4.2. Representation of Solution on an Interval and Reduced Response Operator
We adapt a representation of
developed in [
27], where only Dirichlet control and boundary conditions were considered. Fix
. We extend
to
as follows: first evenly with respect to
, and then periodically. Thus
for all positive integers
k.
Define
to be the solution to the Goursat-type problem
A proof of solvability of this problem can be found in [
33].
Consider the IBVP on the interval
:
Let
Then, the solution to (
17)–(
20) on
can be written as
In what follows, we only consider for some finite T, so all sums will be finite.
Let us now change the condition (
20) to
. In this case, the solution becomes
To represent the solution of the wave equation on the edge
in a star graph, we must account for the control at
. Thus it will also be useful to represent the solution of a wave equation on an interval when the control is on the right end. Consider the IBVP:
Set
, and extend
to
by
. Define
to be the solution to the Goursat-type problem
Let
. One can then verify that
We now show that the system (
10)–(
15) is well-posed. Recall that
was defined in (
9), and
.
Theorem 2.
(a) If , then there exists a unique solution solving the system (10)–(15), and mapping is in . (b) For each , the mapping is a continuous mapping .
Proof. On with , the wave will be generated by the “control” , whereas on the wave is generated by the two controls , . We assume first that .
Here,
p is independent of
j by (
12). We have that
is given by
Note that
has already been explicitly determined in (
23). Thus, by (
21), we have an explicit solution for
if we can solve for
p. We now prove the existence, uniqueness, and regularity of
P.
By (
21) and (
26), we have for
,
For
, we have by (
21), (
24), and (
26) that
We remark that at the moment, we have not yet solved for either
P or
for any
j. Let
We solve for
P with an iterative argument using steps of length
. The iterations are necessary because the upper limits in the sums in (
27), (
28) increase with time due to reflections of the wave at the various vertices. In what follows, we label by
various terms that we have already solved for, which by (
24), includes
. For
we have by unit wave speed that
. Suppose now
. Then,
and hence
for
, for all
j with
. By (
13), we have
and hence from (
27) and (
28) we get
It is easy to show that this is a Volterra equation of the second kind (VESK), and so admits a unique solution P with . Furthermore, by differentiating this equation we get .
Having solved for
P on
, we now suppose
. Thus for any
j and any
, we have
, so all terms in (
27), (
28) involving
, with
and
, are known. Thus by (
27), (
28), and
We can solve this VESK to determine uniquely
for
, with the estimate
holding. Iterating this process, we solve for the unique
for
as desired. The case for
is then obtained by continuity. Part (a) of the Theorem follows easily from (
21),(
23), and (
26). Part (b) of the theorem follows from Part (a) and (
27) and (
28). □
Proposition 1. For one can determine , , and N.
Proof. Let
, so
. From (
24) and (
29) one has, for
,
Thus, we have
, where
denotes various continuous functions. We have by (
22)
Clearly, the discontinuity at
gives us
and
N. That
determines
is proven in [
16]. □
Define the “reduced response operator” on
, with
, by
associated to the IBVP (
17)–(
20). From (
21), we immediately obtain
Lemma 1. For , and any , we havewithwith If T is finite, the sums above are finite. Proof. Using (
25), it is easy to see that
The lemma now follows easily from (
21). □
In what follows, we refer to
as the “reduced response function”. For
, we denote the solution to the system (
10)–(
15) as
. We also use the following.
Lemma 2. Let . For , we have Here , and for , and .
This result holds from the proof of Theorem 2, the unit speed of wave propagation, and the properties of wave reflections off
, see [
33]. The details are left to the reader.
The following result follows from (
22), (
25), and Lemma 2. The details of the proof are left to the reader.
Corollary 1. Let . For , we have Here , and for , .
4.3. Solution of Inverse Problem
Here, we establish some notation. We recall the following notation: for we list the incident edges by . Here, is chosen to be the edge lying on the path from to , and the remaining edges are labeled randomly.
Now let
be some fixed interior vertex, and let
satisfy
. Denote by
the unique subtree of
having
as root with incident edge
, and by
the set of its vertices, see
Figure 4.
We define an associated response operator as follows. Let
be the boundary vertices on
. Suppose
solves the IBVP
Then we define an associated reduced response operator
with associated response function
Suppose we determined . It would follow from Proposition 1 that one could recover the following data: , , and , where is the vertex adjacent to in . In this section we will present an iterative method to determine the operator from the -tuple of operators, , which we know by hypothesis for some . An important ingredient is the following generalization to a tree of Corollary 1.
Lemma 3. Let , and let be associated with (33)–(38), defined by (7) and (8). The response function for has the form Here, , and the sequence is positive and strictly increasing. If T is finite then the sums are finite.
Proof. The proof follows from the proof of Corollary 1, together with the transmission and reflection properties of waves at interior vertices, and reflection properties at boundary vertices. □
Fix . The rest of this section shows how to recover from .
