   Chapter 14.3, Problem 77E

Chapter
Section
Textbook Problem

Verify that the function u = 1 / x 2 + y 2 + z 2 is a solution of the three-dimensional Laplace equation uxx + uyy + uzz = 0.

To determine

To verify: The function u=1x2+y2+z2 satisfies the three-dimensional equation uxx+uyy+uzz=0 .

Explanation

Given:

The function is, u=1x2+y2+z2 .

Verification:

Take the partial derivative of the given function with respect to x and obtain ux .

ux=x(1x2+y2+z2)=(12)(x2+y2+z2)32(2x+0+0)=x(x2+y2+z2)32

Thus, ux=x(x2+y2+z2)32 . (1)

Take the partial derivative the equation (1) with respect to x and obtain uxx .

2ux2=x(x(x2+y2+z2)32)=[(x2+y2+z2)32(1)x(32)(x2+y2+z2)52(2x)]=(x2+y2+z2)32+3x2(x2+y2+z2)52=1(x2+y2+z2)32+3x2(x2+y2+z2)52

On further simplification, the value of 2ux2 is obtained as follows.

2ux2=1(x2+y2+z2)32+3x2(x2+y2+z2)52=(x2+y2+z2)+3x2(x2+y2+z2)52=x2y2z2+3x2(x2+y2+z2)52=2x2y2z2(x2+y2+z2)52

Thus, the partial derivative, 2ux2=2x2y2z2(x2+y2+z2)52 .

Take the partial derivative of the given function with respect to y and obtain uy .

uy=y(1x2+y2+z2)=(12)(x2+y2+z2)32(0+2y+0)=y(x2+y2+z2)32

Thus, uy=y(x2+y2+z2)32 . (2)

Take the partial derivative the equation (1) with respect to y and obtain uyy .

2uy2=y(y(x2+y2+z2)32)=[(x2+y2+z2)32(1)+y(32)(x2+y2+z2)52(2y)]=(x2+y2+z2)323y2(x2+y2+z2)52=1(x2+y2+z2)323y2(x2+y2+z2)52

On further simplification, the value of 2uy2 is obtained as follows.

2uy2=1(x2+y2+z2)32+3y2(x2+y2+z2)52=(x2+y2+z2)+3y2(x2+y2+z2)52=x2y2z2+3y2(x2+y2+z2)52=2y2x2z2(x2+y2+z2)52

Thus, the partial derivative, 2uy2=2y2x2z2(x2+y2+z2)52

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