1. Introduction
Consider the following non-homogeneous fractional evolution boundary value problem:
where
,
,
,
denotes the (time) fractional derivative in the sense of Caputo, with
, and
;
, is the (space) fractional derivative, in the sense of Riemann–Liouville, with
; and
denotes the length of the time interval.
Let us fix the framework: When the initial data
and the source terms
are given, the direct problem consists of looking for the unknown function
u in an adequate functional space. On the other hand, when the initial data
and only the spatial part
p are given, the inverse problem consists of looking for the unknown functions
via, for example, the supplementary data:
where
is a given function of
.
Notice that the fractional operator
is unbounded, non-self-adjoint with compact resolvent admitting a countable complete basis of eigenfunctions, in
[
1]. This will allow us to look for classical solutions via the biorthogonal representation with respect to
x. Indeed, the uniform convergence of the resulting series in
(and of their derivatives) and the uniqueness of the corresponding limits, with respect to the norms
and
, will lead to the existence of classical solutions of the direct problem. For the inverse problems, we will make use of the Schauder fixed point theorem in an adequate convex and bounded subset.
Before we present and prove our results, let us dwell on the existing literature. An analytical and numerical study of inverse problems for fractional diffusion equations has undergone intensive development over the present decade. Kirane and Malik [
2] considered at some later date a one-dimensional fractional diffusion equation with a non-local, non-self-adjoint boundary condition. They developed a solution and source function using a bi-orthogonal method and obtained the uniqueness. In addition, they have prove an existence result when both the initial and the final conditions are smooth enough. In [
3], the problem in [
2] is generalized. Furati et al. used the same bi-orthogonal spatial system to obtain a separable formal solution and source function and to show their uniqueness. They then used the asymptotic expansion of the generalized Mittag–Leffler function to achieve a result of existence under certain smoothness criteria on the initial and final conditions. Aleroev et al. [
1] consider a linear heat equation, with a non-local boundary condition, involving a fractional derivative in time. They establish a space independent source term and the distribution of temperature for a problem with integral type over determining condition. They proved the solution’s existence and uniqueness, and its continuous dependence on the data. Among other papers devoted to determining unknown source terms, we may cite [
4,
5,
6].
As was mentioned in the abstract, an incorrect use of the estimates in the generalized Mittag–Leffler functions is commonly performed in the literature on this topic. This leads to false proofs of the Fourier series’ convergence to recover the equation satisfied by the solution, the initial data or the boundary conditions. Remark 1 explains this fairly widespread mistake. In the present work, the correct framework to recover the decay of fractional Fourier coefficients is established; this allows one to correctly recover the initial data, the boundary conditions and the partial differential equation satisfied within the space-time domain.
The manuscript is organized as follows: In
Section 2, fundamental concepts of fractional calculus are recalled. In
Section 3, we obtain results for the direct problem using Fourier’s method; we determine the existence, uniqueness and continuous dependence on the data. In
Section 4, the inverse problem associated with the direct problem is considered, and we show the results of the existence, uniqueness and continuous dependence of the solution on the data.
3. Direct Problem
In this section, we will study the forward problem:
where
p,
f and
are given functions such that
and
.
We are interested in classical solutions to Problem (10), that is, solutions which are continuous on and satisfy and , for any .
Example 1. Consider the case and , with and . We fix the initial condition , the spatial and temporal source terms and for any and as the following:The exact solution is then , since it is known that Since the problem (10) is linear, we can split its solution as follows:
where
is the solution of the homogeneous problem:
and
is the solution of the problem:
For every direction
,
, the spectral problem associated with Problem (12) is:
The eigenfunctions of (14) are given by [
9]:
where
are the associated eigenvalues. Notice that these eigenvalues form a countable set that is denoted by
. Moreover, they satisfy the following:
- (P1)
There are only finitely many real eigenvalues and the rest appears as complex conjugate pairs;
- (P2)
as ;
- (P3)
for n sufficiently large and as .
