Determination of an unknown source term and the
temperature distribution for the linear heat equation
involving fractional derivative in time
Mokhtar Kirane, Salman Amin Malik
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Mokhtar Kirane, Salman Amin Malik. Determination of an unknown source term and the
temperature distribution for the linear heat equation involving fractional derivative in time.
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Determination of an unknown source term and the temperature distribution for
the linear heat equation involving fractional derivative in time
Mokhtar Kirane, Salman A. Malik
Laboratoire de Mathématiques, Image et Applications, Université de La Rochelle, Avenue M. Crépeau, 17042 La Rochelle Cedex, France.
Abstract
We consider the inverse problem of finding the temperature distribution and the heat source whenever the temperatures
at the initial time and the final time are given. The problem considered is one dimensional and the unknown heat source
is supposed to be space dependent only. The existence and uniqueness results are proved.
Key words: Inverse problem, fractional derivative, heat equation, integral equations, biorthogonal system of
functions, Fourier series
2010 MSC: 80A23, 65N21, 26A33, 45J05, 34K37, 42A16.
1. Introduction
In this paper, we are concerned with the problem of finding u(x, t) (the temperature distribution) and f (x) (the
source term) in the domain QT = (0, 1) × (0, T ) for the following system
Dα0+ (u(x, t) − u(x, 0)) − u xx (x, t) =
u(x, 0) = φ(x) u(x, T ) =
u(1, t) = 0, u x (0, t) =
f (x),
(x, t) ∈ QT ,
(1)
ψ(x),
u x (1, t),
x ∈ [0, 1],
t ∈ [0, T ],
(2)
(3)
where Dα0+ stands for the Riemann-Liouville fractional derivative of order 0 < α < 1, φ(x) and ψ(x) are the initial and
final temperatures respectively. Our choice of the term Dα0+ (u(x, .) − u(x, 0))(t) rather then the usual term Dα0+ u(x, .)(t)
is to avoid not only the singularity at zero but also to impose meaningful initial condition.
The problem of determination of temperature at interior points of a region when the initial and boundary conditions
along with heat source term are specified are known as direct heat conduction problems. In many physical problems,
determination of coefficients or right hand side (the source term, in case of the heat equation) in a differential equation
from some available information is required; these problems are known as inverse problems. These kind of problems
are ill posed in the sense of Hadamard. A number of articles address the solvability problem of the inverse problems
(see [1], [2], [3], [4], [5], [6], [7], [8] and references therein).
From last few decades fractional calculus grabbed great attention of not only mathematicians and engineers but
also of many scientists from all fields. Indeed fractional calculus tools have numerous applications in nanotechnology,
control theory, viscoplasticity flow, biology, signal and image processing etc, see the latest monographs, [9], [10], [11],
[12], [13] articles [14], [15] and reference therein. The mathematical analysis of initial and boundary value problems
(linear or nonlinear) of fractional differential equations has been studied extensively by many authors, we refer to [16],
[17] , [18], [19], [20] and references therein.
In the next section we recall some definitions and notations from fractional calculus. Section 3 is devoted to our
main results; we provide the expressions for the temperature distribution and the source term for the problem (1)-(3).
Furthermore, we proved the existence and uniqueness of the inverse problem.
Email addresses: [email protected] (Mokhtar Kirane), [email protected] (Salman A. Malik)
Preprint submitted to Elsevier
December 21, 2010
2. Preliminaries and notations
In this section, we recall basic definitions, notations from fractional calculus (see [11], [12], [13]).
For an integrable function f : R+ → R the left sided Riemann-Liouville fractional integral of order 0 < α < 1 is
defined by
∫t
1
f (τ)
J0α+ f (t) :=
dτ, t > 0,
(4)
Γ(α)
(t − τ)1−α
0
where Γ(α) is the Euler Gamma function. The integral (4) can be written as a convolution
J0α+ f (t) = (ϕα ⋆ f )(t),
{
where
ϕα :=
tα−1 /Γ(α),
0
(5)
t > 0,
t ≤ 0.
