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Abstract
In this article, an inverse nonlinear convection–diffusion problem is considered for the identification of an unknown solely time-dependent diffusion coefficient in a subregion of a bounded domain in . The missing data are compensated by boundary observations on a part of the surface of the subdomain: the total flux through that surface or the values of the solution at that surface are measured. Two solution methods are discussed. In both cases, the solvability of the problem is proved using coefficient to data mappings. More specific, a nonlinear numerical algorithm based on Rothe’s method is designed and the convergence of approximations towards the weak solution in suitable function spaces is shown. In the proofs, also the monotonicity methods and the Minty–Browder argument are employed. The results of numerical experiments are discussed.
1 Introduction
Recovery of missing parameters in partial differential equations from overspecified data plays an important role in inverse problems arising in engineering and physics. These problems are widely encountered in the modelling of interesting phenomena, e.g. heat conduction and hydrology. Another challenge of mathematical modelling is to determine what additional information is necessary and/or sufficient to ensure the (unique) solvability of an unknown physical parameter of a given process.
This contribution is devoted to these subjects. More specific, the purpose of this paper is a study on recovery of a time-dependent diffusion coefficient in nonlinear parabolic problems. It is assumed to have some overposed nonlocal data as a side condition. Similar but steady-state settings can be found in the so-called ‘Spontaneous potential well-logging’, which is an important technique to detect parameters of the formation in petroleum exploitation, cf. [Citation1–Citation4]. The determination of a time-dependent diffusivity in parabolic equations has been considered in papers [Citation5, Citation6]. Nonlinear problems have been studied in [Citation7–Citation10]. The problems studied in these papers are all one-dimensional in space. However, the analysis made in this article is valid for every dimension .
The mathematical setting is the following. Let be a bounded domain with a Lipschitz continuous boundary
. The domain
is split into two nonoverlapping parts
and
with the assumption that
cannot surround
. A transient convection–diffusion process is considered in
. The diffusion coefficient
takes the form
for a known
and
for
and
for
. The boundary
is split into three nonoverlapping parts, namely
(Neumann part),
(Dirichlet part) and
, on which beside a Dirichlet boundary condition (BC) also the total flux through this part is prescribed. It is assumed that
with
, where
is the Lebesgue measure. The purpose of this article is to study the following nonlinear parabolic initial boundary value problem (IBVP) (Equation1
1
1 )–(Equation2
2
2 ): find a couple
such that (
fixed)
1
1 and such that the following boundary measurements are satisfied
2
2 More specific, the inverse problem of determining the unknown coefficient
from the measured data (Equation2
2
2 ) is considered. It is determined under which assumptions on the data this inverse problem has a weak solution
. Also, a nonlinear numerical algorithm based on backward Euler method is designed to approximate the solution
and the convergence of approximations towards the weak solution in suitable function spaces is shown. An easier case with
and
is studied in [Citation11].
The remainder of this paper is organized as follows. Section 2 summarizes the mathematical tools and the assumptions on the data that are needed. Two different solution methods are presented in detail in Section 3. The existence of a solution to (Equation11
1 ) and (Equation2
2
2 ) for each solution method is shown in Section 4. Finally, some numerical experiments are developed in Section 5.
2 Variational setting and assumptions
First, some standard notations are introduced. The euclidian norm of a vector in
is denoted by
. To increase the readability of the text, it is assumed without loss of generality that
and
. The suitable choice of a test space is
which is clearly a Hilbert space with norm
, where
represents the norm in
. Also a subspace of
is considered, namely
Due to the homogeneous Dirichlet boundary condition (BC) on the boundary
, the following Friedrichs inequality holds for every function
3
3 Consequently, the norm in the space
is equivalent with the seminorm
. The norm in the trace space
, with
, is written by
. The dual space of
is denoted by
.
What now follows is a list of assumptions that are necessary to prove the existence of a weak solution to the nonlinear problem (Equation11
1 ) and (Equation2
2
2 ) on a single time step. The continuous function
(diffusion term) has to satisfy
4
4
5
5 Moreover,
6
6
Remark 2.1
The condition in (Equation6
6
6 ) means that
does not depend on the solution
, but only on time and position, i.e.
