# An Existence Result to a Strongly Coupled Degenerated System

Authors: L. Hadjadj; K. Hamdache; D. Hamroun
Journal: Abstract and Applied Analysis
 (2008)

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## Abstract

We consider a mathematical model to describe the growth of a vascular tumor including
tumor cells, macrophages, and blood vessels. The resulting system of equations is reduced to a
strongly2×2 coupled nonlinear parabolic system of degenerate type. Assuming the initial data
are far enough from 0, we prove existence of a global weak solution with finite entropy to the
problem by using an approximation procedure and a time discrete scheme.


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## Body

## 1. Introduction

The theory of mixtures is used to
develop a mathematical model that governs the interactions of macrophages,
tumor cells, and blood vessels within a vascular tumor, focusing on  the ability
of macrophages to both lyse tumor cells and stimulate angiogenesis. In recent
years, a variety of macroscopic continuum models have been derived
by [1–6]. In their wake, we present
hereafter a simplified model.The vascular tumor is viewed as a mixture of three
constituents: tumor cells, tumor-associated
macrophages (abbreviated by TAMs), and blood vessels. We denote their
respective volume fractions byα,β,
and γ,
and we assume that the mixture is saturated, so we take(1.1)α+β+γ=1.We suppose that the tumor
undergoes one dimensional and one side growth, parallel to the x-axis, by
occupying the region 0≤x≤L(t).
Each phase is associated with velocity, pressure, and spatial stress denoted,
respectively, by v1,P1,σ1 for the tumor cells, v2,P2,σ2 for the TAMs and v3,P3,σ3 for the blood vessels. Formulating
conservation of mass for the three volume fractions, under the assumption that
each phase has the same constant density, we get(1.2)αt+(αv1)x=q1,βt+(βv2)x=q2,γt+(γv3)x=q3,where the indices t and x are set for partial derivatives and q1,q2, and q3
are the rates of production related to each
phase, satisfying q1+q2+q3=0.
We suppose that the volume fraction of tumor cells increase by proliferation
and decrease by apoptosis, necrosis or lysis, the volume fraction of TAMs
increase by proliferation and by influx from capillaries and decrease by
natural death or after lysing tumor cells. Thus, we write(1.3)q1(α,β,γ)=k1αγ−k2α−k3(1−γ)α−k4αβ,(1.4)q2(α,β,γ)=k5β(1−σ(α+γ))−k6β(1−α)−k7αβ,where σ and ki,i=1,…,7 are nonnegative constants. Assuming that the
momentum is conserved and the motions of cells and blood vessels are so slow
that inertial terms can be neglected, we can write(1.5)(ασ1)x+F1=0,(βσ2)x+F2=0,(γσ3)x+F3=0,where(1.6)F1=Pαx+dαβ(v2−v1)+dαγ(v3−v1),(1.7)F2=Pβx−dαβ(v2−v1)+dβγ(v3−v2),(1.8)F3=Pγx−dαγ(v3−v1)−dβγ(v3−v2), represent the momentum supply
related to each phase, d is a positive constant and P is assumed to be a common pressure. When
neglecting viscous effect, the partial stress tensors are given
by(1.9)σi=−Pi=−(P+Σi),i=1,2,3,where(1.10)Σ1=λα,Σ2=μβ(1+θα),Σ3=P0represent the pressures due to
cell-cell interactions exerted on tumor cells, macrophages and blood vessels
respectively. λ,μ and θ are nonnegative constants and P0>0 is constant.The following initial and boundary conditions are
considered:(1.11)L=l>0,α=α0≥0,β=β0≥0,α0+β0≤1att=0,(1.12)α=αb≥0,β=βb≥0,αb+βb≤1atx=0.We impose the
no flux boundary condition at the free boundary that we suppose moving at the
same velocity as the tumor cells, so(1.13)v1=v2=v3=0atx=L(t),(1.14)dLdt=v1|x=L(t)=0thatisL(t)=l∀t≥0.Adding the three continuity
equations (1.2) and the three momentum equations (1.5), we get using
(1.13)(1.15)αv1+βv2+γv3=0,(ασ1)x+(βσ2)x+(γσ3)x=0.The last equality and (1.9)
imply(1.16)Px=−(αΣ1+βΣ2+γΣ3)x.Using (1.6), (1.9), and the first
relation in (1.15), the first equation of (1.5) reduces to give either α=0,
which we reject because it can be only transient, or
(1.17)Px+1α(αΣ1)x=−dv1,
which together with (1.16) gives(1.18)v1=1d((αΣ1+βΣ2+γΣ3)x−1α(αΣ1)x).Using (1.10) and the fact that P0 is constant, (1.18) can be rewritten as
follows:(1.19)v1=1d(αx(2λα+μθβ2−P0−2λ)+βx(2μβ(1+θα)−P0)).Similarly, the second equation
of (1.5) simplifies into(1.20)v2=1d(αx(2λα+μθβ2−P0−μθβ)+βx(2μ(β−1)(1+θα)−P0)).Thus, substituting the relations
(1.19)-(1.20) into (1.2), the equations of α and β become(1.21)αt−1d((2λα(1−α)−μθαβ2+P0α)αx+(−2μβα(1+θα)+P0α)βx)x=q1βt−1d((−2λαβ+μθβ2(1−β)+P0β)αx+(2μβ(1−β)(1+θα)+P0β)βx)x=q2.Having regard to the saturation
condition (1.1), we omit the equation of γ.
The resulting problem (1.21) is strongly coupled with full diffusion matrix which
is generally not positive definite. To simplify it, we reduce the number of
biological parameters by setting(1.22)λ=μ=P0,θ=0.In this case, (1.13),
(1.14), (1.19)
and (1.20) reduce the boundary condition at x=l to(1.23)αx=βx=0.Without loss of generality, we
set l=1,2λ/d=1 and for technical reasons, to get the maximum
principle (see Lemma 3.7), we need to take σ=1 in (1.4) and(1.24)k5≤k6.
In summary, denoting u=(α,β),u0=(α0,β0),ub=(αb,βb) and q=(q1,q2),
the problem (1.21) simplifies into(1.25)ut−(A(u)ux)x=q(u),with A and q given by(1.26)A(α,β)=(α(32−α)α(12−β)β(12−α)β(32−β)),q1(α,β)=k1α(1−α−β)−k2α−k3α(α+β)−k4αβ,q2(α,β)=k5β2−k6β(1−α)−k7αβ.The system (1.25) is complemented
with the boundary and initial conditions(1.27)u(t,0)=ub(t),ux(t,1)=0,u(0,x)=u0(x),and has to be solved in ℝ+×(0,1).In recent years, cross-diffusion systems have drawn a
great deal attention. For example in [7], the global existence was established, as well as the
existence of a global attractor in a case of triangular positive definite
diffusion matrix. In [8], the well-posedness and the properties of steady
states for a degenerate parabolic system with triangular positive (semi)
definite matrix, modeling the chemotaxis movement of cells, were investigated.
In [9, 10], the existence of global
weak solution was shown for a nonlinear problem with full diffusion matrix. The
proof was based on a symmetrization of the problem via an exponential
transformation of variables, backward Euler approximation of the time
derivative and an entropy functional. Here, we use analogous arguments, but in
our case, after the transformation of variables the resulting matrix B is not positive definite. To overcome this
difficulty, we approximate B by positive definite matrices Bτ which tend towards B as τ→0,
if the condition 0≤α,β,α+β≤1 is satisfied. This needs to prove that the set {(α,β)∈L∞(Ω)×L∞(Ω),0≤α,β,α+β≤1} is time invariant.Throughout this paper, we use the following notations:
letT,τ be positive real numbers, we will denote by C all the positive constants which are
independent of τ.
We set  Ω=(0,1),QT=(0,T)×Ω and s+=max(s,0) the positive part of the real number s.
We write ux:=∂xu and ut:=∂tu  for partial derivatives of a real-valued
function u=u(t,x).
Moreover we will use the Sobolev space HD1(Ω)={u∈H1(Ω);u(0)=0} equipped with the norm of H1 and we denote as usual, by (H1)′ the dual of H1.
In the case of vectorial functions, we designate the corresponding Lebesgue and
Sobolev spaces, respectively, by 𝕃2,𝕃∞,ℍ1,ℍD1.
Finally we set ln(r,s)=(lnr,lns) for r,s>0 and e(r,s)=(er,es) for r,s∈ℝ.The remainder of this paper is organized as follows.
In Section2, we introduce the weak formulation of the problem and state our
main existence result in Theorem 2.2. In proving this
theorem, we define and
solve in Section 3 an auxiliary problem which will be useful further. Then, in
Section 4, we formulate a semidiscrete version in time of the problem, using a backward
Euler approximation, combined with a perturbation of the diffusion matrix.
This leads to a recursive sequence of elliptic problems depending on the small
parameter τ.
Performing the limit as τ→0,
with the help of Aubin compactness lemma and the Sobolev embedding H1(0,1)↪L∞(0,1),
we get a weak solution to our problem. Finally, an appendix is devoted to the
proof of a technical lemma.


