$ functions, there is no way to add multiples of the functions $P_0=1$, $P_2=\frac12(3\cos^2\theta-1)$, $P_3$, ... together to get something proportional to $P_1=\cos\theta$.]

Step 3: Now that you know something about the $A_l$'s, Write out $V(r=R)$ and apply the boundary condition $V(r=R)=0$ to decide which $B_l$'s are possible, and which are not.

$$\begineq 0&=V(r=R,\theta)\\ &=-E_0R\cos\theta + B_0 R^{-1} + B_1 R^{-2}\cos \theta+ B_2 R^{-3}\frac 12(3 \cos^2\theta+1)+... \endeq$$ The only way to get this expression to vanish for all $\theta$ is if $B_{l\neq 1}=0$ and $B_1=E_0 R^3$.

The potential on the surface of the sphere is constant ($V=0$) so that: $$\begineq V(r=R,\theta) &= 0=\left[-E_0 r \cos \theta+\sum_{l=0}^\infty B_lr^{-(l+1)}P_l(\cos\theta)\right]^{r=R}\\ &=-E_0 R \cos \theta+B_0\frac 1R +B_1\frac1{R^2}\cos\theta +B_2\frac{1}{R^3}\frac12(3\cos^2\theta-1)+....\endeq$$

Notice that the $B_1$ involves a $\cos\theta$ factor. If $B_1=+E_0R^3$ then this term will cancel out with $-E_0R\cos\theta$. There is no sum of the other factors with non-zero $B_l$s that could add up to 0, so we conclude that all the other $B_l$s are zero.

$\color{#faa}{ A_0=0\\} \color{#f00}{A_1=-E_0}\\ \color{#faa}{A_2=0\\ A_3=0\\A_4=0\\ A_5=0\\A_6=0\\ ...\\ }$
$\color{#aaf}{ B_0=0\\} \color{#00f}{ B_1=+E_0R^3\\\\} \color{#aaf}{B_2=0\\ B_3=0\\B_4=0\\ B_5=0\\B_6=0\\ ...\\ }$

So, the solution which matches both boundary conditions is apparently

$$\begineq V(r,\theta)&=-E_0r\cos\theta+E_0\frac{R^3}{r^2}\cos\theta\\ &=-E_0r\cos\theta\left(1-\frac{R^3}{r^3}\right). \endeq$$

Does the electric field behave as expected?

For $r\gg R$, we have $V(r\gg R,\theta)\to -E_0r\cos\theta=-E_0 z$, and therefore the electric field is $-\myv \grad V=\uv z \frac{\del V}{\del z}= E_0 \uv z$.

For $r=R$, the electric field should be perpendicular to the surface of the conducting sphere, that is, it should have a component in the $\uv r$ direction, but no $\theta$ or $\phi$ component.

We'll use the Spherical-Polar form of the gradient: $$\begineq \myv E(r,\theta, \phi)&=-\myv\grad V\\ &=-\frac{\del V}{\del r} \uv r-\frac 1r\frac{\del V}{\del \theta}\uv \theta -\frac{1}{r\sin \theta} \frac{\del V}{\del \phi}\uv \phi \\ \endeq $$

Taking the terms in turn:

$\phi$ component: $V$ does not depend on $\phi$, so $\frac{\del V}{\del \phi}=0$, and therefore $E_\phi=0$.

$\theta$ component: $$\begineq E_\theta &=-\frac 1r\frac{\del V}{\del \theta}\\ &=\frac 1r E_0 r(-\sin\theta)\left(1-\frac{R^3}{r^3}\right)\\ &=-E_0\sin\theta\left(1-\frac{R^3}{r^3}\right)\\ \endeq $$ From this expression, we can tell that $E_\theta(r=R)=0$. So the field is perpendicular to the surface at $r=R$.

$r$ component: $$\begineq E_r&=-\frac{\del V}{\del r}\\ &=-\frac{\del }{\del r}\left[-E_0\cos\theta\left(r-\frac{R^3}{r^2}\right)\right] \\ &=E_0\cos\theta\left(1+2\frac{R^3}{r^3}\right)\\ \endeq $$

Charge density, $\sigma$: On the surface of a conductor, the charge density $\sigma$ is related to the electric field just outside the conductor: $$\begineq \sigma(r=R,\theta)&=\epsilon_0 E=\epsilon_0 E_r(r=R)\\ &=\epsilon_03E_0\cos\theta\\ \endeq $$