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Prove Green's First Identity for twice differentiable scalar-valued functions \(u\) and \(v\) defined on a region \(D\) : $$\iiint_{D}\left(u \nabla^{2} v+\nabla u \cdot \nabla v\right) d V=\iint_{S} u \nabla v \cdot \mathbf{n} d S$$ where \(\nabla^{2} v=\nabla \cdot \nabla v .\) You may apply Gauss' Formula in Exercise 48 to \(\mathbf{F}=\nabla v\) or apply the Divergence Theorem to \(\mathbf{F}=u \nabla v\)

Short Answer

Expert verified
Question: Prove Green's first identity for twice differentiable scalar-valued functions \(u\) and \(v\) defined on a region \(D\). Solution: Using the steps above, we applied the Divergence Theorem to \(\mathbf{F}=u \nabla v\) and obtained the equation $$\iiint_{D} (u \nabla^{2} v + \nabla u \cdot \nabla v) dV = \iint_{S} (u \nabla v \cdot \mathbf{n}) dS$$ which is Green's first identity for the given functions \(u\) and \(v\).

Step by step solution

01

Compute the divergence of \(\mathbf{F}\)

Let \(\mathbf{F} = u \nabla v\). We need to compute the divergence of \(\mathbf{F}\): $$\nabla \cdot \mathbf{F} = \nabla \cdot (u \nabla v)$$
02

Apply the product rule for divergence

The product rule for divergence states that: $$\nabla \cdot (u \nabla v) = u \nabla^{2} v + \nabla u \cdot \nabla v$$
03

Use the Divergence Theorem

By Divergence Theorem, we know that: $$\iiint_{D} (\nabla \cdot \mathbf{F}) dV = \iint_{S} (\mathbf{F} \cdot \mathbf{n}) dS$$ We substitute the expression for divergence (from step 2) and \(\mathbf{F}\): $$\iiint_{D} (u \nabla^{2} v + \nabla u \cdot \nabla v) dV = \iint_{S} (u \nabla v \cdot \mathbf{n}) dS$$
04

Verify Green's First Identity

We've shown that $$\iiint_{D} (u \nabla^{2} v + \nabla u \cdot \nabla v) dV = \iint_{S} (u \nabla v \cdot \mathbf{n}) dS$$ which is the Green's first identity for twice differentiable functions \(u\) and \(v\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Divergence Theorem
The Divergence Theorem is a fundamental tool in vector calculus, connecting the flow of a vector field across a closed surface to the behavior of the vector field inside the volume encased by that surface. Essentially, it translates volume integrals into surface integrals. For a vector field \( \mathbf{F} \) defined over a region \( D \) with boundary \( S \), the Divergence Theorem states:
\[\iiint_{D} (abla \cdot \mathbf{F}) \, dV = \iint_{S} (\mathbf{F} \cdot \mathbf{n}) \, dS\]
where \( \mathbf{n} \) is the outward unit normal on the surface \( S \). This theorem is crucial in the context of Green's First Identity as it allows the conversion of the divergence of a vector field inside a volume into a flux through the enclosing surface. It simplifies calculations and provides deep insights into the field's global properties. Remember, without the Divergence Theorem, connecting the integral over the surface to that over the volume would be significantly harder.
To apply it effectively, ensure:
  • The vector field \( \mathbf{F} \) is well-defined and smooth.
  • The region \( D \) is closed and bounded with a piecewise smooth boundary.
Product Rule for Divergence
The product rule for divergence is similar to the product rule in differentiation, adapted for vector calculus. It gives a method to find the divergence of the product of a scalar field and a vector field. For this identity, it states:
\[abla \cdot (u abla v) = u abla^2 v + abla u \cdot abla v\]
Here, \( u \) is a scalar function and \( abla v \) is a vector function derived from another scalar function \( v \). This rule is foundational when proving Green's First Identity, as it breaks down complex divergence calculations into more manageable components.
Key points to remember include:
  • \( abla u \cdot abla v \) represents the dot product of gradients, highlighting how each function changes in space.
  • \( abla^2 v \) involves the Laplacian, showing how \( v \) diverges or converges.

