Laplace's equation

Laplace's equation is a second-order linear partial differential equation

∇2ψ=0\nabla^{2}\psi=0

where ∇2\nabla^{2} is the Laplacian. A solution to this equation is said to be a harmonic function. It is a particular case of the Poisson's equation, where f=0f=0.

It is useful to solve the Laplace equation in coordinate systems other than Cartesian, for example, in spherical coordinates where the Spherical harmonics appear. In many coordinate systems, it is possible to solve the equation using separation of variables.

One-dimensional case

In one dimension, using Cartesian coordinates, the equation simplifies to a ordinary differential equation:

d2ψdx2=0\frac{ d^{2} \psi }{ d x^{2} } =0

and the general solution is simply

ψ(x)=mx+b\psi(x)=mx+b

which is just a straight line. The constants are determined by the boundary conditions. From this, we can derive a couple of useful results that carry over to the two- and three-dimensional cases:

  1. Given a certain aa, ψ(x)\psi(x) is the average of ψ(x+a)\psi(x+a) and ψ(x−a)\psi(x-a):ψ(x)=12[ψ(x+a)+ψ(x−a)]\psi(x)=\frac{1}{2}[\psi(x+a)+\psi(x-a)]and the harmonics that solve the equation fit the endpoints exactly. In a way, Laplace's equation is a rule on how to average a function.
  2. There are no local critical points within the harmonic. The only critical points are at the edges of the function. This is a direct consequence of point 1., as if there were maxima and minima within the function, then there would be cases in which the solution would not be an average (as both sides would be bigger or smaller).

Two-dimensional case

In two dimensions, using Cartesian coordinates, the equation is

∂2ψ∂x2+∂2ψ∂y2=0\frac{ \partial ^{2} \psi }{ \partial x^{2} }+\frac{ \partial ^{2} \psi }{ \partial y^{2} } =0

Since it is a partial differential equation, it has no closed-form general solution. Nevertheless, the properties from the one-dimensional case still apply:

  1. The value of ψ(x,y)\psi(x,y) is the average of all surrounding points, given by a closed integral over a circle of radius RR around the point:$$\psi(x,y)=\frac{1}{2\pi R}\underset{ \text{circle} }{ \oint }\psi\ dl
2. The maxima and minima are all at the boundary. The surface of the function is perfectly smooth, with no hills or valleys. #### Solution by separation of variables Separation of variables instructs us to look for solutions of the form

\psi(x,y)=\phi(x)\theta(y)

sothatourequationbecomesso that our equation becomes

\theta \frac{ \partial ^{2} \phi }{ \partial x^{2} }+\phi \frac{ \partial ^{2}\theta }{ \partial y^{2} } =0

Dividing through by $\psi$ we get

\frac{1}{\phi}\frac{ \partial ^{2}\phi }{ \partial x^{2} } + \frac{1}{\theta}\frac{ \partial ^{2}\theta }{ \partial y^{2} } =0

Inotherterms,In other terms,

\frac{1}{\phi}\frac{ \partial ^{2}\phi }{ \partial x^{2} } =- \frac{1}{\theta}\frac{ \partial ^{2}\theta }{ \partial y^{2} }

These two terms are dependent only on $x$ and $y$ respectively, but they are also equal. This means that if one changes (say, $x$ varies and thus $\phi$), than so does the other. The issue is that $\theta$ is not dependent on $x$, but by equality it must change to mimic the other term. In substance, the $\theta$ would be changed by $x$ despite being independent of it. This makes no sense and the only scenario in which this does not happen is if both terms are actually independent of *all* variables, that is, they are constant. This permits us to write

\frac{1}{\phi}\frac{ \partial ^{2}\phi }{ \partial x^{2} } =C_{1},\qquad \frac{1}{\theta}\frac{ \partial ^{2} \theta }{ \partial y^{2} } =C_{2}

and thus $C_{1}=-C_{2}$. Renaming them to $k^{2}$, we get

\frac{ \partial ^{2}\phi }{ \partial x^{2} } =k^{2}\phi,\qquad \frac{ \partial ^{2}\theta }{ \partial y^{2} } =-k^{2}\theta

Sincethesearejust[[Harmonicoscillator∣harmonicoscillators]],wehavethesolutionSince these are just [[Harmonic oscillator|harmonic oscillators]], we have the solution

\psi(x,y)=(Ae^{kx}+Be^{-kx})(C\sin ky+D\cos ky)

