Mandrik, P. A.

English

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Differential Equations, Vol. 37, No. 2, 2001, pp. 257–260. Translated from Differentsial’nye Uravneniya, Vol. 37, No. 2, 2001, pp. 238–241.
Original Russian Text Copyright c© 2001 by Mandrik.
PARTIAL
DIFFERENTIAL EQUATIONS
Solution of the Heat Equation
with Mixed Boundary Conditions
on the Surface of an Isotropic Half-Space
P. A. Mandrik
Belarussian State University, Belarus
Received November 16, 1999
Consider the nonstationary heat equation
Trr(r, z, τ) + r−1Tr(r, z, τ) + Tzz(r, z, τ) = a−1Tτ (r, z, τ) (1)
in cylindrical coordinates (r > 0, z > 0, τ > 0) in the case of axial symmetry under the initial
condition T (r, z, 0) = 0 and the mixed boundary conditions
−Tz(r, 0, τ) = λ−1q1(τ)q2(r), 0 < r < R, τ > 0, (2)
T (r, 0, τ) = 0, R < r <∞, τ > 0. (3)
The mathematical statement of this problem describes, in particular, a nonstationary temper-
ature field under a local heating of an isotropic half-space (with temperature conductivity a > 0
and thermal conductivity λ > 0) through a circular domain of radius R on the surface z = 0;
moreover, the surface of the half-space is thermally insulated outside this circular domain. In this
case, obviously, the homogeneous initial condition and the boundary condition (3) do not affect the
generality of the statement of the problem.
Applying the Laplace transform f¯(s) =
∫∞
0
f(τ) exp(−sτ)dx, Re s > 0, to problem (1)–(3),
for the Laplace transform of the temperature we have
T¯rr(r, z, s) + r−1T¯r(r, z, s) + T¯zz(r, z, s) = σT¯ (r, z, s), σ = s/a, (4)
−T¯z(r, 0, s) = λ−1q¯1(s)q2(r), 0 < r < R, (5)
T¯ (r, 0, s) = 0, R < r <∞. (6)
From now on, for brevity, we omit the obvious constraint Re s > 0 on the parameter s.
Taking into account the boundedness of T¯ (r, z, s) as
√
r2 + z2 → ∞ and using separation of
variables or the Hankel transform
H
[
T¯ (r, z, s)
]
= T¯H(p, z, s) =
∞∫
0
T¯ (r, z, s)J0(pr)r dr, p > 0,
we can write out the solution of Eq. (4) in the form
T¯ (r, z, s) =
∞∫
0
C¯(p, s) exp
(
−z
√
p2 + σ
)
J0(pr)dp (7)
and represent the normal derivative on the surface z = 0 as
−T¯z(r, z, s) =
∞∫
0
C¯(p, s)
√
p2 + σ exp
(
−z
√
p2 + σ
)
J0(pr)dp, (8)
0012-2661/01/3702-0257$25.00 c© 2001 MAIK “Nauka/Interperiodica”
258 MANDRIK
where C¯(p, s) is the unknown transform to be determined and J0(pr) is the Bessel function of the
first kind and zero order.
Taking into account the mixed boundary conditions (5) and (6), for C¯(p, s), we obtain the pair
of integral equations
∞∫
0
C¯(p, s)
√
p2 + σJ0(pr)dp = λ−1q¯1(s)q2(r), 0 < r < R, (9)
∞∫
0
C¯(p, s)J0(pr)dp = 0, R < r <∞, (10)
in the domain of L-transforms.
By choosing the functions C¯(p, s) in the form [1]
C¯(p, s) =
p√
p2 + σ
R∫
0
ϕ¯(t, s) sin
(
t
√
p2 + σ
)
dt, (11)
where ϕ¯(t, s) is some new unknown analytic function, we guarantee the validity of (10), since
[2, p. 203]
∞∫
0
sin
(
t
√
p2 + σ
) pJ0(pr)√
p2 + σ
dp =
{ 0 for r > t
(t2 − r2)−1/2 cos
(√
(t2 − r2) σ
)
for r < t.
Substituting (11) into (9) and following the lines of [1], for the transform ϕ¯(t, s), we obtain the
integral equation
r∫
0
tϕ¯(t, s)√
r2 − t2 exp
(
−
√
(r2 − t2) σ
)
dt+
R∫
0
ϕ¯(t, s) sin
(
t
√
σ
)
dt −
R∫
r
tϕ¯(t, s)√
t2 − r2 sin
(√
(t2 − r2)σ
)
dt
=
q¯1(s)
λ
r∫
0
q2(%)% d%, 0 < r < R.
Applying the integrating factor 2µ cos
(√
(r2 − µ2)σ
)
/
√
r2 − µ2 and performing manipulations
similar to the solution of the Abel equation in [3, p. 46], we arrive at the integral equation
ϕ¯(t, s)− 1
pi
R∫
0
ϕ¯(%, s)
[
sin ((%− t)√σ )
%− t −
sin ((%+ t)
√
σ )
%+ t
]
d%
=
2q¯1(s)
piλ
t∫
0
q2(µ) cos
(√
(t2 − µ2) σ
)
µ√
t2 − µ2 dµ, 0 < t < R,
or
ψ¯(t, s)− 1
pi
R∫
0
ψ¯(%, s)
[
sin ((%− t)√σ )
%− t −
sin ((%+ t)
√
σ )
%+ t
]
d%
=
2
piλs
t∫
0
q2(µ) cos
(√