. If
, then
is depending on
, for instance
with
.
The vector-valued function (convection term) is also continuous and
7
7
8
8
Remark 2.2
If , then the right-hand side (RHS)
of (Equation1
1
1 ) can be redefined as
. From now on, it is assumed that
such that the
-term in problem (Equation1
1
1 ) and (Equation2
2
2 ) can be cancelled out of the equations if
, see also equation (Equation14
14
14 ). Therefore, assumptions (Equation7
7
7 ) and (Equation8
8
8 ) are only necessary if
.
Remark 2.3
The source function in problem (Equation1
1
1 ) and (Equation2
2
2 ) depends only on the time and space variable. However, our numerical procedure can be generalized to a nonlinear
, assuming
if
. The purpose of this paper is to keep the regularity as low as possible during the analysis. This is the reason why is started with the lowest plausible assumptions. Further, during the analysis, more assumptions are necessary on the function
.
The following assumptions on the other data functions are adopted9
9
10
10
11
11
12
12
13
13
14
14
The initial datum satisfies15
15 Note that the data function
can be prolonged into the whole domain
by a function
in such a way that [Citation12, Lemma 5.1]
16
16 Finally, some useful (in)equalities are stated, which can easily be derived:
and
Remark 2.4
In this contribution, the values are generic and positive constants independent of the discretization parameter
, see Section 3. They can be different from place to place. The value
is small and
. To reduce the number of arbitrary constants, we use the notation
if there exists a positive constant
such that
.
3 A single time step
Rothe’s method [Citation13] is applied to prove the existence of a weak solution to problem (Equation11
1 ) and (Equation2
2
2 ). An equidistant time-partitioning is used with time step
for any
. Let us introduce the notations
and for any function
The following recursive approximation scheme is proposed for
: find the unknown couple
on each time step that satisfies
17
17 where
. In this section, the existence of
for any
is shown in two ways: two different methods for solving (Equation17
17
17 ) are presented.
In the first one, it is assumed that is given and a solution of the direct problem
18
18 is looked for. Afterwards, it is shown that the trace of
on
depends continuously on
. Then,
is determined such that
. This solution method is discussed in detail in Section 3.1.
In the second solution method, the following direct problem is solved19
19 for a given
. It is proved that the total flux through
,
depends continuously on
. At the end,
is found such that
. The reader can find more details in Section 3.2. Only the differences with the previous solution method are pointed out.
Finally, in Section 3.3, a lemma is stated about the existence of a solution on a single time step of problem (Equation1717
17 ). This lemma is based on the results of Sections 3.1 and 3.2.
3.1 First solution method
First, the variational formulation of (Equation1818
18 ) is considered
20
20 In the following lemma, the theory of monotone operators is applied to prove the existence of a weak solution to (Equation20
20
20 ) for given
. The interested reader is referred to [Citation14] for further information.
Lemma 3.1
(Unicity) Assume (Equation44
4 )–(Equation15
15
15 ). For any given
,
, there exist a
and an uniquely determined
solving (Equation18
18
18 ) for
.
Proof
We consider the nonlinear operators defined as
together with the linear functionals
such that
We proof that
is strictly monotone, coercive and demicontinuous. In the case of
, it follows from the monotonicity of
that
The strict monotonicity of
follows by an application of the Friedrichs inequality (Equation3
3
3 ). Secondly, we do the case that
. Using the mean value theorem and the Young’s inequality, we have
Fixing a sufficiently small
and
gives the strict monotonicity of the operator
. The operator
is coercive if
This is always satisfied. If
, then the following lower bound is valid for sufficiently small
In the case
, the second constant in this lower bound disappears
The demicontinuity of
follows from the continuity of
and
. The functionals
belong to
if
due to
, the boundedness of
, the trace theorem (cf. [Citation15, Theorem 3.9]) and the growth condition on
Consequently, since
, the equation
admits a unique solution for
.
In the subsequent Lemma, the existence of an uniform bound for independent of
is proved.
