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Chapter 9: Problem 26

Let \(C\) be a closed curve and \(D\) the enclosed region. Prove Green's identities in two dimensions. a. First prove $$ \int_{D}(v \nabla \cdot \mathbf{F}+\mathbf{F} \cdot \nabla v) d A=\int_{C}(v \mathbf{F}) \cdot d \mathbf{s} $$ b. Let \(\mathbf{F}=\nabla u\) and obtain Green's first identity, $$ \int_{D}\left(v \nabla^{2} u+\nabla u \cdot \nabla v\right) d A=\int_{C}(v \nabla u) \cdot d \mathbf{s} $$ c. Use Green's first identity to prove Green's second identity, $$ \int_{D}\left(u \nabla^{2} v-v \nabla^{2} u\right) d A=\int_{C}(u \nabla v-v \nabla u) \cdot d \mathbf{s} $$

Short Answer

Expert verified
Green's identities in two dimensions are a consequence of the divergence theorem and Stokes' theorem, and are obtained by appropriate substitutions and simplifications.

Step by step solution

01

Proving the First Identity

Start with the left-hand side (LHS) of the provided equation: \( \int_{D}(v \nabla \cdot \mathbf{F}+\mathbf{F} \cdot \nabla v) d A \). Divide it into two integrals: \( \int_{D}v \nabla \cdot \mathbf{F} d A + \int_{D}\mathbf{F} \cdot \nabla v d A \). Use the divergence theorem on the first integral and the Stokes' theorem on the second one. Combine both results to verify the RHS of the equation.
02

Obtain Green's First Identity

Substitute \( \mathbf{F} \) with \( \nabla u \) in the previously proven equation. Work through the same calculations to obtain Green's first identity: \( \int_{D}\left(v \nabla^{2} u+\nabla u \cdot \nabla v\right) d A=\int_{C}(v \nabla u) \cdot d \mathbf{s} \).
03

Using Green's First Identity to Prove Second One

Green's second identity can be derived from the first one. Form the second identity's LHS by subtracting the Green's identities for \( u \nabla^{2} v \) and \( v \nabla^{2} u \). Verify that the LHS equals to the RHS: \( \int_{C}(u \nabla v-v \nabla u) \cdot d \mathbf{s} \).

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Most popular questions from this chapter

Consider a constant electric dipole moment \(\mathrm{p}\) at the origin. It produces an electric potential of \(\phi=\frac{\mathrm{p} \cdot \mathrm{r}}{4 \pi \epsilon_{0} r^{3}}\) outside the dipole. Noting that \(\mathbf{E}=-\nabla \phi\), find the electric field at \(\mathbf{r}\).

For spherical coordinates, $$ \begin{aligned} x &=\rho \sin \theta \cos \phi \\ y &=\rho \sin \theta \sin \phi \\ z &=\rho \cos \theta \end{aligned} $$ find the scale factors and derive the following expressions: $$ \begin{gathered} \nabla f=\frac{\partial f}{\partial \rho} \hat{\mathbf{e}}_{\rho}+\frac{1}{\rho} \frac{\partial f}{\partial \theta} \hat{\mathbf{e}}_{\theta}+\frac{1}{\rho \sin \theta} \frac{\partial f}{\partial \phi} \hat{\mathbf{e}}_{\phi} \\ \nabla \cdot \mathbf{F}=\frac{1}{\rho^{2}} \frac{\partial\left(\rho^{2} F_{\rho}\right)}{\partial \rho}+\frac{1}{\rho \sin \theta} \frac{\partial\left(\sin \theta F_{\theta}\right)}{\partial \theta}+\frac{1}{\rho \sin \theta} \frac{\partial F_{\phi}}{\partial \phi}. \end{gathered} $$ $$ \begin{aligned} \nabla \times \mathbf{F}=& \frac{1}{\rho \sin \theta}\left(\frac{\partial\left(\sin \theta F_{\phi}\right)}{\partial \theta}-\frac{\partial F_{\theta}}{\partial \phi}\right) \hat{\mathbf{e}}_{\rho}+\frac{1}{\rho}\left(\frac{1}{\sin \theta} \frac{\partial F_{\rho}}{\partial \phi}-\frac{\partial\left(\rho F_{\phi}\right)}{\partial \rho}\right) \\ &+\frac{1}{\rho}\left(\frac{\partial\left(\rho F_{\theta}\right)}{\partial \rho}-\frac{\partial F_{\rho}}{\partial \theta}\right) \hat{\mathbf{e}}_{\phi} \end{aligned} $$ $$ \nabla^{2} f=\frac{1}{\rho^{2}} \frac{\partial}{\partial \rho}\left(\rho^{2} \frac{\partial f}{\partial \rho}\right)+\frac{1}{\rho^{2} \sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial f}{\partial \theta}\right)+\frac{1}{\rho^{2} \sin ^{2} \theta} \frac{\partial^{2} f}{\partial \phi^{2}}. $$

Use Stokes' Theorem to evaluate the integral $$ \int_{C}-y^{3} d x+x^{3} d y-z^{3} d z $$ for \(C\) the (positively oriented) curve of intersection between the cylinder \(x^{2}+y^{2}=1\) and the plane \(x+y+z=1\)

Compute the following integrals: a. \(\int_{C}\left(x^{2}+y\right) d x+\left(3 x+y^{3}\right) d y\) for \(C\) the ellipse \(x^{2}+4 y^{2}=4\). b. \(\int_{S}(x-y) d y d z+\left(y^{2}+z^{2}\right) d z d x+\left(y-x^{2}\right) d x d y\) for \(S\) the positively oriented unit sphere. c. \(\int_{C}(y-z) d x+(3 x+z) d y+(x+2 y) d z\), where \(C\) is the curve of intersection between \(z=4-x^{2}-y^{2}\) and the plane \(x+y+z=0\). d. \(\int_{C} x^{2} y d x-x y^{2} d y\) for \(C\) a circle of radius 2 centered about the origin. e. \(\int_{S} x^{2} y d y d z+3 y^{2} d z d x-2 x z^{2} d x d y\), where \(S\) is the surface of the cube \([-1,1] \times[-1,1] \times[-1,1]\).

Find the lengths of the following curves: a. \(y(x)=x\) for \(x \in[0,2]\). b. \((x, y, z)=(t, \ln t, 2 \sqrt{2} t)\) for \(1 \leq t \leq 2\). c. \(y(x)=\cosh x, x \in[-2,2]\). (Recall the hanging chain example from classical dynamics.)
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