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Chapter 14: Problem 51

Prove Green's Second Identity for scalar-valued functions \(u\) and \(v\) defined on a region \(D\) : $$\iiint_{D}\left(u \nabla^{2} v-v \nabla^{2} u\right) d V=\iint_{S}(u \nabla v-v \nabla u) \cdot \mathbf{n} d S$$ (Hint: Reverse the roles of \(u\) and \(v\) in Green's First Identity.)

Short Answer

Expert verified
Question: Prove Green's Second Identity for scalar-valued functions \(u\) and \(v\) defined on a region \(D\). Answer: Green's Second Identity can be proved by starting with Green's First Identity, reversing the roles of \(u\) and \(v\), subtracting the reversed Green's First Identity from the original one, applying the divergence theorem, and simplifying the expression. The resulting Green's Second Identity is: $$\iiint_{D}\left(u \nabla^{2} v-v \nabla^{2} u\right) d V=\iint_{S}(u \nabla v-v \nabla u) \cdot \mathbf{n} d S$$

Step by step solution

01

Reverse the roles of \(u\) and \(v\) in Green's First Identity

Green's First Identity states that for scalar-valued functions \(u\) and \(v\) defined on a region \(D\): $$\iiint_D (\nabla u \cdot \nabla v) dV = \iint_S (u \nabla v) \cdot \mathbf{n} dS - \iiint_D (u \nabla^2 v) dV$$ Now, reverse the roles of \(u\) and \(v\): $$\iiint_D (\nabla v \cdot \nabla u) dV = \iint_S (v \nabla u) \cdot \mathbf{n} dS - \iiint_D (v \nabla^2 u) dV$$
02

Subtract the reversed Green's First Identity from the original one

Now, subtract the reversed identity from the original one: $$\iiint_D (\nabla u \cdot \nabla v - \nabla v \cdot \nabla u) dV = \iint_S (u \nabla v - \nodeValue 0.00v \nabla u) \cdot \mathbf{n} dS - \iiint_D (u \nabla^2 v - v \nabla^2 u) dV$$ Since \(\nabla u \cdot \nabla v = \nabla v \cdot \nabla u\), the left-hand side becomes: $$\iiint_D 0 dV = \iint_S (u \nabla v - v \nabla u) \cdot \mathbf{n} dS - \iiint_D (u \nabla^2 v - v \nabla^2 u) dV$$
03

Apply the divergence theorem

The left-hand side of the equation is zero, so we can write: $$0 = \iint_S (u \nabla v - v \nabla u) \cdot \mathbf{n} dS - \iiint_D (u \nabla^2 v - v \nabla^2 u) dV$$
04

Simplify the expression and obtain Green's Second Identity

Isolate \(-\iiint_D (u \nabla^2 v - v \nabla^2 u) dV\) on one side of the equation: $$\iiint_D (u \nabla^2 v - v \nabla^2 u) dV = \iint_S (u \nabla v - v \nabla u) \cdot \mathbf{n} dS$$ This is the Green's Second Identity: $$\iiint_{D}\left(u \nabla^{2} v-v \nabla^{2} u\right) d V=\iint_{S}(u \nabla v-v \nabla u) \cdot \mathbf{n} d S$$

