The Laplace Equation
The Laplace Equation and Harmonic Functions
Laplace's equation is
$$ \nabla^2 V = 0 $$In three-dimensional Cartesian coordinates, this becomes
$$ \frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} + \frac{\partial^2 V}{\partial z^2} = 0 $$Functions satisfying Laplace's equation are called harmonic functions. They appear naturally in electrostatics, gravitation, fluid flow, heat flow, and many other areas of physics.
One-Dimensional Example
In one dimension, Laplace's equation reduces to
$$ \frac{d^2V}{dx^2} = 0 $$Integrating twice gives
$$ V(x) = mx + b $$which is simply a straight line.
A linear function has the property that its value at a point is equal to the average of the values on either side:
$$ V(x) = \frac{1}{2} \left( \, V(x+a) + V(x-a) \, \right) $$This averaging property is one of the most important features of Laplace's equation. Roughly speaking, the equation says that the value of the function at a point is determined entirely by the surrounding values.
Laplace's equation therefore acts like a smoothing condition: harmonic functions do not contain sharp spikes, isolated peaks, or abrupt changes.
No Local Maxima or Minima
Another important property is that harmonic functions cannot possess local maxima or minima in the interior of a region. Any extreme values must occur on the boundary.
Intuitively, this follows from the averaging property. If the value at a point must equal the average of nearby values, then it cannot be strictly larger or smaller than all surrounding points.
In two dimensions, Laplace's equation becomes
$$ \frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} = 0 $$Unlike the one-dimensional case, there is no simple general closed-form solution. However, harmonic functions still obey a powerful averaging property.
If we draw a circle of radius \(R\) centred at the point \((x,y)\), then the value of the function at the centre equals the average value around the circle:
$$ V(x,y) = \frac{1}{2\pi R} \oint_{\text{circle}} V \, dl $$Thus harmonic functions are completely determined by surrounding values. Laplace's equation selects the smoothest function compatible with the boundary conditions.
In three dimensions, the same idea holds with spherical averages:
$$ V(\vec{r}) = \frac{1}{4\pi R^2} \oint_{\text{sphere}} V \, da $$where the sphere has radius \(R\) and is centred at the point \(\vec r\).
Again, the value at a point equals the average over the surrounding sphere. This averaging behaviour is one of the defining features of harmonic functions.
Boundary Conditions and Uniqueness
Laplace's equation by itself does not determine a unique solution. Boundary conditions must also be specified.
Suppose \(V_1\) and \(V_2\) both satisfy Laplace's equation in some region:
$$ \nabla^2 V_1 = 0, \qquad \nabla^2 V_2 = 0 $$and suppose they agree on the boundary. Define
$$ V_3 = V_1 - V_2 $$Then
$$ \nabla^2 V_3 = \nabla^2 V_1 - \nabla^2 V_2 = 0 $$and \(V_3 = 0\) on the boundary.
But harmonic functions cannot possess interior maxima or minima. Since the maximum and minimum values on the boundary are both zero, the function must vanish everywhere:
$$ V_3 = 0 $$Therefore
$$ V_1 = V_2 $$This proves the uniqueness theorem:
If a function satisfies Laplace's equation and has the correct values on the boundary, then the solution is unique.
Electrostatics Example
Consider a configuration that is independent of \(z\), so that
$$ V(\vec r) \rightarrow V(x,y) $$Laplace's equation then becomes
$$ \frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} = 0 $$Suppose we have the following boundary conditions:
$$ V(x,0) = V(x,a) = 0 $$ $$ V(0,y) = V_0(y) $$ $$ V \rightarrow 0 \; \text{as} \; x \rightarrow \infty $$Physically, this corresponds to an infinite strip bounded by grounded conducting plates at \(y=0\) and \(y=a\), while the boundary at \(x=0\) is maintained at some specified potential \(V_0(y)\).
Since the potential is specified on all boundaries, the uniqueness theorem guarantees that the solution is uniquely determined.