## 2. Main Result

We set the
following assumptions:(H1)
α0,β0∈H1(Ω),0<α0,β0,α0+β0≤1,lnα0,lnβ0∈L∞(Ω),(H2)
αb,βb∈ℝ,0<αb,βb,αb+βb≤1.
The matrix A(α,β) is not positive even if 0≤α,β,α+β≤1,
so the problem (1.25)–(1.27) has no classical solution in general. A weak
solution is defined as follows.Definition 2.1.
Let (H1)-(H2) be satisfied ant letT>0.u=(α,β)  is said to be a weak solution of problem
(1.25)–(1.27) on QT if
(1)
u∈L2(0,T;ℍ1(Ω))∩H1(0,T;(ℍ1(Ω))′)∩𝕃∞(QT) with 0≤α,β,α+β≤1,(2)
u(0,x)=u0(x) a.e. in Ω, u(t,0)=ub a.e. in (0,T),(3)
∫0T〈ut,φ〉dt+∫QTA(u)ux⋅φxdxdt=∫QTq(u)⋅φdxdt,
for all φ∈L2(0,T;ℍD1(Ω)), where 〈⋅,⋅〉 is the dual product between (ℍ1(Ω))′ and ℍ1(Ω).Our main result is the following.Theorem 2.2.
Assume (H1)-(H2) are satisfied. Then for everyT>0,
there exists (at least) a weak solution u=(α,β) on QT to the system (1.25)–(1.27), satisfying the
entropy inequality(2.1)∫Ω(G1(α(t))+G2(β(t)))dx+14∫0t∫Ω(|αx|2+|βx|2)dxds≤∫Ω(G1(α0)+G2(β0))dx+C,where C>0 depends on T,αb,βb,ki,i=1,…,7,G1,G2  being positive functions defined on ℝ+ by(2.2)G1(s)=s(lns−lnαb−1)+αb,G2(s)=s(lns−lnβb−1)+βb.The proof of this existence result is based on the entropy inequality (2.1),
which is formally obtained by testing (1.25) with (ln(α/αb),ln(β/βb)),
and integrating by parts (see Section 3.3 for details).This estimate suggests to use the change of unknownU=ln(α,β),
which transforms the problems (1.25)–(1.27) into the following one(2.3)(eU)t−(B(U)Ux)x=Q(U)inQT,U(t,0)=Ub(t),Ux(t,1)=0in(0,T),U(0,x)=U0(x)inΩ,with Ub=ln(αb,βb),U0=ln(α0,β0),Q(U)=q(eU) and the new diffusion matrix is B(U)=A(eU)diag(eU1,eU2) and takes the form(2.4)B(r,s)=(e2r(32−er)er+s(12−es)er+s(12−er)e2s(32−es)).The matrix B(r,s) resulting of this transformation is still not
positive definite. Nevertheless, in the case er+es≤1 which is under interest, B(r,s) is positive definite. Moreover, this change of
variables leads to nonnegative solutions, without using maximum principle,
since α=eU1 and β=eU2.


## 3. Auxiliary Problems

We will use a
time discretization scheme to study (1.25)–(1.27). In order to prove global existence for the
resulting stationary problem, it may be useful to introduce an artificial
perturbation of the diffusion matrix A of type εI,
where ε>0 and I is the identity matrix (see the proof of
Proposition 3.3 below and Lemma 3.6). Nevertheless, the choice of the parameter ε is technical and cannot be done independently
of the time discretization parameter τ.
Here, we take ε=τ,
this choice being dictated by the sake of coherency of the discretization
scheme proposed in Section 4. Indeed, in the case where ε is independent of τ,
this procedure is seriously compromised. More details on this question are
given in Remark 3.8, at the end of this section. In summary, we need to solve the
following problem:(3.1)1τ(u−u˜)−((A(u)+τI)ux)x=q(u)inΩ,u(0)=ub,ux(1)=0,where τ>0 is a small parameter and u˜=u˜(x) is a fixed function. Before giving the
existence result for this problem, let us define the solutions we deal
with.Definition 3.1.
u=(α,β)∈ℍ1(Ω) is said to be a weak solution of problem (3.1) if u(0)=ub,0≤α,β,α+β≤1inΩ and if for every φ∈ℍD1(Ω),
it holds(3.2)1τ∫Ω(u−u˜)⋅φdx+∫Ω(A(u)+τI)ux⋅φxdx=∫Ωq(u)⋅φdx.We
have the following result.Theorem 3.2.
Letub=(αb,βb)∈ℝ2 satisfy assumption (H2), u˜=(α˜,β˜)∈𝕃∞(Ω) such that 0<α˜,β˜,α˜+β˜≤1 a.e. in Ω and lnu˜∈𝕃∞(Ω).
Then for all 0<τ<1,
there exists a weak solution uτ=(ατ,βτ)∈ℍ1(Ω) of the problem (3.1). Moreover, lnuτ∈ℍ1(Ω) and it holds(3.3)∫Ω(G1(ατ)+G2(βτ))dx+τ4∫Ω(|ατx|2+|βτx|2)dx≤∫Ω(G1(α˜)+G2(β˜))dx+Cτ,where G1 and G2 are defined by (2.2).For the proof, according to the change of unknown introduced
in Section2, we will consider the following stationary problem(3.4)1τ(eU−eU˜)−(Bτ(U)Ux)x=Q+(U)inΩ,U(0)=Ub,Ux(1)=0,where U˜=lnu˜,Bτ is the matrix defined by(3.5)Bτ=B++Dτ,B+ is given by(3.6)B+(r,s)=(e2r(32−min(er,1−es))er+s(12−es)er+s(12−er)e2s(32−min(es,1−er))),and Dτ is a diagonal matrix with(3.7)D11τ(r,s)=D22τ(s,r)=(e2r+s+5er(er+es))(er+es−1)++τer.The vector field Q+ is defined by its components(3.8)Q1+(r,s)=k1er(1−er−es)+−min(er,1)(k2+k3min(er+es,1)+k4min(es,1)),Q2+(r,s)=min(es,1)(k5min(es,(1−er)+)−k6(1−er)+−k7min(er,1))Clearly if er+es≤1,
then B+(r,s)=B(r,s),D11τ(r,s)=D22τ(s,r)=τer→0 when τ→0 and Q+(r,s)=Q(r,s).
In addition, we have for i=1,2
that
(3.9)|Qi+(r1,r2)|≤C,|Qi+(r1,r2)|≤Ceri,∀(r1,r2)∈ℝ2,where C>0 depends only on ki,i=1,…,7. We will prove the following
result.Proposition 3.3.
Assume thatU˜∈𝕃∞(Ω) is such that U˜1,U˜2≤0 a.e. in Ω.
Then for all 0<τ<1,
there exists a weak solution Uτ∈H1(Ω) to problem (3.4).The proof of this result relies on the Leray-Schauder fixed point
theorem, so we start by studying the linear problems associated with (3.4).


