Existence and incompressible limit of a tissue growth model with autophagy
aa r X i v : . [ m a t h . A P ] F e b EXISTENCE AND INCOMPRESSIBLE LIMIT OF A TISSUE GROWTHMODEL WITH AUTOPHAGY
Jian-Guo Liu and Xiangsheng Xu
Department of Physics and Department of MathematicsDuke UniversityDurham, NC 27708, USA andDepartment of Mathematics & StatisticsMississippi State UniversityMississippi State, MS 39762, USA
Abstract.
In this paper we study a cross-diffusion system whose coefficient matrix is non-symmetricand degenerate. The system arises in the study of tissue growth with autophagy. The existence ofa weak solution is established. We also investigate the limiting behavior of solutions as the pressuregets stiff. The so-called incompressible limit is a free boundary problem of Hele-Shaw type. Ourkey new discovery is that the usual energy estimate still holds as long as the time variable staysaway from 0. Introduction
Let Ω be a bounded domain in R N with Lipschitz boundary ∂ Ω and T any positive number. Weconsider the initial boundary value problem ∂ t n − div ( n ∇ p ) = G ( d ) n − K ( d ) n + K ( d ) n ≡ R in Ω T ≡ Ω × (0 , T ),(1.1) ∂ t n − div ( n ∇ p ) = ( G ( d ) − D ) n + K ( d ) n − K ( d ) n ≡ R in Ω T ,(1.2) b∂ t d − ∆ d = − ψ ( d ) n + an in Ω T ,(1.3) n ∇ p · n = n ∇ p · n = 0 on Σ T ≡ ∂ Ω × (0 , T ),(1.4) d = d b on Σ T ,(1.5) ( n ( x, , n ( x, , d ( x, n (0)1 ( x ) , n (0)2 ( x ) , d (0) ( x )) on Ω,(1.6)where n is the unit outward normal to ∂ Ω and n = n + n , p = n γ , γ ≥ . (1.7)This problem was proposed as a tissue growth model with autophagy in [8]. In the model, cellsare classified into two phases: normal cells and autophagic cells, and n , n are their respectivedensities. The third unknown function d represents the concentration of nutrients. We assume thatboth cells have the same birth rate. Their death rates are different because autophagic cells havean extra death rate D due to the “self-eating” mechanism. Thus if G ( d ) is the net growth rate ofnormal cells then G ( d ) − D gives the net growth rate for autophagic cells. Two types of cells canchange from one to another. The transition rates are denoted by K ( d ) , K ( d ), respectively. Sinceautophagy is a reversible process, we have(1.8) K ( d ) ≥ , K ( d ) ≥ . Mathematics Subject Classification.
Primary: 35B45, 35B65, 35Q92, 35K51.
Key words and phrases.
Autophagy; existence; incompressible limit; tissue growth models.
Both cells consume nutrients with the consumption rate ψ ( d ). However, autophagic cells alsoprovide nutrients by degrading its own constituents with a supply rate a . We assume(1.9) D, a ∈ (0 , ∞ ) . Moreover,(1.10) ψ (0) = 0 , ψ ( d ) is increasing, and there is d > ψ ( d ) = a. The first condition in (1.10) means that when there is no nutrient the consumption rate should bezero. The number d is the so-called critical nutrient concentration. When d < d autophagic cellssupplies more nutrients than they consume, while d > d indicates that autophagic cells consumesmore nutrients than they supply.For the spatial motion of cells, we take a fluid mechanical point of view. That is, it is driven bya velocity field equals to the negative gradient of the pressure (Darcy’s law) [14]. And the pressurearises from mechanical contact between cells. Denote by p the pressure. Then we can assume that(1.7), (1.1), and (1.2) hold.One can also model tissue growth as free boundary problems [9]. They are also called geometricor incompressible models and describe tissue as a moving domain (see [6] and the references therein).Building a link between these two classes of models has attracted the attention of many researchersin recent years. The first result in this direction was obtained in [14] for a purely mechanical model.It indicates that the limit of the mechanical model gives rise to a free boundary problem as thepressure becomes stiff. Since then the same result has been achieved for a variety of models, whichincluded active motion [15], viscosity [17], different laws of state [7] and more than one species ofcells [4]. In each case the limit model turns out to be a free boundary model of Hele-Shaw type.The objective of this paper is to study the existence assertion for (1.1)-(1.6) and the limitingbehavior of solutions as γ → ∞ .We largely follow the approach adopted in [18] for the existence assertion. To understand thenature of the limiting model for our problem, we define a family of maximal monotone graphs [2]in R × R by ϕ γ ( s ) = (cid:0) s + (cid:1) γ +1 = (cid:26) s γ +1 if s ≥ s < . Obviously,(1.11) ϕ γ ( s ) → ϕ ∞ ( s ) ≡ (cid:26) [0 , ∞ ) if s = 1,0 if s < γ → ∞ [2]. The total density n = n ( γ ) satisfies the problem ∂ t n ( γ ) − γγ + 1 ∆ v ( γ ) = G ( d ( γ ) ) n ( γ ) − Dn ( γ )2 ≡ R ( γ ) in Ω T ,v ( γ ) = (cid:16) n ( γ ) (cid:17) γ +1 a.e. on Ω T ,(1.12) ∇ v ( γ ) · n = 0 on Σ T , n ( γ ) ( x,
0) = n (0) ≡ n (0)1 + n (0)2 on Ω.Thus if we formally take γ → ∞ , we expect to arrive at the following problem ∂ t n ( ∞ ) − ∆ v ( ∞ ) = G ( d ( ∞ ) ) n ( ∞ ) − Dn ( ∞ )2 ≡ R ( ∞ ) in Ω T , (1.13) v ( ∞ ) ∈ ϕ ∞ ( n ( ∞ ) ) a.e. on Ω T ,(1.14) ∇ v ( ∞ ) · n = 0 on Σ T ,(1.15) n ( ∞ ) ( x,
0) = n (0) on Ω.(1.16) TISSUE GROWTH MODEL WITH AUTOPHAGY 3 If n (0) ≤ n ( ∞ ) and lim γ →∞ n ( γ ) = n ( ∞ ) in L (0 , T ; L (Ω)) (also see [22] for related results). If n (0) > ϕ ∞ and theresulting problem (1.13)-(1.16) becomes singular. Thus identifying the limit of the sequence { n ( γ ) } is an interesting issue. When R ( γ ) ≡
0, this problem was solved in [5] through an application ofthe Aronson-B´enilan inequality [1](1.17) ∂ t n ( γ ) ≥ − n ( γ ) γt . The precise result there is: If Ω = R N , n (0) ( x ) has a star-shaped profile, and R ( γ ) =0, then n ( ∞ ) ≡ lim γ →∞ n ( γ ) exists and is given by n ( ∞ ) ( x ) = (cid:26) x ∈ A , n (0) ( x ) if x / ∈ A ,where A is the coincident set of the solution of the following variational inequalities − ∆ w ≥ n (0) − , w ≥ , (cid:16) ∆ w + n (0) − (cid:17) w = 0 in R N .A remarkable fact is that the limit n ( ∞ ) is a function of x only. A similar result was establishedfor hyperbolic conservation laws in [22]. However, if R ( γ ) changes sign, inequalities of the Aronson-B´enilan type no longer hold [16]. To circumvent this difficulty, the authors of [6] established aweaker version of (1.17) along with an L estimate for the gradient of the pressure. Our problemhere does not quite fit the framework developed in [6]. This forces us to take a totally differentapproach. It seems more convenient for us to work with v ( γ ) = (cid:0) n ( γ ) (cid:1) γ +1 instead of the pressure.Our key estimate is: Z Tτ Z Ω (cid:16) v ( γ ) (cid:17) dxdt + Z Tτ Z Ω (cid:12)(cid:12)(cid:12) ∇ v ( γ ) (cid:12)(cid:12)(cid:12) dxdt ≤ cτ for all γ ≥ τ ∈ (0 , T ).Here and in what follows the letter c denotes a generic positive constant whose value is determinedby the given data. That is, the sequence { v ( γ ) } is bounded in L ( τ, T ; W , (Ω)) for each τ ∈ (0 , T ).Before we introduce our complete results, we state the definition of a weak solution. Definition 1.1.
