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Articles

Determination of a time-dependent diffusivity in a nonlinear parabolic problem

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Pages 307-330 | Received 20 Nov 2013, Accepted 19 Feb 2014, Published online: 10 Apr 2014

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 Rd,d1. 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.

AMS Subject Classifications:

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. [Citation1Citation4]. The determination of a time-dependent diffusivity in parabolic equations has been considered in papers [Citation5, Citation6]. Nonlinear problems have been studied in [Citation7Citation10]. The problems studied in these papers are all one-dimensional in space. However, the analysis made in this article is valid for every dimension d1.

The mathematical setting is the following. Let ΩRd be a bounded domain with a Lipschitz continuous boundary Γ. The domain Ω is split into two nonoverlapping parts Ω0 and Ω\Ω¯0 with the assumption that Ω\Ω¯0 cannot surround Ω0. A transient convection–diffusion process is considered in Ω. The diffusion coefficient K takes the form K=k(t,x)κ(t,x) for a known κ and k(t,x)=1 for xΩ\Ω¯0 and k(t,x)=k(t) for xΩ0. The boundary Γ is split into three nonoverlapping parts, namely ΓN (Neumann part), ΓD (Dirichlet part) and Γ0ΓΩ0, on which beside a Dirichlet boundary condition (BC) also the total flux through this part is prescribed. It is assumed that Γ¯DΓ¯0= with μ(Γ0)>0, where μ is the Lebesgue measure. The purpose of this article is to study the following nonlinear parabolic initial boundary value problem (IBVP) (Equation1)–(Equation2): find a couple (K,u) such that (T>0 fixed)1 tθ(u)-·Ku+a(u)=finQT:=(0,T)×Ω;u=gDin(0,T)×ΓD;-Ku-a(u)·ν=gNin(0,T)×ΓN;u(0)=u0inΩ;1 and such that the following boundary measurements are satisfied2 Γ0-Ku-a(u)·ν=h(t)in(0,T);u=U(t)on(0,T)×Γ0.2 More specific, the inverse problem of determining the unknown coefficient K from the measured data (Equation2) is considered. It is determined under which assumptions on the data this inverse problem has a weak solution (K,u). Also, a nonlinear numerical algorithm based on backward Euler method is designed to approximate the solution (K,u) and the convergence of approximations towards the weak solution in suitable function spaces is shown. An easier case with θ(u)=u and a0 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 (Equation1) and (Equation2) 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 v in Rd is denoted by |v|. To increase the readability of the text, it is assumed without loss of generality that gD=0 and gN=0. The suitable choice of a test space isV:=φH1(Ω);φ|ΓD=0,φ|Γ0=constant,which is clearly a Hilbert space with norm uV2=u2+u2, where · represents the norm in L2(Ω). Also a subspace of V is considered, namelyW:=φH1(Ω);φ|Γ0ΓD=0.Due to the homogeneous Dirichlet boundary condition (BC) on the boundary ΓD, the following Friedrichs inequality holds for every function uV3 uVCu.3 Consequently, the norm in the space V is equivalent with the seminorm u. The norm in the trace space L2(γ), with γΓ, is written by ·L2(γ). The dual space of V is denoted by V.

What now follows is a list of assumptions that are necessary to prove the existence of a weak solution to the nonlinear problem (Equation1) and (Equation2) on a single time step. The continuous function θ:RR (diffusion term) has to satisfy4 θ(0)=0;4 5 |θ(s)|C(1+|s|)a.e. inR.5 Moreover,6 θ(s)0a.e. inRifa(u)=a(aindependent ofu);θ(s)θ0>0a.e. inRifa(u)a(adepends onu).6

Remark 2.1

The condition a(u(t,x))=a(t,x) in (Equation6) means that a does not depend on the solution u, but only on time and position, i.e. (t,x)QT. If a(u)a, then a is depending on u, for instance a(u)=bu with b:QTRd.

The vector-valued function a:RRd (convection term) is also continuous and7 |a(s)|=|a1(s),a2(s),,ad(s)|Ca.e. inR;7 8 |a(s)|=|a1(s),a2(s),,ad(s)|Ca.e. inR.8

Remark 2.2

If a(u)=a, then the right-hand side (RHS) f of (Equation1) can be redefined as f+·a. From now on, it is assumed that ·taL2((0,T),L2(Ω)) such that the a(u)-term in problem (Equation1) and (Equation2) can be cancelled out of the equations if a(u)=a, see also equation (Equation14). Therefore, assumptions (Equation7) and (Equation8) are only necessary if a(u)a.

Remark 2.3

The source function f in problem (Equation1) and (Equation2) depends only on the time and space variable. However, our numerical procedure can be generalized to a nonlinear f(u), assuming θθ0>0 if a(u)=a. 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 θ.

