Existence and uniqueness solution of an inverse problems
for fractional evolution equations
Mahmoud M. El-borai & Osama L. Mostafa & Hoda A. Foad m−ml−elborai@yahoo.com & &hoda−rg@yahoo.com Faculty of Science, Alexandria University, Alexandria, Egypt
Khadug S. Sharnana & khadog−[email protected] Faculty of Science, Al-Marqab University, Al-khums, Libya
Abstract
In this paper we concerned with study an inverse problem in a Hilbert space H for fractional abstract differential equation of the form
dαu(t)
dtα =Au(t) +γ(t)u(t), (1)
with the initial condition
u(0) =u0 ∈ H (2)
and the overdetermination condition
(u(t), v) =w(t) (3)
where (.,.) is the inner product in H,γis a real an unknown function,w is a given real function,u0and v are given elements in H, 0 < α ≤ 1,u is unknown,and A is a linear
closed operator defined on a dense subset D(A) in H into H.
It assumed that A generates an analytic semigroup Q(t).This condition implies kQ(t)k ≤β
for allt ≥ 0,β is a positive constant.
Key words: Initial and overdetermination conditions; Solutions in Hilbert space; Frac-tional differential equations.
1
Introduction
classical mathematics(computational algebra,differential and integral equations,partial dif-ferential equations,functional analysis) can be classified as inverse.and they are among the most complicated ones ( since they are unstable and usually nonlinear).
At the same time,the inverse problems began to be studied and applied systematically in physics,geophysics,medicine,astronomy,and all other areas of knowledge where mathemati-cal methods are used.The reason is that solutions to inverse problems describe important properties of media under study,such as density and velocity of wave propagation,elasticity parameters,conductivity,dielectric permittivity and magnetic permeability,and properties and location of inhomogeneities in inaccessible areas,etc.
Several authors [1,2,7,8] studied the exist and unique solvability of the inverse prob-lem of determining a pair of functions{u, γ}.
In a Hilbert space H with the inner product h., .i and the norm k.k,they consider the following problem ( see [8] )
du(t)
dt +Au(t) =γ(t)g(t), (4)
with the initial condition
u(0) =u0 ∈ H (5)
and the overdetermination condition
(u(t), v) =w(t) (6)
where g and ω are given functions, u0 and v are given elements of H, A is not necessarily bounded and self-adjoint operator,{u(t), γ(t)}is unknown functions, there existsµ >0 such that
kAuk ≤ µ2 kuk ∀ u ∈ D(A), (7)
that is A is a positive-define operator.
Our aim in this paper we study the exist and unique solution of an inverse problem in a Hilbert space H for fractional abstract differential equation of the form
dαu(t)
dtα =Au(t) +γ(t)u(t), (8)
with the initial condition
u(0) =u0 ∈ H (9)
and the overdetermination condition
where (.,.) is the inner product in H, γ is a real an unknown function,w is a given real function, u0 and v are given elements in H, 0 < α ≤ 1,u is unknown,and A is a linear
closed operator defined on a dense subset D(A) in H into H.
2
Abstract inverse problem
definition 2.1 A pair of function {u, γ} is said to be a strictly solution of the inverse problem (1)-(3) if
u ∈ D(A),dαu(t)
dtα ∈ H (11)
for eacht ∈ (0, T],γ ∈ C(J) and the relations (1)-(3) are satisfied. In this case we say that the inverse problem (1)-(3) is solvable.
We need the following
conditions:-(A1) u0, v ∈ D(A),for allt ∈ J, J=[0,T].
(A2) |w(t)| ≥ c,t ∈ J,,and c is a positive constant,
(A3) d αw
dtα ∈ C(J),where C(J) the set of continuous functions defined on J.
remark 1
To prove that{u, γ}is a strictly solution of the inverse problem (1)-(3),we shall prove that the equivalence between the inverse problem (1)-(3) and the equation
γ(t) =h(t) +pγ(t) (12)
Where
h(t) = 1
w(t)
dαw(t)
dtα
and p is a linear operator defined on C(J) with
values:-(pγ)(t) =− 1
w(t)(Au(t), v) (13)
Theorem 2.1 Suppose that the conditions (A1-A3) are satisfied.Then the following asser-tions are
valid:-(i) If the inverse problem (1) is solvable, then so equation (12) has a solutionγ ∈ C(J),
(ii) If equation (12) has a solution γ ∈ C(J) and the compatibility condition
(u0, v) =w(0) (14)
holds, then the inverse problem (1)-(3) is solvable.
Proof: Assume that the inverse problem (1)-(3) is solvable.
