##### The Laplace Equation
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In regions of space where there is no charge density, the potential must solve Laplace's equation. Let us discuss how solutions to this equation look, in increasingly complicated situations.

###### The Laplace Equation in One Dimension

In one dimension, the potential is a single-variable function \(\phi (x)\) and the Laplace equation reads

\[ \frac{d^2 \phi(x)}{dx^2} = 0. \tag{Lap_1d}\label{Lap_1d} \]

The solution to this is

- Gr (3.6)

\[ \phi(x) = a x + b \tag{Lap_1d_sol}\label{Lap_1d_sol} \]

Properties:

- Balance: \(\phi(x)\) is the average of \(\phi(x + dx)\) and \(\phi(x - dx)\) for any \(dx\) (with \(x \pm dx\) still being in the region where Laplace is satisfied, of course).
- No extrema: \(\phi(x)\) has no local extrema. Max/min values must occur at boundaries.

In a particular problem, to fix the solution (said otherwise: to fix the parameters \(a\) and \(b\) in Lap_1d_sol), we need to appeal to boundary conditions. Concretely, for a finite segment, a solution exists and is unique if one is provided with any of these possibilities:

- \(\phi\) at both boundaries
- \(\phi\) and \(\frac{d\phi}{dx}\) at one boundary
- \(\phi\) at one boundary, \(\frac{d\phi}{dx}\) at the other.

Specifying \(\frac{d\phi}{dx}\) at both boundaries provides insufficient information, since you get an inconsistency if the derivatives don't match.

###### The Laplace Equation in Two Dimensions

In two dimensions, the potential becomes a function of two variables (here: \(x\) and \(y\)), so Laplace's equation now reads

\begin{equation*} \frac{\partial^2 \phi (x,y)}{\partial x^2} + \frac{\partial^2 \phi (x,y)}{\partial y^2} = 0. \tag{Lap_2d}\label{Lap_2d} \end{equation*}

Properties:

- Balance: \(\phi(x,y)\) equals the average value around the point:

\[ \phi(x,y) = \frac{1}{2\pi R} \oint dl ~\phi \]

- No extrema: \(\phi\) has no local maxima or minima. All extrema occur at the boundaries.

###### The Laplace Equation in Three Dimensions

In three dimensions, we will write the potential as a function of a 3-dimensional vector, \(\phi({\bf r})\). The Laplace equation is (we repeat)

\[ {\boldsymbol \nabla}^2 \phi ({\bf r}) = 0 \]

Theorem: if \(\phi\) satisfies Laplace, then its value at a point equals its value averaged over a sphere \(S_R({\bf r})\) of any radius \(R\) centered on this point (and of course not containing any charges),

\[ \phi({\bf r}) = \frac{1}{4\pi R^2} \oint_{S_R({\bf r})} da' ~\phi ({\bf r}') \tag{p_ball_avg}\label{p_ball_avg} \]

Physicist's proof

Consider a sphere of radius \(R\) centered at the origin carrying charge \(q\) spread with a uniform surface charge density over its surface. Bring in a point charge \(q'\) from infinity up to a distance \(R'\) (with \(R' > R\)) from the center of the sphere.

We know that the field created by the sphere coincides with that of a point charge \(q\) at the origin. The work required to bring the \(q'\) charge into position is thus simply \(W = \frac{q q'}{4 \pi \varepsilon_0 R'}\) by Wab.

We can however proceed the other way: fixing \(q'\) in place, and then bringing the charged sphere into position; the work (energy) has to coincide with our previous result. But this energy is now given by the integral of the potential \(\phi_{q', {\bf r'}}\) created by \(q'\) (sitting at \({\bf r'}\)) over the sphere times the surface charge density on the sphere, namely

\[ W = \oint_{S_R} da ~\sigma ~\phi_{q', {\bf r}'} ({\bf r}) \]

But \(\sigma = q/4\pi R^2\) and is a constant over the sphere, so \(W = q \times \frac{1}{4\pi R^2} \oint da ~\phi_{q', {\bf r}'} ({\bf r})\).

