1. Introduction
Under consideration is the parabolic equation
where
,
G is a domain in
with boundary
, and
. The Equation (
1) is furnished with the initial-boundary conditions
where
, with
being the outward unit normal to
, and with the overdetermination conditions
where
is a collection of points lying in
G. Assuming that
for some known functions
, the problem consists in recovering both a solution to (
1) under (
2) and (
3) and functions
,
, characterizing
g. Note that any function can be approximated by the sums of this form for a suitable choice of basis functions
.
Inverse problems of recovering the boundary regimes are classical. They arise in many different problems of mathematical physics, in particular, in the heat and mass transfer theory, diffusion, filtration (see [
1,
2]), and ecology [
3,
4,
5,
6,
7].
A particular attention is payed to numerical solution of the problems (
1)–(
3) and close to them. Most of the methods are based on reducing the problems to optimal control ones and minimization of the corresponding quadratic functionals (see, for instance, [
8,
9,
10,
11,
12,
13,
14]). However, the problem is that these functionals can have several local minima (see Section 3.3 in [
15]). First, we describe some articles, where pointwise measurements are employed as additional data. Numerical determination of constant fluxes in the case of
is described in [
9]. Similar results are presented in [
16] for
. The three-dimensional problem of recovering constant fluxes of green house gases is discussed in [
3], but numerical results are presented only in the one-dimensional case. In [
4] (see also [
5]) the method of recovering a constant surface flux relying on the approach developed in [
17] is described, where special solutions to the adjoint problem are employed (see also [
6,
7]). The surface fluxes depending on
t are recovered in [
12,
18,
19,
20] in the case of
, and in [
11,
21,
22] in the case of
. The flux depending on time and spatial variables is reconstructed in [
14,
23].
In literature, there are results in the case in which additional Dirichlet data are given on a part of the boundary and the flux is reconstructed with the use of these data on another part of the boundary (see [
24]). The article [
13] is devoted to the recovering of the flux
(the function
is unknown) with the use of final or integral overdetermination data. The existence and uniqueness theorems for solutions to the inverse problems of recovering the surface flux with the use of integral data are presented in [
25,
26].
There is a limited number of theoretical results devoted to the problem (
1)–(
3). We refer the reader to the article [
27] (see also [
28]), where, in the case of
,
, and
, the existence and uniqueness theorems of classical solutions to the problem (
1)–(
3) are established. In contrast to our case, the problem is well-posed in the Hadamard sense. If the points
are interior points of
G then the problem becomes ill-posed and this fact was observed in many articles (see [
29]). In this article we describe a new approach to the existence theory of solutions to this problem and establish the corresponding existence and uniqueness theorems. We hope that these results can be used in developing new numerical algorithms for solving the problem.
2. Preliminaries
Let
E be a Banach space. By
(
G is a domain in
), we mean the space of
E-valued measurable functions such that
[
30]. The symbols
and
stand for the Sobolev spaces (see the definitions in [
30,
31]). If
or
then the latter spaces is denoted by
. The definitions of the Hölder spaces
can be found in [
32]. By the norm of a vector, we mean the sum of the norms of coordinates. Given an interval
, put
and, respectively,
. Denote by
the inner product in
. Let
designate the distance between the sets
. In this case,
is the distance from a point
x to
. Denote by
the ball of radius
centered at
x.
We say that a boundary
of a domain
G belongs to
,
(see the definition in Chapter 1 in [
32]) if, for each point
, there exists a neighborhood
about
and a coordinate system
y (the local coordinate system) obtained from the initial one by the translation of the origin and rotation such that the axis
is directed as the interior normal to
at
and the equation of the part
of the boundary is of the form
,
,
; moreover,
(where
),
, and
. The smoothness of
, with
an open subset of
, is defined similarly. The numbers
for a given
G are fixed and we can assume without loss of generality that
, with
M the Lipschitz constant of the function
. We employ the straightening of the boundary, i.e., the transformation
,
,
, with
y the local coordinate system at a given point
b.