Step 1
For the first step, let be the vertex adjacent to the root , with associated edge labeled . By Proposition 1, we can use to recover , , .
Step 2
Consider
. In Step 2, we show how to solve for
, see
Figure 5.
Since
is the root of
, the following equation is essentially a restatement of Lemma 1 to trees; the details of its proof are left to the reader.
Here, , and . In what follows in Step 2, for readability, we rewrite as .
Since we know
and
, we can solve the wave equation on
with known boundary data. We identify
as the interval
with
corresponding to
. Then
, restricted to
, solves the following Cauchy problem, where we view
x as the “time” variable:
Since the function is known, we can thus uniquely determine and . Thus is determined.
We now show how
p and
can be used to determine
. The following equation follows from the definition of the response operators for any
:
In what follows, it is convenient to extend
as zero for
. By Lemma 1 and by an adaptation of Lemma 2 to general trees, we have the following expansions:
Here,
and
, and
and
are positive and increasing. Clearly
, can all be determined by
and
, whereas for now
and the sets
are unknown. Inserting (
39),(
41) and (
42) into (
40), we get
Here all sums have 1 as lower limit of summation.
Lemma 4. The sets can be determined by and .
Proof. We mimic an iterative argument in [
26]. Differentiating (
43) and then matching the delta singularities, we get
Since the sequences
are all strictly increasing, clearly we have
, so that
, and so
and
. We represent that the set
determines the set
by
We now match the term
with its counterpart on the right hand side of (
44). There are three possible cases.
Case 1:.
In this case, we must have
Case 2a: and
Note that the last inequality can be verified by an observer at this stage. Then
and
and hence
Case 2b: and Then . Note we have not yet solved for . In this case, we now repeat the matching coefficient argument just used with .
Again there are three cases:
Case 2bi: . Note all of these terms are known, so this inequality can be verified. In this case, , so and .
Case 2bii: and . Then , and . Thus and .
Case 2biii: and . Then , and we need to continue our procedure with .
Repeating this procedure as necessary, say for a total of
times, we solve for
. We represent this process as
We must have
finite by (
44) and the finiteness of the graph.
Iterating this procedure, suppose for
we have
Here is chosen to be minimal, and so . We wish to solve for .
We can again distinguish three cases:
Case 1: Note that we know and , so these inequalities are verifiable. In this case, we must have and , so we have determined in this case.
Case 2: There exists an integer
Q and pairs
, with
, such that
Note that all the numbers
have been determined, so these equations can be all verified. We can assume all pairs
satisfying (
45) with
are listed. In this case, we have either
Case 2i:
It follows then that
, and
We thus solve for .
Case 2ii: It follows then that , and we have to repeat this process with .
Repeating the reasoning in Case 2ii as often as necessary, we eventually solve for
. Thus,
Hence, we can solve for for any positive integer L given knowledge of for sufficiently large. □
It remains to solve for
. In what follows, we set
for
. We use
to denote various functions that we have already established to be determined by
and
. Having already solved for
, we can eliminate from (
43) the Heavyside functions to get, recalling
,
We solve this with an iterative argument. Let
. For
, we have for
that
so
. Hence
Letting
, we get
We solve this VESK to determine
. Now for
, we have for
that
, and so those terms in (
46) with
can be absorbed in
G to again give
We solve this VESK to determine . Iterating this procedure, we solve for for any finite s.
Step 3 Because
are determined by assumption for
, the functions
are determined. In Step 2, we showed
is also determined. Hence by (
4),
is also determined. We can now carry out the argument in Step 2 on the remaining edges
incident on
to determine
for all
j.
Step 4 For each , we use Proposition 1, to find the associated together with the valence of the vertex adjacent to . Careful reading of Steps 2 and 3 shows that we can use and for any .
Step 5 Let
be the vertices adjacent to
, other than
. We now iterate Steps 2–4 for the each of these vertices. Choose for instance
. If it were a boundary vertex, this fact would be determined in Step 4, and then this algorithm goes to the next vertex, which we, for convenience, still label
. We can thus assume
is an interior vertex. Let us label an incident edge (other than
) as
, see
Figure 6.
We wish to determine
. Mimicking Step 2, let
solve (
1)–(
6), let
. We have the following formula holding by the definition of response operators:
Of course is assumed to be known. We determine b as follows. We have, from Step 2, that is known. We identify as the interval with corresponding to .
Then
arises as a solution to the following Cauchy problem on
, where we view
x as the “time” variable:
Since and are all known, we can thus determine .
The rest of the argument here is a straightforward adaptation of Steps 2–4 above. The details are left to the reader.
Step 6 Arguing as in Step 5, we determine for all other vertices adjacent to and their associated edges. The details are left to the reader.
Steps above 6 Clearly this procedure can be iterated until all edges of our finite graph have been covered.