22 Solution of the Laplace equation in spherical coordinates
Let \(u: \mathbb R^3 \mapsto \mathbb R\) be a function defined in an open subset \(\Omega \subset \mathbb R^3\).
Then \(u\) is a solution to the Laplace equation, if \[ \Delta u = 0 \] holds in \(\Omega\).
We refer to \(\Delta\) as the Laplace operator.
In different coordinate systems, \(\Delta\) has different representations.
22.1 Representations of the Laplace operator
22.1.1 Cartesian coordinates
\[ \frac{\partial^2 u}{\partial x^2} +\frac{\partial^2 u}{\partial y^2} +\frac{\partial^2 u}{\partial z^2} = 0 \tag{22.1}\]
22.1.2 Spherical coordinates
We transform from a Cartesian to a spherical system with \[ \begin{align} x & = r \cos \phi \sin \theta \\ y & = r \sin \phi \sin \theta \\ z & = r \cos \theta \\ r^2 & = x^2 + y^2 + z^2 \end{align} \]
\[ \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial u}{\partial r} \right) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta {\partial u \over \partial \theta} \right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 u}{\partial \varphi^2} = 0 \tag{22.2}\]
22.1.3 Cylindrical coordinates
\[ \frac{1}{\rho} \frac{\partial}{\partial \rho} \left( \rho \frac{\partial u}{\partial \rho} \right) + \frac{1}{\rho^2} \frac{\partial^2 u}{\partial \varphi^2} + \frac{\partial^2 u}{\partial z^2 } = 0 \tag{22.3}\]
\(\rho^2 = x^2 + y^2\)
22.1.4 Polar coordinates
\[ \frac{1}{\rho} \frac{\partial}{\partial \rho} \left( \rho \frac{\partial u}{\partial \rho} \right) + \frac{1}{\rho^2} \frac{\partial^2 u}{\partial \varphi^2} =0 \tag{22.4}\]
\(\rho^2 = x^2 + y^2\)
22.2 The radial part of the Laplace equation
We consider the Laplace equation in spherical coordinates as defined in 22.2.
For now, we assume that \(u = u(r)\) is a function of the radial distance.
Thus, the Laplace equation reduces to an ordinary differential equation
\[ \frac{\mathrm d}{\mathrm d r} \left( r^2 \frac{\mathrm d u}{\mathrm d r} \right) = 2r \frac{\mathrm d u}{\mathrm d r} + r^2 \frac{\mathrm d^2 u}{\mathrm d r^2} = 0. \]
This equation is referred to as Euler’s differential equation.
We solve this ODE with the ansatz \[ u(r) = r^\alpha, \quad u'(r) = \alpha r ^{\alpha - 1}, \quad u''(r) = (\alpha-1)\alpha r^{\alpha - 2}. \]
It follows \[ \begin{align} 2r \alpha r^{\alpha-1} + r^2 (\alpha-1)\alpha r^{\alpha - 2} &= 0 \\ 2 \alpha u + (\alpha-1)\alpha u &= 0 \\ \alpha(\alpha+1)u &= 0 \quad\text{(characteristic equation)} \end{align} \]
The characteristic equation has the solution \[ u(r) = C_0 r^{-1} + C_1 r^0. \]
At this point, we introduce the
We call a function homogeneous of degree \(\lambda\), if under a proportional scaling of the variables \(x\) by a factor \(t\) the function scales with a factor \(t^\lambda\):
\[ f(t \mathbf x ) = t^\lambda f(\mathbf x) \]
Example:
\(f(x,y) = x^2 + y^2\) is a homogeneous function.
We replace \(x\) and \(y\) by \(tx\) and \(ty\), factor out powers of \(t\), i.e., \[ f(tx, ty) = (tx)^2 + (ty)^2 = t^2(x^2 + y^2) \] and recognize that \(f(x,y)\) is a homogeneous function of degree \(2\).
Now we make the following ansatz:
We assume that \(u\) can be defined as a homogeneous function \(u_n\) of degree \(n\). This is a valid choice, since homogeneous functions are solutions to Laplace’s equation.
We set \[ \begin{align} u_n(x,y,z) & = u_n(r \cos\phi \sin\theta, r \sin\phi \sin\theta, r \cos\theta) \\ & = r^n u_n(\cos\phi \sin\theta, \sin\phi \sin\theta, \cos\theta) \\ & = r^n S_n(\phi, \theta) \end{align} \]
\(u_n = r^n S_n(\phi, \theta)\) is called a solid spherical harmonic.
