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) (Equation11 1 )–(Equation22 2 ): find a couple such that ( fixed)1 1 and such that the following boundary measurements are satisfied2 2 More specific, the inverse problem of determining the unknown coefficient from the measured data (Equation22 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 (Equation22 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 iswhich is clearly a Hilbert space with norm , where represents the norm in . Also a subspace of is considered, namelyDue 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 (Equation22 2 ) on a single time step. The continuous function (diffusion term) has to satisfy4 4 5 5 Moreover,6 6
Remark 2.1
The condition in (Equation66 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 and7 7 8 8
Remark 2.2
If , then the right-hand side (RHS) of (Equation11 1 ) can be redefined as . From now on, it is assumed that such that the -term in problem (Equation11 1 ) and (Equation22 2 ) can be cancelled out of the equations if , see also equation (Equation1414 14 ). Therefore, assumptions (Equation77 7 ) and (Equation88 8 ) are only necessary if .
Remark 2.3
The source function in problem (Equation11 1 ) and (Equation22 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 (Equation22 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 satisfies17 17 where . In this section, the existence of for any is shown in two ways: two different methods for solving (Equation1717 17 ) are presented.
In the first one, it is assumed that is given and a solution of the direct problem18 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 considered20 20 In the following lemma, the theory of monotone operators is applied to prove the existence of a weak solution to (Equation2020 20 ) for given . The interested reader is referred to [Citation14] for further information.
Lemma 3.1
(Unicity) Assume (Equation44 4 )–(Equation1515 15 ). For any given , , there exist a and an uniquely determined solving (Equation1818 18 ) for .
Proof
We consider the nonlinear operators defined astogether with the linear functionals such thatWe proof that is strictly monotone, coercive and demicontinuous. In the case of , it follows from the monotonicity of thatThe strict monotonicity of follows by an application of the Friedrichs inequality (Equation33 3 ). Secondly, we do the case that . Using the mean value theorem and the Young’s inequality, we haveFixing a sufficiently small and gives the strict monotonicity of the operator . The operator is coercive ifThis is always satisfied. If , then the following lower bound is valid for sufficiently small In the case , the second constant in this lower bound disappearsThe 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 (Equation44 4 )–(Equation1515 15 ). There exists a positive constant such that
Proof
We consider as a parameter. We set into (Equation2020 20 ) and getRemark 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 (Equation33 3 )Using the monotonicity of and the uniform bounds (Equation99 9 ) and (Equation1010 10 ) one can easily getAnalogue, in the case , we can readily obtainand thereforeFixing 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 mapThen, the inverse problem with the measured output data can be formulated as the following operator equation: search such thatThe 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 (Equation1717 17 ), see Lemma 3.7.
Lemma 3.3
( depends continuously on ) Assume (Equation44 4 )–(Equation1515 15 ). There exists a such that the function is continuous for .
Proof
Subtract (Equation2020 20 ) for from (Equation2020 20 ) for and set to getThe first term in the RHS can be bounded using the Cauchy’s and Young’s inequality, Lemma 3.2 and the uniform bound (Equation1010 10 )The second term in the RHS disappears if . An obvious calculation employing the monotonicity of gives21 21 We use the mean value theorem in the case of to getTherefore, we can derive the following estimate22 22 We fix a sufficiently small and to conclude that in both cases23 23 Using the trace theorem and the Friedrichs inequality, we deduce in both cases that24 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 that25 25 To get rid of the nonhomogeneous Dirichlet BC on , the solution of (Equation1919 19 ) is prescribed as , where is unknown. Next, define the functions and by settingThanks 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 (Equation1919 19 ) can be rewritten (for given ) for all as26 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 )–(Equation1616 16 ). For any given , , there exist a and an uniquely determined solving (Equation2323 23 ) for . Moreover, the function is solving (Equation1919 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 (Equation44 4 )–(Equation1616 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 byNow, the inverse problem with the measured output data can be formulated as the following operator equation: search such thatIn the following lemma, the continuity of the map is proved.
Lemma 3.6
( depends continuously on ) Assume (Equation44 4 )–(Equation1616 16 ). There exists a such that the function is continuous for .
Proof
Subtract (Equation2323 23 ) for from (Equation2323 23 ) for and set to getwhich implies using Lemma 3.528 28 Recall that . Hence, the existence of a function such that and is guaranteed by Friedman [Citation12, Lemma 5.1]. Using problem (Equation1919 19 ), we have thatTherefore, using the mean value theorem we deduce thatThis completes the proof.
3.3 Solvability of (Equation1717 17 )
The following Lemma is a consequence of Sections 3.1 and 3.2.
