(mathematics) Formulas, obtained from Green's theorem, which relate the volume integral of a function and its gradient to a surface integral of the function and its partial derivatives.
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(mathematics) Formulas, obtained from Green's theorem, which relate the volume integral of a function and its gradient to a surface integral of the function and its partial derivatives.
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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.
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This identity is derived from the divergence theorem applied to the vector field
: Let φ and ψ be scalar functions defined on some region U in R3, and suppose that φ is twice continuously differentiable, and ψ is once continuously differentiable. Then

where
is the Laplace operator,
is the boundary of region U and n is the outward pointing unit normal of surface element dS.
If φ and ψ are both twice continuously differentiable on U in R3, and ε is once continuously differentiable:
![\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 n} - \varphi {\partial \psi \over \partial n}\right)\, dS.](http://wpcontent.answers.com/math/a/7/6/a76546f5aca45a041bdd386401217d98.png)
For the special case of ε = 1 all across U in R3 then:

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:

Green's third identity derives from the second identity by choosing
, where G is a fundamental solution, or Green's function, of the Laplace equation. This means that:

For example in
, the fundamental solution has the form:

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},\eta) \nabla^2 \psi(\mathbf{y})\right]\, dV_\mathbf{y} - \psi(\eta)= \oint_{\partial U} \left[ G(\mathbf{y},\eta) {\partial \psi \over \partial n} (\mathbf{y}) - \psi(\mathbf{y}) {\partial G(\mathbf{y},\eta) \over \partial n} \right]\, dS_\mathbf{y}.](http://wpcontent.answers.com/math/2/6/e/26e9882918d66577afd8c3af95bf1a93.png)
A further simplification arises if ψ is itself a harmonic function, i.e. a solution to the Laplace equation. Then
and the identity simplifies to:
![\psi(\eta)= \oint_{\partial U} \left[\psi(\mathbf{y}) {\partial G(\mathbf{y},\eta) \over \partial n} - G(\mathbf{y},\eta) {\partial \psi \over \partial n} (\mathbf{y}) \right]\, dS_\mathbf{y}.](http://wpcontent.answers.com/math/3/8/d/38d712b20944b1dec6305ca07deb6da0.png)
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