(6)
The Riemann-Liouville fractional derivative of order 0 < α < 1 is defined by
Dα0+
d
d
1
f (t) := J01−α
f (t) =
+
dt
Γ(1 − α) dt
∫t
0
f (τ)
dτ.
(t − τ)α
(7)
In particular, when α = 0 and α = 1 then
f (t) = f (t) and
f (t) = f ′ (t) respectively; notice that the RiemannLiouville fractional derivative of a constant is not equal to zero.
The Laplace transform of the Riemann-Liouville integral of order 0 < α < 1 of a function with at most exponentially growth is
L{J0α+ f (t) : s} = L{ f (t) : s}/sα .
D00+
Also, for 0 < α < 1 we have
D10+
(
)
J0α+ Dα0+ f (t) − f (0) = f (t) − f (0).
(8)
The Mittag-Leffler function plays an important role in the theory of fractional differential equations, for any z ∈ C
the Mittag-Leffler function with parameter ξ is
Eξ (z) =
+∞
∑
k=0
zk
Γ(ξk + 1)
(Re(ξ) > 0).
(9)
In particular, E1 (z) = ez
The Mittag-Leffler function of two parameters Eξ,β (z) which is a generalization of (9) is defined by
Eξ,β (z) =
+∞
∑
k=0
zk
Γ(ξk + β)
(z, β ∈ C; Re(ξ) > 0).
(10)
3. Main Results
Let eξ (t, µ) := Eξ (−µtξ ) where Eξ (t) is the Mittag-Leffler function with one parameter ξ as defined in (9) and µ is
a positive real number. Let us state and prove the
Lemma 3.1. The solution of equation
v(t) + µJ0α+ v(t) = g(t)
(11)
for µ ∈ R satisfies the integral equation
∫t
v(t) =
g′ (t − τ)eα (τ, µ)dτ + g(0)eα (t, µ),
0
2
(12)
Proof. Taking the Laplace transform of both sides of the equation (11), it follows
L{v(t) : s} =
sα L{g(t) : s}
.
sα + µ
(13)
There are different ways for getting the solution from the equation (13) by the inverse Laplace transform.
Writing the equation (13) as
(
)
sα−1
sα−1
sL{g(t) : s} − g(0) + α
g(0);
L{v(t); s} = α
s +µ
s +µ
and using the inverse Laplace transform, we obtain
∫t
v(t) =
g′ (t − τ)eα (τ, µ)dτ + g(0)eα (t, µ),
0
which is (12).
3.1. Solution of the inverse problem
The key factor in determination of the temperature distribution and the unknown source term for the system (1)-(3)
is the specially chosen basis for the space L2 (0, 1) which is
{2(1 − x),
{4(1 − x) cos 2πnx}∞
n=1 ,
{4 sin 2πnx}∞
n=1 }.
(14)
In [21], it is proved that the set of functions in (14) forms a Riesz basis for the space L2 (0, 1), hence the set is
closed, minimal and spans the space L2 (0, 1).
The set of functions (14) is not orthogonal and in [22] a biorthogonal set of functions given by
{1,
{cos 2πnx}∞
n=1 ,
{x sin 2πnx}∞
n=1 },
(15)
is constructed.
The solution of the inverse problem for the linear system (1)-(3) can be written in the form
∞
∑
u(x, t) = 2u0 (t)(1 − x) +
u1n (t)4(1 − x) cos 2πnx +
n=1
f (x) = 2 f0 (1 − x) +
∞
∑
∞
∑
u2n (t)4 sin 2πnx
(16)
n=1
f1n 4(1 − x) cos 2πnx +
n=1
∞
∑
f2n 4 sin 2πnx
(17)
n=1
where u0 (t), u1n (t), u2n (t), f0 , f1n and f2n are to be determined. By plugging the expressions of u(x, t) and f (x) from
(16) and (17) into the equation (1), the following system of fractional differential equations is obtained
Dα0+ (u2n (t) − u2n (0)) − 4πnu1n (t) + 4π2 n2 u2n (t) =
Dα0+ (u1n (t)
− u1n (0)) + 4π n u1n (t) =
Dα0+ (u0 (t) − u0 (0)) =
2 2
f2n ,
(18)
f1n ,
f0 .