We say that ( n , n , d ) is a weak solution to (1.1) - (1.6) if: (D1) n , n , d are all non-negative and bounded with (1.18) ∂ t n , ∂ t n , ∂ t d ∈ L (0 , T ; (cid:0) W , (Ω) (cid:1) ∗ ) , n γ +12 , d ∈ L (0 , T ; W , (Ω)) , where n is given as in (1.7) and (cid:0) W , (Ω) (cid:1) ∗ denotes the dual space of W , (Ω) ; JIAN-GUO LIU AND XIANGSHENG XU (D2)
There hold − Z Ω T n ∂ t ξ dxdt + Z Ω T n ∇ n γ · ∇ ξ dxdt = Z Ω T R ξ dxdt − h n ( · , T ) , ξ ( · , T ) i + Z Ω n (0)1 ( x ) ξ ( x, dx for each ξ ∈ H (0 , T ; W , (Ω)) , − Z Ω T n ∂ t ξ dxdt + Z Ω T n ∇ n γ · ∇ ξ dxdt = Z Ω T R ξ dxdt − h n ( · , T ) , ξ ( · , T ) i + Z Ω n (0)2 ( x ) ξ ( x, dx for each ξ ∈ H (0 , T ; W , (Ω)) , and − b Z Ω T d∂ t ζdxdt + Z Ω T ∇ d · ∇ ζdxdt = Z Ω T ( − ψ ( d ) n + an ) ζdxdt − b h d ( · , T ) , ζ ( · , T ) i + b Z Ω d (0) ( x ) ζ ( x, dx for each ζ ∈ H (0 , T ; W , (Ω)) , where h· , ·i denotes the duality pairing between W , (Ω) and (cid:0) W , (Ω) (cid:1) ∗ and H (0 , T ; W , (Ω)) = { v ∈ L (0 , T ; W , (Ω)) : ∂ t v ∈ L (0 , T ; W , (Ω)) } ; (D3) (1.5) is satisfied. To see that the two equations in (D2) make sense, we can conclude from (D1) that n , n , d ∈ C ([0 , T ]; (cid:0) W , (Ω) (cid:1) ∗ ). Since n is bounded and γ ≥ γ +12 , we also have n γ ∈ L (0 , T ; W , (Ω)). Theorem 1.2.
Assume: (H1)
G, K , K , ψ are all continuous functions; (H2) (1.8) , (1.9) , and (1.10) hold; (H3) b ∈ (0 , ∞ ) and ∂ Ω is Lipschitz; (H4) n (0)1 , n (0)2 ∈ W , (Ω) ∩ L ∞ (Ω) , d (0) ∈ L ∞ (Ω) , and d b ∈ L (0 , T ; W , (Ω)) ∩ L ∞ (Ω T ) .Then there is a weak solution to (1.1) - (1.6) . Set L = max {k d b k ∞ , Σ T , k d (0) k ∞ , Ω , d } , (1.19) G = max s ∈ [0 ,L ] G ( s ) . (1.20) Theorem 1.3.
Let the assumptions of Theorem 1.2 hold. Assume: (H5) G ′ ( s ) is bounded; (H6) d b ∈ W ,s (Ω T ) for some s > N + 2 and d (0) ∈ W , ∞ (Ω) ; (H7) (cid:12)(cid:12)(cid:8) n (0) ( x ) ≥ σ (cid:9)(cid:12)(cid:12) ≤ e G T k n (0) k ∞ , Ω | Ω | for some σ ∈ (cid:0) , e − G T (cid:1) ; (H8) ∂ Ω is C , . TISSUE GROWTH MODEL WITH AUTOPHAGY 5
Denote by ( n ( γ ) , n ( γ )1 , n ( γ )2 , d ( γ ) ) the solution obtained in Theorem 1.2. Then as γ → ∞ , we have ( n ( γ ) , n ( γ )1 , n ( γ )2 ) → ( n ( ∞ ) , n ( ∞ )1 , n ( ∞ )2 ) weak ∗ in ( L ∞ (Ω T )) and strongly in (cid:0) C ([ τ, T ]; (cid:0) W , (Ω) (cid:1) ∗ ) (cid:1) for each τ ∈ (0 , T ) , (1.21) v ( γ ) → v ( ∞ ) weakly in L ( τ, T ; W , (Ω)) for each τ ∈ (0 , T ) , (1.22) ∇ v ( γ ) → ∇ v ( ∞ ) strongly in L ( τ, T ; ( L (Ω)) N ) for each τ ∈ (0 , T ) , (1.23) n ( γ )2 n ( γ ) → η ( ∞ ) weak ∗ in L ∞ (Ω T ) , (1.24) d ( γ ) → d ( ∞ ) weak ∗ in L ∞ (0 , T ; W , ∞ (Ω)) and strongly in L (Ω T ) . (1.25) The limit ( n ( ∞ ) , v ( ∞ ) , n ( ∞ )1 , n ( ∞ )2 , η ( ∞ ) , d ( ∞ ) ) satisfies − Z Ω T n ( ∞ ) ∂ t ξ dxdt + Z Ω T ∇ v ( ∞ ) · ∇ ξ dxdt = Z Ω T R ( ∞ ) ξ dxdt, − Z Ω T n ( ∞ )1 ∂ t ξ dxdt + Z Ω T (cid:16) − η ( ∞ ) (cid:17) ∇ v ( ∞ ) · ∇ ξ dxdt = Z Ω T R ( ∞ )1 ξ dxdt, − Z Ω T n ( ∞ )2 ∂ t ξ dxdt + Z Ω T η ( ∞ ) ∇ v ( ∞ ) · ∇ ξ dxdt = Z Ω T R ( ∞ )2 ξ dxdt, and − b Z Ω T d ( ∞ ) ∂ t ξ dxdt + Z Ω T ∇ d ( ∞ ) · ∇ ξ dxdt = Z Ω T ( − ψ ( d ( ∞ ) ) n ( ∞ ) + an ( ∞ )2 ) ξ dxdt − b h d ( ∞ ) ( · , T ) , ξ ( · , T ) i + b Z Ω d (0) ( x ) ξ ( x, dx for each ( ξ , ξ , ξ ) ∈ (cid:0) H (0 , T ; W , (Ω)) (cid:1) with ( ξ , ξ , ξ ) = 0 near t = 0 and ( ξ , ξ , ξ ) | t = T = 0 and each ξ ∈ H (0 , T ; W , (Ω)) , where R ( ∞ ) is given as in (1.13) and R ( ∞ )1 = G ( d ( ∞ ) ) n ( ∞ )1 − K ( d ( ∞ ) ) n ( ∞ )1 + K ( d ( ∞ ) ) n ( ∞ )2 ,R ( ∞ )2 = (cid:16) G ( d ( ∞ ) ) − D (cid:17) n ( ∞ )2 + K ( d ( ∞ ) ) n ( ∞ )1 − K ( d ( ∞ ) ) n ( ∞ )2 . Moreover, (1.14) holds and (1.26) v ( ∞ ) (cid:16) ∆ v ( ∞ ) + R ( ∞ ) (cid:17) = 0 . If we compare the equations in (D2) with the ones here, two pieces are missing. One is that weare no longer able to identify the initial conditions for ( n ( ∞ ) , n ( ∞ )1 , n ( ∞ )2 ). This is to be expecteddue to the fact that ϕ ∞ is not defined on the set { n (0) > } . A redeeming feature is that we canview (1.26), the so-called complementary condition, as some kind of compensation for this lack ofinitial conditions. More significantly, this condition connects our limits to the geometric form ofthe Hele-Shaw problem [6]. At least formally, it says − ∆ v ( ∞ ) = R ( ∞ ) on Ω( t ) ≡ { v ( ∞ ) ( x, t ) > } .The second one is that we have not been able to show(1.27) η ( ∞ ) = n ( ∞ )2 n ( ∞ ) . This can be derived from the precompactness of { n ( γ ) } in some L q (Ω T ) space with q ∈ [1 , ∞ ) (seethe proof of (2.59) in Section 2 below). Unfortunately, this result is not available to us because inthe generality considered here the sequence {∇ n ( γ ) } cannot be shown to be bounded in a function JIAN-GUO LIU AND XIANGSHENG XU space. Furthermore, it does not seem to be possible to obtain any estimates on ∂ t v ( γ ) that areuniform in γ . As a result, the precompactness of { v ( γ ) } in some L q (Ω T ) space is also an issue. Thisis so in spite of the fact that we have (1.23).We can easily see that (1.14) implies n ( ∞ ) ≤ T and(1.28) (cid:16) − n ( ∞ ) (cid:17) v ( ∞ ) = 0 on Ω T .(1.29)Obviously, we can no longer expect n ( ∞ ) to be independent of t due to the presence of R ( ∞ ) . Theterm ∆ v ( ∞ ) may be a pure distribution. We define v ( ∞ ) ∆ v ( ∞ ) = div (cid:16) v ( ∞ ) ∇ v ( ∞ ) (cid:17) − (cid:12)(cid:12)(cid:12) ∇ v ( ∞ ) (cid:12)(cid:12)(cid:12) in the sense of distributions.Also note that the assumption (H7) implies that n (0) is close to 0 on a large set. The smaller T is,the easier it is for (H7) to hold.The remainder of the paper is devoted to the proof of the above two theorems. To be specific,Section 2 contains the proof of Theorem 1.2, while Theorem 1.3 is established in Section 3.2. Existence of a global weak solution and Proof of Theorem 1.2