### 3.1. Linear Problems Associated with (3.4)

In the sequel,
we let once for all0<τ<1 and U˜∈𝕃∞(Ω) fixed. Let U¯∈𝕃∞(Ω) be given; we consider the
following linear problem: find U∈ℍ1(Ω) satisfying(3.10)U(0)=Uband∀φ∈ℍD1(Ω),1τ∫Ω(eU¯−eU˜)⋅φdx+∫ΩBτ(U¯)Ux⋅φxdx=∫ΩQ+(U¯)⋅φdx.
We have the following
result.Lemma 3.4.
For everyU¯∈𝕃∞(Ω),
problem (3.10) has a unique solution U∈ℍ1(Ω).Proof.
We will
apply the Lax-Milgram lemma. We setV=U−Ub,
so (3.10) goes over into the following equivalent problem(3.11)V(0)=0and∀φ∈ℍD1(Ω),∫ΩBτ(U¯)Vx⋅φxdx=−1τ∫Ω(eU¯−eU˜)⋅φdx+∫ΩQ+(U¯)⋅φdx.Next, we define a bilinear form 𝒜 on ℍD1(Ω)×ℍD1(Ω) and a linear form ℒ on ℍD1(Ω) by setting(3.12)𝒜(V,φ)=∫ΩBτ(U¯)Vx⋅φxdx,ℒ(φ)=−1τ∫Ω(eU¯−eU˜)⋅φdx+∫ΩQ+(U¯)⋅φdx.The continuity of 𝒜 and ℒ follows from the boundedness of U¯ and U˜.
For the coerciveness of 𝒜,
it is sufficient to prove that the matrix D defined by(3.13)D(r,s)=B+(r,s)+er+s(er+es−1)+diag(er,es)is positive definite. Let us
compute D(r,s)ξ⋅ξ for (r,s)∈ℝ2 and ξ=(ξ1,ξ2)∈ℝ2.
We consider first the case where er+es≤1 so that(3.14)D(r,s)ξ⋅ξ=e2r(32−er)ξ12+e2s(32−es)ξ22−er+s(er+es−1)ξ1ξ2.The
elementary inequality(3.15)1−a−b≤32−a32−bif0≤a,b,a+b≤1and Young inequality lead to(3.16)−er+s(er+es−1)ξ1ξ2≥−12(e2r(32−er)ξ12+e2s(32−es)ξ22),so(3.17)D(r,s)ξ⋅ξ≥12(e2r(32−er)ξ12+e2s(32−es)ξ22)≥14min(e2r,e2s)∥ξ∥2.In the case er+es>1,
we have(3.18)D(r,s)ξ⋅ξ=e2r(12+es(er+es))ξ12+e2s(12+er(er+es))ξ22−er+s(er+es−1)ξ1ξ2.From the
inequality(3.19)(1−er−es)2≤(er+es)2≤94(12+es(er+es))(12+er(er+es)),we deduce that(3.20)er+es−1≤3212+es(er+es)12+er(er+es).
So, thanks to Young inequality we get(3.21)D(r,s)ξ⋅ξ≥14(e2r(12+es(er+es))ξ12+e2s(12+er(er+es))ξ22)≥18min(e2r,e2s)∥ξ∥2.Hence, we infer that for all ξ∈ℝ2,U¯=(U¯1,U¯2)∈𝕃∞(Ω),(3.22)Bτ(U¯)ξ⋅ξ≥D(U¯)ξ⋅ξ≥18min(e−2∥U¯1∥L∞(Ω),e−2∥U¯2∥L∞(Ω))∥ξ∥2,so Lax-Milgram lemma implies the
existence of a unique solution V∈ℍD1(Ω) of problem (3.11). Consequently, U=V+Ub is the unique solution of (3.10).