The following assumptions on the other data functions are adopted9 0<C0kC1;9 10 0<D0κD1;10 11 κC([0,T],H1(Ω));11 12 UC([0,T]);12 13 hL2((0,T))hC([0,T]);13 14 tfL2((0,T),L2(Ω))fC([0,T],L2(Ω)).14

The initial datum satisfies15 u0H1(Ω).15 Note that the data function U can be prolonged into the whole domain Ω by a function U~ in such a way that [Citation12, Lemma 5.1]16 U~C[0,T],H1(Ω),U~=0on0,T×ΓD;Uon0,T×Γ0.16 Finally, some useful (in)equalities are stated, which can easily be derived:2i=1n(ai-ai-1)ai=an2-a02+i=1n(ai-ai-1)2,aiR;Abel's summation ruleandabεa2+Cεb2,a,bR,ε>0.Young's inequality

Remark 2.4

In this contribution, the values C,ε,Cε are generic and positive constants independent of the discretization parameter τ, see Section 3. They can be different from place to place. The value ε is small and Cε=Cε-1. To reduce the number of arbitrary constants, we use the notation ab if there exists a positive constant C such that aCb.

3 A single time step

Rothe’s method [Citation13] is applied to prove the existence of a weak solution to problem (Equation1) and (Equation2). An equidistant time-partitioning is used with time step τ=T/n<1, for any nN. Let us introduce the notations ti=iτ and for any function zzi=z(ti),δzi=zi-zi-1τ.The following recursive approximation scheme is proposed for i=1,,n: find the unknown couple (ki,ui)R+×V on each time step that satisfies17 δθ(ui)-·Kiui+a(ui)=fiinΩ;ui=0onΓD;-Kiui-a(ui)·ν=0onΓN;Γ0-Kiui-a(ui)·ν=hiui=UionΓ0,17 where Ki=kiκi. In this section, the existence of (Ki,ui) for any i=1,,n is shown in two ways: two different methods for solving (Equation17) are presented.

In the first one, it is assumed that ki is given and a solution of the direct problem18 δθ(ui)-·Kiui+a(ui)=fiinΩ;ui=0onΓD;-Kiui-a(ui)·ν=0onΓN;Γ0-Kiui-a(ui)·ν=hi18 is looked for. Afterwards, it is shown that the trace of ui on Γ0 depends continuously on ki. Then, ki is determined such that ui|Γ0=Ui. This solution method is discussed in detail in Section 3.1.

In the second solution method, the following direct problem is solved19 δθ(ui)-·Kiui+a(ui)=fiinΩ;ui=0onΓD;-Kiui-a(ui)·ν=0onΓN;ui=UionΓ019 for a given ki. It is proved that the total flux through Γ0, Γ0-Kiui-a(ui)·ν, depends continuously on ki. At the end, ki is found such that Γ0-Kiui-a(ui)·ν=hi. 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 (Equation17). This lemma is based on the results of Sections 3.1 and 3.2.

3.1 First solution method

First, the variational formulation of (Equation18) is considered20 1τθ(ui),φ+Kiui+a(ui),φ=fi,φ-hiφ|Γ0+1τθ(ui-1),φ,(φV).20 In the following lemma, the theory of monotone operators is applied to prove the existence of a weak solution to (Equation20) for given Ki. The interested reader is referred to [Citation14] for further information.

Lemma 3.1

(Unicity)    Assume (Equation4)–(Equation15). For any given ki>0, i=1,,n, there exist a τ0>0 and an uniquely determined ukiV solving (Equation18) for τ<τ0.

Proof

We consider the nonlinear operators Ai:VV,i=1,,n, defined asAi(u),φ:=1τθ(u),φ+Kiu+a(u),φ,together with the linear functionals Fi:VR,i=1,,n, such thatFi,φ:=fi,φ-hiφ|Γ0+1τθ(ui-1),φ.We proof that Ai is strictly monotone, coercive and demicontinuous. In the case of a(u)=a, it follows from the monotonicity of θ thatAi(u)-Ai(v),u-v=1τθ(u)-θ(v),u-v+Ki(u-v),(u-v)C0D0(u-v)2.The strict monotonicity of Ai follows by an application of the Friedrichs inequality (Equation3). Secondly, we do the case that a(u)a. Using the mean value theorem and the Young’s inequality, we haveAi(u)-Ai(v),u-v=1τθ(ξ1)[u-v],u-v+Ki(u-v),(u-v)+a(ξ2)[u-v],(u-v)θ0τu-v2+C0D0(u-v)2-Cεa(ξ2)[u-v]2-ε(u-v)2θ0τ-Cεu-v2+C0D0-ε(u-v)2.Fixing a sufficiently small ε>0 and τ(ε)>0 gives the strict monotonicity of the operator Ai. The operator Ai is coercive iflimuV+Ai(u),uuV=+.This is always satisfied. If a(u)a, then the following lower bound is valid for sufficiently small εAi(u),u=1τθ(u),u+Kiu+a(u),u(4)θ0τu2+C0D0u2-Cεa(u)2-εu2(7)CuV2-C.In the case a(u)=a, the second constant in this lower bound disappearsAi(u),uC0D0u2(3)CuV2.The demicontinuity of Ai follows from the continuity of θ and a. The functionals Fi belong to V if ui-1L2(Ω) due to fiL2(Ω), the boundedness of h, the trace theorem (cf. [Citation15, Theorem 3.9]) and the growth condition on θ|Fi,φ|fiφ+|hi|φL2(Γ0)|Γ0|+1τθ(ui-1)φ(1+ui-1)φV.Consequently, since u0L2(Ω), the equation Ai(u)=Fi admits a unique solution for i=1,,n.