The proof will be divided into several lemmas. Before we begin, we state the following three wellknown results.
Lemma 2.1.
Let h ( s ) be a convex and lower semi-continuous function on R [12] . Assume that (C1) f ∈ W (0 , T ) ≡ (cid:8) ϕ ∈ L (0 , T ; W , (Ω)) : ∂ t ϕ ∈ L (cid:0) , T ; (cid:0) W , (Ω) (cid:1) ∗ (cid:1)(cid:9) ; (C2) g ∈ L (0 , T ; W , (Ω)) with the property g ( x, t ) ∈ ∂h ( f ( x, t )) for a.e ( x, t ) ∈ Ω T , where ∂h is the subgradient of h .Then the function t R Ω h ( f ( x, t )) dx is absolutely continuous on [0 , T ] and (2.1) ddt Z Ω h ( f ) dx = h ∂ t f, g i . If h ( s ) = s , this lemma is a special case of the well known Lions-Magenes lemma ([20],p.176–177). Formula (2.1) is trivial if f is smooth. The general case can be established by suitableapproximation. See ([12], p. 101) for the details. Lemma 2.2 (Lions-Aubin) . Let X , X and X be three Banach spaces with X ⊆ X ⊆ X . Supposethat X is compactly embedded in X and that X is continuously embedded in X . For ≤ p, q ≤ ∞ ,let W p,q (0 , T ) = { u ∈ L p ([0 , T ]; X ) : ∂ t u ∈ L q ([0 , T ]; X ) } . Then: (i) If p < ∞ , then the embedding of W p,q (0 , T ) into L p ([0 , T ]; X ) is compact. (ii) If p = ∞ and q > , then the embedding of W p,q (0 , T ) into C ([0 , T ]; X ) is compact. The proof of this lemma can be found in [19]. We mention in passing that Lemmas 2.1 and 2.2imply that W (0 , T ) is contained in C ([0 , T ]; L (Ω)). Lemma 2.3.
Let Ω be a bounded domain in R N with Lipschitz boundary and ≤ p < N . Thenthere is a positive number c = c ( N ) such that k u − u S k p ∗ ≤ cd N +1 − pN | S | p k∇ u k p for each u ∈ W ,p (Ω) ,where S is any measurable subset of Ω with | S | > , u S = | S | R S udx , and d is the diameter of Ω . TISSUE GROWTH MODEL WITH AUTOPHAGY 7
This lemma can be inferred from Lemma 7.16 in [11].Our approximate problems are similar to those in [18]. For each ε >
0, we consider ∂ t n − ε ∆ n = γ div ( n γ ∇ n ) + G ( d ) n + ( G ( d ) − D ) n in Ω T ,(2.2) ∂ t n − ε ∆ n = γ div (cid:0) n n γ − ∇ n (cid:1) + G ( d ) n − K ( d ) n + K ( d ) n in Ω T ,(2.3) ∂ t n − ε ∆ n = γ div (cid:0) n n γ − ∇ n (cid:1) + ( G ( d ) − D ) n + K ( d ) n − K ( d ) n in Ω T ,(2.4) b∂ t d − ∆ d = − ψ ( d ) n + an in Ω T ,(2.5) ∇ n · n = ∇ n · n = ∇ n · n = 0 on Σ T ,(2.6) d = d b on Σ T ,(2.7) ( n, n , n , d ) | t =0 = (cid:16) n (0) ( x ) , n (0)1 ( x ) , n (0)2 ( x ) , d (0) ( x ) (cid:17) on Ω.(2.8) Lemma 2.4.
Assume that (H1) - (H4) hold. Then for each fixed ε > there exists a quadruplet ( n, n , n , d ) in the function space ( W (0 , T )) ∩ ( L ∞ (Ω T )) such that (2.2) - (2.8) are all satisfied inthe sense of Definition 1.1.Proof. This lemma will be established via the Leray-Schauder fixed point theorem ([11], p.280).For this purpose, we introduce a cut-off function(2.9) θ ℓ ( s ) = s ≤ s if 0 < s < ℓ , ℓ if s ≥ ℓ ,where ℓ > M from (cid:0) L (Ω T ) (cid:1) into itself asfollows: Let ( w, v , v , u ) ∈ (cid:0) L (Ω T ) (cid:1) . We first consider the initial boundary value problem ∂ t n − div h ε + γ ( θ ℓ ( v ) + θ ℓ ( v )) θ γ − ℓ ( w ) ∇ n i = θ ℓ ( v ) G ( θ ℓ ( u ))+ ( G ( θ ℓ ( u )) − D ) θ ℓ ( v ) in Ω T ,(2.10) ∇ n · n = 0 on Σ T , n ( x,
0) = n (0) ( x ) on Ω.(2.11)For given ( w, v , v , u ) the above problem for n is linear and uniformly parabolic. Thus we canconclude from the classical result ([13], Chap. III) that there is a unique weak solution n to (2.10)-(2.11) in the space W (0 , T ). Use the function n so obtained to form the following two initialboundary problems ∂ t n − ε ∆ n = γ div h θ ℓ ( v ) θ γ − ℓ ( w ) ∇ n i + ( G ( θ ℓ ( u )) − K ( θ ℓ ( u ))) θ ℓ ( v )+ θ ℓ ( v ) K ( θ ℓ ( u )) in Ω T ,(2.12) ∇ n · n = 0 on Σ T , n ( x,
0) = n (0)1 ( x ) on Ω, ∂ t n − ε ∆ n = γ div h θ ℓ ( v ) θ γ − ℓ ( w ) ∇ n i + ( G ( θ ℓ ( u )) − K ( θ ℓ ( u )) − D ) θ ℓ ( v )+ θ ℓ ( v ) K ( θ ℓ ( u )) in Ω T ,(2.13) ∇ n · n = 0 on Σ T ,(2.14) n ( x,
0) = n (0)2 ( x ) on Ω.(2.15) JIAN-GUO LIU AND XIANGSHENG XU
Each of the two problems here has a unique solution in W (0 , T ). Then we solve the followinglinear problem b∂ t d − ∆ d = − ( ψ ( θ ℓ ( u )) − a ) θ ℓ ( w ) − aθ ℓ ( v ) in Ω T , d = d b on Σ T , d ( x,
0) = d (0) ( x ) on Ω.We define ( n, n , n , d ) = M ( w, v , v , u ). Evidently, M is well-defined. Claim 2.5.