We look for solutions of the form
$$ V(x,y) = X(x) Y(y) $$Substituting into Laplace's equation gives
$$ X''(x)Y(y) + X(x)Y''(y) = 0 $$Dividing by \(X(x) \, Y(y)\) gives
$$ \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = 0 $$ $$ \frac{X''(x)}{X(x)} = - \frac{Y''(y)}{Y(y)} $$The left-hand side depends only on \(x\), while the right-hand side depends only on \(y\). Since the two sides must remain equal for all values of \(x\) and \(y\), they must both equal a constant:
$$ \frac{X''(x)}{X(x)} = - \frac{Y''(y)}{Y(y)} = \lambda $$This produces two ordinary differential equations:
$$ X''(x)=\lambda X(x) $$ $$ Y''(y)=-\lambda Y(y) $$Let's solve the \( Y \) equation first. The boundary conditions are
$$ Y(0) = Y(a) = 0 $$We now examine the possible signs of \( \lambda \):
\( i) \lambda < 0 \)
Let
$$ \lambda=-\mu^2 $$Then
$$ Y''(y)=\mu^2Y(y) $$whose general solution is
$$ Y(y) = c_1 e^{\mu y} + c_2 e^{-\mu y} $$ $$ Y(0) = c_1 + c_2 \quad \Rightarrow \quad c_1 = - c_2 $$ $$ Y(a) = c_1 (e^{\mu a} - e^{- \mu a}) = 0 $$\( \Leftrightarrow \; c_1 = 0 \; \) or \( \; e^{2 \mu a} = 1 \Rightarrow 2 \mu a = \ln1 = 0 \)
So applying the boundary conditions gives only the trivial solution \( Y(y)=0 \). Therefore no nonzero solutions exist for \( \lambda < 0 \).
\( ii) \lambda = 0 \)
In this case,
$$ Y''(y)=0 $$giving
$$ Y(y) = c_1y + c_2 $$ $$ Y(0) = c_2 = 0 $$ $$ Y(a) = c_1a = 0 $$Applying the boundary conditions again forces \( Y(y) = 0 \), so no nontrivial solution exists here either.
\( iii) \lambda > 0 \)
Let \( \lambda = \mu^2 \) where \( \mu \ge 0 \)
Then
$$ Y''(y) = -\mu^2Y(y) $$whose general solution is
$$ Y(y) = c_1 e^{i\mu y} + c_2 e^{-i\mu y} $$ $$ Y(0) = c_1 + c_2 \quad \Rightarrow \quad c_1 = - c_2 $$ $$ Y(a) = c_1 (e^{i \mu a} - e^{- i \mu a}) = 0 $$\( \Leftrightarrow c_1 = 0 \, \) or \( \, e^{2 i \mu a} = 1 \)
$$ \Rightarrow \quad 2 \mu a = 2 \pi n, \quad n = 0, 1, 2, ... $$ $$ \mu = \frac{n \pi}{a}, \quad \lambda_n = \mu^2 = \left( \frac{n \pi}{a} \right)^2 $$ $$ Y(y) = c_1 (e^{i \mu y} - e^{- i \mu y}) = 2 c_1 i \sin(\mu y) $$ $$ = A \sin \left( \frac{n \pi y}{a} \right) $$Equivalently, using trig functions
$$ Y(y) = c_1 \sin(\mu y) + c_2 \cos(\mu y) $$ $$ Y(0) = c_2 = 0 $$ $$ Y(a) = c_1 \sin(\mu a) = 0 $$ $$ \Rightarrow \quad \mu a = \pi n, \quad n = 0, 1, 2, ... $$ $$ \mu = \frac{n \pi}{a}, \quad \lambda_n = \mu^2 = \left( \frac{n \pi}{a} \right)^2 $$And so
$$ Y_n(y) = A_n \sin \left( \frac{n \pi y}{a} \right) $$The equation for \(X(x)\) becomes
$$ X''(x) = \mu^2 X(x) $$whose general solution is a linear combination of \( e^{\mu x} \) and \( e^{- \mu x} \). Therefore
$$ V_n(x,y) = (c_n e^{\mu x} + d_n e^{- \mu x}) \sin(\mu y) $$Since the potential must vanish as \(x\rightarrow\infty\), the exponentially growing solution is discarded:
$$ c_n = 0 $$Therefore
$$ V_n(x,y) = d_n \sin \left( \frac{n\pi y}{a} \right) \, e^{- n \pi x / a} $$By linearity, the general solution is a superposition of all allowed modes:
$$ V(x,y) = \sum_{n=1}^{\infty} d_n \sin \left( \frac{n\pi y}{a} \right) \, e^{- n \pi x / a} $$Each sine mode on the boundary extends into the interior while decaying exponentially with increasing \(x\). Higher-frequency modes decay more rapidly, so fine spatial structure on the boundary disappears more quickly deeper into the region.