Lemma 3.2
(Uniform bound for ) Assume (Equation4
4
4 )–(Equation15
15
15 ). There exists a positive constant
such that
Proof
We consider as a parameter. We set
into (Equation20
20
20 ) and get
Remark that the solution on the previous time step,
, belongs to
. First, we consider the case
. Using the Cauchy’s and Young’s inequalities together with the growth condition on
, we obtain the trace theorem and the Friedrichs inequality (Equation3
3
3 )
Using the monotonicity of
and the uniform bounds (Equation9
9
9 ) and (Equation10
10
10 ) one can easily get
Analogue, in the case
, we can readily obtain
and therefore
Fixing a sufficiently small positive
in both cases and applying the Friedrichs inequality if
, we conclude the proof.
To prove that the trace of on
depends continuously on
, it is necessary to define the following coefficient to data map
Then, the inverse problem with the measured output data
can be formulated as the following operator equation: search
such that
The continuity of this input–output map
is investigated in the following lemma. It is this property that leads to the existence of a solution to problem (Equation17
17
17 ), see Lemma 3.7.
Lemma 3.3
( depends continuously on
) Assume (Equation4
4
4 )–(Equation15
15
15 ). There exists a
such that the function
is continuous for
.
Proof
Subtract (Equation2020
20 ) for
from (Equation20
20
20 ) for
and set
to get
The first term in the RHS can be bounded using the Cauchy’s and Young’s inequality, Lemma 3.2 and the uniform bound (Equation10
10
10 )
The second term in the RHS disappears if
. An obvious calculation employing the monotonicity of
gives
21
21 We use the mean value theorem in the case of
to get
Therefore, we can derive the following estimate
22
22 We fix a sufficiently small
and
to conclude that in both cases
23
23 Using the trace theorem and the Friedrichs inequality, we deduce in both cases that
24
24
3.2 Second solution method
In contrast to the first solution method, the Lipschitz continuity of is needed in the second solution method, i.e. there exists a real constant
such that
25
25 To get rid of the nonhomogeneous Dirichlet BC on
, the solution of (Equation19
19
19 ) is prescribed as
, where
is unknown. Next, define the functions
and
by setting
Thanks to the properties of
and
, the function
is monotonically increasing if
is independent of
and strict monotonically increasing if
depends on
. Using the preceding assumptions, the variational formulation of (Equation19
19
19 ) can be rewritten (for given
) for all
as
26
26 The following growth condition on
is satisfied due to the growth condition on
Analogue as in the previous subsection one can state three lemmas.
Lemma 3.4
(Unicity) Assume (Equation44
4 )–(Equation16
16
16 ). For any given
,
, there exist a
and an uniquely determined
solving (Equation23
23
23 ) for
. Moreover, the function
is solving (Equation19
19
19 ).
Proof
This is almost an exact analogue of the proof of Lemma 3.1, except for the appearance of the following lower bound that is valid for each 27
27
Lemma 3.5
(Uniform bound for ) Assume (Equation4
4
4 )–(Equation16
16
16 ). There exists a positive constant
such that for
Proof
The proof follows very closely the proof of Lemma 3.2 using the lower bound (Equation2424
24 ).
At this point, define the coefficient to data map by
Now, the inverse problem with the measured output data
can be formulated as the following operator equation: search
such that
In the following lemma, the continuity of the map
is proved.
Lemma 3.6
( depends continuously on
) Assume (Equation4
4
4 )–(Equation16
16
16 ). There exists a
such that the function
is continuous for
.
Proof
Subtract (Equation2323
23 ) for
from (Equation23
23
23 ) for
and set
to get
which implies using Lemma 3.5
28
28 Recall that
. Hence, the existence of a function
such that
and
is guaranteed by Friedman [Citation12, Lemma 5.1]. Using problem (Equation19
19
19 ), we have that
Therefore, using the mean value theorem we deduce that
This completes the proof.
3.3 Solvability of (Equation17
17
17 )
The following Lemma is a consequence of Sections 3.1 and 3.2.