In the subsequent Lemma, the existence of an uniform bound for uki independent of ki is proved.

Lemma 3.2

(Uniform bound for uki)    Assume (Equation4)–(Equation15). There exists a positive constant C such thatukiV2Cτ2forC0kiC1,i=1,,n.

Proof

We consider ki>0 as a parameter. We set φ=uki into (Equation20) and get1τθ(uki),uki+Kiuki,uki=fi,uki-hiuki|Γ0+1τθ(ui-1),uki-a(uki),uki.Remark that the solution on the previous time step, ui-1, belongs to L2(Ω). First, we consider the case a(u)=a. Using the Cauchy’s and Young’s inequalities together with the growth condition on θ, we obtain the trace theorem and the Friedrichs inequality (Equation3)fi,ukiCεfi2+εuki2Cε+εuki2,hiuki|Γ0Cεhi2+εukiL2(Γ0)2|Γ0|Cε+εuki2,1τθ(ui-1),ukiCετ2θ(ui-1)2+εuki2Cετ2+εuki2.Using the monotonicity of θ and the uniform bounds (Equation9) and (Equation10) one can easily getC0D0-εuki2Cετ2.Analogue, in the case a(u)a, we can readily obtaina(uki),ukiCεa(uki)2+εuki2(7)Cε+εuki2and thereforeθ0τuki2+C0D0-εuki2Cετ2.Fixing a sufficiently small positive ε in both cases and applying the Friedrichs inequality if a(u)=a, we conclude the proof.

To prove that the trace of ui on Γ0 depends continuously on ki, it is necessary to define the following coefficient to data mapT:[C0,C1]R:kiT(ki):=uki|Γ0.Then, the inverse problem with the measured output data U(t) can be formulated as the following operator equation: search ki such thatT(ki)=Ui.The continuity of this input–output map T is investigated in the following lemma. It is this property that leads to the existence of a solution to problem (Equation17), see Lemma 3.7.

Lemma 3.3

(uki depends continuously on ki)   Assume (Equation4)–(Equation15). There exists a τ0>0 such that the function T is continuous for τ<τ0.

Proof

Subtract (Equation20) for ki=β from (Equation20) for ki=α and set φ=uα-uβ to get1τθ(uα)-θ(uβ),uα-uβ+ακi(uα-uβ),(uα-uβ)=(β-α)κiuβ,(uα-uβ)-a(uα)-a(uβ),(uα-uβ).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)(β-α)κiuβ,(uα-uβ)Cε(β-α)κiuβ2+ε(uα-uβ)2Cετ2(β-α)2+ε(uα-uβ)2.The second term in the RHS disappears if a(u)=a. An obvious calculation employing the monotonicity of θ gives21 (αD0-ε)(uα-uβ)2Cετ2(α-β)2.21 We use the mean value theorem in the case of a(u)a to geta(uα)-a(uβ),(uα-uβ)Cεa(uα)-a(uβ)2+ε(uα-uβ)2Cεuα-uβ2+ε(uα-uβ)2.Therefore, we can derive the following estimate22 θ0τ-Cεuα-uβ2+(αD0-ε)(uα-uβ)2Cετ2(α-β)2.22 We fix a sufficiently small ε and τ(ε) to conclude that in both cases23 (uα-uβ)2(α-β)2τ2.23 Using the trace theorem and the Friedrichs inequality, we deduce in both cases that24 T(α)-T(β)=uα-uβL2(Γ0)|Γ0|(uα-uβ)(21)1τ|α-β|.24

3.2 Second solution method

In contrast to the first solution method, the Lipschitz continuity of θ is needed in the second solution method, i.e. there exists a real constant θ1 such that25 θ(s)θ1a.e. inR.25 To get rid of the nonhomogeneous Dirichlet BC on Γ0, the solution of (Equation19) is prescribed as ui:=vi+U~i,i=1,,n, where vi is unknown. Next, define the functions ϑ:RR and b:RRd by settingϑ(s)=θ(s+U~i)andb(s)=a(s+U~i).Thanks to the properties of θ and a, the function ϑ is monotonically increasing if b is independent of vi and strict monotonically increasing if b depends on vi. Using the preceding assumptions, the variational formulation of (Equation19) can be rewritten (for given Ki) for all φW as26 1τϑ(vi),φ+Kivi+b(vi),φ=fi,φ-KiU~i,φ+1τϑ(vi-1),φ.26 The following growth condition on ϑ is satisfied due to the growth condition on θϑ(v)1+v+U~i(16)1+v,v:ΩR.Analogue as in the previous subsection one can state three lemmas.