Multiplying both sides of (1) by v scalarly in H, we obtain the relation
dα(u(t), v)
dtα = (Au(t), v) +γ(t)(u(t), v) (15)
dαw(t)
dtα = (Au(t), v) +γ(t)w(t) (16)
From (12) and (16) , one gets 1
w(t)
dαw(t)
dtα = −pγ(t) +γ(t)
This means that γ solves equation (12).
Now: We prove if equation (12) has solution γ ∈ C(J) and the compatibility condi-tion
(u0, v) =w(0)
holds,then the inverse problem (1) and (3) is solvable.
When inserting this function in (1), the resulting problem (1),(2) can be treated as a direct problem having a unique solution u .
Such that the problem (1) and (2) are equivalent to the integral equation
u(t) = u0+ Γ(1α)
Z t
0 (t − θ)
α −1 Au(θ) dθ+ 1
Γ(α)
Z t
0 (t − θ)
α −1 γ(θ)u(θ)dθ
Using theorem 1.2 we have the solution u is given by
u(t) =
Z ∞
0 ξα(θ)Q(t αθ)u
0 dθ
+ α Z t
0
Z ∞
0 θ(t−s)
α−1ξ
Let us prove now that u satisfies the overdetermination condition (3). In this case u andγ
are known, consequently (16) will represent an identity
γ(t)w(t) = dαw(t)
dtα −(Au(t), v), (18)
subtracting equation (16) from (18) , one gets
dαw(t)
dtα =
dα(u(t), v)
dtα
Applying the fractional integral of orderαand taking into account the compatibility condi-tion (14),we find out that u satisfies the overdeterminacondi-tion condicondi-tion (3) and that the pair
{u, γ} is a strictly solution of the inverse problem (1)-(3).
This complete the proof of the theorem.
remark 2
We shall suppose that adjoint operatorA∗ of the closed operator A exists and that if
dαφ(t)
dtα = ψ(t),
then
φ(t) = φ(0) + 1 Γ(α)
Z t
0 (t − s)
α −1 ψ(s) ds
where Γ(α) is the gamma function, 0 < α < 1 , φ , ψ are abstract function of t with values in H and the integral is taken in Bochner,s sense (see [3]).
Theorem 2.2 Let the conditions (A1-A3) and the compatibility condition (3) hold,then
there exists a unique strictly solution of the inverse problem (1)-(3).
Proof: Using (12),(13) and (17), one gets
Since
γ(t) =h(t) +pγ(t),
(pγ)(t) =− 1
w(t)(Au(t), v) =− 1
w(t)(u(t), A
and
u(t) =
Z ∞
0 ξα(θ)Q(t αθ)u
0 dθ
+ α Z t
0
Z ∞
0 θ(t − s)
α− 1 ξ
α(θ) Q((t−s)αθ) γ(s)u(s) dθ ds
then
γ(t) = h(t)− 1
w(t)[(
Z ∞
0 ξα(θ) Q(t
αθ)u
0 dθ
+ α Z t
0
Z ∞
0 θ(t − s)
α− 1 ξ
α(θ) Q((t−s)αθ) γ(s)u(s) dθ ds, A∗v)]
= h(t)− 1
w(t)
Z ∞
0 (ξα(θ) Q(t αθ)u
0, A∗v) dθ
− α
w(t)
Z t
0
Z ∞
0 (θ(t − s)
α −1 ξ
α(θ) Q((t−s)αθ) u(s), A∗v)γ(s)dθ
then
γ(t) =ψ(t)−
Z t
0 (t − s)
α −1 K(t, s) γ(s) ds (19)
where
ψ(t) = h(t)− 1
w(t)
Z ∞
0 ξα(θ) (Q(t αθ)u
0, A∗v) dθ
and
K(t, s) = α
w(t)
Z ∞
0
θ ξα(θ) (Q((t−s)αθ) u(s) , A∗v)dθ
According to conditions (A1-A2)and (A3),functions w−1(t) and h(t) are continuous on
J.
We shall prove now that the functionψ is continuous on J . In fact ,
|
Z ∞
0 ξα(θ) (
Z t2
t1
d dtQ(t
αθ)u
0 dt, A∗v) dθ|
=|
Z ∞
0 ξα(θ) (
Z t2
t1
α tα−1 θ Q(tαθ) A u0 dt, A∗v) dθ|
We shall prove that equation has a unique solutionγ ∈ C(J).
Using the method of successive approximations,we set
γn+1(t) =ψ(t)−
Z t
0 (t − s)
α −1 K(t, s) γ
n(s) ds, γ0(t) = 0,
for allt ∈ J, n= 1,2, .....