(19)
(20)
Since J0α+ 1 = tα /Γ(1 + α) and for 0 < α < 1, J0α+ Dα0+ (u0 (t) − u0 (0)) = u0 (t) − u0 (0) then the solution of equation (20) is
u0 (t) =
f0 t α
+ u0 (0).
Γ(1 + α)
Setting λ := 4π2 n2 , F(t) := f1n tα /Γ(1 + α) + u1n (0) then equation (19) can be written as
u1n (t) + λJ0α+ u1n (t) = F(t),
3
(21)
and using lemma (3.1), the solution of the equation (19) is given by
f1n α
u1n (t) =
Γ(1 + α)
∫t
(t − τ)α−1 eα (τ, λ)dτ + u1n (0)eα (t, λ).
(22)
0
Noticing that equation (18) can be written as
u2n (t) + λJ0α+ u2n (t) = h(t)
(23)
(
)
where h(t) := J0α+ f2n + 4πnu1n (t) + u2n (0), so its solution by virtue of lemma (3.1) is
∫t
u2n (t) =
h′ (t − τ)eα (τ, λ)dτ + h(0)eα (t, λ).
(24)
0
As h(0) = u2n (0) and
′
h (t) =
=
[
]
d α
J ( f2n + 4πnu1n (t)) − u2n (0)
dt 0+
(
)
α
D0+ f2n + 4πnu1n (t) .
From equation (22), we have
Dα0+ u1n (t) =
f1n α
Dα tα−1 ⋆ eα (t, λ) + u1n (0)Dα0+ eα (t, λ),
Γ(1 + α) 0+
since Dα0+ tα−1 = 0 and Dα0+ eα (t, λ) = eα,1−α (t, λ), where eα,β (t, λ) := tβ−1 Eα,β (−λtα ); consequently the equation (24)
becomes
∫t
u2n (t) =
s(t − τ)eα (τ, λ)dτ + u2n (0)eα (t, λ).
(25)
0
where we have set
t−α
.
Γ(1 − α)
and f0 . In order to get these unknowns, we use the initial
s(t) := 4πnu1n (0)eα,1−α (t, λ) + f2n
The unknown constants are u2n (0), u1n (0), u0 (0), f2n , f1n
and final temperatures as given in (2)
2u0 (0)(1 − x) +
+
∞
∑
n=1
∞
∑
u1n (0)4(1 − x) cos 2πnx +
+
∞
∑
n=1
∞
∑
u2n (0)4 sin 2πnx = 2φ0 (1 − x)
n=1
φ1n 4(1 − x) cos 2πnx +
n=1
2u0 (T )(1 − x) +
∞
∑
∞
∑
φ2n 4 sin 2πnx,
n=1
u1n (T )4(1 − x) cos 2πnx +
∞
∑
u2n (T )4 sin 2πnx = 2ψ0 (1 − x)
n=1
ψ1n 4(1 − x) cos 2πnx +
n=1
∞
∑
ψ2n 4 sin 2πnx.
n=1
By identification we get
u2n (0) = φ2n ,
u1n (0) = φ1n ,
4
u0 (0) = φ0 ,
(26)
u2n (T ) = ψ2n ,
u1n (T ) = ψ1n ,
u0 (T ) = ψ0 ,
(27)
where {φ0 ,φ1n , φ2n } and {ψ0 ,ψ1n , ψ2n } are the coefficients of the series expansion in the basis (14)of the functions φ(x)
and ψ(x), respectively. In terms of the biorthogonal basis (15) these are
∫1
φ0 =
∫1
φ(x)dx,
φ1n =
∫1
φ(x) cos 2πndx,
φ2n =
φ(x)x sin 2πndx,
0
0
0
∫1
∫1
∫1
ψ0 =
ψ(x)dx,
φ1n =
0
ψ(x) cos 2πndx,
ψ2n =
0
ψ(x)x sin 2πndx.