For each fixed pair ε > and ℓ > , the operator M is continuous and its range isprecompact.Proof. The key observation here is that each initial boundary value problem in the definition of M is linear and uniformly parabolic. This together with (H1) implies that M is continuous. Onecan easily verify that the range of M is bounded in ( W (0 , T )) , which is compactly embedded in (cid:0) L (Ω T ) (cid:1) . It is similar to the proof of Lemma 2.4 in [18]. We shall omit the details. (cid:3) Now we are in a position to apply Corollary 11.2 in ([11], p.280), thereby obtaining that M hasa fixed point. That is, there is a ( n, n , n , d ) in ( W (0 , T )) such that ∂ t n − ε ∆ n = γ div h ( θ ℓ ( n ) + θ ℓ ( n )) θ γ − ℓ ( n ) ∇ n i + θ ℓ ( n ) G ( θ ℓ ( d ))+ ( G ( θ ℓ ( d )) − D ) θ ℓ ( n ) in Ω T ,(2.16) ∇ n · n = 0 on Σ T , n ( x,
0) = n (0) ( x ) on Ω, ∂ t n − ε ∆ n = γ div h θ ℓ ( n ) θ γ − ℓ ( n ) ∇ n i + ( G ( θ ℓ ( d )) − K ( θ ℓ ( d ))) θ ℓ ( n )+ θ ℓ ( n ) K ( θ ℓ ( d )) in Ω T ,(2.17) ∇ n · n = 0 on Σ T , n ( x,
0) = n (0)1 ( x ) on Ω, ∂ t n − ε ∆ n = γ div h θ ℓ ( n ) θ γ − ℓ ( n ) ∇ n i + ( G ( θ ℓ ( d )) − K ( θ ℓ ( d )) − D ) θ ℓ ( n )+ θ ℓ ( n ) K ( θ ℓ ( d )) in Ω T ,(2.18) ∇ n · n = 0 on Σ T ,(2.19) n ( x,
0) = n (0)2 ( x ) on Ω,(2.20) b∂ t d − ∆ d = − ( ψ ( θ ℓ ( d )) − a ) θ ℓ ( n ) − aθ ℓ ( n ) in Ω T ,(2.21) d = d b on Σ T , d ( x,
0) = d (0) ( x ) on Ω.(2.22)Now we pick(2.23) ℓ ≥ L, where L is given as in (1.19). Note that θ ℓ ( d ) = min { d, ℓ } . On account of (1.10), we have( ψ ( θ ℓ ( d )) − a )( d − L ) + = ( ψ ( θ ℓ ( d )) − ψ ( d ))( d − L ) + ≥ T . TISSUE GROWTH MODEL WITH AUTOPHAGY 9
With this in mind, we use ( d − L ) + as a test function in (2.21) to derive b ddt Z Ω (cid:2) ( d − L ) + (cid:3) dx + Z Ω (cid:12)(cid:12) ∇ ( d − L ) + (cid:12)(cid:12) dx = Z Ω [ − ( ψ ( θ ℓ ( d )) − a ) θ ℓ ( n ) − aθ ℓ ( n )] ( d − L ) + dx ≤ . Integrate to obtain(2.24) d ≤ L in Ω T .Note that θ ℓ ( n ) = 0 in { n ≤ } . With this in mind, we use n − as a test function in (2.12) to derive − ddt Z Ω (cid:0) n − (cid:1) dx − ε Z Ω |∇ n − | dx = Z Ω θ ℓ ( n ) K ( θ ℓ ( d )) n − dx ≥ . Consequently, n ≥ . By the same token, n ≥ . Use d − as a test function in (2.21) to get − b ddt Z Ω (cid:0) d − (cid:1) dx − Z Ω (cid:12)(cid:12) ∇ d − (cid:12)(cid:12) dx = Z Ω [ − ( ψ ( θ ℓ ( d )) − a ) θ ℓ ( n ) − aθ ℓ ( n )] d − dx = a Z Ω [ θ ℓ ( n ) − θ ℓ ( n )] d − dx ≥ . Here we have used the fact that ψ (0) = 0. Integrate to obtain(2.25) d ≥ T .This together with (2.24) implies(2.26) θ ℓ ( d ) = d. Add (2.17) to (2.18) and subtract the resulting equation from (2.16) to derive ∂ t ( n − ( n + n )) − ε ∆( n − ( n + n )) = 0 in Ω T .Recall the initial boundary conditions for ( n − ( n + n )) to deduce(2.27) n = n + n . Let λ ∈ (0 , ∞ ), and define(2.28) w = e − λt n. We easily check that w satisfies ∂ t w + λw − ε ∆ w = γ div h ( θ ℓ ( n ) + θ ℓ ( n )) θ γ − ℓ ( e λt w ) ∇ w i + e − λt θ ℓ ( n ) G ( d )+ e − λt ( G ( d ) − D ) θ ℓ ( n ) in Ω T ,(2.29) ∇ w · n = 0 on Σ T , w ( x,