Lemma 3.4

(Unicity)    Assume (Equation4)–(Equation16). For any given ki>0, i=1,,n, there exist a τ0>0 and an uniquely determined vkiW solving (Equation23) for τ<τ0. Moreover, the function uki=vki+U~iV is solving (Equation19).

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 vW27 ϑ(v),v=θ(v+U~i),v+U~i-θ(v+U~i),U~i-θ(v+U~i),U~i-C1+v+U~iU~i(3)-Cε-εv2.27

Lemma 3.5

(Uniform bound for uki)    Assume (Equation4)–(Equation16). There exists a positive constant C such that for i=1,,nvkiV2Cτ2forC0kiC1,i=1,,n.

Proof

The proof follows very closely the proof of Lemma 3.2 using the lower bound (Equation24).

At this point, define the coefficient to data map Ψ byΨ:[C0,C1]R:kiΨ(ki):=-Γ0kiκiuki+a(uki)·ν.Now, the inverse problem with the measured output data h(t) can be formulated as the following operator equation: search ki such thatΨ(ki)=hi.In the following lemma, the continuity of the map T is proved.

Lemma 3.6

(uki depends continuously on ki)   Assume (Equation4)–(Equation16). There exists a τ0>0 such that the function Ψ is continuous for τ<τ0.

Proof

Subtract (Equation23) for ki=β from (Equation23) for ki=α and set φ=vα-vβ to get1τϑ(vα)-ϑ(vβ),vα-vβ+ακi(vα-vβ),(vα-vβ)=(β-α)κivβ,(vα-vβ)+(β-α)κiU~i,(vα-vβ)-b(vα)-b(vβ),(vα-vβ),which implies using Lemma 3.528 (vα-vβ)2(α-β)2τ2.28 Recall that Γ¯DΓ¯0=. Hence, the existence of a function ΦC(Ω¯) such that Φ|ΓD=0 and Φ|Γ0=1 is guaranteed by Friedman [Citation12, Lemma 5.1]. Using problem (Equation19), we have thatΨ(ki)=-kiκiuki+a(uki)·ν,1Γ0=fi,Φ-1τθ(uki),Φ-kiκiuki+a(uki),Φ+1τθ(ui-1),Φ.Therefore, using the mean value theorem we deduce that|Ψ(α)-Ψ(β)|=|τ-1θ(uα)-θ(uβ),Φ+a(uα)-a(uβ),Φ+ακi(uα-uβ),Φ+(α-β)κiuβ,Φ|(22)θ1τvα-vβΦ+vα-vβΦ+(vα-vβ)Φ+|α-β|vβ+U~iΦ(3)1τ(vα-vβ)+|α-β|(25)|α-β|τ2.This completes the proof.

3.3 Solvability of (Equation17)

The following Lemma is a consequence of Sections 3.1 and 3.2.

Lemma 3.7

Assume (Equation4)–(Equation16). If U(t)T([C0,C1])t[0,T], then there exist a τ0>0 and a couple (ki,ui)R+×V that solves (Equation17) for τ<τ0. If θ is Lipschitz continuous and h(t)Ψ([C0,C1])t[0,T], then there exist a τ0>0 and a couple (ki,ui)R+×V that solves (Equation17) for τ<τ0.

4 Convergence

In this section, the stability estimates are derived. Afterwards, the limit for n is passed to get the existence of a solution to problem (Equation1)–(Equation2). Again, the two different solution methods are considered.

4.1 First solution method

The variational formulation of (Equation18) on time step ti reads as29 δθ(ui),φ+Kiui+a(ui),φ+hiφ|Γ0=fi,φφVui|Γ0=Ui.29 The formulation (Equation26) has a solution on ti according to Lemma 3.7. The next step is the stability analysis. First, two functions are introduced, which simplifies the proofs. Let γ be any monotone increasing real function with γ(0)=0. A primitive function of γ is denoted by Φγ(z):=0zγ(s)ds. The function Φγ(z) is convex because Φγ(z)=γ(z)0. One can check that30 γ(z1)(z2-z1)Φγ(z2)-Φγ(z1)γ(z2)(z2-z1),z1,z2R.30 According to γ(0)=0 and (Equation27), a function Φ~γ(z) can be defined such thatΦ~γ(z):=zγ(z)-Φγ(z)0,zR.Some estimates on the function ui 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 C such that

(i)

i=1nuiV2τC;

(ii)

max1inuiV2+i=1nui-ui-1V2C        if a(u)=a and i=1nδui2τ+max1inuiV2+i=1nui-ui-1V2C        if a(u)a;

