   Chapter 16.5, Problem 35E

Chapter
Section
Textbook Problem

Recall from Section 14.3 that a function g is called harmonic on D if it satisfies Laplace’s equation, that is, ∇2g = 0 on D. Use Green’s first identity (with the same hypotheses as in Exercise 33) to show that if g is harmonic on D, then ∮ C D n   g   d s = 0 . Here Dn g is the normal derivative of g defined in Exercise 33.

To determine

To show: The CDngds=0 .

Explanation

Consider the expression of second vector form of Green’s Theorem.

CFnds=DdivF(x,y)dA (1)

As the gn occurs in the line integral, then the equation (1) can be modified as follows.

Cf(g)nds=Ddivf(g)dA=D[fdiv(g)+gf]dA {div(fF)=fdivF+Ff}=D[f2g+gf]dA {div(g)=2g}=Df2gdA+DgfdA

Rearrange the equation.

Df2gdA=Cf(g)ndsDgfdA (2)

Consider f(x,y)=1 , therefore f=0

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