It is easy to see that
|γ2(t)−γ1(t)| ≤ M tα,
where
M = β
u0
supku(t)k kA∗vk
By induction,one gets
|γn+1(t)−γn(t)| ≤ M
n tnα (Γ(α))n
Γ(nα+ 1) (20)
It can be prove that all the functionsγn+1(t)−γn(t) are continuous on J, (see [4], [6]).
Using (20),we see that the series P∞k=1[γk(t) −γk−1(t)] uniformly converges on J to a
continuous functionγ(t),which represents the unique solution of (19).
According to theorem (1) this confirms that the inverse problem (1)-(3) is solvable.
To prove the uniqueness of this solution,we assume to the contrary that there were two different solutions{u1, γ1} and {u2, γ2} of the inverse problem (1)-(3).
We claim that in this case γ1 6= γ2 for all points of J. In fact if γ1 = γ2 on J then applying the uniqueness theorem to the corresponding direct problem (1)-(2), we would haveu1= u2.
Since both pairs satisfy identity (8),the functions γ1 and γ2 give two different solutions
of equation (19).But this contradicts the uniqueness of solutions to the Volterra integral equation (19).
This complete the proof of the theorem.
Example of an inverse problems
Let a functionu(x, t) satisfy the equation
∂ u(x, t)
∂ t −
∂2u(x, t)
and the boundary conditions
u(0, t) =g1(t) = 0, u(L, t) =g2(t) = 0, t >0 (22)
It is required to determine the value of u(x,t) at the initial instant t=0
u(x,0) =γ(x) (23)
given the values of u(x,t) at a fixed instant of timet=T >0
u(x, T) =f(x), 0≤ x≤ L (24)
This problem is inverse to the problem of finding a function u(x,t) satisfying (21)-(23),prove that the functionγ(x) is given.
solution :
Set
v(x, t) =u(x, t)−µ(x, t) where
µ(x, t) = x
Lg2(t) + (1− x L)g1(t)
then
v(x, t) =u(x, t)− x
L(g2(t)−g1(t))−g1(t) (25)
We notice that
v(0, t) =g1(t)−g1(t) = 0,
v(L, t) =g2(t)−g2(t) = 0
Now
∂ v ∂ t =
∂ u ∂ t −
x L(g
0 2(t)−g
0
1(t))−g 0
1(t) (26)
and
∂2v
∂ x2 =
∂2u
∂ x2 (27)
subtraction (27) from (26)
∂ v ∂ t −
∂2v
∂ x2 =
∂ u ∂ t −
∂2u
∂ x2 −
x L(g
0 2(t)−g
0
1(t))−g 0 1(t)
= h(x, t)− x
L(g
0 2(t)−g
0
1(t))−g 0 1(t)
= h∗(x, t). (28)
and
v(x, T) = u(x, T)− x
L(g2(T)−g1(T))−g1(T)
= f(x)− x
L(g2(T)−g1(T))−g1(T)
= f∗(x). (30)
The solution to the direct problem (28)-(29) is given by the formula
v(x, t) =
∞ X
n=1
An(t) sin (nπ x
L ) (31)
where{An} are the Fourier coefficient of A(t):
An(t) = L2
Z L
0 v(x, t) sin(
nπ x L )dx.
