# Green's identities

In mathematics, Green's identities are a set of three identities in vector calculus. They are named after the mathematician George Green, who discovered Green's theorem.

## Green's first identity

This identity is derived from the divergence theorem applied to the vector field F = ψφ: Let φ and ψ be scalar functions defined on some region UR3, and suppose that φ is twice continuously differentiable, and ψ is once continuously differentiable. Then[1]

$\int_U \left( \psi \Delta \varphi + \nabla \varphi \cdot \nabla \psi \right)\, dV = \oint_{\partial U} \psi \left( \nabla \varphi \cdot \bold{n} \right)\, dS=\oint_{\partial U}\psi\nabla\varphi\cdot d\mathbf{S}$

where $\Delta$ is the Laplace operator, U is the boundary of region U, n is the outward pointing unit normal of surface element dS and dS is the oriented surface element. This theorem is a special case of the divergence theorem, and is essentially the higher dimensional equivalent of integration by parts with ψ and the gradient of φ replacing u and v.

Note that Green's first identity above is a special case of the more general identity derived from the divergence theorem by substituting F = ψΓ:

$\int_U \left( \psi \nabla \cdot \mathbf{\Gamma} + \mathbf{\Gamma} \cdot \nabla \psi\right)\, dV = \oint_{\partial U} \psi \left( \mathbf{\Gamma} \cdot \bold{n} \right)\, dS=\oint_{\partial U}\psi\mathbf{\Gamma}\cdot d\mathbf{S}.$

## Green's second identity

If φ and ψ are both twice continuously differentiable on UR3, and ε is once continuously differentiable, we can choose F = ψεφφεψ and obtain:

$\int_U \left[ \psi \nabla \cdot \left( \epsilon \nabla \varphi \right) - \varphi \nabla \cdot \left( \epsilon \nabla \psi \right) \right]\, dV = \oint_{\partial U} \epsilon \left( \psi {\partial \varphi \over \partial \mathbf{n}} - \varphi {\partial \psi \over \partial \mathbf{n}}\right)\, dS.$

For the special case of ε = 1 all across UR3 then:

$\int_U \left( \psi \Delta \varphi - \varphi \Delta \psi\right)\, dV = \oint_{\partial U} \left( \psi {\partial \varphi \over \partial \mathbf{n}} - \varphi {\partial \psi \over \partial \mathbf{n}}\right)\, dS.$

In the equation above φ/∂n is the directional derivative of φ in the direction of the outward pointing normal n to the surface element dS:

${\partial \varphi \over \partial \mathbf{n}} = \nabla \varphi \cdot \mathbf{n}=\nabla_\mathbf{n}\varphi.$

In particular, this demonstrates that the Laplacian is self-adjoint in the L2 inner product for functions vanishing on the boundary.

## Green's third identity

Green's third identity derives from the second identity by choosing φ = G, where the Green's function G is taken to be a fundamental solution of the Laplace operator. This means that:

$\Delta G(\mathbf{x},\mathbf{\eta}) = \delta(\mathbf{x} - \mathbf{\eta}).$

For example in R3, a solution has the form:

$G(\mathbf{x},\mathbf{\eta})= \frac{-1}{4 \pi \|\mathbf{x} - \mathbf{\eta} \|}.$

Green's third identity states that if ψ is a function that is twice continuously differentiable on U, then

$\int_U \left[ G(\mathbf{y},\mathbf{\eta}) \Delta \psi(\mathbf{y})\right]\, dV_\mathbf{y} - \psi(\mathbf{\eta})= \oint_{\partial U} \left[ G(\mathbf{y},\mathbf{\eta}) {\partial \psi \over \partial \mathbf{n}} (\mathbf{y}) - \psi(\mathbf{y}) {\partial G(\mathbf{y},\mathbf{\eta}) \over \partial \mathbf{n}} \right]\, dS_\mathbf{y}.$

A simplification arises if ψ is itself a harmonic function, i.e. a solution to the Laplace equation. Then 2ψ = 0 and the identity simplifies to:

$\psi(\mathbf{\eta})= \oint_{\partial U} \left[\psi(\mathbf{y}) \frac{\partial G(\mathbf{y},\mathbf{\eta})}{\partial \mathbf{n}} - G(\mathbf{y},\mathbf{\eta}) \frac{\partial \psi}{\partial \mathbf{n}} (\mathbf{y}) \right]\, dS_\mathbf{y}.$

The second term in the integral above can be eliminated if we choose G to be the Green's function for the boundary of the region U where the problem is posed (Dirichlet boundary condition):

$\psi(\mathbf{\eta}) = \oint_{\partial U} \psi(\mathbf{y}) \frac{\partial G(\mathbf{y},\mathbf{\eta})}{\partial \mathbf{n}} \, dS_\mathbf{y}.$

This form is used to construct solutions to Dirichlet boundary condition problems. To find solutions for Neumann boundary condition problems, the Green's function with vanishing normal gradient on the boundary is used instead.

