1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

332 Chapter 5 Higher Dimensions and Other Coordinates

Thus in order to satisfy Eq. (20), we must haveB=0. It is possible to show
that

J 1 / 2 (λρ)=^2

π

sin√(λρ)

λρ

, Y 1 / 2 (λρ)=−^2

π

cos√(λρ)

λρ

.

Our solution to Eqs. (18) and (20) is, therefore,

R(ρ)=sin(λρ)

ρ

, (21)

and Eq. (19) is satisfied ifλ^2 n=(nπ/a)^2. The solution of the problem of
Eqs. (13)–(16) can be written in the form

u(ρ,t)=

∑∞

n= 1

sin(λnρ)

ρ

[

ancos(λnct)+bnsin(λnct)

]

. (22)

Thea’s andb’s are, as usual, chosen so that the initial conditions Eqs. (15)
and (16) are satisfied.

C. Pressure in a Bearing

The pressure in the lubricant inside a plane-pad bearing satisfies the problem

∂

∂x

(

x^3 ∂p

∂x

)

+x^3 ∂

(^2) p
∂y^2
=− 1 , a<x<b, −c<y<c,( 23 )
p(a,y)= 0 , p(b,y)= 0 , −c<y<c,( 24 )
p(x,−c)= 0 , p(x,c)= 0 , a<x<b.( 25 )
(Hereaandcare positive constants andb=a+1.) Equation (23) is ellip-
tic and nonhomogeneous. To reduce this equation to a more familiar one, let
p(x,y)=v(x)+u(x,y),wherev(x)satisfies the problem
(
x^3 v′

)′

=− 1 , a<x<b, (26)

v(a)= 0 ,v(b)= 0. (27)

Then, whenvis found,umust be the solution of the problem

∂

∂x

(

x^3

∂u

∂x

)

+x^3

∂^2 u

∂y^2 =^0 , a<x<b, −c<y<c,(^28 )

u(a,y)= 0 , u(b,y)= 0 , −c<y<c,( 29 )

u(x,±c)=−v(x), a<x<b.( 30 )

If we now assume thatu(x,y)=X(x)Y(y), the variables can be separated: