1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

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5.6 Temperature in a Cylinder 317


of the angular coordinateθ.(Wewritev(r,t)then.) As a consequence of this
assumption, the two-dimensional Laplacian operator becomes


∇^2 v=

1

r


∂r

(

r

∂v
∂r

)

.

Suppose that the temperaturev(r,t)in a large cylinder (radiusa)satisfies
the problem


1
r


∂r

(

r∂v∂r

)

=^1 k∂v∂t, 0 <r<a, 0 <t, (1)

v(a,t)= 0 , 0 <t, (2)
v(r, 0 )=f(r), 0 <r<a. (3)

Because the differential equation (1) and boundary condition (2) are ho-
mogeneous, we may start the separation of variables by assumingv(r,t)=
φ(r)T(t). Using this form forv, we find that the partial differential equa-
tion (1) becomes


1
r

(rφ′)′T=^1
k

φT′.

After dividing through this equation byφT, we arrive at the equality


(rφ′(r))′
rφ(r) =

T′(t)
kT(t). (4)

The two members of this equation must both be constant; call their mutual
value−λ^2. Then we have two linked ordinary differential equations,


T′+λ^2 kT= 0 , 0 <t, (5)
(rφ′)′+λ^2 rφ= 0 , 0 <r<a. (6)

The boundary condition, Eq. (2), becomesφ(a)T(t)=0, 0<t.Itwillbesat-
isfied by requiring that


φ(a)= 0. (7)
We can recognize Eq. (6) as Bessel’s equation withμ=0. (See Summary,
Section 5.5.) The general solution, therefore, has the form


φ(r)=AJ 0 (λr)+BY 0 (λr).

IfB=0,φ(r)must become infinite asrapproaches zero. The physical impli-
cations of this possibility are unacceptable, so we require thatB=0. In effect
we have added the boundedness condition
∣∣
v(r,t)


∣∣

bounded atr= 0 , (8)

which we shall employ frequently.

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