Lemma 3.7
Assume (Equation44
4 )–(Equation16
16
16 ). If
, then there exist a
and a couple
that solves (Equation17
17
17 ) for
. If
is Lipschitz continuous and
, then there exist a
and a couple
that solves (Equation17
17
17 ) for
.
4 Convergence
In this section, the stability estimates are derived. Afterwards, the limit for is passed to get the existence of a solution to problem (Equation1
1
1 )–(Equation2
2
2 ). Again, the two different solution methods are considered.
4.1 First solution method
The variational formulation of (Equation1818
18 ) on time step
reads as
29
29 The formulation (Equation26
26
26 ) has a solution on
according to Lemma 3.7. The next step is the stability analysis. First, two functions are introduced, which simplifies the proofs. Let
be any monotone increasing real function with
. A primitive function of
is denoted by
. The function
is convex because
. One can check that
30
30 According to
and (Equation27
27
27 ), a function
can be defined such that
Some estimates on the function
are deduced in the following lemma. These a priori estimates will serve as uniform bounds in order to prove convergence.
Lemma 4.1
Let the assumptions of Lemma 3.7 be fulfilled. Then there exists a positive constant such that
(i) | |||||
(ii) |
| ||||
(iii) | |||||
(iv) |
|
Proof
(i) Setting into (Equation26
26
26 ), multiplying by
and summing it up for
we have
According to Abel’s summation rule, the assumption that
is monotonically nondecreasing, the growth condition on
and
, we can derive a lower bound for the first term on the left-hand side (LHS) of the above equation:
On the first term of the right-hand side, we apply the Cauchy and Young inequalities and the Friedrichs inequality to obtain
In the same way, using the trace theorem, one can prove that
After fixing a sufficiently small positive
, an application of the Friedrichs inequality concludes the proof.
(ii) Choosing into (Equation26
26
26 ), multiplying by
and summing up over
, one can get
Using the monotonicity of
and the mean value theorem, the first term on the LHS can be estimated by
The second term on the LHS can be rewritten using Abel’s summation rule, namely
We apply Abel’s summation rule, the Cauchy and Young inequalities, the trace theorem, (Equation13
13
13 ), the Friedrichs inequality (Equation3
3
3 ) and Lemma 4.1(i) on the second term of the RHS to get
when
. Analogue one can prove that for
, it holds that
If
, collecting all considerations above results in
Secondly, we discuss the case
. Also, the following partial summation formula is satisfied
Employing the mean value theorem, assumption (Equation8
8
8 ) and Lemma 4.1(i), we get for
We arrive at
Fixing a sufficiently small
and applying the Friedrichs inequality in both cases give the estimates.
(iii) This follows immediately from Lemma 4.1(ii) coupled with the growth condition on .
(iv) The relation (Equation2626
26 ) can be rewritten for
as
A standard argumentation yields
which implies
31
31 An application of Lemma 4.1(ii) gives the first inequality. Taking the second power in (Equation28
28
28 ), multiplying the inequality by
, summing it up for
, and applying Lemma 4.1(i), we get the second inequality.
The existence of a weak solution will be proved using Rothe’s method. The variational formulation of (Equation11
1 ) and (Equation2
2
2 ) reads as: find
such that
32a
32a
32b
32b
Now, let us introduce the following piecewise linear in time functions and
and the step functions
and
Similarly, the functions
and
are defined. The variational formulation (Equation26
26
26 ) can be rewritten as (
and
)
33a
33a
33b
33b
Now, (29) follows after passage to the limit as in (30). In the case that
is independent of
, the assumptions made so far are not strong enough to prove convergence. Namely, to establish convergence, the strong convergence of
in
is needed. The preceding observation leads to the assumption
Corollary 4.2
(i) There exists a with
(i.e.
is a.e. differentiable in
). Moreover, there exists a subsequence of
(denoted by
again) such that (
)
(ii) Let
. There exists a
with
(i.e.
is a.e. differentiable in
). Moreover, there exists a subsequence of
(denoted by
again) such that (
)
Proof
Thanks to the Rellich-Kondrachov Compactness Theorem [Citation16, Theorem 1, p.272], we have that
(i) | Applying Lemma 4.1(iii) and (iv), we get |
(ii) | Applying Lemma 4.1(ii), we have |
Finally, the theorem to be proved is the following.