Below, we assume that
or
G is a domain with compact boundary of the class
. The coefficients of the Equation (
1) are assumed to be real. We consider an elliptic operator
L, i.e., there exists a constant
such that
Assign
for
and
for
. The symbol
stands for an inner product in
. Let
and assume that
Moreover, we suppose that the functions
admits extensions to the whole
such that the condition (
5) is valid in
. If
G is a domain with compact boundary of the class
such an extension always exists (see Theorem 1 in Subsection 4.3.6 of Section Remarks in [
33]). Consider the equation
where the operator
is a formal adjoint to
L. Its coefficients also satisfy (
5). Let
. Introduce the functions
,
. It follows from Theorems 3.5 and 3.1 in [
34] and Theorem 3.3 in [
35] that
Theorem 1. Assume that and the conditions (5) hold. Fix . Then there exists a number such that, for all λ with , there exists a unique solution to the Equation (6) decreasing at ∞ such that for all , and for all , . In every domain a solution admits the representation In what follows, we denote by a solution obtained in Theorem 1 for a given j.
Consider the problem
where
or
or
G is a domain with compact boundary of the class
.
Theorem 2. Let , , , and . Then there exists a number such that, for all λ with , there exists a unique solution to the problem (
11).
The theorem results from Theorem 5.7 for
, Theorem 7.11 for
and Theorem 8.2 in the case of a domain with compact boundary in [
31].
The following Green formula holds.
Lemma 1. Let the conditions (5) hold and let , where is chosen so that Theorem2
is valid for . If is a solution to the problem (
11)
with from the class specified in Theorem2
then If and in some neighborhood about , then Proof. The proof is conventional. It suffices to approximate the functions
by sequences of smooth functions in the corresponding norms, to write out the above Formulas (
12) and (
13) for these approximations, and pass to the limit. □
Assume that or G is a domain with compact boundary of the class . Given a collection of points , construct the points such that . Denote by the set of these points. Let . For , there exists a local coordinate system y such that the axes agree with the principal directions on the surface at , in this case, where are the principal curvatures of the surface at 0. In the case of , the equation of the boundary in some neighborhood about b is of the form and is the curvature of the curve at b.
Lemma 2. Assume that, for every , the set consists of finitely many points and, for every , we havewhere are principal curvatures of Γ for and κ is the curvature of Γ for at b. Then there are constants , such that for every and all , . Remark 1. For , the condition (14) can be reformulated as follows. There exists a constant such that where y is a local coordinate system at . The claim follows from the fact that there exists an orthogonal transformation of coordinates such that the new axes agree with the principal directions on the surface Γ at . Proof. Take
. We prove the claim in the case of
. If
then the proof is simpler and we omit it. Let
y be a local coordinate system at
b. Since
is a superposition of an orthogonal transformation and a translation, the distances between points and their images are the same. We have
,
,
,
,
,
, and
Remark 1 implies that
in some neighborhood about 0. Fix a parameter
such that
. In this case there exists
such that
for
. Therefore, we obtain
The converse inequality follows directly from the definition of the quantity J.
Below, we preserve the notations of Lemma 2. Take
. We can define the transformations
and
. For
, put
where the parameter
is chosen below. The map
takes
onto
. Similar notations are used in the case of
, i.e.,
Below, we assume that, for every
, the set
consists of finitely many points and
where
are the principal curvatures of
for
and, respectively,
is the curvature of
for
.
Let
be a solution to the Equation (
6). Given
, construct the point
lying on the straight line joining
and
b and such that
,
. The point
is symmetric to
with respect to the surface
. Let
be a solution to the Equation (
6), where the point
is replaced with
. Denote by
the functions defined by the equality (
4), where
is replaced with
. In what follows, we assume that the closures of coordinate neighborhoods about the points
are disjoint, otherwise, we can always reduce them. Fix a point
. The quantity
is positive (it depends on
and the angles between the vectors
and
). Let
(where
is the coordinate neighborhood about
b). Without loss of generality, we can also assume that the constant
is positive for all
and all
j, otherwise, we decrease the parameter
of the coordinate neighborhoods
. Denote by
a constant smaller than the minimum of these constants. Theorem 1 for
and
yields
where
and
are constants independent of
j,
, and
such that
. □
Lemma 3. Assume that the conditions (5) and (15) hold, , andfor some . Then there exists a number such that, for , we have the representation Proof. Consider the case of
. We have
Theorem 1 implies that
where
. We can assume that
for all such
and
j. To estimate the second integral
on the right-hand side of (
22) from above, we derive that
In view of the definitions, there exists a constant
such that
for all
and, thereby,
for some constant
. For the first summand
on the right-hand side of (
22), we have
Consider the last integral in (
25) that is multiplied by
. This quantity is written as
where