(iii)

max1inθ(ui)V2C;

(iv)

max1inδθ(ui)VC        and       i=1nδθ(ui)V2τC;

Proof

(i) Setting φ=ui into (Equation26), multiplying by τ and summing it up for i=1,,j we havei=1jδθ(ui),uiτ+i=1jKiui,uiτ=i=1jfi,uiτ-i=1jhiui|Γ0τ-i=1ja(ui),uiτ.According to Abel’s summation rule, the assumption that θ is monotonically nondecreasing, the growth condition on θ and u0L2(Ω), we can derive a lower bound for the first term on the left-hand side (LHS) of the above equation:i=1jθ(ui)-θ(ui-1),ui=θ(uj),uj-θ(u0),u0-i=1jθ(ui-1),ui-ui-1(27)θ(uj),uj-θ(u0),u0-i=1jΩΦθ(ui)-Φθ(ui-1)=θ(uj),uj-ΩΦθ(uj)-θ(u0),u0-ΩΦθ(u0)=ΩΦ~θ(uj)-ΩΦ~θ(u0)-Ωu0θ(u0)-C.On the first term of the right-hand side, we apply the Cauchy and Young inequalities and the Friedrichs inequality to obtaini=1jfi,uiτCεi=1jfi2τ+εi=1jui2τCε+εi=1jui2τ.In the same way, using the trace theorem, one can prove thati=1jhiui|Γ0τCεi=1j|hi|2τ+εi=1juiL2(Γ0)2τCε+εi=1jui2τ;i=1ja(ui),uiτCεi=1ja(ui)2τ+εi=1jui2τCε+εi=1jui2τ.After fixing a sufficiently small positive ε, an application of the Friedrichs inequality concludes the proof.

(ii) Choosing φ=δui into (Equation26), multiplying by τ and summing up over i=1,,j, one can geti=1jδθ(ui),δuiτ+i=1jKiui,δuiτMYAMP]=i=1jfi,δuiτ-i=1jhiδui|Γ0τ-i=1ja(ui),δuiτ.Using the monotonicity of θ and the mean value theorem, the first term on the LHS can be estimated byi=1jδθ(ui),δuiτ=i=1j1τθ(ui)-θ(ui-1),ui-ui-10ifa(u)=a;i=1jδθ(ui),δuiτθ0τi=1jui-ui-12=θ0i=1jδui2τifa(u)a.The second term on the LHS can be rewritten using Abel’s summation rule, namelyi=1jKiui,δuiτC0D02uj2-u02+i=1j(ui-ui-1)2.We apply Abel’s summation rule, the Cauchy and Young inequalities, the trace theorem, (Equation13), the Friedrichs inequality (Equation3) and Lemma 4.1(i) on the second term of the RHS to geti=1jhiδui|Γ0τ=uj|Γ0hj-u0|Γ0h0-i=1jδhiui-1|Γ0τεujL2(Γ0)2+Cε|hj|2+C+Ci=1j|δhi|2τ+Ci=1jui-1L2(Γ0)2τCε+εuj2+Cu02τ+Ci=1jui2τCε+εuj2,when τ<τ0. Analogue one can prove that for τ<τ0, it holds thati=1jfi,δuiτ=fj,uj-f0,u0-i=1jδfi,ui-1τCε+εuj2.If a(u)=a, collecting all considerations above results inC0D02uj2-u02+i=1j(ui-ui-1)2Cε+εuj2.Secondly, we discuss the case a(u)a. Also, the following partial summation formula is satisfiedi=1ja(ui),δuiτ=a(uj),uj-a(u0),u0-i=1jδa(ui),ui-1τ.Employing the mean value theorem, assumption (Equation8) and Lemma 4.1(i), we get for τ<τ0i=1ja(ui),δuiτCε+εuj2+εi=1jδui2τ.We arrive atθ0i=1jδui2τ+C0D02uj2+i=1j(ui-ui-1)2Cε+εuj2+εi=1jδui2τ.Fixing a sufficiently small ε 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 θ.

(iv) The relation (Equation26) can be rewritten for φV asδθ(ui),φ=fi,φ-Kiui,φ-a(ui),φ-hiφ|Γ0.A standard argumentation yields|δθ(ui),φ|1+fi+|hi|+uiφV,which implies31 δθ(ui)V=supφVφV1|δθ(ui),φ|1+fi+|hi|+ui.31 An application of Lemma 4.1(ii) gives the first inequality. Taking the second power in (Equation28), multiplying the inequality by τ, summing it up for i=1,,j, 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 (Equation1) and (Equation2) reads as: find (K,u) such that32a tθ(u),φ+Ku+a(u),φ+hφ|Γ0=f,φφV32a 32b u|Γ0=U.32b