Now, consider a real number
such that:
and this is possible since
and consequently the interval
is not empty. Therefore, the property (P3) implies that
Since
, then applying Theorem 1, we obtain the estimate:
Remark 1. In the published papers on this topic, the mistake commonly made is due to a misapplication of Theorem 1. Indeed, the following estimatecannot hold true for all the eigenvalues , , since . Consequently, there is no real number such thatTherefore, the conditions for applying Theorem 1 are not satisfied to obtain (18). Based on [
11], for every
, the family
is a basis of
, which is not orthogonal, because the operator
is not self-adjoint. Consequently, for each
, we introduce the family:
corresponding to the eigenfunctions of the adjoint operator of
, with respect to the inner product in
:
where
denotes the conjugate of the complex number
. In the sequel, the inner products in
and
will be simply denoted by
and
, respectively.
Notice that the adjoint operator of
is the right-sided Riemann–Liouville fractional derivative of order
defined by:
Moreover, the family
is biorthogonal to
in
and satisfies
On the other hand, for any integers
, the resulting problem in the variable
t is:
whose solution is given by:
where:
since
The solution of Problem (12) is then formally given by:
In the same way, we can look for the classical solution of the problem (13) as:
where the coefficients
are to be found. To this end, if
, then:
with
Substituting the solution
w given by (25) in Equation (13), we find formally:
Therefore, for each
, the coefficient
satisfies the equation:
Equation (28) is supplemented with the initial condition:
Following [
12], the solution of (28) is:
Therefore, the solution of problem (13) is written as:
In what follows, c is a constant that might change from line to line.
Theorem 2. Let , and such that , and . Then, there exists a unique classical solution of Problem (10).
Proof. Existence of a solution:
To lighten the proof without loss of generality, we present the case
and explain below the general case
. In this situation, we consider the problem:
with
and
. We need to prove the convergence of series (24), (31) and the series associated with their fractional derivatives in space and time. We set
. Then:
where
Then, there is a constant
such that, for every
,
and
n are sufficiently large such that:
The series of general terms is convergent since . Using the M-test of Weierstrass, the series (24) is normally convergent on . Consequently, the function defined by (24) is continuous on . The same idea can be used for the series (31) by setting . Hence, the functions v and w verify the initial and the boundary conditions of (12) and (13), respectively.
Finally, to prove the convergence of the series associated to the fractional derivatives of
v and
w, it suffices to prove the convergence of series associated with:
,
,
and
. To this end, let
and
so
and
From above, for any
,
and
, one has:
and
Therefore, the series and converges normally on , for every .
Applying the same ideas, we obtain the normal convergence of the series and on , for every . Whence, we obtain the desired convergence on the whole set by Lemma 1, which achieves the claim.
- –
Uniqueness of the solution:
Let
be two classical solutions of Problem (10) and set
. Then,
satisfies the following problem:
Proceeding as above with
we obtain that
, for every
, which implies that
. □
Remark 2. The generalization of the previous proof to the case proceeds as follows: We set . Then,Direct computations give: Theorem 3. Under the same conditions as in Theorem 2, the solution of the direct problem (10) depends continuously on the given data.
Proof. The arguments developed in Remark 2 allow, without loss of generality, to restrict ourselves to the case .
Let
u and
be the solutions to the direct problem, corresponding to the data
and
, respectively. Using (11), we find
where
and
correspond to the data
and
, respectively. The same arguments as applied to (24) before lead to
where
c is a positive constant. On the other hand, a direct computation gives
It holds that
Therefore, there is a constant
such that
which achieves the proof. □
4. The Inverse Problem
Let us consider the problem:
where the functions
p,
and
are given, and the functions
u and
f are unknown.
Example 2. As above, consider the case and , with and . We fix the initial condition , the spatial source term for any and the “mean value” , of the solution on Ω:The unknown temporal source term is given by and the exact solution . To solve Problem (35), we start by expanding
and
, using systems (15) and obtain:
with
and
By substitution in (35), we formally obtain:
It follows that
with the initial condition
and
is defined by (23). Using (36) and (6) we obtain:
Using (41), we formally find:
According to [
12], the solution of (41) is given by:
Replacing
given by (44) in (43), we then find the integral equation satisfied by the unknown function
:
where
and
Theorem 4. Let
such that and .
