2. Inverse Source Problem for
In this section, we studied ISPs (
2), (
3) and (
2), (
4) using a variational approach. We used the following lemmas (see [
2] (Lemma 2.1), [
2] (Corollary 3.1), and [
2] (Theorem 6.1)) to study the uniqueness of the solution to the ISPs.
Lemma 1 (Strongly positive definiteness)
. For all , for , the kernel satisfies:with Lemma 2 (Energy estimate)
. For any satisfying:it holds for all that there exists a strict positive constant such that: Lemma 3 (Existence of a solution to the forward problem for
)
. Let the following assumptions be fulfilled:
AS-1: satisfying and is uniformly elliptic;
AS-2: satisfying for a.a. ;
AS-3: and ;
AS-4: or for and ;
AS-5:.
Then, there exists a unique weak solution u to Problem (2), with and . From this lemma, we were not able to conclude that . We first considered the time-average condition for , which was well-defined as an element of .
Theorem 1 (Uniqueness if
with
using measurement (3))
. Let the conditions of Lemma 3 be satisfied. Assume that and . Moreover, assume that satisfies:Then, there exists, at most, one couple with , such that Problem (2), together with Condition (3), is satisfied. Proof. We supposed that there were two solutions
and
satisfying (
2) and (
3). Then, the differences
and
satisfied the weak form:
for all
with
and
in
. Next, we integrated (
7) in time over
. Using [
21] (Lemma 7.3),
almost everywhere (a.e.) in
and
, we obtained:
First, we considered the case
for all
, i.e.,
for all
. We multiplied (
8) by the reciprocal function
, chose
as the test function, and integrated in time over
to obtain:
Using integration by parts, we obtained for the first term on the left-hand side (LHS) of (
9) that:
It was clear that
for all
. Therefore, with Lemma 1 (
is strongly positive definite), we obtained:
The second term on the LHS of (
9) could be handled by employing
(note that a unique symmetric positive definite square root
exists, since the symmetric matrix
is positive, definite in space and time—cf. [
22] (Theorem 1)—and integration by parts as follows:
Therefore, from (
9), we obtained:
Hence, we obtained that
a.e. in
. This was a Volterra equation of the first kind. From [
23] (Theorem 3.5 on p. 44), it followed that
a.e. in
. Hence, from (
7), it followed that
for all
An application of [
24] (Proposition 18.2) implied that
in
. The proof was similar if
for all
, then (
10) became:
and the remainder of the proof proceeded as before. □
If we assumed that solution u satisfied additionally
, then we could show the uniqueness of a solution to ISP (
2) and (
3), with time-dependent coefficients in operator
L.
Theorem 2 (Uniqueness if
using measurement (3))
. Assume . Moreover, assume that satisfies:or, either,Then, there exists at most one couple with , such that Problem (2), together with Condition (3), is satisfied. Proof. Let
with
for all
. We placed
with
in (
7) and integrated over time
to obtain:
Employing integration in parts, we obtained:
Note that
for all
. From Lemma 2 for
, it followed that:
Therefore, and, thus, a.e. in and a.e. in . □
If
satisfies
, then the uniqueness of a solution to ISP (
2) and (
4) with
was established in [
12] (Theorem 3.1). If
, then
can be uniquely determined from (
4) for general operator
assuming the monotonicity condition on
and
or
—see [
13] (Theorem 3.1). Unfortunately, this assumption on
is typically not satisfied as mentioned before. The purpose of the next theorem was to show that [
13] (Theorem 3.1) stayed valid if solution
u to Problem (
2) satisfies
and
.
Theorem 3 (Uniqueness if
using measurement (4))
. Assume that . Moreover, assume that satisfies:or, either,and assume that: Then, there exists at most one couple with and , such that Problem (2), together with Condition (4), is satisfied. Proof. Since
for a.a.
, we could consider
as a test function, as conducted in [
13] (Theorem 3.1). Lemma 1 stayed valid if
and
. Hence, we could apply Lemma 1 for
with
, and all other steps stayed valid in the proof of [
13] (Theorem 3.1) (note that all integrals stayed finite), i.e., we could conclude the current proof. □
Remark 1. The assumptions on solution u in the previous theorem were fulfilled if the conditions (5) were satisfied. Indeed, we obtained:andas for and . Remark 2. A counterexample (if h changed its sign) violating the uniqueness of a solution to the inverse problem considered in Theorem 3 could be found in [13] (Example 3.1). Whether the monotonicity conditions on h can be relaxed in case of a non-autonomous operator L, as conducted in [7] for a classical diffusion equation, is still an open problem. Moreover, the recovery of from (4) if without additional assumptions on the solution is also still an open problem. 3. Inverse Source Problem for
In this section, we studied ISPs (
2), (
3) and (
2), (
4) for
. First, we mentioned the well-posedness of the forward problem [
25] (Theorem 4.1).