Theorem 4.3
Let the assumptions of Lemma 3.7 be fulfilled. Suppose that there exists a positive real constant such that
. Then there exists a weak solution to (29).
Proof
Take any and integrate (30a) on
to get for each
34
34 We have to pass to the limit for
in (Equation31
31
31 ). Each term in (Equation31
31
31 ) is considered separately. The Rothe sequence
is bounded in the space
thanks to Lemma 4.1(i), indeed
Therefore, the reflexivity of
implies (for a subsequence denoted by
again) the following useful result
35
35 Thus
. Firstly, we apply the Minty–Browder’s trick [Citation16] to prove that
. It holds that for
Hence, using Lemma 4.1(iv), we obtain for
Therefore, according to Corollary 4.2(i)
36
36 Thanks to the monotonicity of the function
, it yields for any fixed
that
Due to (Equation32
32a
32a ) and (Equation33
32b
32b ), we have that, for
,
Firstly, suppose that
with
and
. We get
Next, dividing this equation by
, taking the limit
and using the continuity of
, we arrive at
Secondly, assume that
with
and
. Then
Combining the previous results give us
Therefore,
. Applying Corollary 4.2(i) gives for all
that
Now, we focus on the second term in the LHS. This term can be split into two parts. Firstly, we consider the second part. The sequences
and
have the same limit in the space
. Employing Lemma 4.1(ii) gives
37
37 Therefore, if
, it holds, thanks to assumption (Equation8
8
8 ) and Corollary 4.2(ii), that for all
Secondly, applying the Green theorem and taking
in the first part, we deduce that
38
38 At this point, we need some auxiliary results. In light of equation (Equation32
32a
32a ) and Lemma 4.1(i), applying the Nečas inequality [Citation17]
implies
Passing to the limit for
and applying Corollary 4.2(ii), we obtain
which is valid for any small
. Hence,
39
39 Repeating this consideration for
instead of
gives
40
40 Due to the construction, we have that
. This yields that
(for a subsequence) in
. Therefore, employing
, (Equation10
10
10 ), (Equation32
32a
32a ), (Equation34
34
34 ) and (Equation36
36
36 ), we get after passage to the limit as
in (Equation35
35
35 ) that
Applying the density argument
, we conclude that
Moreover, due the continuity of
and
one can deduce that, for any
Collecting all considerations above and passing to the limit for
in (Equation31
31
31 ), we arrive at
This is valid for any
. Differentiation with respect to
gives (29a). Taking the limit
in (30b) and using the continuity of
we get (29b), which concludes the proof.
4.2 Second solution method
Consider the notations in Section 3.2. The variational formulation of (Equation1919
19 ) on time step
for all
is
41
41 with
. Using analogous notations as in the previous Section 4.1, the variational formulation (Equation37
37
37 ) can be rewritten as
42a
42a where
and
. The variational formulation of (Equation1
1
1 ) and (Equation2
2
2 ) for all
reads as: find
with
such that
43a
43a
Analysis similar to that in the previous Section 4.1 shows that the limit in (38a) results into (39a). There is an extra term to take under consideration, namely (
and
)
The convergence of (38b)–(39b) for any
follows from the continuity of
.
This subsection concludes with an analogue of Theorem 4.3.
Theorem 4.4
Let the assumptions of Lemma 3.7 be fulfilled. Moreover, suppose that there exist real constants and
such that
. Then there exists a weak solution to (39).
5 Numerical experiments
The aim of the simulations is to analyse both algorithms proposed in Sections 3 and 4. The 2D Finite Elements code Freefem++ is used.
The domain under consideration is the rectangle , with
in
. The time interval is
, i.e.
. The boundary
is split into three nonoverlapping parts, namely
(right),
(top and bottom) and
(left part of
), see Figure .