Understanding this rule allows for deeper insights into the structural properties of the functions involved, enabling efficient simplification of expressions often encountered in physics and engineering.
Twice Differentiable Functions
Twice differentiable functions are essential in the context of calculus and differential equations. A function is twice differentiable if it is differentiable, and its first derivative is also differentiable. For Green's First Identity, having both functions \( u \) and \( v \) be twice differentiable is a necessary condition, which ensures that the second derivatives exist and are continuous.
Properties of Twice Differentiable Functions include:
  • Smoothness: Such functions are typically smoother, allowing for accurate gradient and divergence calculations.
  • Existence of Laplacian: The Laplacian \( abla^2 v \) is well-defined, revealing how a function spreads out from a point.
  • Stability: Solutions to differential equations involving such functions are often stable and physically meaningful.

The requirement for twice differentiable functions ensures the robustness of Green's First Identity, guaranteeing that all necessary derivatives exist and behave predictably. This makes mathematical operations involving these functions more reliable across various fields of study, from theoretical physics to applied engineering. Understanding the behavior and constraints of twice differentiable functions is vital for students aiming to tackle advanced problems in mathematics.

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Most popular questions from this chapter

Prove the following identities. Assume that \(\varphi\) is \(a\) differentiable scalar-valued function and \(\mathbf{F}\) and \(\mathbf{G}\) are differentiable vector fields, all defined on a region of \(\mathbb{R}^{3}\). $$\nabla \times(\nabla \times \mathbf{F})=\nabla(\nabla \cdot \mathbf{F})-(\nabla \cdot \nabla) \mathbf{F}$$

\(A\) scalar-valued function \(\varphi\) is harmonic on a region \(D\) if \(\nabla^{2} \varphi=\nabla \cdot \nabla \varphi=0\) at all points of \(D\) Show that the potential function \(\varphi(x, y, z)=|\mathbf{r}|^{-p}\) is harmonic provided \(p=0\) or \(p=1,\) where \(\mathbf{r}=\langle x, y, z\rangle .\) To what vector fields do these potentials correspond?

The heat flow vector field for conducting objects is \(\mathbf{F}=-k \nabla T,\) where \(T(x, y, z)\) is the temperature in the object and \(k>0\) is a constant that depends on the material. Compute the outward flux of \(\mathbf{F}\) across the following surfaces S for the given temperature distributions. Assume \(k=1\). \(T(x, y, z)=-\ln \left(x^{2}+y^{2}+z^{2}\right) ; S\) is the sphere \(x^{2}+y^{2}+z^{2}=a^{2}\).

Consider the potential function \(\varphi(x, y, z)=G(\rho),\) where \(G\) is any twice differentiable function and \(\rho=\sqrt{x^{2}+y^{2}+z^{2}} ;\) therefore, \(G\) depends only on the distance from the origin. a. Show that the gradient vector field associated with \(\varphi\) is \(\mathbf{F}=\nabla \varphi=G^{\prime}(\rho) \frac{\mathbf{r}}{\rho},\) where \(\mathbf{r}=\langle x, y, z\rangle\) and \(\rho=|\mathbf{r}|\) b. Let \(S\) be the sphere of radius \(a\) centered at the origin and let \(D\) be the region enclosed by \(S\). Show that the flux of \(\mathbf{F}\) across \(S\) is $$\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=4 \pi a^{2} G^{\prime}(a) $$ c. Show that \(\nabla \cdot \mathbf{F}=\nabla \cdot \nabla \varphi=\frac{2 G^{\prime}(\rho)}{\rho}+G^{\prime \prime}(\rho)\) d. Use part (c) to show that the flux across \(S\) (as given in part (b)) is also obtained by the volume integral \(\iiint_{D} \nabla \cdot \mathbf{F} d V\). (Hint: use spherical coordinates and integrate by parts.)

a. Let \(\mathbf{a}=\langle 0,1,0\rangle, \mathbf{r}=\langle x, y, z\rangle,\) and consider the rotation field \(\mathbf{F}=\mathbf{a} \times \mathbf{r} .\) Use the right-hand rule for cross products to find the direction of \(\mathbf{F}\) at the points (0,1,1),(1,1,0),(0,1,-1), and (-1,1,0). b. With \(\mathbf{a}=\langle 0,1,0\rangle,\) explain why the rotation field \(\mathbf{F}=\mathbf{a} \times \mathbf{r}\) circles the \(y\) -axis in the counterclockwise direction looking along a from head to tail (that is, in the negative \(y\) -direction).

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