0
By virtue of (21), (22), (25), and conditions (27) we have the expressions of unknowns f0 , f1n and f2n as
[
]
ψ 0 − φ0
f0 = Γ(1 + α)
,
Tα
[
]
ψ1n − φ1n eα (T, λ)
,
f1n = Γ(1 + α) T
∫
α (T − τ)α−1 eα (τ, λ)dτ
(28)
(29)
0
[ ψ2n − φ2n eα (T, λ) −
f2n
=
∫T
φ1n eα,1−α (T − τ, λ)eα (τ, λ)dτ ]
0
Γ(1 − α)
∫T
(T −
.
(30)
τ)−α eα (τ, λ)dτ
0
The unknown source term and the temperature distribution for the problem (1)-(3) are given by the series (17) and
(16), where the unknowns u0 (0), u1n (0), u2n (0) are calculated from (26), while f0 , f1n , f2n are given by (28)-(30). In
the next subsection we will show the uniqueness of the solution.
3.2. Uniqueness of the solution
Suppose {u1 (x, t), f1 (x)}, {u2 (x, t), f2 (x)} are two solution sets of the inverse problem for the system (1)-(3). Define
u(x, t) = u1 − u2 and f (x) = f1 − f2 then the function u(x, t) satisfies the following equation and boundary conditions
Dα0+ u(x, t) − u xx (x, t) =
u(x, 0) = 0 u(x, T ) =
u(1, t) = 0, u x (0, t) =
Using the biorthogonal basis (15), we have
(x, t) ∈ QT ,
x ∈ [0, 1],
t ∈ [0, T ].
f (x),
0,
u x (1, t),
(31)
(32)
(33)
∫1
u0 (t) =
u(x, t)dx,
(34)
0
∫1
u1n (t) =
u(x, t) cos 2πnxdx,
(n = 1, 2, ...),
(35)
u(x, t)x sin 2πnxdx
(n = 1, 2, ...).
(36)
0
and
∫1
u2n (t) =
0
5
Alike,
∫1
f0 =
f (x)dx,
(37)
0
∫1
f 1n =
f (x) cos 2πnxdx,
(n = 1, 2, ...),
(38)
f (x)x sin 2πnxdx
(n = 1, 2, ...).
(39)
0
and
∫1
f 2n =
0
Taking the fractional derivative
we get
Dα0+
under the integral sign of equation (34), using (31), (37) and integration by parts,
Dα0+ u0 (t) = f 0 ,
(40)
with the boundary conditions
u0 (0) = 0,
The solution of the problem (40) is
u0 (t) =
u0 (T ) = 0.
(41)
tα
f ;
Γ(1 + α) 0
taking into account the boundary conditions (41), we get
f 0 = 0,
⇒
u0 (t) = 0.
(42)
Taking the fractional derivative Dα0+ under the integral sign of equation (35) then by using (31), (32), (38) and
integration by parts, we obtain
Dα0+ u1n (t) + λu1n (t) = f 1n ;
(43)
the associated boundary conditions are
u1n (0) = 0,
u1n (T ) = 0.
(44)
The solution of the problem (43) is
u1n (t) = eα,α (t, λ) ⋆ f 1n .
Using (44) we have
f 1n = 0,
⇒
u1n (t) = 0,
(n = 1, 2, ...).
In the same manner from (36), we obtain the equation
C
Dα0+ u2n (t) + 4π2 n2 u2n (t) = 4πnu1n (t) + f 2n ;
(45)
the associated boundary conditions are
u2n (0) = 0,
u2n (T ) = 0.
(46)
Taking account of u1n (t) = 0 then the solution of the equation (45) satisfies
u2n (t) = eα,α (t, λ) ⋆ f 2n ;
using (46) we have
f 2n = 0,
⇒
u2n (t) = 0,
(n = 1, 2, ...).
For every t ∈ [0, T ], the functions u(x, t) and f (x) are orthogonal to the system of functions given in (15), which
form a basis of the space L2 (0, 1), consequently,
u(x, t) = 0,
f (x) = 0.