Now, let us introduce the following piecewise linear in time functions θn and unθn(0)=θ(u0)θn(t)=θ(ui-1)+(t-ti-1)δθ(ui)fort(ti-1,ti];un(0)=u0un(t)=ui-1+(t-ti-1)δuifort(ti-1,ti];and the step functions u¯n and θ¯nu¯n(0)=u0,u¯n(t)=ui,fort(ti-1,ti];θ¯n(0)=θ(u0),θ¯n(t)=θ(ui),fort(ti-1,ti].Similarly, the functions K¯n,h¯n,U¯n and f¯n are defined. The variational formulation (Equation26) can be rewritten as (t(0,T) and φV)33a tθn(t),φ+K¯n(t)u¯n(t)+a(u¯n(t)),φ+h¯n(t)φ|Γ0=f¯n(t),φφV33a 33b u¯n|Γ0=U¯n.33b

Now, (29) follows after passage to the limit as τ0 in (30). In the case that a is independent of u, the assumptions made so far are not strong enough to prove convergence. Namely, to establish convergence, the strong convergence of u¯n in L2((0,T),L2(Ω)) is needed. The preceding observation leads to the assumptionθ(s)θ0>0a.e. inR.

Corollary 4.2

(i) There exists a wC([0,T],V)L((0,T),L2(Ω)) with twL2((0,T),V) (i.e. w is a.e. differentiable in (0,T)). Moreover, there exists a subsequence of θn (denoted by θn again) such that (t(0,T))θnwinC([0,T],V),θ¯n(t)w(t)inL2(Ω),θn(t)w(t)inL2(Ω),tθntwinL2((0,T),V).(ii) Let θθ0>0. There exists a uC([0,T],L2(Ω))L((0,T),L2(Ω)) with tuL2((0,T),L2(Ω)) (i.e. u is a.e. differentiable in (0,T)). Moreover, there exists a subsequence of un (denoted by un again) such that (t(0,T))unuinC([0,T],L2(Ω)),u¯n(t)u(t)inV,un(t)u(t)inV,tuntuinL2((0,T),L2(Ω)).

Proof

Thanks to the Rellich-Kondrachov Compactness Theorem [Citation16, Theorem 1, p.272], we have thatVL2(Ω)(L2(Ω))V.

(i)

Applying Lemma 4.1(iii) and (iv), we getmaxt[0,T]θ¯n(t)2+0TtθnV2C.

Consequently, an application of [Citation13, Lemma 1.3.13] gives the proof.
(ii)

Applying Lemma 4.1(ii), we havemaxt[0,T]u¯n(t)V2+0Ttun2C.

Hence, an application of [Citation13, Lemma 1.3.13] concludes the proof.

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 θ0 such that θθ0>0. Then there exists a weak solution to (29).