In the experiments, the exact solution is prescribed as follows
The function
is the indicator function of the subset
of
. Moreover, it is assumed that
Some simple calculations give the exact data for the numerical experiment
This section is split into two subsections. The first two are devoted to a different solution method. The results are summarized in the third subsection. The purpose is the recovery of
For the time discretization, an equidistant time partitioning is chosen with time step
and for the space discretization, a fixed uniform mesh is used consisting of
triangles.
5.1 First solution method
An uncorrelated noise is added to the additional condition in order to simulate the errors present in real measurements. The noise is generated randomly with given magnitude
and
. Applying the backward Euler difference scheme into (Equation20
20
20 ), a recurrent system of nonlinear elliptic BVPs for
and
have to be solved
44
44 with
45
45 On every time level, Newton’s method is applied to deal with the nonlinearity. More precisely, given a solution
from iteration
, a perturbation
is searched such that
fulfills the nonlinear problem (Equation40
40
40 ). The perturbation
is zero at the boundaries where the Dirichlet conditions are applied. Inserting
in the BVP (Equation40
40
40 ), the
term in the LHS is linearized as
As initial guess
, the solution of the linear problem with
and
is taken. The iterations are stopped when
or when the number of iterations exceed the limit
. The unknowns
are determined by the nonlinear conjugate gradient method. On each time step
the functional
is minimized. The starting point for this algorithm on the first time step is set as
. The starting point on the following time steps is
, the minimizing value of the functional in the previous time step. The algorithm stops after maximum
iterations with the prescribed error tolerance
.
5.2 Second solution method
Again, an uncorrelated noise with magnitude and
is added to the additional condition
. Utilizing the backward Euler difference scheme into (Equation23
23
23 ), a recurrent system of linear elliptic BVPs for
and
has to be solved
46
46 where
is as in (Equation41
41
41 ) and
on
. The nonlinearity is handled in the same way as in the first solution method. Now, on each time step
the functional
is minimized.
5.3 Results
At each time step, the resulting elliptic BVP’s (Equation4040
40 ) and (Equation42
40
40 ) are solved numerically by the finite element method (FEM) using second-order (P2-FEM) Lagrange polynomials. The results from the recovery of
using both solution methods for the different values of the amplitude
are shown in Figures –. The exact
is denoted by a solid line and the approximations
by linespoints;
. The evolution of the absolute
-error for the different time steps is shown in Figure .
Figure 4. Noise : numerical value of
using the first solution method (a) and the second solution method (b);
![Figure 4. Noise e=1%: numerical value of k~i using the first solution method (a) and the second solution method (b); i=1,…,50.](/cms/asset/d96dc01a-4e00-434c-8ce4-87e46548939a/gipe_a_900615_f0004_oc.gif)
Figure 5. The absolute -error using the first solution method (a) and using the second solution method (b) for the different noise levels;
![Figure 5. The absolute k~i-error using the first solution method (a) and using the second solution method (b) for the different noise levels; i=1,…,50.](/cms/asset/a055de22-9834-4885-b80f-e58eb61f50d8/gipe_a_900615_f0005_oc.gif)
The experiments demonstrate that the approximation becomes less accurate with increasing magnitude when the number of time discretization intervals and the number of triangles in the space discretization is fixed. In this experiment, the second solution method is more accurate.
6 Conclusion
In this contribution, a nonlinear parabolic problem of second order with an unknown diffusion coefficient in a subregion is considered. In this subregion, the diffusion coefficient is only time dependent. Two different solution methods are considered. In the first solution method, an additional Dirichlet condition is prescribed on a part of the surface of the region with unknown coefficient. In the second solution method, an additional total flux condition is prescribed through the same surface. First, for both solution methods, the existence of a solution on a single time step is proved using coefficient to data mappings. Afterwards, a numerical algorithm based on Rothe’s method is established and the convergence of this scheme is shown. No uniqueness of the solution can be assured. The convergence of the numerical algorithm is illustrated by a numerical experiment.
Acknowledgments
The authors thank the reviewers for closely reading the manuscript and for their useful suggestions.
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