Chapter 4 477
21.u(r,θ)=
∑∞
1bn(r
c)n/ 2
sin(nθ/ 2 ),bn=^1
π∫ 2 π0f(θ )sin(nθ/ 2 )dθ.23.u(x,y)=
∑
cnsinh(λny)sin(λnx),λn=( 2 n− 1 )π/ 2 a,
cn=2sin(λna)/(aλ^2 nsinh(λnb)).25.wsatisfies the potential equation in the rectangle with boundary condi-
tions
w( 0 ,y)=0, wx(a,y)=ay/b,0<y<b,
w(x, 0 )=0, w(x,b)=0, 0 <x<a.
w(x,y)=∑∞
n= 1bnsin(λny)cosh(λnx),λn=nπ/b,bn= 2 a(− 1 )n+^1 /n^2 π^2 cosh(λna).- The equations become
∂^2 φ
∂y∂x=∂^2 φ
∂x∂y,(
1 −M^2
)∂^2 φ
∂x^2 +∂^2 φ
∂y^2 =^0.29.φ(x,y)=
∫∞
0(
A(α)cos(αx)+B(α)sin(αx))
e−βydα+c,whereβ=α√
1 −M^2 ,cis an arbitrary constant, andA(α)
B(α)}
=−U^0
βπ∫∞
−∞f′(x){
cos(αx)
sin(αx)}
dx.- If(x(s),y(s))is the parametric representation for the boundary curveC,
then the vectory′i−x′jis normal toC,and
∫
C∂u
∂nds=∫
C∂u
∂xdy−∂u
∂ydx.By Green’s theorem,
∫C∂u
∂xdy−∂u
∂ydx=∫∫
R(
∂^2 u
∂x^2+∂
(^2) u
∂y^2
)
dA,which is 0, sinceusatisfies the potential equation inR.- Substitute directly.
35.−∇u=−(xi+yj)/(x^2 +y^2 ).