Further, \[ \begin{align} 2r \frac{ \partial u_{n} }{ \partial r } + r^2 \frac{ \partial^{2} u_{n} }{ \partial r^{2} } & = 2 r n r^{n-1} S_n + r^2 n(n-1)r^{n-2} S_n \\ & = 2 n u_n + n(n-1) u_n \\ & = n(n+1) u_n \end{align} \]
We recognize that \(f(r) = u_n(r)\) and \(\lambda = n(n+1)\) are eigenfunctions and eigenvalues of the eigenvalue problem
\[ \Delta f(r) = \lambda f(r). \]
We see that the radial part of the Laplace equation for \(u_n\) can be equivalently replaced by \[ n (n+1) u_n, \] which we insert back into the full Laplace equation 22.2 to obtain
\[ n(n+1) S_n + \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta {\partial S_n \over \partial \theta} \right) + \frac{1}{\sin^2 \theta} \frac{\partial^2 S_n}{\partial \varphi^2} = 0. \]
Note that the equation has no dependence on \(r\).
22.3 Zonal spherical harmonics
Now we seek solutions \(S_n(\theta)\). These functions are referred to as zonal spherical harmonics. On a surface of a sphere they only vary with the polar angle \(\theta\). Their zeros are parallel to the equator of the sphere.
The Laplace equation 22.2 reduces to \[ \frac{1}{\sin \theta} \frac{\mathrm d}{\mathrm d \theta} \left( \sin \theta {\mathrm d S_n \over \mathrm d \theta} \right) + n(n+1) S_n = 0 \]
We substitute \(\cos\theta\) and set \(S_n(\theta)=P_n(t)\):
\[ \begin{align} t & = \cos\theta \\ \mathrm{d}t & = -\sin \theta \,\mathrm{d}\theta \\ S_n(\theta) & = P_n(\cos\theta) = P_n(t) \end{align} \] With the chain rule we obtain \[ \begin{align} \frac{\mathrm d P_n(t)}{\mathrm d \theta} & = \frac{\mathrm d P_n(t)}{\mathrm d t} \frac{\mathrm d t}{\mathrm d \theta} \\ & = -\frac{\mathrm d P_n(t)}{\mathrm d t} \sin\theta \\ \sin^2\theta & = 1 - t^2 \end{align} \]
Useful steps: \[ \frac{1}{\sin \theta} \frac{\mathrm{d}}{\mathrm{d}\theta} = \frac{1}{\sin \theta} \left( -\frac{\sin \theta}{\mathrm{d}t} \right) = -\frac{\mathrm{d}}{\mathrm{d}t} \]
\[ \sin \theta \frac{\mathrm{d}}{\mathrm{d}\theta} = \sin \theta \left( -\sin \theta \frac{\mathrm{d}}{\mathrm{d}t} \right) = -\sin ^{2}\theta \frac{\mathrm{d}}{\mathrm{d}t} \]
Finally, we end at the Legendre differential equation
\[ \frac{\mathrm d}{\mathrm d t} \left[ (1-t^2) \frac{\mathrm d P_n(t)}{\mathrm d t} \right] + n(n+1) P_n(t) = 0 \]
The functions \(P_n(t) = P_n(\cos\theta)\) are the Legendre polynomials.
When a pure dependency of \(\theta\) is assumed, then \[ r^{n} S_{n}(\theta) = r^{n} P_{n}(\cos \theta) \] and \[ r^{-(n+1)}S_n = r^{-(n+1)}P_{n}(\cos \theta) \] are solid spherical harmonics. Note that we have arrived at these two equations by replacing \(n\) with \(-(n+1)\).
22.4 The angular component of the Laplace equation
Previously, we have encountered zonal spherical harmonics or Legendre polynomials as solutions to the Laplace equation.
Formally, the zonal spherical harmonics \(S_n(\theta) = P_n(\cos\theta)\) correspond to the Legendre polynomials.