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Key Concepts

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

Vector Calculus
Vector calculus is an essential mathematical tool used to analyze physical quantities that have both magnitude and direction. In the context of Green's Second Identity, vector calculus allows us to understand the behavior of vector fields related to the scalar-valued functions u and v. The key operations in vector calculus are the gradient (denoted as \(abla\)), which measures the rate and direction of change in a scalar field, and the divergence (denoted as \(abla^2\) or simply \(\text{div}\)), which represents the extent to which a vector field is spreading out or converging at a point.In our exercise, we make use of these concepts to establish a relationship between the volume integral of certain expressions involving these functions and their respective surface integrals over the boundary of the volume. Understanding the vector calculus operations is imperative to solving problems like Green's Second Identity and many others in physics and engineering.
Divergence Theorem
The Divergence Theorem is a powerful statement in vector calculus that relates the flux of a vector field through a closed surface to the divergence of the field inside the volume enclosed by that surface. Mathematically, it can be expressed as\[\iint_{S} \mathbf{F} \cdot \mathbf{n} dS = \iiint_{V} abla \cdot \mathbf{F} dV\]where \(S\) is the closed surface, \(V\) is the volume enclosed by \(S\), \(\mathbf{F}\) is a vector field, and \(\mathbf{n}\) is the outward unit normal vector to the surface. The Divergence Theorem is used in the final steps of proving Green's Second Identity to simplify the mathematical expressions by recognizing that the volume integral of the divergence of a vector field is equal to the surface integral of the field over the volume's boundary.
Triple Integral
Triple integrals extend the concept of integrals to three-dimensional space, allowing us to calculate the volume under a surface in \(\mathbb{R}^3\). In the context of our problem, triple integrals are used to evaluate the integral of a scalar or vector quantity throughout a volume \(D\). The notation for a triple integral is written as\[\iiint_D f(x, y, z) dV\]where \(f(x, y, z)\) is the function to be integrated and \(dV\) represents an infinitesimal volume element. In Green's Second Identity, we use triple integrals to calculate the total of the quantities involving \(u abla^2v\) and \(v abla^2u\) throughout the region \(D\).
Surface Integral
A surface integral allows us to integrate over a two-dimensional surface in three-dimensional space, which is often required when working with vector fields and flux calculations. The surface integral can be represented as\[\iint_{S} \mathbf{F} \cdot d\mathbf{S}\]where \(\mathbf{F}\) is a vector field and \(d\mathbf{S}\) is a differential element of the surface \(S\), often written as \(\mathbf{n} dS\) where \(\mathbf{n}\) is the normal vector to the surface at that point. Green's Second Identity involves surface integrals that relate the behavior of the functions \(u\) and \(v\) on the boundary surface of the volume.
Scalar-valued functions
Scalar-valued functions are functions that assign a single real number to each point in a space. In multivariable calculus and specifically in Green's Second Identity, we deal with scalar-valued functions like \(u\) and \(v\) that depend on the three spatial coordinates \((x, y, z)\). The functions can represent physical quantities such as temperature, pressure, or potential at various points in space.Understanding scalar-valued functions and how they interact through gradient and Laplacian operations is foundational for proving identities like Green's Second Identity, which involves both the functions themselves and their derivatives.

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Most popular questions from this chapter

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.)

Use Green's Theorem to evaluate the following line integrals. Unless stated otherwise, assume all curves are oriented counterclockwise. \(\oint\left(2 x+e^{y^{2}}\right) d y-\left(4 y^{2}+e^{x^{2}}\right) d x,\) where \(C\) is the boundary of the square with vertices \((0,0),(1,0),(1,1),\) and (0,1)

\(\mathbb{R}^{2}\) Assume that the vector field \(\mathbf{F}\) is conservative in \(\mathbb{R}^{2}\), so that the line integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) is independent of path. Use the following procedure to construct a potential function \(\varphi\) for the vector field \(\mathbf{F}=\langle f, g\rangle=\langle 2 x-y,-x+2 y\rangle\) a. Let \(A\) be (0,0) and let \(B\) be an arbitrary point \((x, y) .\) Define \(\varphi(x, y)\) to be the work required to move an object from \(A\) to \(B\) where \(\varphi(A)=0 .\) Let \(C_{1}\) be the path from \(A\) to \((x, 0)\) to \(B\) and let \(C_{2}\) be the path from \(A\) to \((0, y)\) to \(B .\) Draw a picture. b. Evaluate \(\int_{C_{1}} \mathbf{F} \cdot d \mathbf{r}=\int_{C_{1}} f d x+g d y\) and conclude that \(\varphi(x, y)=x^{2}-x y+y^{2}\) c. Verify that the same potential function is obtained by evaluating the line integral over \(C_{2}\)

Let S be the disk enclosed by the curve \(C: \mathbf{r}(t)=\langle\cos \varphi \cos t, \sin t, \sin \varphi \cos t\rangle,\)for \(0 \leq t \leq 2 \pi,\) where \(0 \leq \varphi \leq \pi / 2\) is a fixed angle. Use Stokes' Theorem and a surface integral to find the circulation on \(C\) of the vector field \(\mathbf{F}=\langle-y, x, 0\rangle\) as a function of \(\varphi .\) For what value of \(\varphi\) is the circulation a maximum?

The Navier-Stokes equation is the fundamental equation of fluid dynamics that models the flow in everything from bathtubs to oceans. In one of its many forms (incompressible, viscous flow), the equation is $$\rho\left(\frac{\partial \mathbf{V}}{\partial t}+(\mathbf{V} \cdot \nabla) \mathbf{V}\right)=-\nabla p+\mu(\nabla \cdot \nabla) \mathbf{V}.$$ In this notation, \(\mathbf{V}=\langle u, v, w\rangle\) is the three-dimensional velocity field, \(p\) is the (scalar) pressure, \(\rho\) is the constant density of the fluid, and \(\mu\) is the constant viscosity. Write out the three component equations of this vector equation. (See Exercise 40 for an interpretation of the operations.)
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