1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

360 Chapter 5 Higher Dimensions and Other Coordinates

Here,DrandDzare the diffusion constants in the radial and axial di-

rections, respectively. The term containing∂C/∂zrepresents physical

movement of particles at speedU.

Show that the change of variablesρ=r/R,ζ=z/L,u(ρ, ζ )=C(r,z)

leads to the equivalent equations

b^1

ρ

∂

∂ρ

(

ρ∂u

∂ρ

)

+∂

(^2) u
∂ζ^2
+p∂u
∂ζ
= 0 , 0 <ρ< 1 , 0 <ζ < 1 ,
∂u
∂ρ
( 1 ,ζ)= 0 , 0 <ζ < 1 ,
and identify the parametersbandp.
32.When a boundedness condition atρ=0isadded,productsolutionsof
the foregoing equation are found to have the formu(ρ, ζ )=R(ρ)Z(ζ ):
R 0 (ρ)= 1 , Z 0 (ζ )=

{

e−pζ

1

,

Rn(ρ)=J 0 (λnρ), Zn(ζ )=

{

em^1 ζ

em^2 ζ

,

wherem 1 < 0 <m 2 are the roots of the equationm^2 +pm−λ^2 nb=0and

λnis chosen to satisfyJ 0 ′(λn)=0.

a.Check the details of the solution.

b.Show that theλ’s also satisfyJ 1 (λn)= 0.

33.The solution of the problem in Exercise 31 has the form

u(ρ, ζ )=a 0 e−pζ+b 0 +

∑∞

n= 1

(

anem^1 ζ+bnem^2 ζ

)

J 0 (λnρ).

The coefficients would normally be found by applying boundary condi-

tions

u(ρ, 0 )=f(ρ), u(ρ, 1 )=g(ρ), 0 <ρ< 1.

In this case, however, information is scarce. The author suggests discard-

ing the solutions that do not approach 0 asζ→∞. The justification is

thatg(ρ)is approximately 0. The solution then becomes

u(ρ, ζ )=a 0 e−pζ+

∑∞

n= 1

anem^1 ζJ 0 (λnρ),