6
3.3. Existence of the solution
Suppose φ(x) ∈ C 4 [0, 1] be such that φ′ (0) = φ′ (1), φ′′ (0) = φ′′ (1). As φ1n is the coefficient of the cosine Fourier
expansion of the function φ(x) with respect to basis (15), from (26) the expression for u1n (0) can be written as
∫1
φ(x) cos 2πnxdx
u1n (0) =
0
which integrated by parts four times gives
u1n (0)
∫1
=
1
16π4 n4
=
1
φ4 ,
16π4 n4 1n
φ4 (x) cos 2πnxdx,
0
where φ41n is the coefficient of the cosine Fourier expansion of the function φ4 (x) with respect to the basis (15).
Alike, we obtain
1
u2n (0) =
φ4 ,
(47)
16π4 n4 2n
where φ42n is the coefficient of sine Fourier transform of the function φ4 with respect to the basis (15).
The set of functions {φ41n , φ42n n = 1, 2, ...} is bounded as by supposition we have φ(x) ∈ C 4 [0, 1], and due to the
inequality
∞
∑
(φ4in )2 ≤ C ∥ φ4 (x) ∥2L2 (0,1) , i = 1, 2
n=1
which is true by the Bessel inequality of the trigonometric series. Similarly, we have for ψ(x) ∈ C 4 [0, 1] the set of
functions {ψ41n , ψ42n n = 1, 2, ...} is bounded and
∞
∑
(ψ4in )2 ≤ C ∥ ψ4 (x) ∥2L2 (0,1) ,
i = 1, 2.
n=1
The Mittag-Leffler functions eα (t; µ) and eα,β (t; µ) for µ > 0 and 0 < α ≤ 1, 0 < α ≤ β ≤ 1 respectively, are
completely monotone functions (see [23] page 268). Furthermore, we have
Eα,β (µtα ) ≤ M,
t ∈ [a, b],
(48)
where [a, b] is a finite interval with a ≥ 0, and
∫t
(t − τ)α−1 Eα,β (λtα )dτ < ∞,
0
(see [24] page 9).
The solution u(x, t) becomes
(
u(x, t) =
+
)
2(1 − x) f0 t /Γ(1 + α) + φ0
∞ (
∑
n=1
+
α
)
φ41n
f1n α α−1
t
⋆ eα (t, λ) +
e
(t,
λ)
4(1 − x) cos 2πnx
α
Γ(1 + α)
16π4 n4
)
∞ (
∑
φ42n
φ41n
t−α
⋆
e
(t,
λ)
+
e
(t,
λ)
+
f
e
(t,
λ)
4 sin 2πnx
α
α,1−α
2n
α
Γ(1 − α)
4π3 n3
16π4 n4
n=1
7
(49)
Setting M = max{M1 , M2 } where
|eα | ≤ M1 ,
|eα,1−α | ≤ M2
t ∈ [0, T ], T is finite.
Taking into account the values of f0 , f1n , f2n from (28)-(30) and the properties of Mittag-Leffler function, the
series (49) is bounded above by
(
)
tα
2(1 − x) |ψ0 | + (1 − α )|φ0 |
T
}
)
({
∞
4
4
∑ ψ − φ M tα
φ41n
1n
1n
M
4(1 − x) cos 2πnx
+
+
T α 16π4 n4
16π4 n4
n=1
{
}
)
∞ ( 4
∑
φ1n M
φ41n M
φ42n M
t1−α ψ42n − φ42n M
+
+
−
+
4 sin 2πnx.
(50)
4π3 n3 T 1−α
16π4 n4
16π4 n4 16π4 n4
n=1
For every (x, t) ∈ QT the series (50) is bounded above by the uniformly convergent series
) ∑
)
∞ ( 4
∞ (
∑
|ψ1n | + |φ41n |2M
(4πn − 1)|φ41n |M |ψ42n | − |φ42n |2M
+
+
.