To arrive at \(S_n(\theta, \phi)\), we use a separation of variables approach of the form:
\[S_n(\theta, \phi) = \Theta(\theta)\Phi(\phi)\]
and write:
\[ \underbrace{ n(n+1)\Theta\Phi + \frac{\Phi}{\sin\theta}\frac{\mathrm d}{\mathrm d \theta}\left(\sin\theta \frac{\mathrm d \Theta}{\mathrm d \theta}\right) }_{ =m^{2} } + \underbrace{ \frac{\Theta}{\sin^2\theta}\frac{\mathrm d^2 \Phi}{\mathrm d \phi^2} }_{ =-m^{2} } =0 \]
or, after multiplying by \(\dfrac{\sin^{2}\theta}{\Theta \Phi}\):
\[ n(n+1) \sin^2\theta + \frac{\sin\theta}{\Theta} \frac{\mathrm d}{\mathrm d \theta} \left(\sin\theta \frac{\mathrm d \Theta}{\mathrm d \theta}\right) = m^2 \]
\[\frac{1}{\Phi} \frac{\mathrm d^2 \Phi}{\mathrm d \phi^2} = -m^2\]
Solutions can be provided for both equations.
First, we consider:
\[\Phi''+m^2 \Phi = 0\]
This is a harmonic differential equation with the solution:
\[ \Phi(\phi) = A_m \cos m \phi + B_m \sin m \phi \]
The other equation is:
\[\dfrac{1}{\sin\theta}\frac{\mathrm d}{\mathrm d \theta} \left(\sin\theta \frac{\mathrm d \Theta}{\mathrm d \theta}\right) + \left[ n(n+1) - \frac{m^2}{\sin^2\theta} \right]\Theta = 0\]
We seek the solution \(\Theta(\theta)\), substitute \(t = \cos\theta\), \(\mathrm d t = -\sin\theta \mathrm d\theta\), and assume \(\Theta(\theta) = P_n^m(\cos\theta) = P_n^m(t)\). Thus:
\[\frac{\mathrm d}{\mathrm d t} \left[ (1-t^2) \frac{\mathrm d P_n^m(t)}{\mathrm d t} \right] + \left[ n(n+1) - \frac{m^2}{1-t^2} \right] P_n^m(t) = 0\]
The desired \(P_n^m(t)\) are associated Legendre functions of order \(m\) and degree \(n\).
The associated Legendre functions can be most easily defined through the derivatives of the Legendre polynomials with \(0 \le m \le n\):
\[ \begin{align} P_n^m(t) & = (1 - t^2)^{\frac{m}{2}} \frac{\mathrm d^m}{\mathrm dt^m}P_n(t) \\ P_n^m(\cos \theta) & = \sin ^{m}\theta P_n^{(m)}(\cos \theta) \end{align} \]
where \(P_{n}^{(m)}\) is the \(m\)-th derivative of \(P_{n}\) with respect to \(\cos \theta\).
22.4.1 Properties
Orthogonality:
\[ \int\limits_{-1}^{+1} P_n^m(t) P_{n'}^m(t) \, \dd t = \begin{cases} 0 & n \ne n' \\ \frac{2 (n+m)!}{(2n+1)(n-m)!} & n = n' \end{cases} \]
Symmetry: Follows from properties of Legendre polynomials \(P_n\), i.e., from \(P_n(-t) = (-1)^n P_n(t)\) we obtain
\[ P_n^m(-t) = (-1)^{n+m} P_n^m(t) \]
For \(m=0\), the associated Legendre functions are Legendre polynomials, i.e.,
\[ P_n^m(t) = P_n(t) \text{ for } m=0. \]
When \(m>0\), the order represents the index \(m\) of the associated azimuthal Fourier coefficient in \(\sin m \varphi\) and \(\cos m \varphi\).
The magnitude of the \(P_n^m\) strongly varies with \(m\), e.g.,
\[ \frac{P_4^1}{P_4^4} = \frac{1}{2016}. \]
22.5 Schmidt semi-normalized Legendre functions
To better reflect the relative weight of the spherical harmonic coefficients, A. Schmidt has introduced a normalization, such that
\[ Y_n^m(t) = \begin{cases} P_n^m(t) & m = 0 \\ \sqrt{2 \frac{(n-m)!}{(n+m)!}} P_n^m(t) & m>0. \end{cases} \]
22.6 Summary
The variations in latitude can be described by associated Legendre functions, while longitudinal variations can be represented by a Fourier series.
We can therefore write
\[ f(\theta, \varphi) = \sum\limits_{n=0}^\infty \sum\limits_{m=0}^n \underbrace{(g_n^m \cos m \varphi + h_n^m \sin m \varphi)}_{\text{Fourier series in longitude}} \underbrace{P_n^m(\cos\theta)}_{\text{assoc. LF in latitude}} \]