0) = n (0) ( x ) on Ω. Set(2.30) M = max { max d ∈ [0 ,L ] | G ( d ) | , max d ∈ [0 ,L ] | G ( d ) − D |} . Then the last two terms in (2.29) can be estimated as follows: (cid:12)(cid:12)(cid:12) e − λt θ ℓ ( n ) G ( d ) + e − λt ( G ( d ) − D ) θ ℓ ( n ) (cid:12)(cid:12)(cid:12) ≤ e − λt θ ℓ ( n ) | G ( d ) | + e − λt | G ( d ) − D | θ ℓ ( n ) ≤ M e − λt ( θ ℓ ( n ) + θ ℓ ( n )) ≤ M e − λt θ ℓ ( n ) ≤ M w. Subsequently, ∂ t w + ( λ − M ) w − div h ε + γ ( θ ℓ ( n ) + θ ℓ ( n )) θ γ − ℓ ( e λt w ) ∇ w i ≤ T .Choose λ = 2 M . Then use ( w − k n (0) k ∞ , Ω ) + as a test function in the above differential inequalityto derive w ≤ k n (0) k ∞ , Ω a.e. in Ω T .This immediately implies(2.31) n ≤ e M T k n (0) k ∞ , Ω a.e. in Ω T .Thus if, in addition to (2.23), we further require ℓ ≥ e M T k n (0) k ∞ , Ω , (2.32)then θ ℓ ( n ) = n, θ ℓ ( n ) = n , θ ℓ ( n ) = n and problem (2.16)-(2.22) reduces to problem (2.2)-(2.8). This completes the proof of Lemma2.4. (cid:3) Let ε ∈ (0 , n (0)1 ( x ) by n (0)1 ( x ) + ε in (2.8) and denote the resulting solution to(2.2)-(2.8) by ( n ( ε ) , n ( ε )1 , n ( ε )2 , d ( ε ) ). That is, we have ∂ t n ( ε ) − ε ∆ n ( ε ) = γ div h(cid:16) n ( ε ) (cid:17) γ ∇ n ( ε ) i + G ( d ( ε ) ) n ( ε )1 +( G ( d ( ε ) ) − D ) n ( ε )2 in Ω T ,(2.33) ∂ t n ( ε )1 − ε ∆ n ( ε )1 = γ div (cid:20) n ( ε )1 (cid:16) n ( ε ) (cid:17) γ − ∇ n ( ε ) (cid:21) + G ( d ( ε ) ) n ( ε )1 − K ( d ( ε ) ) n ( ε )1 + K ( d ( ε ) ) n ( ε )2 in Ω T ,(2.34) ∂ t n ( ε )2 − ε ∆ n ( ε )2 = γ div (cid:20) n ( ε )2 (cid:16) n ( ε ) (cid:17) γ − ∇ n ( ε ) (cid:21) + ( G ( d ( ε ) ) − D ) n ( ε )2 + K ( d ( ε ) ) n ( ε )1 − K ( d ( ε ) ) n ( ε )2 in Ω T ,(2.35) b∂ t d ( ε ) − ∆ d ( ε ) = − ψ ( d ( ε ) ) n ( ε ) + an ( ε )2 in Ω T ,(2.36) ∇ n ( ε ) · n = ∇ n ( ε )1 · n = ∇ n ( ε )2 · n = 0 on Σ T ,(2.37) d ( ε ) = d b on Σ T ,(2.38) ( n ( ε ) , n ( ε )1 , n ( ε )2 , d ( ε ) ) (cid:12)(cid:12)(cid:12) t =0 = ( n (0) ( x ) + ε, n (0)1 ( x ) + ε, n (0)2 ( x ) , d (0) ( x )) on Ω.(2.39) TISSUE GROWTH MODEL WITH AUTOPHAGY 11
In addition, we have n ( ε )1 ≥ , n ( ε )2 ≥ , n ( ε ) = n ( ε )1 + n ( ε )2 ≤ c, ≤ d ( ε ) ≤ c. (2.40)Here and in what follows the letter c is independent of ε . As we shall see, the addition of ε in (2.39)is to ensure that n ( ε ) stays away from 0 below. Lemma 2.6.
We have Z Ω T (cid:12)(cid:12)(cid:12)(cid:12) ∇ (cid:16) n ( ε ) (cid:17) γ +12 (cid:12)(cid:12)(cid:12)(cid:12) dxdt + ε Z Ω T (cid:12)(cid:12)(cid:12)(cid:12) ∇ q n ( ε )1 (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) ∇ q n ( ε )2 (cid:12)(cid:12)(cid:12)(cid:12) ! dxdt ≤ c. Proof.
Pick τ >
0. Use ln( n ( ε )1 + τ ) as a test function in (2.34) to derive ddt Z Ω (cid:16) ( n ( ε )1 + τ ) ln( n ( ε )1 + τ ) − n ( ε )1 (cid:17) dx + Z Ω n ( ε )1 n ( ε )1 + τ ∇ (cid:16) n ( ε ) (cid:17) γ ∇ n ( ε )1 dx + ε Z Ω n ( ε )1 + τ |∇ n ( ε )1 | = Z Ω (cid:16) G ( d ( ε ) ) n ( ε )1 − K ( d ( ε ) ) n ( ε )1 + K ( d ( ε ) ) n ( ε )2 (cid:17) ln( n ( ε )1 + τ ) dx ≤ Z Ω (cid:12)(cid:12)(cid:12)(cid:16) G ( d ( ε ) ) − K ( d ( ε ) ) (cid:17) n ( ε )1 ln( n ( ε )1 + τ ) (cid:12)(cid:12)(cid:12) dx + Z { n ( ε )1 + τ ≥ } K ( d ( ε ) ) n ( ε )2 ln( n ( ε )1 + τ ) dx ≤ C Z Ω n ( ε ) ( n ( ε )1 + τ ) dx + 2 C Z { n ( ε )1 + τ ≤ } | n ( ε )1 ln n ( ε )1 | dx ≤ c. Here(2.41) C = max { max d ∈ [0 ,L ] | G ( d ) | , max d ∈ [0 ,L ] K ( d ) , max d ∈ [0 ,L ] K ( d ) } . Integrate and take τ → Z Ω T ∇ (cid:16) n ( ε ) (cid:17) γ · ∇ n ( ε )1 dxdt + 4 ε Z Ω T (cid:12)(cid:12)(cid:12)(cid:12) ∇ q n ( ε )1 (cid:12)(cid:12)(cid:12)(cid:12) dxdt ≤ c. Similarly, Z Ω T ∇ (cid:16) n ( ε ) (cid:17) γ · ∇ n ( ε )2 dxdt + 4 ε Z Ω T (cid:12)(cid:12)(cid:12)(cid:12) ∇ q n ( ε )2 (cid:12)(cid:12)(cid:12)(cid:12) dxdt ≤ c. Add up the two preceding inequalities to obtain the desired result. (cid:3)
Lemma 2.7.
