Fundamental solution A
fundamental solution of Laplace's equation satisfies \Delta u = u_{xx} + u_{yy} + u_{zz} = -\delta(x-x',y-y',z-z'), where the
Dirac delta function denotes a unit source concentrated at the point . No function has this property: in fact it is a
distribution rather than a function; but it can be thought of as a limit of functions whose integrals over space are unity, and whose support (the region where the function is non-zero) shrinks to a point (see
weak solution). It is common to take a different sign convention for this equation than one typically does when defining fundamental solutions. This choice of sign is often convenient to work with because −Δ is a
positive operator. The definition of the fundamental solution thus implies that, if the Laplacian of is integrated over any volume that encloses the source point, then \iiint_V \nabla \cdot \nabla u \, dV =-1. The Laplace equation is unchanged under a rotation of coordinates, and hence we can expect that a fundamental solution may be obtained among solutions that only depend upon the distance from the source point. If we choose the volume to be a ball of radius around the source point, then Gauss's
divergence theorem implies that -1= \iiint_V \nabla \cdot \nabla u \, dV = \iint_S \frac{du}{dr} \, dS = \left.4\pi a^2 \frac{du}{dr}\right|_{r=a}. It follows that \frac{du}{dr} = -\frac{1}{4\pi r^2}, on a sphere of radius that is centered on the source point, and hence u = \frac{1}{4\pi r}. Note that, with the opposite sign convention (used in
physics), this is the
potential generated by a
point particle, for an
inverse-square law force, arising in the solution of the
Poisson equation. A similar argument shows that in two dimensions u = -\frac{\log(r)}{2\pi}. where denotes the
natural logarithm. Note that, with the opposite sign convention, this is the
potential generated by a pointlike
sink (see
point particle), which is the solution of the
Euler equations in two-dimensional
incompressible flow.
Green's function A
Green's function is a fundamental solution that also satisfies a suitable condition on the boundary of a volume . For instance, G(x,y,z;x',y',z') may satisfy \nabla \cdot \nabla G = -\delta(x-x',y-y',z-z') \qquad \text{in } V, G = 0 \quad \text{if} \quad (x,y,z) \qquad \text{on } S. Now if is any solution of the Poisson equation in : \nabla \cdot \nabla u = -f, and assumes the boundary values on , then we may apply
Green's identity, (a consequence of the divergence theorem) which states that \iiint_V \left[ G \, \nabla \cdot \nabla u - u \, \nabla \cdot \nabla G \right]\, dV = \iiint_V \nabla \cdot \left[ G \nabla u - u \nabla G \right]\, dV = \iint_S \left[ G u_n -u G_n \right] \, dS. \, The notations
un and
Gn denote normal derivatives on . In view of the conditions satisfied by and , this result simplifies to u(x',y',z') = \iiint_V G f \, dV - \iint_S G_n g \, dS. \, Thus the Green's function describes the influence at of the data and . For the case of the interior of a sphere of radius , the Green's function may be obtained by means of a reflection : the source point at distance from the center of the sphere is reflected along its radial line to a point
P' that is at a distance \rho' = \frac{a^2}{\rho}. \, Note that if is inside the sphere, then
P′ will be outside the sphere. The Green's function is then given by \frac{1}{4 \pi R} - \frac{a}{4 \pi \rho R'}, \, where denotes the distance to the source point and denotes the distance to the reflected point
P′. A consequence of this expression for the Green's function is the
Poisson integral formula. Let , , and be
spherical coordinates for the source point . Here denotes the angle with the vertical axis, which is contrary to the usual American mathematical notation, but agrees with standard European and physical practice. Then the solution of the Laplace equation with Dirichlet boundary values inside the sphere is given by u(P) =\frac{1}{4\pi} a^3\left(1-\frac{\rho^2}{a^2}\right) \int_0^{2\pi}\int_0^{\pi} \frac{g(\theta',\varphi') \sin \theta'}{(a^2 + \rho^2 - 2 a \rho \cos \Theta)^{\frac{3}{2}}} d\theta' \, d\varphi' where \cos \Theta = \cos \theta \cos \theta' + \sin\theta \sin\theta'\cos(\varphi -\varphi') is the cosine of the angle between and . A simple consequence of this formula is that if is a harmonic function, then the value of at the center of the sphere is the mean value of its values on the sphere. This mean value property immediately implies that a non-constant harmonic function cannot assume its maximum value at an interior point.
