Editing Laplace's equation (section)

==In three dimensions==

===Fundamental solution===
A [[fundamental solution]] of Laplace's equation satisfies
<math display="block"> \Delta u = u_{xx} + u_{yy} + u_{zz} = -\delta(x-x',y-y',z-z'),</math>
where the [[Dirac delta function]] {{math|''δ''}} denotes a unit source concentrated at the point {{math|(''x''′, ''y''′, ''z''′)}}. No function has this property: in fact it is a [[Distribution (mathematics)|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 {{math|''u''}} is integrated over any volume that encloses the source point, then
<math display="block"> \iiint_V \nabla \cdot \nabla u \, dV =-1.</math>

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 {{mvar|r}} from the source point. If we choose the volume to be a ball of radius {{mvar|a}} around the source point, then Gauss's [[divergence theorem]] implies that
<math display="block"> -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}.</math>

It follows that 
<math display="block"> \frac{du}{dr} = -\frac{1}{4\pi r^2},</math>
on a sphere of radius {{mvar|r}} that is centered on the source point, and hence
<math display="block"> u = \frac{1}{4\pi r}.</math>

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 [[Poisson equation]]. A similar argument shows that in two dimensions
<math display="block"> u = -\frac{\log(r)}{2\pi}.</math>
where {{math|log(''r'')}} denotes the [[natural logarithm]]. Note that, with the opposite sign convention, this is the [[potential]] generated by a pointlike [[Potential flow|sink]] (see [[point particle]]), which is the solution of the [[Euler equations (fluid dynamics)|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 {{mvar|S}} of a volume {{mvar|V}}. For instance, 
<math display="block">G(x,y,z;x',y',z')</math>
may satisfy
<math display="block"> \nabla \cdot \nabla G = -\delta(x-x',y-y',z-z') \qquad \text{in } V,</math> 
<math display="block"> G = 0 \quad \text{if} \quad (x,y,z) \qquad \text{on } S.</math>

Now if {{math|''u''}} is any solution of the Poisson equation in {{mvar|V}}:
<math display="block"> \nabla \cdot \nabla u = -f,</math>

and {{math|''u''}} assumes the boundary values {{math|''g''}} on {{mvar|S}}, then we may apply [[Green's identities|Green's identity]], (a consequence of the divergence theorem) which states that

<math display="block"> \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. \,</math>

The notations ''u<sub>n</sub>'' and ''G<sub>n</sub>'' denote normal derivatives on {{math|''S''}}. In view of the conditions satisfied by {{math|''u''}} and {{math|''G''}}, this result simplifies to

<math display="block"> u(x',y',z') =  \iiint_V G f \, dV - \iint_S G_n g \, dS. \,</math>

Thus the Green's function describes the influence at {{math|(''x''′, ''y''′, ''z''′)}} of the data {{math|''f''}} and {{math|''g''}}. For the case of the interior of a sphere of radius {{math|''a''}}, the Green's function may be obtained by means of a reflection {{harv| Sommerfeld| 1949}}: the source point {{math|''P''}} at distance {{math|''ρ''}} from the center of the sphere is reflected along its radial line to a point ''P''' that is at a distance

<math display="block"> \rho' = \frac{a^2}{\rho}. \,</math>

Note that if {{math|''P''}} is inside the sphere, then ''P&prime;'' will be outside the sphere. The Green's function is then given by
<math display="block"> \frac{1}{4 \pi R} - \frac{a}{4 \pi \rho R'}, \,</math>
where {{mvar|R}} denotes the distance to the source point {{mvar|''P''}} and {{math|''R''′}} denotes the distance to the reflected point ''P''′. A consequence of this expression for the Green's function is the '''[[Poisson integral formula]]'''. Let {{mvar|ρ}}, {{mvar|θ}}, and {{mvar|φ}} be [[spherical coordinates]] for the source point {{math|''P''}}.  Here {{mvar|θ}} 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 {{math|''g''}} inside the sphere is given by {{harv|Zachmanoglou|Thoe|1986|loc=p. 228}}
<math display="block">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'</math> 
where
<math display="block"> \cos \Theta = \cos \theta \cos \theta' + \sin\theta \sin\theta'\cos(\varphi -\varphi')</math>
is the cosine of the angle between {{math|(''θ'', ''φ'')}} and {{math|(''θ''′, ''φ''′)}}.  A simple consequence of this formula is that if {{math|''u''}} is a harmonic function, then the value of {{math|''u''}} 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 ===
{{Main|Spherical harmonics#Laplace's spherical harmonics}}
[[File:Rotating_spherical_harmonics.gif|right|thumb|Real (Laplace) spherical harmonics {{math|''Y<sub>ℓ</sub><sup>m</sup>''}} for {{math|1=''ℓ'' = 0, ..., 4}} (top to bottom) and {{math|1=''m'' = 0, ..., ''ℓ''}} (left to right). Zonal, sectoral, and tesseral harmonics are depicted along the left-most column, the main diagonal, and elsewhere, respectively. (The negative order harmonics <math>Y_{\ell}^{-m}</math> would be shown rotated about the {{math|''z''}} axis by <math>90^\circ/m</math> with respect to the positive order ones.)]]
Laplace's equation in [[Spherical coordinate system|spherical coordinates]] is:<ref>The approach to spherical harmonics taken here is found in {{harv|Courant|Hilbert|1962|loc=§V.8, §VII.5}}.</ref>