Lemma 3.7
Assume (Equation44 4 )–(Equation1616 16 ). If , then there exist a and a couple that solves (Equation1717 17 ) for . If is Lipschitz continuous and , then there exist a and a couple that solves (Equation1717 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 (Equation11 1 )–(Equation22 2 ). Again, the two different solution methods are considered.
4.1 First solution method
The variational formulation of (Equation1818 18 ) on time step reads as29 29 The formulation (Equation2626 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 that30 30 According to and (Equation2727 27 ), a function can be defined such thatSome 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) | if and if | ||||
(iii) | |||||
(iv) | and |
Proof
(i) Setting into (Equation2626 26 ), multiplying by and summing it up for we haveAccording 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 obtainIn the same way, using the trace theorem, one can prove thatAfter fixing a sufficiently small positive , an application of the Friedrichs inequality concludes the proof.
(ii) Choosing into (Equation2626 26 ), multiplying by and summing up over , one can getUsing the monotonicity of and the mean value theorem, the first term on the LHS can be estimated byThe second term on the LHS can be rewritten using Abel’s summation rule, namelyWe apply Abel’s summation rule, the Cauchy and Young inequalities, the trace theorem, (Equation1313 13 ), the Friedrichs inequality (Equation33 3 ) and Lemma 4.1(i) on the second term of the RHS to getwhen . Analogue one can prove that for , it holds thatIf , collecting all considerations above results inSecondly, we discuss the case . Also, the following partial summation formula is satisfiedEmploying the mean value theorem, assumption (Equation88 8 ) and Lemma 4.1(i), we get for We arrive atFixing 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 asA standard argumentation yieldswhich implies31 31 An application of Lemma 4.1(ii) gives the first inequality. Taking the second power in (Equation2828 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 (Equation22 2 ) reads as: find such that32a 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 (Equation2626 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 (Equation3131 31 ). Each term in (Equation3131 31 ) is considered separately. The Rothe sequence is bounded in the space thanks to Lemma 4.1(i), indeedTherefore, the reflexivity of implies (for a subsequence denoted by again) the following useful result35 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 thatDue to (Equation3232a 32a ) and (Equation3332b 32b ), we have that, for ,Firstly, suppose that with and . We getNext, dividing this equation by , taking the limit and using the continuity of , we arrive atSecondly, assume that with and . ThenCombining the previous results give usTherefore, . Applying Corollary 4.2(i) gives for all thatNow, 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) gives37 37 Therefore, if , it holds, thanks to assumption (Equation88 8 ) and Corollary 4.2(ii), that for all Secondly, applying the Green theorem and taking in the first part, we deduce that38 38 At this point, we need some auxiliary results. In light of equation (Equation3232a 32a ) and Lemma 4.1(i), applying the Nečas inequality [Citation17]impliesPassing to the limit for and applying Corollary 4.2(ii), we obtainwhich is valid for any small . Hence,39 39 Repeating this consideration for instead of gives40 40 Due to the construction, we have that . This yields that (for a subsequence) in . Therefore, employing , (Equation1010 10 ), (Equation3232a 32a ), (Equation3434 34 ) and (Equation3636 36 ), we get after passage to the limit as in (Equation3535 35 ) thatApplying the density argument , we conclude thatMoreover, due the continuity of and one can deduce that, for any Collecting all considerations above and passing to the limit for in (Equation3131 31 ), we arrive atThis 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 is41 41 with . Using analogous notations as in the previous Section 4.1, the variational formulation (Equation3737 37 ) can be rewritten as42a 42a where and . The variational formulation of (Equation11 1 ) and (Equation22 2 ) for all reads as: find with such that43a 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 followsThe function is the indicator function of the subset of . Moreover, it is assumed thatSome simple calculations give the exact data for the numerical experimentThis 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 ofFor 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 (Equation2020 20 ), a recurrent system of nonlinear elliptic BVPs for and have to be solved44 44 with45 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 thatfulfills the nonlinear problem (Equation4040 40 ). The perturbation is zero at the boundaries where the Dirichlet conditions are applied. Inserting in the BVP (Equation4040 40 ), the term in the LHS is linearized asAs 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 functionalis 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 (Equation2323 23 ), a recurrent system of linear elliptic BVPs for and has to be solved46 46 where is as in (Equation4141 41 ) and on . The nonlinearity is handled in the same way as in the first solution method. Now, on each time step the functionalis minimized.
5.3 Results
At each time step, the resulting elliptic BVP’s (Equation4040 40 ) and (Equation4240 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 .
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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