is the point
written in the coordinate system
y. Consider the integral
We can assume that the axes of the local coordinate system
y are directed as the principal directions on
at
b. In this case (see Lemma 2) we obtain that
where
is a
-function in some neighborhood about 0. Make the change of variables
in
. We obtain that
Introducing the polar coordinate system, we arrive at the expression
Integrating by parts yields
The last integral here admits the estimate
The second integral on the right-hand side is estimated as
where
is a positive constant. Thus, we establish the representation
Introducing the polar coordinate system, we infer
Making the change of variables
, we obtain the estimate
This inequality and (
24) imply that
where the constant
is independent of
. In this case the last integral
on the right-hand side of (
26) admits the estimate
In view of (
28), the previous integral
in (
26) (
) is estimated as follows:
Finally, the second summand on the right-hand side of (
25) is representable as
In view of our conditions on the coefficients,
for every compact set
, and thereby,
. Involving the condition of the lemma and (
28), we can estimate the integral
on the right-hand side of (
25) by
The representation (
29) and the estimate (
30) validate the equality (
18). The equality (
19) is proven analogously and the former equalities in (
20) and (
21) are consequences of (
18) and (
19). The proof in the case of
is simpler. Display the asymptotics of the main integral
where
,
is the local coordinate system at
b, and
is the equation of the curve
. To reduce arguments, we take
, where the parameter
is defined in Lemma 2. Theorem 1 implies that
As before, we have
. We have the asymptotic formula (see
Section 1 , Chapter 2 in [
36])
where
is a point in which
S reaches its maximum. Applying this formula to the first integral on the right-hand side of (
31) and estimating the second integral by
, we obtain
All other arguments are similar. The proof in the case of is even simpler and we omit it.
It remains to prove the latter inequalities in (
19) and (
20). As before, take
. The asymptotics from Theorem 1 ensure that
where
If
then we have
This equality and the previous arguments validate the claim. □
Remark 2. Let . Then the condition (15) holds if for all j. We consider the problem (
11), where
, i.e., the problem
and we obtain some estimates of its solution. Fix
j and take
. In Lemma 4 below, we use functions
such that
on the set
and
. The condition
ensures the inclusion
. The map
,
takes a neighborhood
onto the set
. Denote
and
.
Lemma 4. Assume that the conditions (5) hold, , and . Then there exists a number such that, for , there exists a unique a solution to the problem (33) and (34) in the space satisfying the estimates If , with φ from the above-described class of functions, then there exist constants such thatwhere is arbitrarily small constant. If additionally andthen for any φ and there exist constants such that Proof. Theorem 2 for
ensures the existence and uniqueness of solutions provided that
for some
. Multiply the Equation (
33) by a function
and integrate the result over
G. Integrating by parts, we infer
Separating the real and imaginary parts, we obtain
Summing (
42) and (
41) and estimating the modules of the right-hand sides
Below, we use the inequality
The last integral is estimated by
Similarly, we have
where
and
are arbitrary positive constants. The embedding theorems and interpolation inequalities (see [
30]) imply that
Estimating the right-hand side of (
43) with the use of (
44) and (
45), we arrive at the inequality
Choosing sufficiently small
and increasing
, if necessary, we derive that
where the constant
is independent of
with
and
can be taken arbitrarily small. Using (
46) and interpolation inequalities we obtain that
and the estimate (
35) is proven. Rewriting (
33) in the coordinate system
y, we obtain the problem
Multiply the equation (
47) by
. The result is the problem
Introduce the coordinate system
z, with
,
. In this case, the function
is a solution to the problem
Multiplying the Equation (
50) by
and integrating the result over
U, we obtain that
Integrating by parts, we rewrite the first summand in the form
Note that
and integrating by parts we obtain the integrals containing third order derivatives. However, the result of integration is easily justified if we employ smooth approximations of functions in
. Similar arguments can be found, for instance, in the proof of Lemma 7.1 of Chapter 3 in [
37]. We also have
Using (
52)–(
54) in (
51), we obtain
As it is seen, the inequality
is valid for some constant
. Next, we infer
where
is the space with the norm
,
,
is arbitrary, and the last summand is estimated by
(see (
46)). Here we rely on the conventional theorems on pointwise multipliers and Proposition 12.1 of Chapter 1 in [
38]. Next, repeating the arguments of the proof of the estimate (
46), we conclude that
To establish (
37), it suffices to prove the estimate
which is justified by repeating of the proof of (
35). To validate the second part of the claim, we first demonstrate the smoothness of a solution
w. Take an arbitrary point
and the set
. Construct a function
such that
. The function
is a solution to the Equation (
48) from the space
satisfying (
49) on
and
Using the conventional theorems on extension of boundary data inside the domain [
30] and Theorem
Section 3 of Chapter 4 in [
39], we conclude that
.