Equating this with the previous results shows that

\[ \frac{q'}{4\pi \varepsilon_0 R'} = \frac{1}{4\pi R^2} \oint_{S_R} da ~\phi_{q', {\bf r}'} ({\bf r}) \] but this also equals the potential at \({\bf r} = 0\) created by the charge \(q'\) at \({\bf r'}\), i.e. \(\phi_{q', {\bf r}'} (0) = \frac{q'}{4\pi \varepsilon_0 R'}\). In other words, we have thus shown that for the potential created by a single point charge \(q'\) at \(R'\), the value at a point (here the origin) coincides with the value averaged over a sphere or an arbitrary radius \(R\) centered on the same point.

By the principle of superposition, this works for an arbitrary distribution of charges outside the sphere, proving the theorem.

Formal proof

Consider a function \(f({\bf r})\) and its average over a ball of radius \(R\) centered on \({\bf r}\):

\[ f_{S_R} ({\bf r}) \equiv \frac{1}{4\pi R^2}\oint_{S_R ({\bf r})} da' ~ f ({\bf r}') \]

In spherical coordinates defined around the point \({\bf r}\), we have \(da' = R^2 sin \theta d\theta d\varphi \equiv R^2 d\Omega\). Differentiating with respect to \(R\),

\[ \frac{d}{dR} f_{S_R} = \frac{1}{4\pi} \oint_{S_R} d\Omega ~\left.\frac{\partial f}{\partial r}\right|_{r=R} \]

with \(f\) differentiated with respect to the radial coordiate. We can rewrite this by noting that \(R^2 d\Omega ~\hat{\bf r}\) is the normal differential surface area \(d{\bf a}\), while \(\left.\frac{\partial f}{\partial r}\right|_{r=R}\) is the radial component of the gradient of \(f\) in spherical coordinates. Thus,

\[ \frac{d}{dR} f_{S_R} = \frac{1}{4\pi R^2} \oint_{S_R} d{\bf a} \cdot ~\nabla f \]

Invoking the divergence theorem and using the definition of the Laplacian operator \(\nabla^2 = \nabla \cdot \nabla\), we get the following general

Theorem:

\[ \frac{d}{dR} f_{S_R} = \frac{1}{4\pi R^2} \int_{V_R} d\tau ~\nabla^2 f \tag{dfdR_intLap}\label{dfdR_intLap} \]

For the electrostatic potential away from charges, we have \[ \nabla^2 \phi = 0 ~\rightarrow \frac{d}{dR} \phi_{S_R} = 0 \] namely the ball average is independent of the ball size. Since the value at the center is simply the average for an infinitesimally small ball, we get the result announced above.

Theorem (Earnshaw, mathematical version): \(\phi\) has no local extrema except at the boundaries.

Proof: write the second derivatives as

\[ \frac{\partial^2 \phi({\bf r})}{\partial x^2} = f_x ({\bf r}), \hspace{5mm} \frac{\partial^2 \phi({\bf r})}{\partial y^2} = f_y ({\bf r}), \hspace{5mm} \frac{\partial^2 \phi({\bf r})}{\partial z^2} = f_z ({\bf r}), \hspace{5mm} f_x + f_y + f_z = 0. \]

The \(f_a ({\bf r})\) represent the three components of the curvature of \(\phi({\bf r})\). An extremum of \(\phi\) at \({\bf r}_e\) would be characterized by \({\boldsymbol \nabla} \phi |_{{\bf r}_e} \cdot \delta{\bf r} = 0\) for any infinitesimal displacement \(\delta{\bf r}\) around the extremum point. For a local minimum, the second derivative form should be greater than zero, \(\sum_{i,j} \frac{\partial^2 \phi}{\partial r_i \partial r_j} \delta r_i \delta r_j > 0\) for any displacement vector. Choosing alternately displacements along the three axes, the form becomes \(f_x (\delta x)^2\), \(f_y (\delta y)^2\) or \(f_z (\delta z)^2\). Since the squared displacements are necessarily positive, we thus require \(f_x > 0\), \(f_y > 0\) and \(f_z > 0\). This is impossible in view of the \(f_x + f_y + f_z = 0\) condition above.