Proof

Take any ξ(0,T) and integrate (30a) on (0,ξ) to get for each φV34 0ξtθn(t),φ+0ξK¯n(t)u¯n(t)+a(u¯n(t)),φ+0ξh¯n(t)φ|Γ0=0ξf¯n(t),φ.34 We have to pass to the limit for n in (Equation31). Each term in (Equation31) is considered separately. The Rothe sequence {u¯n}nN is bounded in the space L2((0,T),V) thanks to Lemma 4.1(i), indeedu¯nL2((0,T),V)2=0Tu¯n(t)V2dt=i=1nti-1tiu¯n(t)V2dt=i=1nuiV2τC.Therefore, the reflexivity of L2((0,T),V) implies (for a subsequence denoted by u¯n again) the following useful result35 u¯nuinL2((0,T),V).35 Thus uL2((0,T),V)C([0,T],L2(Ω)). Firstly, we apply the Minty–Browder’s trick [Citation16] to prove that w=θ(u). It holds that for t(ti-1,ti]θn-θ¯n(t)=θ(ui-1)+t-ti-1τθ(ui)-θ(ui-1)-θ(ui)=(t-ti)tθn(t).Hence, using Lemma 4.1(iv), we obtain for t(ti-1,ti]θn-θ¯n(t)Vτtθn(t)Vτ.Therefore, according to Corollary 4.2(i)36 θ¯nwinC([0,T],V).36 Thanks to the monotonicity of the function θ, it yields for any fixed vL2((0,T),L2(Ω)) that0Tθ(u¯n)-θ(v),u¯n-v0.Due to (Equation32) and (Equation33), we have that, for τ0,0Tw-θ(v),u-v0.Firstly, suppose that v=u+εz with ε>0 and zL2((0,T),L2(Ω)). We get0Tw-θ(u+εz),-εz0.Next, dividing this equation by -ε, taking the limit ε0 and using the continuity of θ, we arrive at0Tw-θ(u),z0.Secondly, assume that v=u-εz with ε>0 and zL2((0,T),L2(Ω)). Then0Tw-θ(u),z0.Combining the previous results give us0Tw-θ(u),z=0,zL2((0,T),L2(Ω)).Therefore, w=θ(u)L2((0,T),L2(Ω)). Applying Corollary 4.2(i) gives for all φV that0ξtθn(t),φ-0ξtθ(u(t)),φ0ifn.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 un and u¯n have the same limit in the space L2((0,T),V). Employing Lemma 4.1(ii) gives37 limnun-u¯nL2((0,T),V)2=limni=1nti-1ti(un-u¯n)(t)V2dt=limni=1nti-1tiui-1+t-ti-1τ(ui-ui-1)-uiV2dt4limni=1nui-ui-1V2τlimnCn=0.37 Therefore, if a(u)a, it holds, thanks to assumption (Equation8) and Corollary 4.2(ii), that for all φV0ξa(u¯n(t))-a(u(t)),φφT0Tu¯n(t)-u(t)20ifn.Secondly, applying the Green theorem and taking φ{ψC(Ω):ψ|ΓD=0,ψ|Γ0=constant} in the first part, we deduce that38 0ξK¯nu¯n,φ=0ξk¯nκ¯nu¯n,φΩ0+0ξκ¯nu¯n,φΩ\Ω0=0ξk¯nu¯n,κ¯nφ·νΩ0-0ξk¯nu¯n,·(κ¯nφ)Ω0+0ξκ¯nu¯n,φΩ\Ω0=0ξk¯nu¯n,κ¯nφ·νΩ0-0ξk¯nu¯n,κ¯n·φ+κ¯nΔφΩ0MYAMP]+0ξκ¯nu¯n,φΩ\Ω0.38 At this point, we need some auxiliary results. In light of equation (Equation32) and Lemma 4.1(i), applying the Nečas inequality [Citation17]zΓ2εz2+Cεz2,zH1(Ω),0<ε<ε0,implies0Tu¯n-uΓ2ε0T(u¯n-u)2+Cε0Tu¯n-u2ε+Cε0Tu¯n-u2.Passing to the limit for τ0 and applying Corollary 4.2(ii), we obtainlimτ00Tu¯n-uΓ2ε,which is valid for any small ε>0. Hence,39 limτ00Tu¯n-uΓ2=0andu¯nua.e. in(0,T)×Γ.39 Repeating this consideration for Ω0 instead of Ω gives40 limτ00Tu¯n-uΩ02=0andu¯nua.e. in(0,T)×Ω0.40 Due to the construction, we have that C0k¯nC1. This yields that k¯nk (for a subsequence) in L2((0,T)). Therefore, employing k¯nk, (Equation10), (Equation32), (Equation34) and (Equation36), we get after passage to the limit as τ0 in (Equation35) thatlimτ00ξK¯nu¯n,φ=0ξku,κφ·νΩ0-0ξku,κ·φ+κΔφΩ0+0ξκu,φΩ\Ω0=0ξku,κφ·νΩ0-0ξku,·(κφ)Ω0+0ξκu,φΩ\Ω0=0ξkκu,φΩ0+0ξκu,φΩ\Ω0=0ξKu,φ.Applying the density argument {ψC(Ω):ψ|ΓD=0,ψ|Γ0=constant}¯=V, we conclude thatlimτ00ξK¯nu¯n,φ=0ξKu,φ,φV.Moreover, due the continuity of h and f, one can deduce that, for any φV,0ξh¯n(t)-h(t)φ|Γ0φV0T|h¯n(t)-h(t)|τ0ξf¯n(t)-f(t),φφ0Tf¯n(t)-f(t)τ.Collecting all considerations above and passing to the limit for τ0 in (Equation31), we arrive at0ξtθ(u),φ+0ξKu+a(u),φ+0ξhφ|Γ0=0ξf,φφV.This is valid for any ξ(0,T). Differentiation with respect to ξ gives (29a). Taking the limit τ0 in (30b) and using the continuity of U we get (29b), which concludes the proof.

4.2 Second solution method

Consider the notations in Section 3.2. The variational formulation of (Equation19) on time step ti for all φW is41 δϑ(vi),φ+Ki(vi+U~i)+b(vi),φ=fi,φ,Γ0-Ki(vi+U~i)-b(vi)·ν=hi,41 with ui=vi+U~i. Using analogous notations as in the previous Section 4.1, the variational formulation (Equation37) can be rewritten as42a tϑn,φ+K¯n(v¯n+U~¯n)+b(v¯n),φ=f¯n,φ,Γ0-K¯n(v¯n+U~¯n)-b(v¯n)·ν=h¯n,42a where φW and u¯n=v¯n+U~¯n. The variational formulation of (Equation1) and (Equation2) for all φW reads as: find (K,u) with u=v+U~ such that43a tϑ(v),φ+K(v+U~)+b(v),φ=f,φ,Γ0-K(v+U~)-b(v)·ν=h.43a

Analysis similar to that in the previous Section 4.1 shows that the limit τ0 in (38a) results into (39a). There is an extra term to take under consideration, namely (ξ(0,T) and φW)0ξk¯nκ¯nU~¯n-kκU~,φ0.The convergence of (38b)–(39b) for any t(0,T) follows from the continuity of h.