Let such that .
such that and .
Then, there exists at least one solution to Problem (35).
Before proving Theorem 4, we introduce the functional framework for the fixed point integral Equation (45). First, applying the same arguments as before, we can show that the functions and defined by (48) and (47) are -differentiable on . Moreover, the condition implies that the real number .
At this stage, we will establish the existence of at least a solution to the inverse problem (35) and that this solution depends continuously on the data. First of all, we will use the Schauder fixed point theorem in Banach spaces with the Arzela–Ascoli compactness result. To this end, we start by defining the following operator:
To prove that the operator
B admits a fixed point, start by showing that
B maps a certain closed convex set into itself, in the space
equipped with the Bielecki norm. For every
, we introduce the Bielecki norm:
The space
is a Banach space, and the two norms
and
are equivalent.
Lemma 2. Under the above notations, there exists a positive constant such that for any , there is a radius such that the closed convex ballis stable by the operator B; that is, . Proof. For any
,
and
, we have:
At this stage, recall that for any
, the function
is integrable on
. Moreover, it holds that
; consequently:
That is, there are two real constants
and
satisfying
and
such that:
Whence, for any , there exists such that the closed ball K of radius R in is stable by the operator B; this achieves the proof. □
Remark 3. Using the Bielecki norm allows one to have no constraint on the maximum value that T can take. On the other hand, if we use the classical infinite norm, then T must be less than a finite quantity depending on the data of the problem.
Lemma 3. The family is equicontinuous, that is: Proof. It suffices to prove that the family
is equicontinuous, where
We introduce the functions
and
, which are clearly integrable on
. Let
t and
in
. We can assume, without loss of generality, that
.
First, for every
, we have
Using the fact that the map
is in
and applying the dominated convergence theorem of Lebesgue, we deduce that
Fixing now an
and
, it follows that
On the other hand, using the fact that the map
is in
and applying the dominated convergence theorem of Lebesgue, we deduce that
That is:
Let us point out that radii
and
depend only on
t and
, but are independent of
f. Whence, for any
and
, there exists
such that
which achieves the proof. □
By Arzela–Ascoli theorem,
is relatively compact. Now, we are able to use the Schauder fixed-point theorem [
13].
Theorem 5 (Schauder fixed-point theorem). Let be a Banach space, and let be convex and closed. Let be a continuous operator such that is relatively compact. Then, has a fixed point in K.
Proof of Theorem 4.
We know from Lemmas 2 and 3 that B is a continuous operator and is relatively compact, with K convex and closed. By Theorem 5, the operator B admits a fixed point in K, so the integral Equation (45) has a solution in .
Once the existence of the source term f is established, the existence of u follows in a similar way with the arguments developed above for the direct problem. □
Theorem 6. Under the conditions of Theorem 4, every solution of the inverse problem (35) depends continuously on the given data φ, p and ψ.
Proof. Let
,
be two solution sets of the inverse problem, corresponding to the data
,
, respectively. Using (45), we have:
and
where
correspond to the data
,
, respectively.
Remark first that
,
a and
depend continuously on
p and
. Indeed, for the function
, using (39) and (48), we have the estimate:
For the real constant
a, using (46) and (39), we have the estimate:
For the function
, using (47) and (23), we have the estimate:
Let
and
be such that
Now,
and consequently,
where
is a positive constant such that
. Similarly, we obtain:
It follows that
where
is a positive constant such that
. Finally, we obtain:
where
is a positive constant such that
and
Similarly, we obtain:
Then,
Therefore, we obtain the estimate:
Combining the last estimate and (55), we deduce the estimate on the source term:
where
is a positive constant depending on
a,
,
T,
,
and
.
Since
there is
such that
holds true. Whence, there is a positive constant
, independent of
f and
such that
That is, the solution
f depends continuously on the data.
The continuity of the part u of the solution follows with classical arguments for linear problems with the norm: . □