### 3.2. Proof of Proposition3.3

Lemma3.4 and the embedding H1(Ω)⊂L∞(Ω) allow us to define the map 𝒮:𝕃∞(Ω)→𝕃∞(Ω), by setting 𝒮(U¯)=U the solution of (3.10). We will establish, using
the theorem of Leray Schauder, that 𝒮 has a fixed point Uτ in 𝕃∞(Ω),
so Uτ∈H1(Ω) is a solution of the nonlinear problem (3.4).First, we prove that𝒮is continuous. Let (U¯n)n be a sequence in 𝕃∞(Ω) such that  U¯n→U¯ strongly in 𝕃∞(Ω) as n→∞ and let 𝒮(U¯n)=Un.
We use the test function φ=Un−Ub∈ℍD1(Ω) in (3.10), estimate (3.22), and Poincaré inequality to get(3.23)1τ∫Ω(eU˜−eU¯n)⋅(Un−Ub)dx+∫ΩQ+(U¯n)⋅(Un−Ub)dx≥C∥Un−Ub∥ℍ1(Ω)2,where C>0 is independent of n.
So taking into account (3.9), we get, thanks to Young
inequality,(3.24)∥Un−Ub∥ℍ1(Ω)2≤C(τ).Thus, Un is bounded in ℍ1(Ω) and from the compactness of the embedding H1(Ω)↪L∞(Ω), we deduce that there exists a subsequence of (Un)n,
still denoted by (Un)n and a function U∈ℍ1(Ω) such that(3.25)Un→Ustronglyin𝕃∞(Ω),Un⇀Uweaklyinℍ1(Ω).This implies the weak
convergence Bτ(U¯n)Unx⇀Bτ(U¯)Ux in 𝕃2(Ω),
hence there exists a subsequence of (Un)n which converges towards 𝒮(U¯).
Moreover thanks to the uniqueness result for the system (3.10), we see that all
the sequence (Un)n converges to 𝒮(U¯) which ends the proof of continuity of 𝒮.The compactness
of𝒮 follows from the compactness of the embedding H1(Ω) into L∞(Ω) and (3.24). Finally, let us check that the setsΛδ={U¯∈𝕃∞(Ω)/U¯=δ𝒮(U¯)}are
uniformly bounded with respect toδ∈[0,1].
Observe that Λ0={0} and if δ≠0 the equation U¯=δ𝒮(U¯) is equivalent to U¯∈ℍ1(Ω),U¯(0)=δUb and for all φ∈ℍD1(Ω),(3.26)1τ∫Ω(eU¯−eU˜)⋅φdx+1δ∫ΩBτ(U¯)U¯x⋅φxdx=∫ΩQ+(U¯)⋅φdx.The remainder of the proof is a
direct consequence of the following lemma.Lemma 3.5.
Under the assumptions of Proposition3.3, there exists a positive constant C(τ) independent of δ such that if U¯∈ℍ1(Ω) satisfies (3.26), then it holds(3.27)∥U¯x∥𝕃2(Ω)≤C(τ).For the proof, we need the following technical result which will be checked in
the appendix.Lemma 3.6.
For all(r,s),(ξ1,ξ2)∈ℝ2,
one has
(3.28)ℬτ((r,s),(ξ1,ξ2))≡B11τ(r,s)(1+2τe−r)ξ12+B22τ(r,s)(1+2τe−s)ξ22+(B12τ(r,s)(1+2τe−r)+B21τ(r,s)(1+2τe−s))ξ1ξ2≥τ2(ξ12+ξ22),where Bτ is the matrix given in (3.5).Proof of Lemma3.5.
Testing the
equation of (3.26) with φ=U¯−δUb+2τ(e−δUb−e−U¯)∈ℍD1(Ω) leads to(3.29)∫Ωℬτ(U¯,U¯x)dx=−δτ∫Ω(eU¯−eU˜)⋅φdx+δ∫ΩQ+(U¯)⋅φdx.The left-hand side is estimated
using (3.28). We write U¯=(U¯1,U¯2),U˜=(U˜1,U˜2),Ub=(Ub1,Ub2);
the convexity of es leads to(3.30)(eU¯i−eU˜i)(U¯i−δUbi)=(eU¯i(U¯i−δUbi−1)+eδUbi)−(eU˜i(U˜i−δUbi−1)+eδUbi)+((eU¯i−eU˜i)−eU˜i(U¯i−U˜i))≥fi(δ,U¯i)−fi(δ,U˜i),where for i=1,2,fi(δ,s)=es(s−δUbi−1)+eδUbi≥0,
for all (δ,s)∈[0,1]×ℝ.
Using the elementary inequality es≥1+s,
valid for all s∈ℝ,
we get(3.31)(eU¯i−eU˜i)(e−δUbi−e−U¯i)=(eU¯i−δUbi−U¯i+δUbi)−(eU˜i−δUbi−U˜i+δUbi)+(eU˜i−U¯i−(U˜i−U¯i)−1)≥gi(δ,U¯i)−gi(δ,U˜i),with gi(δ,s)=es−δUbi−s+δUbi≥1,
for all (δ,s)∈[0,1]×ℝ.
Combining (3.30) and (3.31), we find(3.32)−δτ∫Ω(eU¯−eU˜)⋅φdx≤1τ∫Ω(f1(δ,U˜1)+f2(δ,U˜2))dx+2∫Ω(g1(δ,U˜1)+g2(δ,U˜2))dx.We recall that a.e. in Ω,−∥U˜i∥∞≤U˜i≤0 for i=1,2,
and since the functions fi,gi are continuous on [0,1]×[−∥U˜i∥∞,0],
we deduce that the right-hand side of (3.32) is uniformly bounded with respect to δ. Now we infer from (3.9) and Poincaré inequality
that(3.33)δ∫ΩQ+(U¯)⋅(U¯−Ub)dx≤C∫Ω|U¯−Ub|dx≤τ22∥U¯x∥𝕃2(Ω)2+C(τ),δ∫ΩQ+(U¯)⋅(e−δUb−e−U¯)dx≤C.
The result follows by combining
all these inequalities.


### 3.3. End of Proof of Theorem3.2

LetUτ be the solution of (3.4) provided by
Proposition 3.3. Recalling that u˜=eU˜ and ub=eUb,
we see that uτ=eUτ satisfies the following
problem:(3.34)1τ(uτ−u˜)−(Aτ(uτ)uτx)x=Q+(lnuτ)inΩ,uτ(0)=ub,uτx(1)=0,where the matrix Aτ is given by Aτ(r,s)=A+(r,s)+hτ(r,s)I with(3.35)A+(r,s)=(r(32−min(r,1−s))r(12−s)s(12−r)s(32−min(s,1−r))),hτ(r,s)=(r+s−1)+(rs+5(r+s))+τ.Note that if r+s≤1 then A+(r,s)=A(r,s) and hτ(r,s)=τ.We will focus on the𝕃∞ and ℍ1 estimates satisfied by the function uτ.
We begin with the following L∞ bounds.Lemma 3.7.
Let the hypotheses of Theorem3.2
hold, and let uτ=(ατ,βτ)∈ℍ1(Ω) satisfy problem (3.34). One
has
(3.36)0<ατ,βτ,ατ+βτ≤1inΩ.Proof.
We write the equation satisfied byατ+βτ and test it with φ=(ατ+βτ−1)+∈HD1(Ω).
We get using (1.24) that(3.37)∫Ωατ+βτ−(α˜+β˜)τφdx+∫Ω(12(ατ+βτ)+hτ(ατ,βτ))|φx|2dx≤0,so ∫Ω(ατ+βτ−(α˜+β˜))φdx≤0.
Consequently, ∫Ωφ2dx≤∫Ω(α˜+β˜−1)φdx≤0,
and hence ατ+βτ≤1.As a consequence of Lemma3.7, we easily see that uτ is a solution to problem (3.1) in the sense of
Definition 3.1. Moreover, since lnuτ=Uτ∈ℍ1(Ω) then lnuτ∈𝕃∞(Ω).
In order to check the entropy inequality (3.3), we test (3.1) with φτ=lnuτ−lnub∈ℍD1(Ω) to get(3.38)1τ∫Ω(uτ−u˜)⋅φτdx+∫Ω(1−ατ−βτ)ατxβτxdx+τ∫Ω(|ατx|2ατ+|βτx|2βτ)dx+∫Ω((32−ατ)|ατx|2+(32−βτ)|βτx|2)dx=∫Ωq(uτ)⋅φτdx.Then, from (3.15) and Young inequality,
we get(3.39)−∫Ω(ατ+βτ−1)ατxβτxdx≥−12∫Ω((32−ατ)|ατx|2+(32−βτ)|βτx|2)dx,so, inserting (3.39) into (3.38) and
using the fact that 3/2−ατ,3/2−βτ≥1/2,
we see that(3.40)1τ∫Ω(uτ−u˜)⋅φτdx+14∫Ω(|ατx|2+|βτx|2)dx≤∫Ωq(uτ)⋅φτdx.Using the boundedness of q(uτ) and the fact that the function sln(s) is bounded in [0,1],
we obtain ∫Ωq(uτ)⋅φτdx≤C, then the convexity of the functions G1 and G2 leads to(3.41)∫Ω(uτ−u˜)⋅φτdx≥∫Ω(G1(ατ)+G2(βτ))dx−∫Ω(G1(α˜)+G2(β˜))dx,which ends the
proof.Remark 3.8.
All the results of this section remain valid if one
consider, instead of (3.1), the following problem:(3.42)1τ(u−u˜)−((A(u)+εI)ux)x=q(u)inΩ,u(0)=ub,ux(1)=0,where τ>0 and ε>0 is independent of τ.
In particular, the solution vτε to (3.42) satisfies the L∞ bounds and the entropy inequality given in
Theorem 3.2, with the constant C>0 independent of ε (and τ). Thus performing the limit as ε→0 in (3.42), we get a function vτ solving the problem(3.43)1τ(u−u˜)−(A(u)ux)x=q(u)inΩ,u(0)=ub,ux(1)=0,which corresponds to the
“natural” time discretization of our problem (1.25)–(1.27). However, from there, the situation becomes
complicated because the obtained solution vτ to problem (3.43) has its components which are
only nonnegative and we no longer have lnvτ∈𝕃∞(Ω) (in fact, we cannot even take the ln of vτ). Therefore, the time discretization scheme
based on (3.43) cannot be solved.