Determining the constants $A$, $B$, $C$ and $D$ requires knowing the boundary conditions, which depends on the problem. The important part is that the solution is not expressible in closed form, as solving this equation gives us an infinite number of possibilities. Each solution individually is likely not the description of the system at hand, but they possess the important property of linearity, that is, the sum of several solutions is still a solution. Thus, it is possible to [[combinazione lineare|linearly combine]] several solutions (an infinite number, actually), to describe the state of the system. As an example, it may take the form

\psi(x,y)=\sum_{n=1}^{\infty} c_{n}e^{-n\pi x/a}\sin\left( \frac{n\pi x}{a} \right)

where $n$ is a nonzero integer and $a$ is a constant. This is a [[Serie di Fourier|Fourier series]], specifically a sine series. From Fourier analysis, we know that any [[Funzione generalmente continua|almost always continuous]] function can be expressed as a Fourier series, so that's the only limitation on $\psi(x,y)$. The coefficients can as usual be derived with a [[Prodotto scalare|dot product]] like

c_{n}=(\psi_{n},\ \psi)=\left( e^{-n\pi x/a}\sin\left( \frac{n\pi x}{a} \right),\ \psi \right)

As is always the case for separable solutions, the set formed by $\{\psi_{n}\}_{n\in \mathbb{N}}$ is a [[Sistema ortonormale completo|complete orthonormal set]] and thus $\psi_{n}\psi_{m}=\delta_{nm}$ with the [[Delta di Kronecker|Kronecker delta]] and $\psi=\sum_{n}^{\infty}c_{n}\psi_{n}$ is always true. We can also find a solution in [[Spherical coordinates]] by starting in three-dimensions and then assuming *azimuthal symmetry*, which states that $\psi$ is independent from the azimuthal angle. This removes one variable, leaving a two-dimensional solution. The most general solution found that is compatible with minimum physical requirements is

\psi(r,\theta)=\sum_{l=0}^{\infty} \left( A_{l}r^{l}+ \frac{B_{l}}{r^{l+1}} \right)P_{l}(\cos \theta)

where $P_{l}(\cos \theta)$ are the [[Polinomi di Legendre|Legendre polynomials]]. ### Three-dimensional case In three dimensions, we get the full equation:

\frac{ \partial ^{2} \psi }{ \partial x^{2} } + \frac{ \partial ^{2} \psi }{ \partial y^{2} } + \frac{ \partial ^{2} \psi }{ \partial z^{2} } =0

In this case, the properties are as follow: 1. The value of $\psi(x,y,z)$ is again the average of all surrounding points, this time determined by a loop integral on a sphere centered on that point:$$\psi(x,y,z)=\frac{1}{4\pi R^{2}}\underset{ \text{sphere} }{ \oint } \psi\ da
  1. The maxima and minima are still at the boundary. There is no easy way to visualize this, but the general property of minimizing the measure of the space spanning the boundary still applies.

The separable variable solution for the three-dimensional case is similar to the two-dimensional case, but becomes more complicated as it ends up requiring a double Fourier series over two sets of solutions. An example is

ψ(x,y,z)=∑n=1∞∑m=1∞cnme−π(n/a)2+(m/b)2 xsin⁡(nπya)sin⁡(mπzb)\psi(x,y,z)=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} c_{nm}e^{-\pi \sqrt{ (n/a)^{2}+(m/b)^{2} }\ x}\sin\left( \frac{n\pi y}{a} \right)\sin\left( \frac{m\pi z}{b} \right)

where nn and mm are the integer indexes and aa and bb are constants.

For a solution in spherical coordinates, see Schrödinger's equation.

Uniqueness theorems

Determining what the boundary conditions are in the one-dimensional case is easy: just provide the values at the edges, or a derivative at an edge and a value at the same edge, or... several other combinations. But we can't provide the derivative at both edges, as that would be either redundant (if they are equal) or nonsense (if they are different).

For the two- and three-dimensional case, such easy logic doesn't hold. The proof that a set of boundary condition provides a unique solution is usually given in the form of a uniqueness theorem. There are several for the Laplace equation.

First uniqueness theorem

This theorem actually also applies to Poisson's equation.

