31 Green’s identities

31.1 Divergence theorem

If \(U\) is harmonic in a domain \(\Omega\), i.e., \(U\) satisfies Laplace’s equation \[ \nabla^{2}U = 0 \] then the surface integral of the normal derivative of \(U\) over the boundary of \(\Omega\) has a special property. Specifically, this integral evaluates to zero under certain conditions.

For a harmonic function \(U\), the surface integral of its normal derivative over the boundary \(\partial\Omega\) is \[ \int _{\partial\Omega} \frac{ \partial U }{ \partial n } \, \dd S = 0 \]

31.1.1 Explanation

This results follows from the divergence theorem (or Gauss’ theorem), from which we obtain \[ \int_{\Omega} \nabla^{2}U \, \dd V = \int_{\partial \Omega} \nabla U \cdot \vb{n} \, \dd S \] Since \(U\) is harmonic, the volume integral on the left-hand side vanishes. Therefore \[ \int _{\partial\Omega} \frac{ \partial U }{ \partial n } \, \dd S = 0 \]

31.1.2 Implications

This result has important consequences in potential theory and physics:

Mass Conservation Analogy: The integral represents the net “flux” of \(U\) through the boundary. For a harmonic function, the flux is zero, analogous to conservation laws.
Boundary Conditions: If \(U\) and \(\frac{\partial U}{\partial n}\) are known on the boundary, they are often constrained by this zero-flux condition.

This property is a fundamental characteristic of harmonic functions and is widely used in solving boundary value problems.

31.2 Green’s identity

Green’s identity can be used to evaluate the value of a potential \(U\) at a point \(\mathbf{r}_0\) in terms of a surface integral over the boundary of a domain. This is commonly referred to as Green’s third identity, which relates the potential \(U\) at \(\mathbf{r}_0\) to its values and normal derivatives on the boundary surface.

Green’s third identity is expressed as: \[ U(\mathbf{r}_0) = \frac{1}{4\pi} \int_{\partial\Omega} \left[ U(\mathbf{r}) \frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial n} - G(\mathbf{r}, \mathbf{r}_0) \frac{\partial U(\mathbf{r})}{\partial n} \right] \, \dd S \] where

\(\Omega\) is the surface bounding the domain
\(U(\mathbf{r})\) is the potential function
\(G(\mathbf{r}, \mathbf{r}_0)\) is the Green’s function, which satisfies the Laplace equation and has a singularity at \(\mathbf{r} = \mathbf{r}_0\)
\(\frac{\partial}{\partial n}\) denotes the derivative in the direction of the outward normal to the surface \(\partial\Omega\)
\(\dd{S}\) is the infinitesimal surface element

31.2.1 Key points

Green’s function: For the Laplace equation, \(G(\mathbf{r}, \mathbf{r}_0)\) is typically chosen as:

\[ G(\mathbf{r}, \mathbf{r}_0) = \frac{1}{|\mathbf{r} - \mathbf{r}_0|}. \]

Interpretation: Green’s third identity provides a way to reconstruct the potential \(U(\mathbf{r}_0)\) inside the domain if \(U\) and its normal derivative \(\frac{\partial U}{\partial n}\) are known on the boundary \(\partial \Omega\) .

This identity is particularly useful in solving boundary value problems in electrostatics, fluid mechanics, and other fields where the Laplace equation governs the potential.

31.3 Mean value theorem

Gauss’ mean value theorem is a fundamental result in potential theory and the study of harmonic functions. It states that the value of a harmonic function at a point is equal to the average of its values over the surface of a sphere centered at that point.

Let \(U\) be a harmonic function in a domain \(\Omega\). For any point \(\mathbf{r}_0\) in \(\Omega\) and any sphere \(S\) of radius \(R\) centered at \(\mathbf{r}_0\) (with the sphere lying entirely within \(\Omega\)), we have:

\[ U(\mathbf{r}_0) = \frac{1}{4\pi R^2} \int_{\partial\Omega} U(\mathbf{r}) \, \dd{S}, \]

where:

\(\mathbf{r}_0\) is the center of the sphere,
\(R\) is the radius of the sphere,
\(\partial \Omega\) is the surface of the sphere,
\(\dd{S}\) is the surface area element on \(\partial \Omega\),
\(U(\mathbf{r})\) is the harmonic function.

