九州大学 システム情報科学府 情報理工学専攻・電気電子工学専攻 2024年1月実施 ベクトル解析

标签：

• Kyushu-University
• Greens-Identities

Author

祭音Myyura (assisted by ChatGPT 5.5)

Description

(1) Prove the following relationship between the volume integral and the surface integral:

$$\iiint_V \left\{ f\nabla^2 g+(\nabla f)\cdot(\nabla g) \right\}\,dV = \iint_S (f\nabla g)\cdot d\mathbf{S}.$$

(2) Prove the following relationship between the volume integral and the surface integral:

$$\iiint_V \left(f\nabla^2 g-g\nabla^2 f\right)\,dV = \iint_S \left(f\nabla g-g\nabla f\right)\cdot d\mathbf{S}.$$

Here,

$$d\mathbf{S}=\mathbf{n}\,dS$$

where $\mathbf{n}$ is the outward unit normal vector to the closed surface $S$.

Kai

(1)

We use the divergence theorem:

$$\iiint_V \nabla\cdot \mathbf{F}\,dV = \iint_S \mathbf{F}\cdot d\mathbf{S}.$$

Choose the vector field

$$\mathbf{F}=f\nabla g.$$

Then

$$\nabla\cdot \mathbf{F} = \nabla\cdot(f\nabla g).$$

Using the product rule for divergence,

$$\nabla\cdot(f\nabla g) = (\nabla f)\cdot(\nabla g)+f\nabla\cdot(\nabla g).$$

Since

$$\nabla\cdot(\nabla g)=\nabla^2 g,$$

we have

$$\nabla\cdot(f\nabla g) = (\nabla f)\cdot(\nabla g)+f\nabla^2 g.$$

Therefore,

$$\iiint_V \nabla\cdot(f\nabla g)\,dV = \iiint_V \left\{ f\nabla^2 g+(\nabla f)\cdot(\nabla g) \right\}\,dV.$$

By the divergence theorem,

$$\iiint_V \nabla\cdot(f\nabla g)\,dV = \iint_S (f\nabla g)\cdot d\mathbf{S}.$$

Hence,

$$\boxed{ \iiint_V \left\{ f\nabla^2 g+(\nabla f)\cdot(\nabla g) \right\}\,dV = \iint_S (f\nabla g)\cdot d\mathbf{S} }$$

as required.

(2)

We again use the divergence theorem.

Choose the vector field

$$\mathbf{F}=f\nabla g-g\nabla f.$$

Then

$$\nabla\cdot \mathbf{F} = \nabla\cdot(f\nabla g-g\nabla f).$$

Using linearity of divergence,

$$\nabla\cdot \mathbf{F} = \nabla\cdot(f\nabla g)-\nabla\cdot(g\nabla f).$$

From the product rule,

$$\nabla\cdot(f\nabla g) = (\nabla f)\cdot(\nabla g)+f\nabla^2 g,$$

and

$$\nabla\cdot(g\nabla f) = (\nabla g)\cdot(\nabla f)+g\nabla^2 f.$$

Thus,

$$\nabla\cdot \mathbf{F} = \left\{(\nabla f)\cdot(\nabla g)+f\nabla^2 g\right\} - \left\{(\nabla g)\cdot(\nabla f)+g\nabla^2 f\right\}.$$

Since the dot product is commutative,

$$(\nabla f)\cdot(\nabla g) = (\nabla g)\cdot(\nabla f),$$

these two terms cancel. Therefore,

$$\nabla\cdot \mathbf{F} = f\nabla^2 g-g\nabla^2 f.$$

Now applying the divergence theorem,

$$\iiint_V \nabla\cdot \mathbf{F}\,dV = \iint_S \mathbf{F}\cdot d\mathbf{S}.$$

Substituting

$$\mathbf{F}=f\nabla g-g\nabla f,$$

we get

$$\iiint_V \left(f\nabla^2 g-g\nabla^2 f\right)\,dV = \iint_S \left(f\nabla g-g\nabla f\right)\cdot d\mathbf{S}.$$

Hence,

$$\boxed{ \iiint_V \left(f\nabla^2 g-g\nabla^2 f\right)\,dV = \iint_S \left(f\nabla g-g\nabla f\right)\cdot d\mathbf{S} }$$

as required.