Orthogonality and Fourier Coefficients
Applying the boundary condition at \(x=0\) gives
$$ V(0,y) = \sum_{n=1}^{\infty} d_n \sin \left( \frac{n\pi y}{d} \right) = V_0(y) $$The functions \( \displaystyle Y_n = \sin \left( \frac{n\pi y}{a} \right) \) are eigenvectors of \( \partial_y^2 \) with eigenvalues \( \displaystyle - \left( \frac{n\pi}{a} \right)^2 \)
\( Y_n \) forms an orthogonal basis \( \Leftrightarrow \) the inner product \( \, \langle Y_n,Y_m\rangle = \delta_{nm} \|Y_n\|^2 \, \), where \( \, \displaystyle \delta_{nm} = \begin{cases} 1 & n = m \\ 0 & n \neq m \end{cases} \)
We can prove \( Y_n \) are orthogonal by showing that \( \partial_y^2 \) is symmetric under the boundary conditions.
The relevant vector space is
$$ \left\{ f : [0,a] \to \mathbb{R} \mid f(0)=f(a)=0 \right\} $$ $$ \left\langle \partial_y^2 f, g \right\rangle = \int_0^a \partial_y^2 f \, g \; dy = \int_0^a f'' g \, dy $$Integrate by parts
$$ \int uv' = uv - \int vu' $$ $$ u = g, \quad u' = g' $$ $$ v = f', \quad v' = f'' $$ $$ = \left[ f'g \right]_0^a - \int_0^a f'g' \, dy $$Integrate by parts again
$$ u = g', \quad u' = g'' $$ $$ v = f, \quad v' = f' $$ $$ = - \left( \left[ f g' \right]_0^a - \int_0^a f g'' \, dy \right) $$ $$ = \int_0^a f \, \partial_y^2 g \; dy = \left\langle f, \partial_y^2 g \right\rangle $$And so \( \partial_y^2 \) is symmetric under the boundary conditions \( Y(0) = Y(a) = 0 \)
$$ \therefore \sum_n c_n Y_n = v \quad \Rightarrow \quad c_n \left\langle Y_n,Y_n \right\rangle = \left\langle Y_n,v \right\rangle $$Therefore the coefficients are obtained by projection:
$$ d_n = \frac{ \langle Y_n,V_0\rangle }{ \|Y_n\|^2 } $$Explicitly,
$$ d_n = \frac{2}{a} \int_0^a V_0(y) \sin \left( \frac{n\pi y}{a} \right) \, dy $$Thus the boundary potential is expanded as a Fourier sine series, and each Fourier mode propagates into the interior with exponential decay.
$$ \sum_{n=1}^{\infty} d_n \int_0^a \sin\left(\frac{n\pi y}{a}\right) \sin\left(\frac{m\pi y}{a}\right) \, dy = \int_0^a V_0(y) \sin\left(\frac{m\pi y}{a}\right) \, dy $$ $$ \int_0^a \sin\left(\frac{n\pi y}{a}\right) \sin\left(\frac{m\pi y}{a}\right) \, dy = \delta_{nm} \left\|Y_n\right\|^2 = \begin{cases} \dfrac{a}{2} & n=m \\[0.4em] \; 0 & n\ne m \end{cases} $$ $$ \therefore \, d_n = \frac{2}{a} \int_0^a V_0(y) \sin\left(\frac{n\pi y}{a}\right) \, dy $$As a concrete example, suppose the strip at \( x=0 \) is a metal plate with constant potential \( V_0 \). Remember, it is insulated from the grounded plates at \( y=0 \) and \( y=a \). Then
$$ d_n = \frac{2V_0}{a} \int_0^a \sin\left(\frac{n\pi y}{a}\right) \, dy = \frac{2V_0}{a} \left[ -\frac{a}{n\pi} \cos\left(\frac{n\pi y}{a}\right) \right]_0^a $$ $$ d_n = \frac{2V_0}{n\pi} \left( 1-\cos n\pi \right) = \begin{cases} \; \; 0 & n \text{ even} \\[0.4em] \dfrac{4V_0}{n\pi} & n \text{ odd} \end{cases} $$ $$ \therefore \, V(x,y) = \frac{4V_0}{\pi} \sum_{n=1,3,5,\ldots}^{\infty} \frac{1}{n} \sin\left(\frac{n\pi y}{a}\right) \, e^{- n \pi x/ a} $$