Green’s First and Second Identity

Section 12.2 Green’s First and Second Identity

We want to derive two useful identities from the divergence theorem. In some sense they are extensions of the formula of `integration by parts’ to higher dimensions. We first introduce the Laplace operator.

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Definition 12.6. Laplace operator.

Suppose that

D \subset R^{N},

and that

u : D \to R

is a function. Then we set

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\begin{aligned} Δ u & := div (grad u) \\ = \sum_{i = 1}^{N} \frac{\partial^{2} u}{\partial x_{i}^{2}} \\ = \frac{\partial^{2} u}{\partial x_{1}^{2}} + \dots + \frac{\partial^{2} u}{\partial x_{N}^{2}}, \end{aligned}

and call

Δ

the Laplace operator or the Laplacian.

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Remark 12.7.

Using the nabla operator from Definition 8.15 and the notation introduced in Remark 11.4 we can formally write

Δ

as the scalar product

\nabla \cdot \nabla .

For this reason one often writes

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Δ u =: \nabla \cdot \nabla u =: \nabla^{2} u

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We need the following product rule.

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Lemma 12.8. Product rule for divergence.

Suppose that

f

is a vector field, and

u

a scalar valued function. Then

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div (u f) = f \cdot grad u + u div f

whenever all derivatives exist.

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Proof.

By the product rule we have

\begin{aligned} div u & = \sum_{i = 1}^{N} \frac{\partial}{\partial x_{i}} (u f_{i}) \\ = \sum_{i = 1}^{N} f_{i} \frac{\partial u}{\partial x_{i}} + u \sum_{i = 1}^{N} \frac{\partial f_{i}}{\partial x_{i}} \\ = f \cdot grad u + u div f \end{aligned}

as required.

Theorem 12.9. Green’s first identity.

Suppose that

D

is a bounded set on which the divergence theorem can be applied. If

u, v : D \to R

have continuous second order partial derivatives on

\overset{―}{D}

then

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\begin{matrix} (12.5) & \int_{D} u Δ v + \nabla u \cdot \nabla v d x = \int_{\partial D} u \nabla v \cdot n d S, \end{matrix}

where

n

is the outward pointing unit normal to

\partial D .

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Proof.

Applying Lemma 12.8 with

f := \nabla v

we get

\begin{aligned} div (u \nabla v) & = div (u f) \\ = f \cdot grad u + u div f \\ = \nabla u \cdot \nabla v + u div (\nabla u) \\ = \nabla u \cdot \nabla v + u Δ u . \end{aligned}

Hence by the Divergence Theorem

\begin{aligned} \int_{D} u Δ u + \nabla u \cdot \nabla v d x & = \int_{D} div (u \nabla v) d x \\ = \int_{\partial D} u \nabla v \cdot n d S, \end{aligned}

completing the proof of the theorem.

Theorem 12.10. Green’s second identity.

Suppose that

D

is a bounded set on which the divergence theorem can be applied. If

u, v : D \to R

have continuous second order partial derivatives on

\overset{―}{D}

then

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\begin{matrix} (12.6) & \int_{D} u Δ v - v Δ u d x = \int_{\partial D} u \nabla v \cdot n - v \nabla u \cdot n d S, \end{matrix}

where

n

is the outward pointing unit normal to

\partial D .

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Proof.

Apply Green’s first identity (12.5) to

u Δ v

and then to

v Δ u

and subtract the two integrals. Then the term

\int_{D} \nabla u \cdot \nabla v d x

cancels as it appears twice with opposite signs.

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