Earnshaw's theorem (physical version)

It is impossible to find a static distribution of charges which generates an electrostatic field displaying a stable equilibrium position in empty space.

Going back to Poisson's equation, we can make a few comments:

- representation 🐟 highlights the 'local' nature of the coupling between electrostatic fields and charges: fields are 'created' where the charges 'sit'. This is also seen by looking at the integrand of p_vcd. If electrostatics was nonlocal, a modified representation like p_vcd would still exist, but not a local differential one like Poisson's equation.
- as written, representations E_vcd and p_vcd require the knowledge of the charge density distribution \(\rho({\bf r})\) throughout space to determine the potential at any given point.
- Poisson's equation 🐟, being purely local, might allow to determine the potential at a specified point, provided we know the charge density distribution around this specified point, and at some set of other reference points (to make the solution unique).

We therefore want to ask the question: under what conditions can an electrostatic problem be fully defined by solving Poisson's equation?

We start by mentioning some cases, and interpreting them thereafter.

Charge density is known throughout space: in this case, the electrostatic potential is uniquely determined by Poisson's equation, which is explicitly solved by p_vcd. One can explicitly verify this:

\[ {\boldsymbol \nabla}^2 \phi ({\bf r}) = \frac{1}{4\pi \varepsilon_0} \int_{\mathbb{R}^3} d\tau' \rho({\bf r}') {\boldsymbol \nabla}^2 \frac{1}{|{\bf r} - {\bf r}'|} = \frac{1}{4\pi \varepsilon_0} \int_{\mathbb{R}^3} d\tau' (-4\pi) \delta ({\bf r} - {\bf r}') = -\frac{\rho ({\bf r})}{\varepsilon_0}. \]

where we have used Lap1or, and the fact that the delta function is always resolved since we integrate over all space. Note: it is implicitly assumed that the integral in p_vcd converges, i.e. that the charge density \(\rho({\bf r})\) is sufficiently well-behaved.

"Known boundary charge" case: charge density in closed volume and boundary surface charge density are both known: the electrostatic potential is uniquely determined in a certain volume \({\cal V}\) bounded by boundary \({\cal S}\), provided the charge density \(\rho ({\bf r})\) is given everywhere within \({\cal V}\), vanishes outside of \({\cal V}\), and the value of the surface charge density \(\sigma\) is given everywhere on the boundary \({\cal S}\). Of course, \({\cal S}\) need not be a connected surface.

"Known boundary potential" case: charge density in closed volume and potential at boundary are both known: the electrostatic potential is uniquely determined in a certain volume \({\cal V}\) bounded by boundary \({\cal S}\), provided the charge density \(\rho ({\bf r})\) is given everywhere within \({\cal V}\), and the value of \(V\) is given everywhere on the boundary \({\cal S}\). Of course, \({\cal S}\) need not be a connected surface.

Here, the logic is quite simple: since the electrostatic potential is known on all the surface enclosing the space \({\cal V}\), and since Poisson's equation is local, we need not consider anything outside of \({\cal V}\) to obtain \(V\) within \({\cal V}\).

Given a solution \(V_1 ({\bf r})\), we can easily show that it is unique. Suppose there was another solution \(V_2 ({\bf r})\). Look at the difference, \(U \equiv V_1 - V_2\). In the bulk, \(U\) obeys the Laplace equation

\[ {\boldsymbol \nabla}^2 U = {\boldsymbol \nabla}^2 V_1 - {\boldsymbol \nabla}^2 V_2 = -\frac{\rho}{\varepsilon_0} + \frac{\rho}{\varepsilon_0} = 0. \]

Moreover, \(U ({\bf r}) = 0\) for \({\bf r} \in {\cal S}\). Since solutions to the Laplace equation take their maximal and minimal value on the boundary, we must have \(U = 0\) \(\forall {\bf r} \in {\cal V}\)

This all feels a bit amateurish and not very systematic. Can we be more precise and general? What kinds of boundary information do we really need to specify the solution uniquely?

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42