## 3.1. Linear Problems Associated with (3.4)

In the sequel,
we let once for all0<τ<1 and U˜∈𝕃∞(Ω) fixed. Let U¯∈𝕃∞(Ω) be given; we consider the
following linear problem: find U∈ℍ1(Ω) satisfying(3.10)U(0)=Uband∀φ∈ℍD1(Ω),1τ∫Ω(eU¯−eU˜)⋅φdx+∫ΩBτ(U¯)Ux⋅φxdx=∫ΩQ+(U¯)⋅φdx.
We have the following
result.Lemma 3.4.
For everyU¯∈𝕃∞(Ω),
problem (3.10) has a unique solution U∈ℍ1(Ω).Proof.
We will
apply the Lax-Milgram lemma. We setV=U−Ub,
so (3.10) goes over into the following equivalent problem(3.11)V(0)=0and∀φ∈ℍD1(Ω),∫ΩBτ(U¯)Vx⋅φxdx=−1τ∫Ω(eU¯−eU˜)⋅φdx+∫ΩQ+(U¯)⋅φdx.Next, we define a bilinear form 𝒜 on ℍD1(Ω)×ℍD1(Ω) and a linear form ℒ on ℍD1(Ω) by setting(3.12)𝒜(V,φ)=∫ΩBτ(U¯)Vx⋅φxdx,ℒ(φ)=−1τ∫Ω(eU¯−eU˜)⋅φdx+∫ΩQ+(U¯)⋅φdx.The continuity of 𝒜 and ℒ follows from the boundedness of U¯ and U˜.
For the coerciveness of 𝒜,
it is sufficient to prove that the matrix D defined by(3.13)D(r,s)=B+(r,s)+er+s(er+es−1)+diag(er,es)is positive definite. Let us
compute D(r,s)ξ⋅ξ for (r,s)∈ℝ2 and ξ=(ξ1,ξ2)∈ℝ2.
We consider first the case where er+es≤1 so that(3.14)D(r,s)ξ⋅ξ=e2r(32−er)ξ12+e2s(32−es)ξ22−er+s(er+es−1)ξ1ξ2.The
elementary inequality(3.15)1−a−b≤32−a32−bif0≤a,b,a+b≤1and Young inequality lead to(3.16)−er+s(er+es−1)ξ1ξ2≥−12(e2r(32−er)ξ12+e2s(32−es)ξ22),so(3.17)D(r,s)ξ⋅ξ≥12(e2r(32−er)ξ12+e2s(32−es)ξ22)≥14min(e2r,e2s)∥ξ∥2.In the case er+es>1,
we have(3.18)D(r,s)ξ⋅ξ=e2r(12+es(er+es))ξ12+e2s(12+er(er+es))ξ22−er+s(er+es−1)ξ1ξ2.From the
inequality(3.19)(1−er−es)2≤(er+es)2≤94(12+es(er+es))(12+er(er+es)),we deduce that(3.20)er+es−1≤3212+es(er+es)12+er(er+es).
So, thanks to Young inequality we get(3.21)D(r,s)ξ⋅ξ≥14(e2r(12+es(er+es))ξ12+e2s(12+er(er+es))ξ22)≥18min(e2r,e2s)∥ξ∥2.Hence, we infer that for all ξ∈ℝ2,U¯=(U¯1,U¯2)∈𝕃∞(Ω),(3.22)Bτ(U¯)ξ⋅ξ≥D(U¯)ξ⋅ξ≥18min(e−2∥U¯1∥L∞(Ω),e−2∥U¯2∥L∞(Ω))∥ξ∥2,so Lax-Milgram lemma implies the
existence of a unique solution V∈ℍD1(Ω) of problem (3.11). Consequently, U=V+Ub is the unique solution of (3.10).


## 3.2. Proof of Proposition3.3

Lemma3.4 and the embedding H1(Ω)⊂L∞(Ω) allow us to define the map 𝒮:𝕃∞(Ω)→𝕃∞(Ω), by setting 𝒮(U¯)=U the solution of (3.10). We will establish, using
the theorem of Leray Schauder, that 𝒮 has a fixed point Uτ in 𝕃∞(Ω),
so Uτ∈H1(Ω) is a solution of the nonlinear problem (3.4).First, we prove that𝒮is continuous. Let (U¯n)n be a sequence in 𝕃∞(Ω) such that  U¯n→U¯ strongly in 𝕃∞(Ω) as n→∞ and let 𝒮(U¯n)=Un.
We use the test function φ=Un−Ub∈ℍD1(Ω) in (3.10), estimate (3.22), and Poincaré inequality to get(3.23)1τ∫Ω(eU˜−eU¯n)⋅(Un−Ub)dx+∫ΩQ+(U¯n)⋅(Un−Ub)dx≥C∥Un−Ub∥ℍ1(Ω)2,where C>0 is independent of n.
So taking into account (3.9), we get, thanks to Young
inequality,(3.24)∥Un−Ub∥ℍ1(Ω)2≤C(τ).Thus, Un is bounded in ℍ1(Ω) and from the compactness of the embedding H1(Ω)↪L∞(Ω), we deduce that there exists a subsequence of (Un)n,
still denoted by (Un)n and a function U∈ℍ1(Ω) such that(3.25)Un→Ustronglyin𝕃∞(Ω),Un⇀Uweaklyinℍ1(Ω).This implies the weak
convergence Bτ(U¯n)Unx⇀Bτ(U¯)Ux in 𝕃2(Ω),
hence there exists a subsequence of (Un)n which converges towards 𝒮(U¯).
Moreover thanks to the uniqueness result for the system (3.10), we see that all
the sequence (Un)n converges to 𝒮(U¯) which ends the proof of continuity of 𝒮.The compactness
of𝒮 follows from the compactness of the embedding H1(Ω) into L∞(Ω) and (3.24). Finally, let us check that the setsΛδ={U¯∈𝕃∞(Ω)/U¯=δ𝒮(U¯)}are
uniformly bounded with respect toδ∈[0,1].
Observe that Λ0={0} and if δ≠0 the equation U¯=δ𝒮(U¯) is equivalent to U¯∈ℍ1(Ω),U¯(0)=δUb and for all φ∈ℍD1(Ω),(3.26)1τ∫Ω(eU¯−eU˜)⋅φdx+1δ∫ΩBτ(U¯)U¯x⋅φxdx=∫ΩQ+(U¯)⋅φdx.The remainder of the proof is a
direct consequence of the following lemma.Lemma 3.5.
Under the assumptions of Proposition3.3, there exists a positive constant C(τ) independent of δ such that if U¯∈ℍ1(Ω) satisfies (3.26), then it holds(3.27)∥U¯x∥𝕃2(Ω)≤C(τ).For the proof, we need the following technical result which will be checked in
the appendix.Lemma 3.6.
For all(r,s),(ξ1,ξ2)∈ℝ2,
one has
(3.28)ℬτ((r,s),(ξ1,ξ2))≡B11τ(r,s)(1+2τe−r)ξ12+B22τ(r,s)(1+2τe−s)ξ22+(B12τ(r,s)(1+2τe−r)+B21τ(r,s)(1+2τe−s))ξ1ξ2≥τ2(ξ12+ξ22),where Bτ is the matrix given in (3.5).Proof of Lemma3.5.
Testing the
equation of (3.26) with φ=U¯−δUb+2τ(e−δUb−e−U¯)∈ℍD1(Ω) leads to(3.29)∫Ωℬτ(U¯,U¯x)dx=−δτ∫Ω(eU¯−eU˜)⋅φdx+δ∫ΩQ+(U¯)⋅φdx.The left-hand side is estimated
using (3.28). We write U¯=(U¯1,U¯2),U˜=(U˜1,U˜2),Ub=(Ub1,Ub2);
the convexity of es leads to(3.30)(eU¯i−eU˜i)(U¯i−δUbi)=(eU¯i(U¯i−δUbi−1)+eδUbi)−(eU˜i(U˜i−δUbi−1)+eδUbi)+((eU¯i−eU˜i)−eU˜i(U¯i−U˜i))≥fi(δ,U¯i)−fi(δ,U˜i),where for i=1,2,fi(δ,s)=es(s−δUbi−1)+eδUbi≥0,
for all (δ,s)∈[0,1]×ℝ.
Using the elementary inequality es≥1+s,
valid for all s∈ℝ,
we get(3.31)(eU¯i−eU˜i)(e−δUbi−e−U¯i)=(eU¯i−δUbi−U¯i+δUbi)−(eU˜i−δUbi−U˜i+δUbi)+(eU˜i−U¯i−(U˜i−U¯i)−1)≥gi(δ,U¯i)−gi(δ,U˜i),with gi(δ,s)=es−δUbi−s+δUbi≥1,
for all (δ,s)∈[0,1]×ℝ.
Combining (3.30) and (3.31), we find(3.32)−δτ∫Ω(eU¯−eU˜)⋅φdx≤1τ∫Ω(f1(δ,U˜1)+f2(δ,U˜2))dx+2∫Ω(g1(δ,U˜1)+g2(δ,U˜2))dx.We recall that a.e. in Ω,−∥U˜i∥∞≤U˜i≤0 for i=1,2,
and since the functions fi,gi are continuous on [0,1]×[−∥U˜i∥∞,0],
we deduce that the right-hand side of (3.32) is uniformly bounded with respect to δ. Now we infer from (3.9) and Poincaré inequality
that(3.33)δ∫ΩQ+(U¯)⋅(U¯−Ub)dx≤C∫Ω|U¯−Ub|dx≤τ22∥U¯x∥𝕃2(Ω)2+C(τ),δ∫ΩQ+(U¯)⋅(e−δUb−e−U¯)dx≤C.
The result follows by combining
all these inequalities.