<math display="block"> \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.</math>

Consider the problem of finding solutions of the form {{math|1=''f''(''r'', ''θ'', ''φ'') = ''R''(''r'') ''Y''(''θ'', ''φ'')}}.  By [[Separation of variables#pde|separation of variables]], two differential equations result by imposing Laplace's equation:

<math display="block">\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.</math>

The second equation can be simplified under the assumption that {{math|''Y''}} has the form {{math|1=''Y''(''θ'', ''φ'') = Θ(''θ'') Φ(''φ'')}}. Applying separation of variables again to the second equation gives way to the pair of differential equations

<math display="block">\frac{1}{\Phi} \frac{d^2 \Phi}{d\varphi^2} = -m^2</math>
<math display="block">\lambda\sin^2\theta + \frac{\sin\theta}{\Theta} \frac{d}{d\theta} \left(\sin\theta \frac{d\Theta}{d\theta}\right) = m^2</math>

for some number {{math|''m''}}. A priori, {{math|''m''}} is a complex constant, but because {{math|Φ}} must be a [[periodic function]] whose period evenly divides {{math|2''π''}}, {{math|''m''}} is necessarily an integer and {{math|Φ}} is a linear combination of the complex exponentials {{math|''e''<sup>±''imφ''</sup>}}. The solution function {{math|''Y''(''θ'', ''φ'')}} is regular at the poles of the sphere, where {{math|1=''θ'' = 0, ''π''}}.  Imposing this regularity in the solution {{math|Θ}} of the second equation at the boundary points of the domain is a [[Sturm–Liouville problem]] that forces the parameter {{math|''λ''}} to be of the form {{math|1=''λ'' = ''ℓ'' (''ℓ'' + 1)}} for some non-negative integer with {{math|''ℓ'' ≥ {{!}}''m''{{!}}}}; this is also explained [[Spherical harmonics#Orbital angular momentum|below]] in terms of the [[Angular momentum operator|orbital angular momentum]].  Furthermore, a change of variables {{math|1=''t'' = cos ''θ''}} transforms this equation into the [[Associated Legendre function|Legendre equation]], whose solution is a multiple of the [[associated Legendre polynomial]] {{math|''P<sub>ℓ</sub><sup>m</sup>''(cos ''θ'')}} .  Finally, the equation for {{math|''R''}} has solutions of the form {{math|1=''R''(''r'') = ''A r<sup>ℓ</sup>'' + ''B r''<sup>−''ℓ'' − 1</sup>}}; requiring the solution to be regular throughout {{math|'''R'''<sup>3</sup>}} forces {{math|1=''B'' = 0}}.<ref group=note>Physical applications often take the solution that vanishes at infinity, making {{math|1=''A'' = 0}}.  This does not affect the angular portion of the spherical harmonics.</ref>

Here the solution was assumed to have the special form {{math|1=''Y''(''θ'', ''φ'') = Θ(''θ'') Φ(''φ'')}}.  For a given value of {{math|''ℓ''}}, there are {{math|2''ℓ'' + 1}} independent solutions of this form, one for each integer {{math|''m''}} with {{math|−''ℓ'' ≤ ''m'' ≤ ''ℓ''}}.  These angular solutions are a product of [[trigonometric function]]s, here represented as a [[Euler's formula|complex exponential]], and associated Legendre polynomials:
<math display="block"> Y_\ell^m (\theta, \varphi ) = N e^{i m \varphi } P_\ell^m (\cos{\theta} )</math>
which fulfill
<math display="block"> r^2\nabla^2 Y_\ell^m (\theta, \varphi ) = -\ell (\ell + 1 ) Y_\ell^m (\theta, \varphi ).</math>

