Editing Green's function (section)

==Green's functions for the Laplacian==
Green's functions for linear differential operators involving the [[Laplacian]] may be readily put to use using the second of [[Green's identities]].

To derive Green's theorem, begin with the [[divergence theorem]] (otherwise known as [[Gauss's theorem]]),
<math display="block">\int_V \nabla \cdot \mathbf A\, dV = \int_S \mathbf A \cdot d\hat\boldsymbol\sigma \,.</math>

Let <math>\mathbf A = \varphi\,\nabla\psi - \psi\,\nabla\varphi</math> and substitute into Gauss' law.

Compute <math>\nabla\cdot\mathbf A</math> and apply the product rule for the ∇ operator,
<math display="block">\begin{align}
 \nabla\cdot\mathbf A &= \nabla\cdot \left(\varphi\,\nabla\psi \;-\; \psi\,\nabla\varphi\right)\\
 &= (\nabla\varphi)\cdot(\nabla\psi) \;+\; \varphi\,\nabla^2\psi \;-\; (\nabla\varphi)\cdot(\nabla\psi) \;-\; \psi\nabla^2\varphi\\
 &= \varphi\,\nabla^2\psi \;-\; \psi\,\nabla^2\varphi.
\end{align}</math>

Plugging this into the divergence theorem produces [[Green's theorem]],
<math display="block">\int_V \left(\varphi\,\nabla^2\psi-\psi\,\nabla^2\varphi\right) dV = \int_S \left(\varphi\,\nabla\psi-\psi\nabla\,\varphi\right) \cdot d\hat\boldsymbol\sigma.</math>

Suppose that the linear differential operator {{mvar|L}} is the [[Laplacian]], ∇<sup>2</sup>, and that there is a Green's function {{mvar|G}} for the Laplacian. The defining property of the Green's function still holds,
<math display="block">L G(\mathbf{x},\mathbf{x}') = \nabla^2 G(\mathbf{x},\mathbf{x}') = \delta(\mathbf{x}-\mathbf{x}').</math>

Let <math>\psi=G</math> in Green's second identity, see [[Green's identities]]. Then,
<math display="block">\int_V \left[ \varphi(\mathbf{x}') \delta(\mathbf{x}-\mathbf{x}') - G(\mathbf{x},\mathbf{x}') \, {\nabla'}^2\,\varphi(\mathbf{x}')\right] d^3\mathbf{x}' = \int_S \left[\varphi(\mathbf{x}')\,{\nabla'} G(\mathbf{x},\mathbf{x}') - G(\mathbf{x},\mathbf{x}') \, {\nabla'}\varphi(\mathbf{x}')\right] \cdot d\hat\boldsymbol\sigma'.</math>

Using this expression, it is possible to solve [[Laplace's equation]] {{math|1=∇<sup>2</sup>''φ''('''x''') = 0}} or [[Poisson's equation]] {{math|1=∇<sup>2</sup>''φ''('''x''') = −''ρ''('''x''')}}, subject to either [[Neumann boundary condition|Neumann]] or [[Dirichlet boundary condition|Dirichlet]] boundary conditions. In other words, we can solve for {{math|''φ''('''x''')}} everywhere inside a volume where either (1) the value of {{math|''φ''('''x''')}} is specified on the bounding surface of the volume (Dirichlet boundary conditions), or (2) the normal derivative of {{math|''φ''('''x''')}} is specified on the bounding surface (Neumann boundary conditions).

Suppose the problem is to solve for {{math|''φ''('''x''')}} inside the region. Then the integral
<math display="block">\int_V \varphi(\mathbf{x}') \, \delta(\mathbf{x}-\mathbf{x}') \, d^3\mathbf{x}'</math>
reduces to simply {{math|''φ''('''x''')}} due to the defining property of the [[Dirac delta function]] and we have
<math display="block">\varphi(\mathbf{x})
= -\int_V G(\mathbf{x},\mathbf{x}') \, \rho(\mathbf{x}')\, d^3\mathbf{x}'
+ \int_S \left[\varphi(\mathbf{x}') \, \nabla' G(\mathbf{x},\mathbf{x}') - G(\mathbf{x},\mathbf{x}') \, \nabla'\varphi(\mathbf{x}')\right] \cdot d\hat\boldsymbol\sigma'.</math>

This form expresses the well-known property of [[harmonic function]]s, that ''if the value or normal derivative is known on a bounding surface, then the value of the function inside the volume is known everywhere''.

In [[electrostatics]], {{math|''φ''('''x''')}} is interpreted as the [[electric potential]], {{math|''ρ''('''x''')}} as [[electric charge]] [[density]], and the normal derivative <math>\nabla\varphi(\mathbf{x}')\cdot d\hat\boldsymbol\sigma'</math> as the normal component of the electric field.

If the problem is to solve a Dirichlet boundary value problem, the Green's function should be chosen such that {{math|''G''(''x'',''x''&prime;)}} vanishes when either {{mvar|x}} or {{mvar|x′}} is on the bounding surface. Thus only one of the two terms in the [[surface integral]] remains. If the problem is to solve a Neumann boundary value problem, it might seem logical to choose Green's function so that its normal derivative vanishes on the bounding surface. However, application of Gauss's theorem to the differential equation defining the Green's function yields
<math display="block">\int_S \nabla' G(\mathbf{x},\mathbf{x}') \cdot d\hat\boldsymbol\sigma'
= \int_V \nabla'^2 G(\mathbf{x},\mathbf{x}') \, d^3\mathbf{x}'
= \int_V \delta (\mathbf{x}-\mathbf{x}')\, d^3\mathbf{x}'
= 1 \,,</math>
meaning the normal derivative of ''G''('''x''','''x'''&prime;) cannot vanish on the surface, because it must integrate to 1 on the surface.<ref>{{cite book |last1=Jackson |first1=John David |title=Classical Electrodynamics |date=1998-08-14 |publisher=John Wiley & Sons |pages=39}}</ref>

The simplest form the normal derivative can take is that of a constant, namely {{math|1/''S''}}, where {{math|''S''}} is the surface area of the surface. The surface term in the solution becomes
<math display="block">\int_S \varphi(\mathbf{x}') \, \nabla' G(\mathbf{x},\mathbf{x}') \cdot d\hat\boldsymbol\sigma' = \langle\varphi\rangle_S </math>
where <math>\langle\varphi\rangle_S </math> is the average value of the potential on the surface. This number is not known in general, but is often unimportant, as the goal is often to obtain the electric field given by the gradient of the potential, rather than the potential itself.

With no boundary conditions, the Green's function for the Laplacian ([[Green's function for the three-variable Laplace equation]]) is
<math display="block">G(\mathbf{x},\mathbf{x}') = -\frac{1}{4 \pi \left|\mathbf{x}-\mathbf{x}'\right|}.</math>

Supposing that the bounding surface goes out to infinity and plugging in this expression for the Green's function finally yields the standard expression for electric potential in terms of electric charge density as

{{Equation box 1
|indent =:
|equation = <math>\varphi(\mathbf{x}) = \int_V \dfrac{\rho(\mathbf{x}')}{4 \pi \varepsilon \left|\mathbf{x} - \mathbf{x}'\right|} \, d^3\mathbf{x}' \, .</math>
|cellpadding= 6
|border
|border colour = #0073CF
|bgcolor=#F9FFF7}}

{{further|Poisson's equation}}