Lemma 4 (Existence of a solution to the forward problem for ). Let the following assumptions be fulfilled:
AS-1,AS-2andAS-3;
AS-6:;
AS-7: and .
Then, there exists a unique weak solution u to Problem (2) with , and . Now, both
and
were well-defined as elements in
. We first considered ISP (
2) and (
3) for
with autonomous operator
L.
Theorem 4 (Uniqueness if
with
using measurement (3))
. Let the conditions of Lemma 4 be satisfied. Assume that . Moreover, assume that satisfies: Then, there exists at most one couple with and , such that Problem (2), together with Condition (3), are satisfied. Proof. We supposed that there were two solutions
and
satisfying (
2) and (
3). Then, the differences
and
satisfied the weak form:
for all
with
,
and
in
. Next, we integrated in time over
. Again, we placed
. We obtained:
We only considered the case
for all
and defined
. We took
in (
13) and integrated in time over
to obtain:
We only pointed out how to handle the first term on the LHS. The other terms can be handled as in Theorem 1. Using integration by parts, we obtained:
Therefore, by Lemma 2 for
, we obtained:
from which we concluded the proof. □
The uniqueness of a solution to ISP (
2) and (
3) for a time-dependent operator
L is still an open problem. When we assumed a higher regularity on the solution of Problem (
2) (in comparison with Lemma 4), we were able to show the following theorem concerning ISP (
2) and (
4). Hence, the recovery of
from Measurement (
4) if
without additional assumptions on the solution is still an open problem.
Theorem 5 (Uniqueness if
using measurement (4))
. Assume that and: Moreover, assume that satisfies:or, either, Then, there exists at most one couple with and , such that Problem (2), together with Condition (4), are fulfilled. Proof. We supposed that there were two solutions
and
satisfying (
2) and (
4). Then, the differences
and
satisfied (
12) for all
with
,
and
in
. Let
with
for all
. Define
. Here, we chose
in (
13) and integrated in time over
to obtain:
The second and third terms on the LHS were positive—see the proof of [
13] (Theorem 3.1). For the first term, we, again, employed integration by parts and Lemma 2 for
to obtain:
It followed that
in
as
in
. We concluded the proof as before. □
Remark 3. The assumptions on solution u in the previous theorem were fulfilled if the conditions (6) were satisfied. Remark 4. Important to note is that no additional damping term in the fractional wave equation was needed in order to obtain the uniqueness to ISP (2) and (4) for , which was the case for the classical wave equations [6]. 4. Conclusions
The purpose of the current study was to determine which type of measurement was suited to recover, in a unique way, a space-dependent source , taking into account the typical behaviour of the solution to FPDEs. The coefficients in the governing operator L might be time-dependent. Case corresponded with a fractional diffusion equation of the space-dependent variable order, whilst case corresponded with the fractional wave analogue.
For a non-autonomous operator
, and under low regularity assumptions (see Lemmas 3 and 4), the time-averaged Measurement (
3) was well-defined as an element in
for
, whilst both Measurements (
3) and (
4) were well-defined for
. This observation suggested that the time-averaged Measurement (
3) was better suited to recover the space-dependent source
if
.
In Theorem 1, we showed that
could be uniquely determined from Measurement (
3) if
assuming that
satisfied either
or
. In the case of a non-autonomous operator
, we needed additional assumptions on the solution and monotonicity conditions on
(i.e.,
or
) to obtain the uniqueness of a solution to ISP (
2) and (
3). No extra conditions on
were needed—see Theorem 2 for more details. In Theorem 3, we extended the result from [
13] (Theorem 3.1) to solutions
u of Problem (
2), satisfying Bound (
5). ISP (
2) and (
4) had a unique solution under appropriate conditions on
u, monotonicity condition on
and
or
. Concerning Theorems 2 and 3, whether the assumptions on
u and/or the monotonicity condition on
h can be further relaxed, is still an open problem. This gives a first direction for future research.
The uniqueness of a solution to ISP (
2) and (
3) for
was only obtained for an autonomous operator
L if
satisfied either
or
—see Theorem 4. Hence, the same problem for a time-dependent operator
is still unsolved. Finally, Theorem 5 concerned the uniqueness of a solution to ISP (
2) and (
4) for
. The monotonicity conditions on
and
h were the same as in Theorem 3. Furthermore, here, the question raised by this study was whether it was possible to relax the conditions on the solution and/or
h. Important to note is that, for this latter ISP, no additional damping term in the governing equation was needed, as was the case for the wave equations with classical derivatives. Moreover, we also addressed that, for the fractional wave equation, the typical behaviour of Solution (
6) was satisfied, and that it might be possible that the final in time Measurement (
4) was more suited to recover the space-dependent source
(
).
Finally, we mentioned that the obtained results were also valid if the order of the fractional derivative was constant.