Page 411 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes It is clear that the temperature u must be a function of r and t only (due to symmetry). So using
cylindrical co-ordinates, we have

u 2 2 u 1 u
h , 0 r a ...(ii)
t r 2 r t
The boundary condition is

u 0 at r a ...(iii)
The initial condition is
u r ,0 f r ...(iv)

To solve eq. (ii), let us assume
u e mt v r ...(v)

Substituting in eq. (ii), we obtain
2 v 1 u
mv r h 2 2 ...(vi)
r r r

Rewriting (vi) in the form
2 v v mr
r 2 2 v 0 ...(vii)
r r h
Let k 2 m /h 2 ...(viii)
and t = kr, we have from (vii)

2 v v
t 2 tv 0 ...(ix)
t t
which has the same form as Bessel s differential equation for n = 0. Hence the general solution of
(ix) is

v AJ kr BY kr ...(x)
0
0
where A and B are arbitrary constants. Now since the temperature must remain finite at r = 0, the
arbitrary constant B in (X) must be equal to zero. We thus have
v AJ kr ...(xi)
0
Since the boundary r = a, of the plate is maintained at zero temperature for all values of t, we
must have

J ka 0 ...(xii)
0
Thus only those values of k are allowed that satisfy equation (xii). Let these values be
k i 1,2,3,... . Equation (viii) gives the following values for m:
i
2
m i k h ...(xiii)
i
A particular solution of (v) that satisfies the boundary condition is

2 2
k th
u A e i J k r
i i 0 i

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