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278 Chapter 4 The Potential Equation

ThesolutionisgivenbyEq.(10),providedthatthecoefficientsarechosen
accordingtoEq.(11).Sincef(θ )is an even function,bn=0, and

a 0 =π^1

∫π

0

f(θ )dθ=^12 ,

an=π^2 cn

∫π

0

f(θ )cos(nθ)dθ=2sinn(πncπ/n^2 ).

Therefore, the solution of the problem is

v(r,θ)=

1

2 +

∑∞

n= 1

2sin(nπ/ 2 )

nπ

rn

cncos(nθ). (12)

The level curves of this function are all arcs of circles that pass through the
boundary pointsr=c,θ=±π/2, wheref(θ )jumps between 0 and 1. Along
thex-axis, the function has the simple closed form 1/ 2 +( 2 /π )tan−^1 (x/c).
The CD has a color graphic of the solution.

Properties of the Solution

Now that we have the form Eq. (10) of the solution of the potential equation,
we can see some important properties of the functionv(r,θ).Inparticular,by
settingr=0weobtain

v( 0 ,θ)=a 0 = 21 π

∫π

−π

f(θ )dθ= 21 π

∫π

−π

v(c,θ)dθ.

This says that the solution of the potential equation at the center of a disk is
equal to the average of its values around the edge of the disk. It is easy to show
also that

v( 0 ,θ)=

1

2 π

∫π

−π

v(r,θ)dθ (13)

for anyrbetween 0 andc! This characteristic of solutions of the potential equa-
tion is called themean value property. From the mean value property, it is just
a step to prove the maximum principle mentioned in Section 4.1, for the mean
value of a function lies between the minimum and the maximum and cannot
equal either unless the function is constant.
An important consequence of the maximum principle — and thus of the
mean value property — is a proof of the uniqueness of the solution of the
Dirichlet problem. Suppose thatuandvare two solutions of the potential
equation in some regionRand that they have the same values on the bound-
ary ofR. Then their difference,w=u−v, is also a solution of the potential
equation inRand has value 0 all along the boundary ofR.Bythemaximum