31.3.1 Interpretation

The theorem says that the value of a harmonic function at the center of a sphere equals the average value of the function over the sphere’s surface.
This property characterizes harmonic functions and is often used to prove uniqueness and regularity results for solutions to Laplace’s equation.

31.3.2 Consequences

Maximum Principle: Since the value at the center of the sphere is an average, a harmonic function cannot have a maximum or minimum inside a domain unless it is constant.
Uniqueness: If two harmonic functions agree on the boundary of a domain, they must agree throughout the domain.
Physical Interpretation: In electrostatics, the theorem reflects the fact that the potential at a point inside a charge-free region depends only on its surrounding values, not on interior contributions.

31.3.3 Uniqueness

We consider a sphere \(\Omega\) of radius \(a\) and a homogeneous charge density \(\rho_{E}\). For any point inside \(\Omega\) the Poisson equation holds \[ \nabla^{2} U(r) = -\frac{\rho_{E}}{\varepsilon_{0}}, \quad r \le a \] On the surface \(\partial\Omega\) we fix the potential and its normal derivative, such that \[ U = 0, \qquad \frac{ \partial U }{ \partial n } = 0 \]

Inside the sphere, we obtain for the potential \[ U(r) = \frac{\rho_{E}}{2\varepsilon_{0}}a^{2} - \frac{\rho_{E}}{6\varepsilon_{0}}r^{2} \] whereas for \(r > a\) we find \[ U(r) = \frac{\rho_{E}a^{3}}{3\varepsilon_{0}r} \]

For \(r=a\) we obtain \[ U(a) = \frac{\rho_{E}}{3\varepsilon_{0}}a^{2} \qquad \frac{ \partial U }{ \partial n } \bigg|_{r=a} = -\frac{\rho_{E}}{3\varepsilon_{0}}a \] which is not consistent with the assumption. It follows that the values at the boundary are determined by \(\nabla^{2}U\) alone. A simultaneous choice of \(U\) and \(\frac{ \partial U }{ \partial n }\) at the boundary overdetermines the problem and typically leaves no room for a solution. Imposing both Dirichlet and Neumann conditions on the same boundary can result in inconsistencies.

31.4 First boundary value problem

31.4.1 How can we eliminate the Neumann term in the Green’s third identity?

We consider the function \(U\) defined on an open domain \(\Omega \subset \mathbb{R}^{3}\) \[ \nabla^{2} U(r) = 0 \] The goal is to eliminate the term \(\dfrac{\partial U}{\partial n}\) from Green’s third identity. \[ U(\mathbf{r}_0) = \frac{1}{4\pi} \int_{\partial\Omega} \left[ U(\mathbf{r}) \frac{\partial }{\partial n} \frac{1}{r} - \frac{1}{r} \frac{\partial U(\mathbf{r})}{\partial n} \right] \, \dd S \]

We introduce a harmonic auxiliary function \(g(\vb{r},\vb{r_{0}})\) such that for any \(\vb{r} \in \Omega\) \[ \nabla^{2} g(\vb{r},\vb{r_{0}}) = 0 \] Further, \[ g(\vb{r},\vb{r_{0}}) = \frac{1}{|\vb{r} - \vb{r_{0}|}} = \frac{1}{r}, \quad \vb{r_{0}}\in \partial\Omega \]

Green’s second identity states \[ \frac{1}{4\pi}\int_\Omega \left( u \nabla^2 v - v \nabla^2 u \right) \,\dd{V} = \frac{1}{4\pi} \int_{\partial \Omega} \left( u \frac{\partial v}{\partial n} - v \frac{\partial u}{\partial n} \right) \,\dd{S} \]

If both \(u\) and \(v\) are harmonic functions, then \[ \frac{1}{4\pi}\int_{\partial \Omega} \left( u \frac{\partial v}{\partial n} - v \frac{\partial u}{\partial n} \right) \,\dd{S} = 0 \] For \(u=U\) and \(v=g\) we obtain \[ \frac{1}{4\pi} \int_{\partial \Omega} \left( U \frac{\partial g}{\partial n} - g \frac{\partial U}{\partial n} \right) \,\dd{S} = 0 \]