The sequences { n ( ε ) } and { d ( ε ) } are precompact in L p (Ω T ) for each p ≥ .Proof. It follows from (2.30) and (2.33) that(2.42) ∂ t n ( ε ) − ε ∆ n ( ε ) ≥ γ div h(cid:16) n ( ε ) (cid:17) γ ∇ n ( ε ) i − M n ( ε ) in Ω T .Let w ( ε ) = e M t n ( ε ) . Then we have(2.43) ∂ t w ( ε ) − ε ∆ w ( ε ) ≥ γ div h(cid:16) n ( ε ) (cid:17) γ ∇ w ( ε ) i in Ω T .Use ( ε − w ( ε ) ) + as a test function in (2.43) to get(2.44) − ddt Z Ω h ( ε − w ( ε ) ) + i dx − γ Z Ω (cid:16) n ( ε ) (cid:17) γ |∇ ( ε − w ( ε ) ) + | dx − ε Z Ω |∇ ( ε − w ( ε ) ) + | dx ≥ . Recall from (2.39) that w ( ε ) ( x,
0) = n ( ε ) ( x, ≥ ε . Integrate to obtain(2.45) n ( ε ) ≥ εe − M T . Subsequently, (cid:0) n ( ε ) (cid:1) r ∈ L (0 , T ; W , (Ω)) for each r ∈ R . We derive from (2.33) that ∂ t (cid:16) n ( ε ) (cid:17) γ +12 = γ + 12 (cid:16) n ( ε ) (cid:17) γ +12 − ∂ t n ( ε ) = γ + 12 div (cid:20)(cid:16) n ( ε ) (cid:17) γ +12 ∇ (cid:16) n ( ε ) (cid:17) γ (cid:21) − γ + 12 ∇ (cid:16) n ( ε ) (cid:17) γ +12 · ∇ (cid:16) n ( ε ) (cid:17) γ + ( γ + 1) ε (cid:20)(cid:16) n ( ε ) (cid:17) γ +12 − ∇ n ( ε ) (cid:21) − ( γ + 1) ε ∇ (cid:16) n ( ε ) (cid:17) γ +12 − · ∇ n ( ε ) + γ + 12 (cid:16) n ( ε ) (cid:17) γ +12 − ( G ( d ( ε ) ) n ( ε )1 + ( G ( d ( ε ) ) − D ) n ( ε )2 )= γ div (cid:20)(cid:16) n ( ε ) (cid:17) γ ∇ (cid:16) n ( ε ) (cid:17) γ +12 (cid:21) − γ ( γ − γ + 1 (cid:16) n ( ε ) (cid:17) γ +12 − (cid:12)(cid:12)(cid:12)(cid:12) ∇ (cid:16) n ( ε ) (cid:17) γ +12 (cid:12)(cid:12)(cid:12)(cid:12) + ε ∆ (cid:16) n ( ε ) (cid:17) γ +12 − ( γ − ε (cid:16) n ( ε ) (cid:17) γ +12 − (cid:12)(cid:12)(cid:12) ∇ p n ( ε ) (cid:12)(cid:12)(cid:12) + γ + 12 (cid:16) n ( ε ) (cid:17) γ +12 − (cid:16) G ( d ( ε ) ) n ( ε )1 + ( G ( d ( ε ) ) − D ) n ( ε )2 (cid:17) . (2.46)Remember that γ +12 − >
0. We can conclude from Lemma 2.6 that the sequence { ∂ t (cid:0) n ( ε ) (cid:1) γ +12 } is bounded in L (cid:0) , T ; (cid:0) W , (Ω) (cid:1) ∗ (cid:1) + L (Ω T ) ≡ { ψ + ψ : ψ ∈ L (cid:0) , T ; (cid:0) W , (Ω) (cid:1) ∗ (cid:1) , ψ ∈ L (Ω T ) } . Now we are in a position to use (i) in Lemma 2.2, thereby obtaining the precompactnessof { (cid:0) n ( ε ) (cid:1) γ +12 } in L (Ω T ).It is easy to see from (2.36) that { d ( ε ) } is bounded in W (0 , T ). The lemma follows from (2.40). (cid:3) We may extract a subsequence of { ( n ( ε ) , n ( ε )1 , n ( ε )2 , d ( ε ) ) } , still denoted by the same notation, suchthat n ( ε ) → n a.e. in Ω T and strongly in L p (Ω T ) for each p ≥ d ( ε ) → d a.e. in Ω T and strongly in L p (Ω T ) for each p ≥ n ( ε )1 → n weak ∗ in L ∞ (Ω T ), n ( ε )2 → n weak ∗ in L ∞ (Ω T ), and (cid:16) n ( ε ) (cid:17) γ +12 → n γ +12 weakly in L (0 , T ; W , (Ω)) as ε → { n ( ε ) } is bounded, we also have (cid:16) n ( ε ) (cid:17) p → n p weakly in L (0 , T ; W , (Ω)) for each p ≥ γ +12 .This combined with (2.43) implies ∂ t n ( ε ) → ∂ t n weakly in L (0 , T ; (cid:0) W , (Ω) (cid:1) ∗ ).Remember that G, K , K , ψ are all continuous functions. We also have G ( n ( ε ) ) → G ( n ) strongly in L p (Ω T ) for each p ≥ ψ ( n ( ε ) ) → ψ ( n ) strongly in L p (Ω T ) for each p ≥
1, and(2.51) K i ( n ( ε ) ) → K i ( n ) strongly in L p (Ω T ) for each p ≥ i = 1 , TISSUE GROWTH MODEL WITH AUTOPHAGY 13
Our key result is the following.
Lemma 2.8.
We have (cid:16) n ( ε ) (cid:17) γ +1 → n γ +1 strongly in L (0 , T ; W , (Ω)) .Proof. We have(2.53) n ( ε ) ∇ ( n ( ε ) ) γ = γγ + 1 ∇ (cid:16) n ( ε ) (cid:17) γ +1 . Thus we can write (2.43) in the form(2.54) ∂ t n ( ε ) − γγ + 1 ∆ w ( ε ) = R ( ε ) , where w ( ε ) = (cid:16) n ( ε ) (cid:17) γ +1 + ε ( γ + 1) γ n ( ε ) ,R ( ε ) = (cid:16) G ( d ( ε ) ) n ( ε )1 + ( G ( d ( ε ) ) − D ) n ( ε )2 (cid:17) . We may assume that n ( ε ) is a classical solution to (2.54) because it can be viewed as the limit of asequence of classical approximate solutions. Use ∂ t w ( ε ) as a test function in (2.54) to derive(2.55) Z Ω ∂ t n ( ε ) ∂ t w ( ε ) dx + γγ + 1 Z Ω ∇ w ( ε ) · ∇ ∂ t w ( ε ) dx = Z Ω R ( ε ) ∂ t w ( ε ) dx We proceed to evaluate each integral in the above equation as follows: Z Ω ∂ t n ( ε ) ∂ t w ( ε ) dx = ( γ + 1) Z Ω (cid:16) n ( ε ) (cid:17) γ (cid:16) ∂ t n ( ε ) (cid:17) dx + ε ( γ + 1) γ Z Ω (cid:16) ∂ t n ( ε ) (cid:17) dx, Z Ω ∇ w ( ε ) · ∇ ∂ t w ( ε ) dx = 12 ddt Z Ω (cid:12)(cid:12)(cid:12) ∇ w ( ε ) (cid:12)(cid:12)(cid:12) dx, Z Ω R ( ε ) ∂ t w ( ε ) dx = ( γ + 1) Z Ω R ( ε ) (cid:16) n ( ε ) (cid:17) γ ∂ t n ( ε ) dx + ε ( γ + 1) γ Z Ω R ( ε ) ∂ t n ( ε ) dx ≤ γ + 12 Z Ω (cid:16) n ( ε ) (cid:17) γ (cid:16) ∂ t n ( ε ) (cid:17) dx + γ + 12 Z Ω (cid:16) n ( ε ) (cid:17) γ (cid:16) R ( ε ) (cid:17) dx + ε ( γ + 1)2 γ Z Ω (cid:16) ∂ t n ( ε ) (cid:17) dx + ε ( γ + 1)2 γ Z Ω (cid:16) R ( ε ) (cid:17) dx. Plug the preceding three results into (2.55) and integrate to derive Z Ω T (cid:18) ∂ t (cid:16) n ( ε ) (cid:17) γ +22 (cid:19) dxdt + ε Z Ω T (cid:16) ∂ t n ( ε ) (cid:17) dxdt + sup ≤ t ≤ T Z Ω (cid:12)(cid:12)(cid:12) ∇ w ( ε ) (cid:12)(cid:12)(cid:12) dx ≤ c. Note ∂ t (cid:16) n ( ε ) (cid:17) γ +1 = 2 (cid:16) n ( ε ) (cid:17) γ +22 ∂ t (cid:16) n ( ε ) (cid:17) γ +22 , ∇ (cid:16) n ( ε ) (cid:17) γ +1 = ( γ + 1) (cid:16) n ( ε ) (cid:17) γ ∇ n ( ε ) . On account of (2.40), { ∂ t (cid:0) n ( ε ) (cid:1) γ +1 } is bounded in L (Ω T ), while { (cid:0) n ( ε ) (cid:1) γ +1 } is bounded in L ∞ (0 , T ; W , (Ω)). By (ii) in Lemma 2.2, the sequence { (cid:0) n ( ε ) (cid:1) γ +1 } is precompact in C ([0 , T ] , L (Ω)).Consequently, { (cid:0) n ( ε ) (cid:1) γ +1 } is precompact in C ([0 , T ] , L p (Ω)) for each p ≥