## 3.3. End of Proof of Theorem3.2

LetUτ be the solution of (3.4) provided by
Proposition 3.3. Recalling that u˜=eU˜ and ub=eUb,
we see that uτ=eUτ satisfies the following
problem:(3.34)1τ(uτ−u˜)−(Aτ(uτ)uτx)x=Q+(lnuτ)inΩ,uτ(0)=ub,uτx(1)=0,where the matrix Aτ is given by Aτ(r,s)=A+(r,s)+hτ(r,s)I with(3.35)A+(r,s)=(r(32−min(r,1−s))r(12−s)s(12−r)s(32−min(s,1−r))),hτ(r,s)=(r+s−1)+(rs+5(r+s))+τ.Note that if r+s≤1 then A+(r,s)=A(r,s) and hτ(r,s)=τ.We will focus on the𝕃∞ and ℍ1 estimates satisfied by the function uτ.
We begin with the following L∞ bounds.Lemma 3.7.
Let the hypotheses of Theorem3.2
hold, and let uτ=(ατ,βτ)∈ℍ1(Ω) satisfy problem (3.34). One
has
(3.36)0<ατ,βτ,ατ+βτ≤1inΩ.Proof.
We write the equation satisfied byατ+βτ and test it with φ=(ατ+βτ−1)+∈HD1(Ω).
We get using (1.24) that(3.37)∫Ωατ+βτ−(α˜+β˜)τφdx+∫Ω(12(ατ+βτ)+hτ(ατ,βτ))|φx|2dx≤0,so ∫Ω(ατ+βτ−(α˜+β˜))φdx≤0.
Consequently, ∫Ωφ2dx≤∫Ω(α˜+β˜−1)φdx≤0,
and hence ατ+βτ≤1.As a consequence of Lemma3.7, we easily see that uτ is a solution to problem (3.1) in the sense of
Definition 3.1. Moreover, since lnuτ=Uτ∈ℍ1(Ω) then lnuτ∈𝕃∞(Ω).
In order to check the entropy inequality (3.3), we test (3.1) with φτ=lnuτ−lnub∈ℍD1(Ω) to get(3.38)1τ∫Ω(uτ−u˜)⋅φτdx+∫Ω(1−ατ−βτ)ατxβτxdx+τ∫Ω(|ατx|2ατ+|βτx|2βτ)dx+∫Ω((32−ατ)|ατx|2+(32−βτ)|βτx|2)dx=∫Ωq(uτ)⋅φτdx.Then, from (3.15) and Young inequality,
we get(3.39)−∫Ω(ατ+βτ−1)ατxβτxdx≥−12∫Ω((32−ατ)|ατx|2+(32−βτ)|βτx|2)dx,so, inserting (3.39) into (3.38) and
using the fact that 3/2−ατ,3/2−βτ≥1/2,
we see that(3.40)1τ∫Ω(uτ−u˜)⋅φτdx+14∫Ω(|ατx|2+|βτx|2)dx≤∫Ωq(uτ)⋅φτdx.Using the boundedness of q(uτ) and the fact that the function sln(s) is bounded in [0,1],
we obtain ∫Ωq(uτ)⋅φτdx≤C, then the convexity of the functions G1 and G2 leads to(3.41)∫Ω(uτ−u˜)⋅φτdx≥∫Ω(G1(ατ)+G2(βτ))dx−∫Ω(G1(α˜)+G2(β˜))dx,which ends the
proof.Remark 3.8.
All the results of this section remain valid if one
consider, instead of (3.1), the following problem:(3.42)1τ(u−u˜)−((A(u)+εI)ux)x=q(u)inΩ,u(0)=ub,ux(1)=0,where τ>0 and ε>0 is independent of τ.
In particular, the solution vτε to (3.42) satisfies the L∞ bounds and the entropy inequality given in
Theorem 3.2, with the constant C>0 independent of ε (and τ). Thus performing the limit as ε→0 in (3.42), we get a function vτ solving the problem(3.43)1τ(u−u˜)−(A(u)ux)x=q(u)inΩ,u(0)=ub,ux(1)=0,which corresponds to the
“natural” time discretization of our problem (1.25)–(1.27). However, from there, the situation becomes
complicated because the obtained solution vτ to problem (3.43) has its components which are
only nonnegative and we no longer have lnvτ∈𝕃∞(Ω) (in fact, we cannot even take the ln of vτ). Therefore, the time discretization scheme
based on (3.43) cannot be solved.