Now
dAn(t)
dt = 2 L Z L 0 ∂ v ∂ t sin (
nπ x L )dx
= 2
L Z L
0 (
∂2v
∂ x2 +h∗(x, t)) sin(
nπ x L )dx
= 2
L[ Z L
0
∂2v ∂ x2 sin(
nπ x L )dx+
Z L
0 h
∗(x, t) sin(nπ x L )dx]
= 2
L[ Z L
0
∂2v
∂ x2 sin(
nπ x
L )dx+rn(x, t)] (32)
where
rn(x, t) =
Z L
0 h
∗(x, t) sin(nπ x L )dx
Now we find
I =
Z L
0
∂2v ∂ x2 sin(
nπ x L )dx
Let
u= sin(nπ x
L ) , dv= ∂2v
∂ x2dx
du= nπ
L cos( nπ x
L ) , v= ∂ v ∂ x
I1 =
2
L{[sin( nπ x
L ) ∂ v ∂ x|
L 0]−[
nπ L Z L 0 cos( nπ x L ) ∂ v ∂ xdx]}
= −2nπ
L2 Z L 0 cos( nπ x L ) ∂ v
∂ x. (33)
Let
u= cos(nπ x
du=−nπ
L sin( nπ x
L ) , v=v
I1 = −2nπ
L2 [cos(
nπ x L )v−v|
L 0 + nπ L Z L 0 sin( nπ x L v dx]
= −2nπ
L2 {[cos(nπ)v−v] +
nπ L
Z L
0 sin (
nπ x L )v dx]
=
½0 n even
4nπ
L2 v n odd +
nπ L
Z L
0 sin (
nπ x L )v dx
= −2
L Z L
0 (
nπ L )
2sin(nπ x
L )v dx , n even
i.e
dAn(t)
dt + ( nπ
L )
2A
n(t) =rn(t), (L.D.E) (34)
Such that
An(0) = L2
Z L
0 γ(x) sin(
nπ x L )dx
= βn (unknown)
To solve equation (34),such that the equation is linear,then
µ(t) = exp
R
(nπ
L)2dt= exp(nπL)2t
Multiplying both sides of (34) by exp(nπL)2
tand integral equation,we obtain the relation
exp (nπ
L )
2t dAn(t)
dt + exp ( nπ
L )
2t(nπ
L )
2 A
n(t) = exp (nπL )2t rn(t)
d
dt[An(t) exp ( nπ
L )
2t] = exp (nπ
L )
2t r n(t)
An(t) exp (nπL )2t|t0=
Z t
0 exp (
nπ L )
2s r n(s)ds
An(t) exp (nπL )2t−An(0) =
Z t
0 exp (
nπ L )
2s r n(s)ds
An(t) = exp−(
nπ L )
2t A
n(0) + exp−(
nπ L )
2t
Z t
0 exp (
nπ L )
2(s−t) r n(s)ds
= exp−(nπ
L )
2tβ
n+ exp−(nπL )2t
Z t
0 exp (
nπ L )
2(s−t) r n(s)ds
Now setting t=T in (31),we get
v(x, T) =
∞ X
n=1
f∗(x) =
∞ X
n=1
An(T) sin(nπ xL ) (35)
To determine constantsAn,we multiply both sides of the equation (35) bysin(mπ x L ) and
integrate from x=0 to x=L,hence
Z L
0 f
∗(x) sin(mπ x L )dx=
Z L
0 (
∞ X
n=1
An(T) sin(nπ x
L ) sin( mπ x
L ))dx
In (35) we interchange the order of integration and summation to
Z L
0 f
∗(x) sin(mπ x L )dx=
∞ X
n=1
(
Z L
0 An(T) sin(
nπ x L ) sin(
mπ x L )dx)
If m=n
Z L
0 f
∗(x) sin(nπ x
L )dx=An(T) L
2 i.e
An(T) = 2
L Z L
0 f
∗(x) sin(nπ x L )dx
Such that
v(x, t) =
∞ X
n=1
An(t) sin(nπ x
L )
=
∞ X
n=1
(exp−(nπ
L )
2t β n
+ exp−(nπ
L )
2tZ t 0 exp (
nπ L )
2(s−t) r
n(s)ds) sin(
nπ x
L ) (36)
Setting t=T,we get
v(x, T) =
∞ X
n=1
An(T) sin(nπ x
L )
=
∞ X
n=1
(exp−(nπ
L )
2T β n
+ exp−(nπ
L )
2T
Z t
0 exp (
nπ L )
2(s−T) r
n(s)ds) sin(
nπ x L )
then
f∗(x) =
∞ X
n=1
(exp−(nπ
L )
2T β n
+ exp−(nπ
L )
2TZ t 0 exp (
nπ L )
2(s−T) r
n(s)ds) sin(
nπ x
To determine the constantsβn,we multiply both sides of the equation (37) by sin(nπ x L ) and
integrate from x=0 to x=L,hence
Z L
0 f
∗(x) sin(nπ x L )dx=
L
2 exp−(
nπ L )
2T β
n+
Z L
0 Msin(
nπ x L )dx
where
M =
∞ X
n=1
(exp−(nπ
L )
2T sin(nπ x
L ) Z t
0 exp (
nπ L )
2(s−T) r
n(s)ds)
then
βn = L2 exp (nπL )2T(
Z L
0 f
∗(x) sin(nπ x L )dx−
Z L
0 Msin(
nπ x L )dx)
= An(T) exp (nπ
L )
2T − 2
Lexp ( nπ
L )
2TZ L 0 Msin(
nπ x L )dx
i.e
An(0) =An(T) exp (nπL )2T −L2 exp (nπL )2T
Z L
0 Msin(
nπ x L )dx
then the solution of (21)-(23) is
u(x, t) =
∞ X
n=1
An(t) sin(nπ x
L ) .
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