1. This asserts(2.56) Z Ω (cid:16) n ( ε ) ( x, t ) (cid:17) q dx → Z Ω n q ( x, t ) dx for each t ∈ [0 , T ] and each q ≥ γ + 1.Take ε → ∂ t n − γγ + 1 ∆ n γ +1 = R ≡ G ( d ) n + ( G ( d ) − D ) n . Subtract this equation from (2.54) and keep (2.53) in mind to get ∂ t ( n ( ε ) − n ) − γγ + 1 ∆ (cid:20)(cid:16) n ( ε ) (cid:17) γ +1 − n γ +1 (cid:21) − ε ∆ n ( ε ) = R ( ε ) − R. (2.57)Use (cid:0) n ( ε ) (cid:1) γ +1 − n γ +1 as a test function in (2.57) to derive γγ + 1 Z Ω T (cid:12)(cid:12)(cid:12)(cid:12) ∇ (cid:20)(cid:16) n ( ε ) (cid:17) γ +1 − n γ +1 (cid:21)(cid:12)(cid:12)(cid:12)(cid:12) dxdt + ε Z Ω T ∇ n ( ε ) · ∇ (cid:20)(cid:16) n ( ε ) (cid:17) γ +1 − n γ +1 (cid:21) dxdt = Z Ω T ( R ( ε ) − R ) (cid:20)(cid:16) n ( ε ) (cid:17) γ +1 − n γ +1 (cid:21) dxdt − Z T (cid:28) ∂ t ( n ( ε ) − n ) , (cid:16) n ( ε ) (cid:17) γ +1 − n γ +1 (cid:29) dt. (2.58)We will show that the last three terms in the above equation all go to 0 as ε →
0. It is easy to seefrom Lemma 2.6 that (cid:12)(cid:12)(cid:12)(cid:12) ε Z Ω T ∇ n ( ε ) · ∇ (cid:20)(cid:16) n ( ε ) (cid:17) γ +1 − n γ +1 (cid:21) dxdt (cid:12)(cid:12)(cid:12)(cid:12) = 4 ε (cid:12)(cid:12)(cid:12)(cid:12)Z Ω T p n ( ε ) ∇ p n ( ε ) · (cid:20)(cid:16) n ( ε ) (cid:17) γ +12 ∇ (cid:16) n ( ε ) (cid:17) γ +12 − n γ +12 ∇ n γ +12 (cid:21) dxdt (cid:12)(cid:12)(cid:12)(cid:12) ≤ c √ ε → ε → Z Ω T ( R ( ε ) − R ) (cid:20)(cid:16) n ( ε ) (cid:17) γ +1 − n γ +1 (cid:21) dxdt → ε → TISSUE GROWTH MODEL WITH AUTOPHAGY 15
Finally, we compute from Lemma 2.1 and (2.56) that Z T (cid:28) ∂ t ( n ( ε ) − n ) , (cid:16) n ( ε ) (cid:17) γ +1 − n γ +1 (cid:29) dt = 1 γ + 2 Z T (cid:20) ddt Z Ω (cid:16) n ( ε ) (cid:17) γ +2 dx + ddt Z Ω n γ +2 dx (cid:21) dt − Z T D ∂ t n ( ε ) , n γ +1 E dt − Z T (cid:28) ∂ t n, (cid:16) n ( ε ) (cid:17) γ +1 (cid:29) dt = 1 γ + 2 (cid:20)Z Ω (cid:16) n ( ε ) ( x, T ) (cid:17) γ +2 dx + Z Ω n γ +2 ( x, T ) dx (cid:21) − γ + 2 Z Ω (cid:16) n (0) ( x ) (cid:17) γ +2 dx − Z T D ∂ t n ( ε ) , n γ +1 E dt − Z T (cid:28) ∂ t n, (cid:16) n ( ε ) (cid:17) γ +1 (cid:29) dt → γ + 2 Z Ω n γ +2 ( x, T ) dx − γ + 2 Z Ω (cid:16) n (0) ( x ) (cid:17) γ +2 dx − Z T (cid:10) ∂ t n, n γ +1 (cid:11) dt = 0 . This completes the proof. (cid:3)
Proof of Theorem 1.2 .
Equipped with the preceding lemmas, we can complete the proof of Theo-rem 1.2. Keeping (2.45) in mind, we can set η ( ε )1 = n ( ε )1 n ( ε ) , η ( ε )2 = n ( ε )2 n ( ε ) . Suppose η ( ε )1 → η , η ( ε )2 → η weak ∗ in L ∞ (Ω T ).We calculate n ( ε )1 ∇ (cid:16) n ( ε ) (cid:17) γ = η ( ε )1 n ( ε ) ∇ (cid:16) n ( ε ) (cid:17) γ = γγ + 1 η ( ε )1 ∇ (cid:16) n ( ε ) (cid:17) γ +1 → γγ + 1 η ∇ n γ +1 = η n ∇ n γ weakly in (cid:0) L (Ω T ) (cid:1) N .We claim that(2.59) η n = n a.e. on Ω T . To see this, for each δ > η ( ε )1 ( n ( ε ) − δ ) + → η ( n − δ ) + weak ∗ in L ∞ (Ω T ).Note that ( n ( ε ) − δ ) + n ( ε ) ≤
1. Subsequently, η ( ε )1 ( n ( ε ) − δ ) + = n ( ε )1 ( n ( ε ) − δ ) + n ( ε ) → n ( n − δ ) + n weak ∗ in L ∞ (Ω T ).We obtain n ( n − δ ) + n = η ( n − δ ) + for each δ > . This implies that n = nη on the set { n > } . If n = 0, then n = 0, and we still have n = nη . This completes the proof of (2.59). Similarly,we can show n ( ε )2 ∇ (cid:16) n ( ε ) (cid:17) γ → n ∇ n γ weakly in (cid:0) L (Ω T ) (cid:1) N .We are ready to pass to the limit in (2.34) and (2.35), thereby finishing the proof of Theorem1.2. (cid:3) The limit as γ → ∞ and proof of theorem 1.3 Once again, the proof will be divided into several lemmas. Now the solution to our problem(1.1)-(1.6) is denoted by ( n ( γ ) , n ( γ )1 , n ( γ )2 , d ( γ ) ). That is, we have ∂ t n ( γ ) − γγ + 1 ∆ (cid:16) n ( γ ) (cid:17) γ +1 = G ( d ( γ ) ) n ( γ ) − Dn ( γ )2 ≡ R ( γ ) in Ω T ,(3.1) ∂ t n ( γ )1 − div (cid:16) n ( γ )1 ∇ (cid:16) n ( γ ) (cid:17) γ (cid:17) = G ( d ( γ ) ) n ( γ )1 − K ( d ( γ ) ) n ( γ )1 + K ( d ( γ ) ) n ( γ )2 ≡ R ( γ )1 in Ω T ,(3.2) ∂ t n ( γ )2 − div (cid:16) n ( γ )2 ∇ (cid:16) n ( γ ) (cid:17) γ (cid:17) = ( G ( d ( γ ) ) − D ) n ( γ )2 + K ( d ( γ ) ) n ( γ )1 − K ( d ( γ ) ) n ( γ )2 ≡ R ( γ )2 in Ω T ,(3.3) b∂ t d ( γ ) − ∆ d ( γ ) = − ψ ( d ( γ ) ) n ( γ ) + an ( γ )2 in Ω T ,(3.4) n ( γ )1 ∇ (cid:16) n ( γ ) (cid:17) γ · n = n ( γ )2 ∇ (cid:16) n ( γ ) (cid:17) γ · n = 0 on Σ T ≡ ∂ Ω × (0 , T ),(3.5) d ( γ ) = d b on Σ T ,(3.6) (cid:16) n ( γ ) , n ( γ )1 , n ( γ )2 , d ( γ ) (cid:17)(cid:12)(cid:12)(cid:12) t =0 = (cid:18) n (0) ( x ) + 1 γ , n (0)1 ( x ) + 1 γ , n (0)2 ( x ) , d (0) ( x ) (cid:19) on Ω.(3.7)As before, the term γ is added in (3.7) to ensure that n ( γ ) stays away from 0 below. Therefore,it possesses enough regularity properties. We wish to find and identify the limit of solutions as γ → ∞ . By our analysis in the preceding section, we have n ( γ )1 ≥ , n ( γ )2 ≥ , n ( γ ) = n ( γ )1 + n ( γ )2 ≤ c, (3.8) 0 ≤ d ( γ ) ≤ L, (3.9)where L is given as in (1.19). In (3.8) and what follows, the generic positive number c is independentof γ . We may assume that(3.10) n ( γ )1 → n ( ∞ )1 , n ( γ )2 → n ( ∞ )2 , n ( γ ) → n ( ∞ ) , d ( γ ) → d ( ∞ ) weak ∗ in L ∞ (Ω T ). Lemma 3.1.