Consider the equation (
50). Multiply (
50) by
and integrate the result over
U. The same arguments as those of the proof of the estimate (
36) can be applied to justify (
37) and (
39). The calculations are rather cumbersome and we omit them. □
Assume that the conditions (
5) and (
15) hold. In this case, for every
j and
, we can consruct the balls
and
. Let
(where
).
Lemma 5. Let the conditions (5) and (15) hold. Then, for every , there exists a function and constants such that for , for , and for all . Proof. In view of (
15), it is not difficult to establish that there exists a parameter
such that
for all
and
for all
. Put
. Obviously,
. Take
. Construct a nonnegative function
such that
,
and the averaged function
where
is the characteristic function of the set
. By construction,
for
and
for
. This function satisfies our conditions. □
The following theorem results from Theorem 7.11 for
and Theorems 8.2 in the case of the domain with compact boundary in [
40].
Theorem 3. Assume that and . Then there exists a constant such that if and the condition (
56)
holds then there exists a unique solution to the problem (
1) and (
2)
such that and Let
E be a Hilbert space. Denote by
the space of functions
u defined on
whose zero extensions
to the negative semiaxis belong to
and
The Laplace transform
is an isomorphism of this space
onto the space
of analytic functions in the domain
such that
If
or
or
(
G is a domain in
) then these properties of the Laplace transform can be found in [
41] (see Theorem 7.1 and Section 8). For
, we similarly define the space
as the subspace of functions in
admitting the zero extensions for
of the same class. This space coincides with
for
and with the space of functions
such that
for
. For
, it coincides with the space of functions in
such that
[
41].
3. Basic Results
We assume here that the conditions (
5), (
15), (
17) are fulfilled. Let
be the matrix with entries
. We assume that
Fix a parameter
greater than the maximum of the parameters defined in Theorem 1 with
, Theorem 2 with
, and Theorem 3. We assume that
By Theorem 3, if the condition (
60) holds for some
, then there exists a unique solution
to the problem (
1) and (
2), where
, such that
. Consider the problem (
1)–(
3). Changing the variables
, we obtain the simpler problem
We assume that
and
where
for
and
for
. For
, the condition (
63) can be rewritten as
For a finite
T, the condition (
63) can be stated as follows: there exists an extension of
on
satisfying (
64). We have
for
and
for
. Here
is the Hankel function. The latter equality is derived in Lemma 1.6.7 in [
42]. The former can be easily obtained if we use the Poisson formula for a solution to the Cauchy problem for the heat equation with the right-hand side equal to the Dirac delta function.
Theorem 4. Assume that and the conditions (
5), (
15), (
58), (
59)
, and (
38)
for hold. Then there exists such that, if and the conditions (60), (63) are fulfilled, then there exists a unique solution to the problem (
1)–(
3)
such that , . Proof. Consider the equivalent problem (
61) and (
62). Assuming that
and applying the Laplace transform to (
61), we arrive at the problem
Next, we use the functions
constructed before Lemma 3. Theorem 1 yields
,
for all
,
. Construct the functions
,
, where the functions
are defined in Lemma 5. The properties of the functions
imply that
. Lemma 1 imply that
where the function
is a solution to the problem (
65). Consider the case of
. The case of
is considered analogously. For the integral on the left-hand side, we have
However, only two summands with
and
are essential on the set
. Indeed, in view of (
16), for
and
, we infer
where
is a constant independent of
. This inequality implies that the remaining integrals decay exponentially. By Lemma 3, we have
Consider the right-hand side in (
67). The integrals over the domain are estimated by means of Lemma 5. On the support of
, Theorem 1 and Lemma 5 ensure the estimate
where the constants
are independent of
. The Hölder inequality yields
Examine the integrals over
in the right-hand side of
. We have
As in the estimate (
69), the last two integrals are estimated by
in view of (
16). Estimate the second and third integrals. In view of Theorem 3 and estimates of Lemma 5 (see (
27)), they admit the estimates
where
is the straightening of the boundary in
. It remains to consider the first integral
Note that
. Lemma 5 ensures the following representation for the first integral
on the right-hand side of (
71):
The second integral on the right-hand side of (
71), in view of Lemma 5, (
28) and (
32), is estimated as follows:
Thus, in view of (
70)–(
72), we have the inequality
with some constant
. Next, we employ Lemma 4. The embedding theorems for
and Lemma 4 imply that
where
is an arbitrarily small constant. Similarly, Lemma 4 ensures that
In view of the conditions on the functions
, there exists a constant
such that
Therefore, we have the estimate
where
is an arbitrarily small constant. We can rewrite (
67) in the form
The left-hand side of this equality is written as
, where the entries of the matrix
are of the form
. The right-hand side is written in the form
where the coordinates of the vectors
are as follows:
Choose
so that the matrix
is invertible for
and the norm of the operator
is bounded by a constant
for all
. It is more convenient to rewrite the system (
74) in the form
Estimate the norm of the operator
. In view of (
73), we have the estimate
Thus, for
, increasing the parameter
if necessary, we can assume that
The norm of the operator
is less than 1/2 in this case and, thereby, the Equation (
75) has a unique solution. Constructing a solution
to the Equation (
75), we can find a solution
to the problem (
61), where
. In view of our conditions, the estimates of Lemma 4 holds. In view of the Equation (
75), a solution
meets the estimates
Hence, we infer
where the constant
is independent of
. The properties of Laplace transform validate the equality
and the previous inequality yields
This inequality ensures that the inverse Laplace transform is defined for the functions
,
, and
Note that the additional smoothness of the functions
ensures the additional smoothness of the functions
. Consider the problem (
61) with the above constructed functions
. By Theorem 3, there exists a unique solution to this problem such that
. We now demonstrate that this function satisfies (
62). Indeed, applying the Laplace transform, we obtain that
is a solution to the problem (
65). Multiplying the equation in (
65) by
and integrating by parts we obtain (
67) with
rather than
. Since
satisfy (
67) with the functions
on the right-hand side, we obtain
.
In the case of
, the arguments are the same. However, in view of another asymtotics of the function
the inequality (
78) can be rewritten as
Uniqueness clearly follows from the above arguments. □
If we state our theorem in the case of a finite interval
, then the condition (
60) looks as follows:
Theorem 5. Assume that and the conditions (
5), (
15), (
58), (
59), (
79), (
63)
and (
38)
for hold. Then there exists a unique solution to the problem (
1)–(
3)
such that , . Proof. Extend the functions
on
as compactly supported functions of the same class. The conditions (
63) are fulfilled for every
. Extend the function
f by zero on
. Theorem 4 ensures existence of a solution to the problem (
1)–(
3). Now we prove uniqueness of solutions. Assume that there are two solutions of the problem from the class pointed out in the statement of the theorem. In this case, their difference
is a solution to the problem
Integrating the equation and the boundary condition with respect to time two times, we obtain that the function
is a solution to the problem
Make the change of variables
(
). We have
Integrating (
82) over
, we obtain that
Let
. Make the change of variables
, with
a solution to the problem
,
, and, respectively,
is a solution to the problem
Note that
and, thereby,
. Since
[
30], we have the estimate (see Theorem 7.11 for
and Theorem 8.2 in the case of a domain with compact boundary in [
31])
Multiply the Equation (
85) by the function
defined in the proof of the previous theorem and integrate over
G. As in the proof of Theorem 4, we obtain the system (see (
75))
where the coordinates of
are written as
. The system can be rewritten as follows
where the right-hand side is analytic for
and we have
where
is independent of
. Thus, every of the quantities
is estimated by
The function
is the Laplace transform of the function
for
and
for
. Fix
and define an additional function
. It is analytic in the right half-plane and is bounded by some constant
on the real semi-axis
. To estimate this function on the on the imaginary axis, we integrate by parts as follows:
For
, we thus have the estimate
In each of the sectors
,
the function
admits the estimate
Applying the Fragment-Lindelef Theorem (see Theorem 5.6.1 in [
43]) we obtain that in each of the sectors
,
the function
admits the estimate
Therefore,
We have equality (
)
and, thereby,
The Parseval identity yields
Since this inequality is true for all , for . Since the parameter is arbitrary, for and for and every j and, therefore, which implies that . □