Laplace's spherical harmonics Laplace's equation in
spherical coordinates is: \nabla^2 f = \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial f}{\partial r}\right) + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial \theta}\left(\sin\theta \frac{\partial f}{\partial \theta}\right) + \frac{1}{r^2 \sin^2\theta} \frac{\partial^2 f}{\partial \varphi^2} = 0. Consider the problem of finding solutions of the form . By
separation of variables, two differential equations result by imposing Laplace's equation: \frac{1}{R}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right) = \lambda,\qquad \frac{1}{Y}\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta \frac{\partial Y}{\partial\theta}\right) + \frac{1}{Y}\frac{1}{\sin^2\theta}\frac{\partial^2Y}{\partial\varphi^2} = -\lambda. The second equation can be simplified under the assumption that has the form . Applying separation of variables again to the second equation gives way to the pair of differential equations \frac{1}{\Phi} \frac{d^2 \Phi}{d\varphi^2} = -m^2 \lambda\sin^2\theta + \frac{\sin\theta}{\Theta} \frac{d}{d\theta} \left(\sin\theta \frac{d\Theta}{d\theta}\right) = m^2 for some number . A priori, is a complex constant, but because must be a
periodic function whose period evenly divides , is necessarily an integer and is a linear combination of the complex exponentials . The solution function is regular at the poles of the sphere, where . Imposing this regularity in the solution of the second equation at the boundary points of the domain is a
Sturm–Liouville problem that forces the parameter to be of the form for some non-negative integer with ; this is also explained
below in terms of the
orbital angular momentum. Furthermore, a change of variables transforms this equation into the
Legendre equation, whose solution is a multiple of the
associated Legendre polynomial . Finally, the equation for has solutions of the form ; requiring the solution to be regular throughout forces . Here the solution was assumed to have the special form . For a given value of , there are independent solutions of this form, one for each integer with . These angular solutions are a product of
trigonometric functions, here represented as a
complex exponential, and associated Legendre polynomials: Y_\ell^m (\theta, \varphi ) = N e^{i m \varphi } P_\ell^m (\cos{\theta} ) which fulfill r^2\nabla^2 Y_\ell^m (\theta, \varphi ) = -\ell (\ell + 1 ) Y_\ell^m (\theta, \varphi ). Here is called a spherical harmonic function of degree and order , is an
associated Legendre polynomial, is a normalization constant, and and represent colatitude and longitude, respectively. In particular, the
colatitude , or polar angle, ranges from at the North Pole, to at the Equator, to at the South Pole, and the
longitude , or
azimuth, may assume all values with . For a fixed integer , every solution of the eigenvalue problem r^2\nabla^2 Y = -\ell (\ell + 1 ) Y is a
linear combination of . In fact, for any such solution, is the expression in spherical coordinates of a
homogeneous polynomial that is harmonic (see
below), and so counting dimensions shows that there are linearly independent such polynomials. The general solution to Laplace's equation in a ball centered at the origin is a
linear combination of the spherical harmonic functions multiplied by the appropriate scale factor , f(r, \theta, \varphi) = \sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell f_\ell^m r^\ell Y_\ell^m (\theta, \varphi ), where the are constants and the factors are known as
solid harmonics. Such an expansion is valid in the
ball r For r > R, the solid harmonics with negative powers of r are chosen instead. In that case, one needs to expand the solution of known regions in
Laurent series (about r=\infty), instead of
Taylor series (about r = 0), to match the terms and find f^m_\ell.
Electrostatics and magnetostatics Let \mathbf{E} be the electric field, \rho be the electric charge density, and \varepsilon_0 be the permittivity of free space. Then
Gauss's law for electricity (Maxwell's first equation) in differential form states \nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}. Now, the electric field can be expressed as the negative gradient of the electric potential V, \mathbf E=-\nabla V, if the field is irrotational, \nabla \times \mathbf{E} = \mathbf{0}. The irrotationality of \mathbf{E} is also known as the electrostatic condition. For the magnetic field, when there is no free current, \nabla\times\mathbf{H} = \mathbf{0}.We can thus define a
magnetic scalar potential, , as \mathbf{H} = -\nabla\psi.With the definition of : \nabla\cdot\mathbf{B} = \mu_{0}\nabla\cdot\left(\mathbf{H} + \mathbf{M}\right) = 0, it follows that \nabla^2 \psi = -\nabla\cdot\mathbf{H} = \nabla\cdot\mathbf{M}. Similar to electrostatics, in a source-free region, \mathbf{M} = 0 and Poisson's equation reduces to Laplace's equation for the magnetic scalar potential , \nabla^2 \psi = 0 A potential that does not satisfy Laplace's equation together with the boundary condition is an invalid electrostatic or magnetic scalar potential. ==Gravitation==