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 θ0 and θ1 such that 0<θ0<θθ1. 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 Ω=-12,1×-1,1, with Ω0=-12,0×-1,1 in R2. The time interval is [0,1], i.e. T=1. The boundary Γ is split into three nonoverlapping parts, namely ΓD (right), ΓN (top and bottom) and Γ0 (left part of Γ), see Figure .

Figure 1. The domain used in the numerical experiments.

Figure 1. The domain used in the numerical experiments.

In the experiments, the exact solution (K,u) is prescribed as followsK(t,x,y)=1+sin(10t)1Ix<0+12;u(t,x,y)=(1+t)sinπ2(1-x).The function 1Ix<0 is the indicator function of the subset (x,y)R2:x<0 of R2. Moreover, it is assumed thatθ(s)=s2+1anda0.Some simple calculations give the exact data for the numerical experimentgD=gN=0;U(t)=1+t2;h(t)=π21+t1.5+sin(10t).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 ofk~(t):=1+sin(10t).For the time discretization, an equidistant time partitioning is chosen with time step τ=0.02 and for the space discretization, a fixed uniform mesh is used consisting of 144528 triangles.

5.1 First solution method

An uncorrelated noise is added to the additional condition U(t) in order to simulate the errors present in real measurements. The noise is generated randomly with given magnitude e=0,0.5 and 1%. Applying the backward Euler difference scheme into (Equation20), a recurrent system of nonlinear elliptic BVPs for (Ki,ui)K(ti),u(ti),i=1,2,,50 and φV have to be solved44 1τθ(ui),φ+Kiui,φ=fi,φ-hiφ|Γ0+1τθ(ui-1),φ;u0=u0;44 with45 fi,φ=1+tisinπ21-x21+ti2sinπ21-x2+1,φ+1.5+sin(10ti)π221+tisinπ2(1-x),φΩ0+0.5π221+tisinπ2(1-x),φΩ\Ω0.45 On every time level, Newton’s method is applied to deal with the nonlinearity. More precisely, given a solution uil from iteration l, a perturbation dui is searched such thatuil+1=uil+duifulfills the nonlinear problem (Equation40). The perturbation dui is zero at the boundaries where the Dirichlet conditions are applied. Inserting uil+1 in the BVP (Equation40), the θ term in the LHS is linearized asθ(uil+1)θ(uil)+θ(uil)dui.As initial guess ui0, the solution of the linear problem with θ(s)=1 and a(s)=0 is taken. The iterations are stopped when uil+1-uilL(Ω)=duiL(Ω)<10-5 or when the number of iterations exceed the limit 25. The unknowns k~ik~(ti),i=1,,50, are determined by the nonlinear conjugate gradient method. On each time step ti,i=1,,50, the functionalJ1(k~i):=ui-U(ti)L2(Γ0)2is minimized. The starting point for this algorithm on the first time step is set as k~10=1. The starting point on the following time steps is k~i-1, the minimizing value of the functional in the previous time step. The algorithm stops after maximum 10 iterations with the prescribed error tolerance 10-6.

Figure 2. Noise e=0%: numerical value of k~i using the first solution method (a) and the second solution method (b); i=1,,50.

Figure 2. Noise e=0%: numerical value of k~i using the first solution method (a) and the second solution method (b); i=1,…,50.

Figure 3. Noise e=0.5%: numerical value of k~i using the first solution method (a) and the second solution method (b); i=1,,50.

Figure 3. Noise e=0.5%: numerical value of k~i using the first solution method (a) and the second solution method (b); i=1,…,50.

5.2 Second solution method

Again, an uncorrelated noise with magnitude e=0,0.5 and 1% is added to the additional condition h(t). Utilizing the backward Euler difference scheme into (Equation23), a recurrent system of linear elliptic BVPs for (Ki,ui)K(ti),u(ti),i=1,2,,50 and φW has to be solved46 1τθ(ui),φ+Kiui,φ=fi,φ+1τθ(ui-1),φ,u0=u0;46 where fi,φ is as in (Equation41) and ui=U~i on Γ0. The nonlinearity is handled in the same way as in the first solution method. Now, on each time step ti,i=1,,50, the functionalJ2(k~i):=-Γ0(k~i+0.5)ui·ν-h(ti)2is minimized.

5.3 Results

At each time step, the resulting elliptic BVP’s (Equation40) and (Equation42) are solved numerically by the finite element method (FEM) using second-order (P2-FEM) Lagrange polynomials. The results from the recovery of k~(t) using both solution methods for the different values of the amplitude e are shown in Figures . The exact k~(ti) is denoted by a solid line and the approximations k~i by linespoints; i=1,,50. The evolution of the absolute k~i-error for the different time steps is shown in Figure .

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.

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.

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.

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.

The experiments demonstrate that the approximation becomes less accurate with increasing magnitude e 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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