We subtract this from equation to \(U(\vb{r_{0}})\): \[ \begin{align} U(\vb{r_{0}}) & = \frac{1}{4\pi} \int_{\partial \Omega} \left( U \frac{\partial}{\partial n}\left( g - \frac{1}{r} \right) - \left( g - \frac{1}{r} \right) \frac{\partial U}{\partial n} \right) \,\dd{S} \\ & = \frac{1}{4\pi} \int_{\partial \Omega} \left( U \frac{\partial}{\partial n}\left( g - \frac{1}{r} \right) \right) \,\dd{S} \end{align} \] We denote \[ G_{1}(\vb{r},\vb{r_{0}})=g(\vb{r},\vb{r_{0}})-\frac{1}{r} \] as Green’s function which satisfies \[ \begin{align} \nabla^{2}G_{1} & = 0 \text{ in } \Omega \\ G_{1} & = 0 \text{ on } \partial\Omega \end{align} \]

Note that for \(\vb{r}=\vb{r_{0}}\), the Green’s function \(G_{1}\) has a singularity.

31.5 Second (exterior) boundary value problem

We restrict \(\dfrac{\partial U}{\partial n}\) at the boundary \(\partial\Omega\). With the Green’s third identity, we are able to find a solution \(U\) in the exterior of \(\Omega\) by construction of an auxiliary function \[ G_{2} = g - \frac{1}{r} \] such that \[ \begin{align} \nabla^{2}G_{2} &= 0 \text{ in } \Omega \\ \frac{ \partial }{ \partial n }G_{2} & = 0 \text{ on } \partial\Omega \end{align} \]

The solution in the exterior is \[ U(\vb{r_{0}}) = \frac{1}{4 \pi} \int _{\partial\Omega } G_{2}(\vb{r, \vb{r_{0}}}) \frac{\partial U}{\partial n} \, \dd S \]

31.6 Second (interior) BVP

For the interior BVP the derivative at the boundary cannot be defined arbitrarily, since both \[ \nabla^{2}U = 0 \text{ in } \Omega \] and \[ \int \frac{\partial U}{\partial n} \, \dd S = 0 \] must be fulfilled. The same restriction holds for \(g\).

To eliminate the boundary values of \(U\) in Green’s third identity, it is required that \[ \frac{\partial g}{\partial n} = \frac{\partial }{\partial n} \frac{1}{r} \]

because of the property of \(G_{2}\) [[#^31751b]]. However, with \(\dd{S} = r^{2}\sin \theta \,\dd{\theta}\,\dd{\varphi}\), \[ \int \frac{\partial }{\partial n} \frac{1}{r} \, \dd S = -\int \frac{\cos(\vb{r},\vb{n})}{r^{2}} \, \dd S = -\int \, \dd \Omega = -4 \pi \]

We have to force that the condition \[ \frac{\partial g}{\partial n} = \frac{\partial }{\partial n} \frac{1}{r} + K \] is fulfilled such that \[ \int \frac{\partial g}{\partial n} \, \dd S = \int \frac{\partial }{\partial n} \frac{1}{r} \, \dd S = 0 \]

It follows \[ K = \frac{4 \pi}{\int \, \dd S } = \frac{4 \pi}{A_{0}} \] and \[ U(\vb{r_{0}}) = \frac{1}{A_{0}} \int U \, \dd S - \frac{1}{4\pi} \int G_{2}(\vb{r},\vb{r_{0}}) \frac{\partial U}{\partial n} \, \dd S \] The solution of the second BVP is \[ U(\vb{r_{0}}) = C - \frac{1}{4\pi} \int G_{2}(\vb{r},\vb{r_{0}}) \frac{\partial U}{\partial n} \, \dd S \] which is uniquely determined up to a constant.

The Green’s function \(G_{2}\) \[ G_{2}(\vb{r, \vb{r_{0}}}) = g_{2}(\vb{r, \vb{r_{0}}})-\frac{1}{r} \] is harmonic for \(\vb{r} \ne \vb{r_{0}}\), such that it holds \[ \begin{align} \nabla^{2}G_{2}(\vb{r},\vb{r_{0}}) & = 0 \text{ for } \vb{r} \ne \vb{r_{0}} \\ \frac{\partial G_{2}}{\partial n} & = \frac{4 \pi}{A_{0}} \end{align} \]