## 4. Proof of Theorem2.2

### 4.1. The Time Discretization Scheme

Let assumptions
(H1)-(H2) hold and letT>0.
We will use the backward Euler approximation of time derivative ut≃(1/τ)(u(tk)−u(tk−1)).
We divide the time interval (0,T) into N subintervals (tk−1,tk] of the same length τ=T/N.
Then, we define recursively uτk,k=1,…,N,
as the weak solution of (3.1) provided by Theorem 3.2 corresponding to the data u˜=uτk−1,
that is,(4.1)1τ(uτk−uτk−1)−((A(uτk)+τI)uτxk)x=q(uτk)inΩ,uτk(0)=ub,uτxk(1)=0,uτ0 being the initial condition u0 of problem (1.25)–(1.27). Let u(τ),u¯(τ)  be the piecewise constant in time
interpolation on (0,T) of uτ1,uτ2,…,uτN  and  uτ0,uτ1,…,uτN−1,
respectively, that is,(4.2)u(τ)(t,x)=uτk(x),u¯(τ)(t,x)=uτk−1(x)on(tk−1,tk]×Ω,k=1,…,N,and let u˜(τ) be the function defined on QT by(4.3)u˜(τ)(t,x)=t−tk−1τ(u(τ)(t,x)−u¯(τ)(t,x))+u¯(τ)(t,x)on(tk−1,tk]×Ω,k=1,…,N.With these notations, we can rewrite (4.1) as(4.4)u˜t(τ)−((A(u(τ))+τI)ux(τ))x=q(u(τ))inΩ,u(τ)(0)=ub,ux(τ)(1)=0.Now, we set for uτk=(ατk,βτk)(4.5)ητk=∫Ω(G1(ατk)+G2(βτk))dx,k=0,…,N,η0=ητ0,η(τ)(t)=ητk,∀t∈(tk−1,tk],k=1,…,N.


### 4.2. Uniform Estimates with Respect toτ

Lemma 4.1.
Letu(τ)=(α(τ),β(τ)) and u¯(τ)=(α¯(τ),β¯(τ)) defined by (4.2). One has(4.6)0<α(τ),β(τ),α(τ)+β(τ)≤1a.e.inQT,0<α¯(τ),β¯(τ),α¯(τ)+β¯(τ)≤1a.e.inQT,and there exists a positive
constant C independent of τ such that(4.7)∥η(τ)∥L∞(0,T),∥u(τ)∥L2(0,T;ℍ1(Ω)),∥u¯(τ)∥L2(0,T;ℍ1(Ω))≤C.Proof.
We apply the results of Theorem3.2,
so the first part is immediate, then (3.3) leads to(4.8)ητk−ητk−1+τ4∫Ω(|ατxk|2+|βτxk|2)dx≤Cτ,k=1,…,N.Summing these inequalities from k=1 to k=m,
for 1≤m≤N,
we get
(4.9)ητm−η0+τ4∑k=1m∫Ω(|ατxk|2+|βτxk|2)dx≤Cmτ.Therefore,(4.10)max1≤m≤Nητm+14∑k=1N∫Ωτ(|ατxk|2+|βτxk|2)dx≤η0+CT,which can be written as(4.11)∥η(τ)∥L∞(0,T)+14∫Ω∑k=1N∫tk−1tk(|ατxk|2+|βτxk|2)dtdx≤η0+CT.This means that(4.12)∥η(τ)∥L∞(0,T)+14∫QT(|αx(τ)|2+|βx(τ)|2)dtdx≤η0+CT.Coming back to (4.9), we deduce
that(4.13)∑k=1N−1∫Ωτ(|ατxk|2+|βτxk|2)dx=∫QT|u¯x(τ)|2dtdx−τ∫Ω|u0x|2dx≤4(η0+CT),and we get the result.We have also the following
estimate.Lemma 4.2.
There exists a positive constantC independent of τ such that(4.14)∥u˜(τ)∥H1(0,T;(ℍ1(Ω))′),∥u˜(τ)∥L2(0,T;ℍ1(Ω))≤C,(4.15)∥u˜(τ)−u(τ)∥L2(0,T;(ℍ1(Ω))′)≤Cτ.Proof.
We use (4.4) and the results of Lemma
4.1 to deduce that ∥u˜t(τ)∥L2(0,T;(ℍ1(Ω))′) is uniformly bounded. Now, since u˜x(τ)=((t−tk−1)/τ)ux(τ)+((tk−t)/τ)u¯x(τ) on (tk−1,tk]×Ω,
then thanks to Lemma 4.1 we deduce that ∥u˜(τ)∥L2(0,T;ℍ1(Ω)) is uniformly bounded. Finally, to check (4.15),
we have from (4.3) that for  t∈(tk−1,tk),∥u˜(τ)−u(τ)∥(ℍ1(Ω))′=(tk−t)∥u˜t(τ)∥(ℍ1(Ω))′≤τ∥u˜t(τ)∥(ℍ1(Ω))′.
This leads to the result by using (4.14).


### 4.3. Passing to the Limit asτ→0: End of Proof of Theorem 2.2

Using (4.14), we
deduce the existence of a function u∈L2(0,T;ℍ1(Ω))∩H1(0,T;(ℍ1(Ω))′) such that as τ→0 at least for some subsequence,(4.16)u˜(τ)⇀uweaklyinL2(0,T;ℍ1(Ω))∩H1(0,T;(ℍ1(Ω))′).Then Aubin compactness lemma and
the compactness of the embedding of H1(Ω) into L∞ lead to the strong convergence(4.17)u˜(τ)→ustronglyinL2(0,T;𝕃∞(Ω)).Moreover, by Lemma 4.1, we infer
the existence of a function v in L2(0,T;ℍ1(Ω)) and a subsequence of uτ such that(4.18)u(τ)⇀vweaklyinL2(0,T;ℍ1(Ω)),and according to (4.15) and (4.16),
we derive that v=u.
Moreover, we have the strong convergence(4.19)u(τ)→ustronglyin𝕃2(QT).Indeed,(4.20)∥u(τ)−u∥L2(QT)≤∥u(τ)−u˜(τ)∥L2(QT)+∥u˜(τ)−u∥L2(QT)≤C∥u˜(τ)−u(τ)∥L2(0,T;ℍ1(Ω))1/2∥u˜(τ)−u(τ)∥L2(0,T;(ℍ1(Ω))′)1/2+∥u˜(τ)−u∥𝕃2(QT),where C>0 is independent of τ.
Using (4.7), (4.14),
(4.15),
and (4.17), it is straightforward to deduce that u(τ)→u strongly in 𝕃2(QT),
and hence a.e. in QT.
Consequently, u=(α,β) satisfies(4.21)0≤α,β,α+β≤1,a.e.inQT.Finally, since H1(0,T;(ℍ1(Ω))′)∩L2(0,T;ℍ1(Ω))⊂𝒞0([0,T];𝕃2(Ω)),
the initial condition is satisfied and thus u is a weak solution of (1.25)–(1.27) in the sense of Definition 2.1.Remark 4.3.
The result proved in this work
remains valid if we replace the conditionλ=μ=P0 by one of these conditions:(4.22)87max(λ,μ)≤P0≤187(λ+μ)or23max(λ,μ)≤P0≤87min(λ,μ).Indeed, in these cases, direct
calculations show that (3.15) becomes(4.23)(−2μb−2λa+2P0)2≤94(2λ(1−a)+P0)(2μ(1−b)+P0)if0≤a,b,a+b≤1.