It can be further verified that the above identity also applies when ψ is a solution to the Helmholtz equation or wave equation and G is the appropriate Green's function. In such a context, this identity is the mathematical expression of the Huygens Principle.

## On manifolds

Green's identities hold on a Riemannian manifold, In this setting, the first two are

\begin{align} \int_M u\Delta v\, dV + \int_M \langle\operatorname{grad}\ u, \operatorname{grad}\ v\rangle\, dV &= \int_{\partial M} u N v d\widetilde{V} \\ \int_M \left (u \Delta v - v \Delta u \right )\, dV &= \int_{\partial M}(u N v - v N u) d \widetilde{V} \end{align}

where u and v are smooth real-valued functions on M, dV is the volume form compatible with the metric, $d\widetilde{V}$ is the induced volume form on the boundary of M, N is oriented unit vector field normal to the boundary, and 2u = div(grad u) is the Laplacian.

## Green's vector identity

Green’s second identity establishes a relationship between second and (the divergence of) first order derivatives of two scalar functions. In differential form

$p_{m}\Delta q_{m}-q_{m}\Delta p_{m}=\nabla\cdot\left(p_{m}\nabla q_{m}-q_{m}\nabla p_{m}\right),$

where pm and qm are two arbitrary twice continuously differentiable scalar fields. This identity is of great importance in physics because continuity equations can thus be established for scalar fields such as mass or energy.[2] Although the second Green’s identity is always presented in vector analysis, only a scalar version is found on textbooks. Even in the specialized literature, a vector version is not easily found. In vector diffraction theory, two versions of Green’s second identity are introduced. One variant invokes the divergence of a cross product [3][4][5] and states a relationship in terms of the curl-curl of the field

$\mathbf{P}\cdot\left(\nabla\times\nabla\times\mathbf{Q}\right)-\mathbf{Q}\cdot\left(\nabla\times\nabla\times\mathbf{P}\right)=\nabla\cdot\left(\mathbf{Q}\times\nabla\times\mathbf{P}-\mathbf{P}\times\nabla\times\mathbf{Q}\right).$

This equation can be written in terms of the Laplacians:

$\mathbf{P}\cdot\Delta \mathbf{Q}-\mathbf{Q}\cdot\Delta \mathbf{P}+\mathbf{Q}\cdot\left[\nabla\left(\nabla\cdot\mathbf{P}\right)\right]-\mathbf{P}\cdot\left[\nabla\left(\nabla\cdot\mathbf{Q}\right)\right]=\nabla\cdot\left(\mathbf{P}\times\nabla\times\mathbf{Q}-\mathbf{Q}\times\nabla\times\mathbf{P}\right).$

However, the terms

$\mathbf{Q}\cdot\left[\nabla\left(\nabla\cdot\mathbf{P}\right)\right]-\mathbf{P}\cdot\left[\nabla\left(\nabla\cdot\mathbf{Q}\right)\right],$

could not be readily written in terms of a divergence. The other approach introduces bi-vectors, this formulation requires a dyadic Green function.[6][7] The derivation presented here avoids these problems.[8]

Consider that the scalar fields in Green's second identity are the Cartesian components of vector fields, i.e.

$\mathbf{P}=\sum_m p_{m}\hat{\mathbf{e}}_{m}, \qquad \mathbf{Q}=\sum_m q_{m}\hat{\mathbf{e}}_{m}.$

Summing up the equation for each component, we obtain

$\sum_m \left[p_m\Delta q_m-q_m\Delta p_m\right]=\sum_m \nabla\cdot\left(p_m\nabla q_m-q_m\nabla p_m \right).$

The LHS according to the definition of the dot product may be written in vector form as

$\sum_m \left[p_{m}\Delta q_{m}-q_{m}\Delta p_{m}\right]=\mathbf{P}\cdot\Delta\mathbf{Q}-\mathbf{Q}\cdot\Delta\mathbf{P}.$

The RHS is a bit more awkward to express in terms of vector operators. Due to the distributivity of the divergence operator over addition, the sum of the divergence is equal to the divergence of the sum, i.e.