16π4 n4
16π4 n4
16π4 n4
n=1
n=1
(51)
By the Weierstrass M-test, the series (49) is uniformly convergent in the domain QT . Hence, the solution u(x, t) is
continuous in the domain QT .
The uniformly convergent series doesn’t provide any information about the convergence of the series obtained
from its term by term differentiation. Now take the Dα0+ derivative from the series expression of u(x, t) given by (49)
∞
∑
Dα0+ Un (x, t) =
n=1
+
∞ (
∑
φ41n
+
eα,1−2α (t, λ) +
4π3 n3
n=1
(
2(1 − x) f0 + φ0
)
t−α
Γ(1 − α)
)
(
∞
∑ φ4
1n
e
(t,
λ)
4(1 − x) cos 2πnx
α,1−α
16π4 n4
n=1
)
φ42n
t−α
f2n
⋆ eα,1−α (t, λ) +
eα,1−α (t, λ) 4 sin 2πnx,
Γ(1 − α)
16π4 n4
(52)
we have used Dα0+ tα−1 = 0, Dα0+ eα (t, λ) = eα,1−α (t, λ), Dα0+ eα,1−α (t, λ) = eα,1−2α (t, λ) and Un (x, t) is defined by the right
hand side of the series (16).
The series (52) is bounded above by the uniformly convergent series
)
∞ ( 4
∑
|ψ2n | + (4πn + 1)|φ41n |M
,
16π4 n4
n=1
hence it converges uniformly, so Dα0+ un (x, t) =
∞
∑
n=1
Dα0+ Un (x, t).
Take the x-derivative of the series expression of u(x, t) from (49), we obtain
∞
∑
(
) ∑
∞ (
α
Un,x (x, t) := −2 f0 t /Γ(1 + α) + φ0 −
n=1
+
)
φ41n
f1n α α−1
t
⋆ eα (t, λ) +
eα (t, λ)
Γ(1 + α)
16π4 n4
n=1
(
) ∑
∞ (
φ41n
8πn sin 2πnx + 4 cos 2πnx −
eα,1−α (t, λ)
4π3 n3
n=1
)
φ42n
t−α
f2n
⋆ eα (t, λ) +
e
(t,
λ)
8πn sin 2πnx.
α
Γ(1 − α)
16π4 n4
8
(53)
∞
∑
Un,xx (x, t) :=
n=1
+
)
φ41n
f1n α α−1
t
⋆ eα (t, λ) +
e
(t,
λ)
α
Γ(1 + α)
16π4 n4
n=1
(
) ∑
∞ (
φ41n
2 2
2 2
16π n sin 2πnx − 16π n (1 − x) cos 2πnx −
eα,1−α (t, λ)
4π3 n3
n=1
)
φ42n
t−α
eα (t, λ) 16π2 n2 sin 2πnx.
f2n
⋆ eα (t, λ) +
Γ(1 − α)
16π4 n4
∞ (
∑
(54)
The series (53) and (54) are bounded above by the uniformly convergent series
)
∞ (
∑
(1 + 2πn)|ψ41n | + 2πn|ψ42n | + (4π2 n2 + 6πn + 2)M|φ41n | + (1 + M)2πn|φ42n |
,
4π4 n4
n=1
)
∞ ( 4
∑
|ψ1n | + 2|ψ42n | + (1 + T M)M|φ41n | + M|φ42n |
,
π2 n2
n=1
respectively, then
∞
∑
n=1
Un,x (x, t) and
∞
∑
n=1
Un,xx (x, t) converges uniformly; consequently
u xx (t, x) =
∞
∑
Un,xx (x, t),
n=1
is uniformly convergent.
Hence the term by term differentiation of the series (49) with respect to time t and x is valid. Similar we can show
that f (x) obtained by series (17) is continuous.
Remark 3.2. The above analysis remains valid if the boundary conditions given by (3) are replaced by
u(0, t) = u(1, t),
u x (0, t)
=
u(1, t) = 0,
u x (0, t)
or
= 0,
u x (1, t),
t ∈ [0, T ],
t ∈ [0, T ].
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