(t2 − µ2) σ
)
µ√
t2 − µ2 dµ, 0 < t < R,
DIFFERENTIAL EQUATIONS Vol. 37 No. 2 2001
SOLUTION OF THE HEAT EQUATION WITH MIXED BOUNDARY CONDITIONS 259
where ψ¯(%, s) := ϕ¯(%, s)/ (sq¯1(s)). There is no ready-to-use method other than the approach of [4]
for the solution of the last integral equation in the domain of L-transforms.
We represent the unknown function ψ¯(t, s) in the form of the series
ψ¯(t, s) = exp
(−R√σ ) ∞∑
n=0
ψn(t)
(√
s
)n−2
=
1
s
exp
(−R√σ )ψ0(t) + exp (−R√σ ) ∞∑
n=0
ψn+1(t)
(√
s
)n−1
,
(12)
which justifies the existence of the inverse Laplace transform L−1
[
ψ¯(t, s)
]
, since
L−1
[
s−1 exp
(−k√s )] = erfc (k/ (2√τ )) ,
L−1
[(√
s
)m−1
exp
(−k√s )] = (exp (−k2/(4τ)) /(2m√piτm+1 ))Hm (k/ (2√τ )) ,
where
erfc(x) =
2√
pi
∞∫
x
exp
(−t2) dt = 1− erf(x) = 1− 2√
pi
∞∑
k=0
(−1)kx2k+1
k!(2k + 1)
,
Hm(x) = m!
∑[m/2]
k=0 (−1)k(2x)m−2k/(k!(m− 2k)!) is the Hermite polynomial [5].
Following [4], we construct a recursion formula for the functions ψn(t) occurring in (12):
ψn(t) =
2
piλ
n∑
j=0
Anj(R)
t∫
0
(√
t2 − µ2
)j−1
q2(µ)µdµ+
1
pi
n∑
m=0
R∫
0
Cm(%, t)ψn−m(%)d%, (13)
ψ0(t) =
2
piλ
t∫
0
q2(µ)µ√
t2 − µ2dµ, Anj(R) =
1
n!
(
1√
a
)n(
n
j
)
Rn−j cos
(
jpi
2
)
,
Cm(%, t) =
1
m!
(
1√
a
)m
sin
(mpi
2
) [
(%− t)m−1 − (%+ t)m−1] ,
where
(
n
j
)
are the binomial coefficients.
Successively substituting the expressions (13) into (12) and the resulting relations into (11) and
finally into (7), we can find the solution of the problem in the domain of L-transforms:
T¯ (r, z, s) = q¯1(s) exp
(−R√σ ) R∫
0
ψ0(t)
∞∫
0
exp
(
−z
√
p2 + σ
)
sin
(
t
√
p2 + σ
) pJ0(pr)√
p2 + σ
dp dt
+ q¯1(s) exp
(−R√σ ) ∞∑
n=0
(√
s
)n+1 n+1∑
j=0
[
2
piλ
An+1 j(R)
R∫
0
t∫
0
(√
t2 − µ2
)j−1
q2(µ)µ
×
∞∫
0
exp
(
−z
√
p2 + σ
)
sin
(
t
√
p2 + σ
) pJ0(pr)√
p2 + σ
dp dµ dt
+
n+1∑
m=0
1
pi
R∫
0
R∫
0
Cm(%, t)ψn+1−m(%)
∞∫
0
exp
(
−z
√
p2 + σ
)
sin
(
t
√
p2 + σ
)
× pJ0(pr)√
p2 + σ
dp d% dt
]
,
(14)
DIFFERENTIAL EQUATIONS Vol. 37 No. 2 2001
260 MANDRIK
where r > 0, z > 0, and Re s > 0. We can readily see that this solution satisfies the mixed
boundary conditions (5) and (6) for z = 0.
In the special case q1(τ)q2(r) = const, the inverse transform T (r, z, τ) of the solution was found
in [1, 4]. For general q1(τ)q2(r), the inverse transform of the solution can be found with the use of
the inverse formula for the Laplace integral [5, p. 807 of the Russian translation; 6, p. 499]:
T (r, z, τ) =
1
2pii
α+i∞∫
α−i∞
exp(τs)T¯ (r, z, s)ds.
Note that the solution (14) satisfies the well-known limit relations as s → 0 (τ → ∞) and as
s→∞ (τ → 0) provided that the limits lims→0 [sq¯1(s)] and lims→∞ [sq¯1(s)] exist.
REFERENCES
1. Kozlov, V.P., Yurchuk, N.I., and Mandrik, P.A., I.F.Zh., 1998, vol. 71, no. 4, pp. 734–743.
2. Prudnikov, A.P., Brychkov, Yu.A., and Marichev, O.I., Integraly i ryady. Spetsial’nye funktsii (Integrals
and Series. Special Functions), Moscow, 1983.
3. Mikhlin, S.G., Lektsii po lineinym integral’nym uravneniyam (Lectures on Linear Integral Equations),
Moscow, 1959.
4. Kozlov, V.P., Yurchuk, N.I., and Mandrik, P.A., Vestn. Bel. univ. Ser. 1 , 1999, no. 2, pp. 37–42.
5. Abramowitz, M. and Stegun, I., Handbook of Mathematical Functions with Formulas, Graphs, and Math-
ematical Tables , Washington, 1964. Translated under the title Spravochnik po spetsial’nym funktsiyam,
Moscow, 1979.
6. Lykov, A.V., Teoriya teploprovodnosti (Theory of Thermal Conduction), Moscow, 1967.
DIFFERENTIAL EQUATIONS Vol. 37 No. 2 2001


Solution of the heat equation with mixed boundary conditions on the surface of an isotropic half-space

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Solution of the heat equation with mixed boundary conditions on the surface of an isotropic half-space

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