$\sum_m \nabla\cdot\left(p_{m}\nabla q_{m}-q_{m}\nabla p_{m}\right)= \nabla\cdot\left(\sum_m p_{m}\nabla q_{m}-\sum_m q_{m}\nabla p_{m}\right).$

Recall the vector identity for the gradient of a dot product

$\nabla\left(\mathbf{P}\cdot\mathbf{Q}\right)=\left(\mathbf{P}\cdot\nabla\right)\mathbf{Q}+\left(\mathbf{Q}\cdot\nabla\right)\mathbf{P}+\mathbf{P}\times\nabla\times\mathbf{Q}+\mathbf{Q}\times\nabla\times\mathbf{P},$

which, written out in vector components is given by

$\nabla\left(\mathbf{P}\cdot\mathbf{Q}\right)=\nabla\sum_m p_{m}q_{m}=\sum_m p_{m}\nabla q_{m}+\sum_m q_{m}\nabla p_{m}.$

This result is similar to what we wish to evince in vector terms 'except' for the minus sign. Since the differential operators in each term act either over one vector (say $p_{m}$’s) or the other ($q_{m}$’s), the contribution to each term must be

$\sum_m p_m \nabla q_m = \left(\mathbf{P}\cdot\nabla\right)\mathbf{Q}+\mathbf{P}\times\nabla\times\mathbf{Q},$
$\sum_m q_m \nabla p_m = \left(\mathbf{Q}\cdot\nabla\right)\mathbf{P}+\mathbf{Q}\times\nabla\times\mathbf{P}.$

These results can be rigorously proven to be correct through evaluation of the vector components. Therefore, the RHS can be written in vector form as

$\sum_m p_{m}\nabla q_{m}-\sum_m q_{m}\nabla p_{m}=\left(\mathbf{P}\cdot\nabla\right)\mathbf{Q}+\mathbf{P}\times\nabla\times\mathbf{Q}-\left(\mathbf{Q}\cdot\nabla\right)\mathbf{P}-\mathbf{Q}\times\nabla\times\mathbf{P}.$

Putting together these two results, a result analogous to Green’s theorem for scalar fields is obtained:

Theorem for vector fields.
$\color{OliveGreen}\mathbf{P} \cdot \Delta \mathbf{Q} - \mathbf{Q} \cdot \Delta \mathbf{P} = \nabla \cdot \left[ \left( \mathbf{P}\cdot \nabla \right) \mathbf{Q} + \mathbf{P} \times \nabla \times \mathbf{Q}-\left(\mathbf{Q}\cdot\nabla\right)\mathbf{P}-\mathbf{Q}\times\nabla\times\mathbf{P}\right].$

The curl of a cross product can be written as

$\nabla\times\left(\mathbf{P}\times\mathbf{Q}\right)=\left(\mathbf{Q}\cdot\nabla\right)\mathbf{P}-\left(\mathbf{P}\cdot\nabla\right)\mathbf{Q}+\mathbf{P}\left(\nabla\cdot\mathbf{Q}\right)-\mathbf{Q}\left(\nabla\cdot\mathbf{P}\right);$

Green’s vector identity can then be rewritten as

$\mathbf{P}\cdot\Delta \mathbf{Q}-\mathbf{Q}\cdot\Delta \mathbf{P}= \nabla\cdot \left[\mathbf{P} \left(\nabla\cdot\mathbf{Q}\right)-\mathbf{Q}\left(\nabla\cdot\mathbf{P}\right)-\nabla\times\left(\mathbf{P}\times\mathbf{Q}\right)+\mathbf{P}\times\nabla\times\mathbf{Q}-\mathbf{Q}\times\nabla\times\mathbf{P}\right].$

Since the divergence of a curl is zero, the third term vanishes and we obtain:

Green's vector identity.
$\color{OliveGreen}\mathbf{P}\cdot\Delta\mathbf{Q}-\mathbf{Q}\cdot\Delta\mathbf{P}=\nabla\cdot\left[\mathbf{P}\left(\nabla\cdot\mathbf{Q}\right)-\mathbf{Q}\left(\nabla\cdot\mathbf{P}\right)+\mathbf{P}\times\nabla\times\mathbf{Q}-\mathbf{Q}\times\nabla\times\mathbf{P}\right].$

With a similar procedure, the Laplacian of the dot product can be expressed in terms of the Laplacians of the factors

$\Delta\left(\mathbf{P}\cdot\mathbf{Q}\right)=\mathbf{P}\cdot\Delta\mathbf{Q}-\mathbf{Q}\cdot\Delta \mathbf{P} +2\nabla\cdot\left[\left(\mathbf{Q}\cdot\nabla\right)\mathbf{P}+\mathbf{Q}\times\nabla\times\mathbf{P}\right].$

As a corollary, the awkward terms can now be written in terms of a divergence by comparison with the vector Green equation

$\mathbf{P}\cdot \left[ \nabla \left(\nabla \cdot \mathbf{Q} \right) \right]-\mathbf{Q} \cdot \left[ \nabla \left( \nabla \cdot \mathbf{P} \right) \right] = \nabla\cdot\left[\mathbf{P}\left(\nabla\cdot\mathbf{Q}\right)-\mathbf{Q}\left(\nabla \cdot \mathbf{P} \right) \right].$

This result can be verified by expanding the divergence of a scalar times a vector on the RHS.