## 4.1. The Time Discretization Scheme

Let assumptions
(H1)-(H2) hold and letT>0.
We will use the backward Euler approximation of time derivative ut≃(1/τ)(u(tk)−u(tk−1)).
We divide the time interval (0,T) into N subintervals (tk−1,tk] of the same length τ=T/N.
Then, we define recursively uτk,k=1,…,N,
as the weak solution of (3.1) provided by Theorem 3.2 corresponding to the data u˜=uτk−1,
that is,(4.1)1τ(uτk−uτk−1)−((A(uτk)+τI)uτxk)x=q(uτk)inΩ,uτk(0)=ub,uτxk(1)=0,uτ0 being the initial condition u0 of problem (1.25)–(1.27). Let u(τ),u¯(τ)  be the piecewise constant in time
interpolation on (0,T) of uτ1,uτ2,…,uτN  and  uτ0,uτ1,…,uτN−1,
respectively, that is,(4.2)u(τ)(t,x)=uτk(x),u¯(τ)(t,x)=uτk−1(x)on(tk−1,tk]×Ω,k=1,…,N,and let u˜(τ) be the function defined on QT by(4.3)u˜(τ)(t,x)=t−tk−1τ(u(τ)(t,x)−u¯(τ)(t,x))+u¯(τ)(t,x)on(tk−1,tk]×Ω,k=1,…,N.With these notations, we can rewrite (4.1) as(4.4)u˜t(τ)−((A(u(τ))+τI)ux(τ))x=q(u(τ))inΩ,u(τ)(0)=ub,ux(τ)(1)=0.Now, we set for uτk=(ατk,βτk)(4.5)ητk=∫Ω(G1(ατk)+G2(βτk))dx,k=0,…,N,η0=ητ0,η(τ)(t)=ητk,∀t∈(tk−1,tk],k=1,…,N.


## 4.2. Uniform Estimates with Respect toτ

Lemma 4.1.
Letu(τ)=(α(τ),β(τ)) and u¯(τ)=(α¯(τ),β¯(τ)) defined by (4.2). One has(4.6)0<α(τ),β(τ),α(τ)+β(τ)≤1a.e.inQT,0<α¯(τ),β¯(τ),α¯(τ)+β¯(τ)≤1a.e.inQT,and there exists a positive
constant C independent of τ such that(4.7)∥η(τ)∥L∞(0,T),∥u(τ)∥L2(0,T;ℍ1(Ω)),∥u¯(τ)∥L2(0,T;ℍ1(Ω))≤C.Proof.
We apply the results of Theorem3.2,
so the first part is immediate, then (3.3) leads to(4.8)ητk−ητk−1+τ4∫Ω(|ατxk|2+|βτxk|2)dx≤Cτ,k=1,…,N.Summing these inequalities from k=1 to k=m,
for 1≤m≤N,
we get
(4.9)ητm−η0+τ4∑k=1m∫Ω(|ατxk|2+|βτxk|2)dx≤Cmτ.Therefore,(4.10)max1≤m≤Nητm+14∑k=1N∫Ωτ(|ατxk|2+|βτxk|2)dx≤η0+CT,which can be written as(4.11)∥η(τ)∥L∞(0,T)+14∫Ω∑k=1N∫tk−1tk(|ατxk|2+|βτxk|2)dtdx≤η0+CT.This means that(4.12)∥η(τ)∥L∞(0,T)+14∫QT(|αx(τ)|2+|βx(τ)|2)dtdx≤η0+CT.Coming back to (4.9), we deduce
that(4.13)∑k=1N−1∫Ωτ(|ατxk|2+|βτxk|2)dx=∫QT|u¯x(τ)|2dtdx−τ∫Ω|u0x|2dx≤4(η0+CT),and we get the result.We have also the following
estimate.Lemma 4.2.
There exists a positive constantC independent of τ such that(4.14)∥u˜(τ)∥H1(0,T;(ℍ1(Ω))′),∥u˜(τ)∥L2(0,T;ℍ1(Ω))≤C,(4.15)∥u˜(τ)−u(τ)∥L2(0,T;(ℍ1(Ω))′)≤Cτ.Proof.
We use (4.4) and the results of Lemma
4.1 to deduce that ∥u˜t(τ)∥L2(0,T;(ℍ1(Ω))′) is uniformly bounded. Now, since u˜x(τ)=((t−tk−1)/τ)ux(τ)+((tk−t)/τ)u¯x(τ) on (tk−1,tk]×Ω,
then thanks to Lemma 4.1 we deduce that ∥u˜(τ)∥L2(0,T;ℍ1(Ω)) is uniformly bounded. Finally, to check (4.15),
we have from (4.3) that for  t∈(tk−1,tk),∥u˜(τ)−u(τ)∥(ℍ1(Ω))′=(tk−t)∥u˜t(τ)∥(ℍ1(Ω))′≤τ∥u˜t(τ)∥(ℍ1(Ω))′.
This leads to the result by using (4.14).


## 4.3. Passing to the Limit asτ→0: End of Proof of Theorem 2.2

Using (4.14), we
deduce the existence of a function u∈L2(0,T;ℍ1(Ω))∩H1(0,T;(ℍ1(Ω))′) such that as τ→0 at least for some subsequence,(4.16)u˜(τ)⇀uweaklyinL2(0,T;ℍ1(Ω))∩H1(0,T;(ℍ1(Ω))′).Then Aubin compactness lemma and
the compactness of the embedding of H1(Ω) into L∞ lead to the strong convergence(4.17)u˜(τ)→ustronglyinL2(0,T;𝕃∞(Ω)).Moreover, by Lemma 4.1, we infer
the existence of a function v in L2(0,T;ℍ1(Ω)) and a subsequence of uτ such that(4.18)u(τ)⇀vweaklyinL2(0,T;ℍ1(Ω)),and according to (4.15) and (4.16),
we derive that v=u.
Moreover, we have the strong convergence(4.19)u(τ)→ustronglyin𝕃2(QT).Indeed,(4.20)∥u(τ)−u∥L2(QT)≤∥u(τ)−u˜(τ)∥L2(QT)+∥u˜(τ)−u∥L2(QT)≤C∥u˜(τ)−u(τ)∥L2(0,T;ℍ1(Ω))1/2∥u˜(τ)−u(τ)∥L2(0,T;(ℍ1(Ω))′)1/2+∥u˜(τ)−u∥𝕃2(QT),where C>0 is independent of τ.
Using (4.7), (4.14),
(4.15),
and (4.17), it is straightforward to deduce that u(τ)→u strongly in 𝕃2(QT),
and hence a.e. in QT.
Consequently, u=(α,β) satisfies(4.21)0≤α,β,α+β≤1,a.e.inQT.Finally, since H1(0,T;(ℍ1(Ω))′)∩L2(0,T;ℍ1(Ω))⊂𝒞0([0,T];𝕃2(Ω)),
the initial condition is satisfied and thus u is a weak solution of (1.25)–(1.27) in the sense of Definition 2.1.Remark 4.3.
The result proved in this work
remains valid if we replace the conditionλ=μ=P0 by one of these conditions:(4.22)87max(λ,μ)≤P0≤187(λ+μ)or23max(λ,μ)≤P0≤87min(λ,μ).Indeed, in these cases, direct
calculations show that (3.15) becomes(4.23)(−2μb−2λa+2P0)2≤94(2λ(1−a)+P0)(2μ(1−b)+P0)if0≤a,b,a+b≤1.



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