Here {{math|''Y<sub>ℓ</sub><sup>m</sup>''}} is called a spherical harmonic function of degree {{mvar|ℓ}} and order {{mvar|m}}, {{math|''P<sub>ℓ</sub><sup>m</sup>''}} is an [[associated Legendre polynomial]], {{math|''N''}} is a normalization constant, and {{mvar|θ}} and {{mvar|φ}} represent colatitude and longitude, respectively. In particular, the [[colatitude]] {{mvar|θ}}, or polar angle, ranges from {{math|0}} at the North Pole, to {{math|''π''/2}} at the Equator, to {{math|''π''}} at the South Pole, and the [[longitude]] {{mvar|φ}}, or [[azimuth]], may assume all values with {{math|0 ≤ ''φ'' < 2''π''}}.  For a fixed integer {{mvar|ℓ}}, every solution {{math|''Y''(''θ'', ''φ'')}} of the eigenvalue problem
<math display="block"> r^2\nabla^2 Y = -\ell (\ell + 1 ) Y</math>
is a [[linear combination]] of {{math|''Y<sub>ℓ</sub><sup>m</sup>''}}.  In fact, for any such solution, {{math|''r<sup>ℓ</sup> Y''(''θ'', ''φ'')}} is the expression in spherical coordinates of a [[homogeneous polynomial]] that is harmonic (see [[Spherical harmonics#Higher dimensions|below]]), and so counting dimensions shows that there are {{math|2''ℓ'' + 1}} 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 {{math|''r<sup>ℓ</sup>''}},
<math display="block"> f(r, \theta, \varphi) = \sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell f_\ell^m r^\ell Y_\ell^m (\theta, \varphi ), </math>
where the {{math|''f<sub>ℓ</sub><sup>m</sup>''}} are constants and the factors {{math|''r<sup>ℓ</sup> Y<sub>ℓ</sub><sup>m</sup>''}} are known as [[solid harmonics]]. Such an expansion is valid in the [[Ball (mathematics)|ball]]
<math display="block"> r < R = \frac{1}{\limsup_{\ell\to\infty} |f_\ell^m|^{{1}/{\ell}}}.</math>

For <math> r > R</math>, the solid harmonics with negative powers of <math>r</math> are chosen instead. In that case, one needs to expand the solution of known regions in [[Laurent series]] (about <math>r=\infty</math>), instead of [[Taylor series]] (about <math>r = 0</math>), to match the terms and find <math>f^m_\ell</math>.

===Electrostatics and magnetostatics===
Let <math>\mathbf{E}</math> be the electric field, <math>\rho</math> be the electric charge density, and <math>\varepsilon_0</math> be the permittivity of free space. Then [[Gauss's law]] for electricity (Maxwell's first equation) in differential form states<ref name="Griffiths-2">Griffiths, David J. ''Introduction to Electrodynamics''. 4th ed., Pearson, 2013. Chapter 2: Electrostatics. p. 83-4. {{ISBN|978-1-108-42041-9}}.</ref>
<math display="block">\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}.</math>

Now, the electric field can be expressed as the negative gradient of the electric potential <math>V</math>,
<math display="block">\mathbf E=-\nabla V,</math>
if the field is irrotational, <math>\nabla \times \mathbf{E} = \mathbf{0}</math>. The irrotationality of <math>\mathbf{E}</math> is also known as the electrostatic condition.<ref name="Griffiths-2"/>

<math display="block">\nabla\cdot\mathbf E = \nabla\cdot(-\nabla V)=-\nabla^2 V</math>
<math display="block">\nabla^2 V = -\nabla\cdot\mathbf E</math>

Plugging this relation into Gauss's law, we obtain Poisson's equation for electricity,<ref name="Griffiths-2"/>
<math display="block">\nabla^2 V = -\frac{\rho}{\varepsilon_0}.</math>

In the particular case of a source-free region, <math>\rho = 0</math> and Poisson's equation reduces to Laplace's equation for the electric potential.<ref name="Griffiths-2"/>

If the electrostatic potential <math>V</math> is specified on the boundary of a region <math>\mathcal{R}</math>, then it is uniquely determined. If <math>\mathcal{R}</math> is surrounded by a conducting material with a specified charge density <math>\rho</math>, and if the total charge <math>Q</math> is known, then <math>V</math> is also unique.<ref name="Griffiths-3">Griffiths, David J. ''Introduction to Electrodynamics''. 4th ed., Pearson, 2013. Chapter 3: Potentials. p. 119-121. {{ISBN|978-1-108-42041-9}}.</ref>

For the magnetic field, when there is no free current, <math display="block">\nabla\times\mathbf{H} = \mathbf{0}.</math>We can thus define a [[magnetic scalar potential]], {{math|''ψ''}}, as 
<math display="block">\mathbf{H} = -\nabla\psi.</math>With the definition of {{math|'''H'''}}:
<math display="block">\nabla\cdot\mathbf{B} = \mu_{0}\nabla\cdot\left(\mathbf{H} + \mathbf{M}\right) = 0,</math>
it follows that
<math display="block">\nabla^2 \psi = -\nabla\cdot\mathbf{H} = \nabla\cdot\mathbf{M}.</math>

Similar to electrostatics, in a source-free region, <math>\mathbf{M} = 0</math> and Poisson's equation reduces to Laplace's equation for the magnetic scalar potential ,
<math display="block">\nabla^2 \psi = 0</math>

A potential that does not satisfy Laplace's equation together with the boundary condition is an invalid electrostatic or magnetic scalar potential.