Proof

Consider a volume VV and let's assume the we know the value of ψ\psi on its surface. It does not need to be solid: there may "holes" within, provided we know ψ\psi on all of those surfaces too. It's also possible for the outer surface to be at infinity provided, again, we know the limit of ψ\psi at infinity. Suppose there are two solutions to the equation:

∇2ψ1=0,∇2ψ2=0\nabla ^{2}\psi_{1}=0, \qquad \nabla ^{2}\psi_{2}=0

both of which have the same value on the surface(s). If we look at their difference, it is

ψ3=ψ1−ψ2\psi_{3}=\psi_{1}-\psi_{2}

which must also obey Laplace's equation:

∇2ψ3=∇2ψ1−∇2ψ2=0\nabla ^{2}\psi_{3}=\nabla ^{2}\psi_{1}-\nabla ^{2}\psi_{2}=0

This takes the value zero on all boundaries, since ψ1\psi_{1} and ψ2\psi_{2} have the same value there. But we know there are no maxima and minima outside of the boundary, thus all other points must be themselves zero, so ψ3=0\psi_{3}=0, which means ψ1=ψ2\psi_{1}=\psi_{2}.

F(r)=−q2r0r4πε0(r2−r02)2F(r)=- \frac{q^{2}r_{0}r}{4\pi \varepsilon_{0}(r^{2}-r_{0}^{2})^{2}}

The same can be proven for Poisson's equation, as the function ff cancels out either way. We can express this as a sort of corollary:

Electrostatic interpretation

This theorem (especially the corollary) is of particular use in electrostatics, where the electric potential obeys Poisson's equation. The corollary can therefore be stated as:

This is useful because the potential is often known, since it comes from artificial batteries that are built provide a specific potential.

Second uniqueness theorem

The second uniqueness theorem is given directly in electromagnetic flavor.

Proof

Suppose there are two fields satisfying the conditions. Both obey Gauss' law:

∇⋅E1=ρε0,∇⋅E2=ρε0\nabla\cdot\mathbf{E}_{1}=\frac{\rho}{\varepsilon_{0}},\qquad \nabla\cdot\mathbf{E}_{2}=\frac{\rho}{\varepsilon_{0}}

also in integral form for a Gaussian form each containing a conductor:

∮i-th conductorE1⋅da=Qiε0,∮i-th conductorE2⋅da=Qiε0\underset{ i\text{-th conductor} }{ \oint }\mathbf{E}_{1}\cdot d\mathbf{a}=\frac{Q_{i}}{\varepsilon_{0}},\qquad \underset{ i\text{-th conductor} }{ \oint }\mathbf{E}_{2}\cdot d\mathbf{a}=\frac{Q_{i}}{\varepsilon_{0}}

and for the outermost surface:

∮outer surfaceE1⋅da=Qtotε0,∮outer surfaceE2⋅da=Qtotε0\underset{ \text{outer surface} }{ \oint }\mathbf{E}_{1}\cdot d\mathbf{a}=\frac{Q_\text{tot}}{\varepsilon_{0}},\qquad\underset{ \text{outer surface} }{ \oint }\mathbf{E}_{2}\cdot d\mathbf{a}=\frac{Q_\text{tot}}{\varepsilon_{0}}

Let's examine the difference

E3=E1−E2\mathbf{E}_{3}=\mathbf{E}_{1}-\mathbf{E}_{2}

which obeys

∇⋅E3=0\nabla\cdot\mathbf{E}_{3}=0

in the region between the conductors and

∮E3⋅da=0\oint \mathbf{E}_{3}\cdot d\mathbf{a}=0

over each boundary surface. Since conductors are equipotential volumes, we know that the potential V3V_{3} associated with E3\mathbf{E}_{3} has to be a constant for all conductor surfaces (it doesn't need to be the same constant for all of them). The trick is to use the product rule

∇⋅(V3E3)=V3(∇⋅E3)+E3⋅(∇V3)=−(E3)2\nabla \cdot(V_{3}\mathbf{E}_{3})=V_{3}(\nabla\cdot\mathbf{E}_{3})+\mathbf{E}_{3}\cdot(\nabla V_{3})=-(E_{3})^{2}

Integrating over V\mathcal{V} and applying the divergence theorem to the left side we get

∫V∇⋅(V3E3) dτ=∮SV3E3⋅da=−∫VE32 dτ\int_{\mathcal{V}}\nabla\cdot (V_{3}\mathbf{E_{3}})\ d\tau=\oint_{\mathcal{S}}V_{3}\mathbf{E}_{3}\cdot d\mathbf{a}=-\int_{\mathcal{V}}E_{3}^{2}\ d\tau

The surface S\mathcal{S} covers all boundaries of the region in question, i.e. the conductors and outer boundary. Since V3V_{3} is constant over each surface, it comes out of the integrals and what remains is zero. Thus we're left with

∫VE32 dτ=0\int_{\mathcal{V}}E_{3}^{2}\ d\tau=0

E32E_{3}^{2} is of course always positive, so the only way for it to vanish is if it's always zero. Therefore E1=E2\mathbf{E}_{1}=\mathbf{E}_{2}.