Assume that (3.11) ∂ t d b ∈ L (0 , T ; W , (Ω)) , d (0) ∈ W , (Ω) . Then we have (3.12) Z Ω T (cid:16) ∂ t d ( γ ) (cid:17) dxdt ≤ c. Furthermore, if (H6) and (H8) hold, then we have (3.13) k∇ d ( γ ) k ∞ , Ω T ≤ c. TISSUE GROWTH MODEL WITH AUTOPHAGY 17
Proof.
Use ∂ t ( d ( γ ) − d b ) as a test function in (3.4) to get b Z Ω (cid:16) ∂ t d ( γ ) (cid:17) dx + 12 ddt Z Ω |∇ d ( γ ) | dx = b Z Ω ∂ t d ( γ ) ∂ t d b dx + Z Ω ∇ d ( γ ) · ∇ ∂ t d b dx + Z Ω (cid:16) − ψ ( d ( γ ) ) n ( γ ) + an ( γ )2 (cid:17) ∂ t ( d ( γ ) − d b ) dx. Integrate to derive(3.14) Z Ω T (cid:16) ∂ t d ( γ ) (cid:17) dxdt + sup ≤ t ≤ T Z Ω |∇ d ( γ ) | dx ≤ c. With the aid of our assumptions (H6) and (H8), we can easily modify the proof of Proposition 2.3in [23] to derive (3.13). The proof is complete. (cid:3)
Clearly, this lemma implies (1.25). Subsequently,(3.15) R ( γ ) → R ( ∞ ) = G ( d ( ∞ ) ) n ( ∞ ) − Dn ( ∞ )2 weak ∗ in L ∞ (Ω T ).The core of our development is the following lemma. Lemma 3.2.
We have (3.16) Z Ω T t (cid:16) v ( γ ) (cid:17) dxdt + Z Ω T t (cid:12)(cid:12)(cid:12) ∇ v ( γ ) (cid:12)(cid:12)(cid:12) dxdt ≤ c. Proof.
Let G be given as in Theorem 1.3. Then(3.17) R ( γ ) ≤ G n ( γ ) . Use this in (3.1) and multiply through the resulting inequality by e − G t to get(3.18) ∂ t w ( γ ) − γe γG t γ + 1 ∆ (cid:16) w ( γ ) (cid:17) γ +1 ≤ T ,where w ( γ ) = e − G t n ( γ ) . For each ε > η ε ( s ) = s > ε , ε s if 0 ≤ s ≤ ε ,0 if s < η ε ( s ) → sgn +0 ( s ) = (cid:26) s > s ≤
0. as ε → σ ∈ (cid:0) , e − G T (cid:1) be given as in (H7). Clearly, η ε (cid:0) w ( γ ) − σ (cid:1) ≥
0. Multiply through (3.18) by thisfunction to get Z Ω Z w ( γ ) ( x,t )0 η ε ( s − σ ) dsdx ≤ Z Ω Z w ( γ ) ( x, η ε ( s − σ ) dsdx. (3.19)Take ε → Z Ω (cid:16) w ( γ ) ( x, t ) − σ (cid:17) + dx ≤ Z Ω (cid:16) w ( γ ) ( x, − σ (cid:17) + dx ≤ (cid:18) k n (0) k ∞ , Ω + 1 γ − σ (cid:19) (cid:12)(cid:12)(cid:12)(cid:12)(cid:26) n (0) ( x ) + 1 γ ≥ σ (cid:27)(cid:12)(cid:12)(cid:12)(cid:12) . Or equivalently, Z Ω (cid:16) n ( γ ) ( x, t ) − σe G t (cid:17) + dx ≤ e G t (cid:18) k n (0) k ∞ , Ω + 1 γ − σ (cid:19) (cid:12)(cid:12)(cid:12)(cid:12)(cid:26) n (0) ( x ) + 1 γ ≥ σ (cid:27)(cid:12)(cid:12)(cid:12)(cid:12) . (3.20)On the other hand, Z Ω (cid:16) n ( γ ) ( x, t ) − σe G t (cid:17) + dx ≥ Z { n ( γ ) ( x,t ) ≥ } (cid:16) n ( γ ) ( x, t ) − σe G t (cid:17) + dx ≥ (1 − σe G t ) (cid:12)(cid:12)(cid:12)n n ( γ ) ( x, t ) ≥ o(cid:12)(cid:12)(cid:12) . This combined with (3.20) implies (cid:12)(cid:12)(cid:12)n n ( γ ) ( x, t ) ≥ o(cid:12)(cid:12)(cid:12) ≤ e G t (cid:16) k n (0) k ∞ , Ω + γ − σ (cid:17) − σe G t (cid:12)(cid:12)(cid:12)(cid:12)(cid:26) n (0) ( x ) + 1 γ ≥ σ (cid:27)(cid:12)(cid:12)(cid:12)(cid:12) → e G t (cid:0) k n (0) k ∞ , Ω − σ (cid:1) − σe G t (cid:12)(cid:12)(cid:12)n n (0) ( x ) ≥ σ o(cid:12)(cid:12)(cid:12) (as γ → ∞ ) ≤ e G t (cid:0) k n (0) k ∞ , Ω − σ (cid:1) − σe G t e G T k n (0) k ∞ , Ω | Ω | . (3.21)The last step is due to our assumption (H7). We easily check e G t (cid:0) k n (0) k ∞ , Ω − σ (cid:1) − σe G t < e G t k n (0) k ∞ , Ω . Hence we can pick a number σ ∈ (cid:18) e G t ( k n (0) k ∞ , Ω − σ ) − σe G t e G T k n (0) k ∞ , Ω , (cid:19) . Consequently,(3.22) sup ≤ t ≤ T (cid:12)(cid:12)(cid:12)n n ( γ ) ( x, t ) ≥ o(cid:12)(cid:12)(cid:12) ≤ σ | Ω | at least for γ sufficiently large . Using (cid:16) w ( γ ) − k n (0) k ∞ , Ω − γ (cid:17) + as a test function in (3.18), we derive the weak maximum principle(3.23) w ( γ ) ≤ k n (0) k ∞ , Ω + 1 γ in Ω T . This together with (3.17) implies(3.24) R ( γ ) ≤ G e G t (cid:18) k n (0) k ∞ , Ω + 1 γ (cid:19) . Let v ( γ ) be given as in (1.12). Use tv ( γ ) as a test function in (3.1) to deduce1 γ + 2 ddt Z Ω t (cid:16) n ( γ ) (cid:17) γ +2 dx + γtγ + 1 Z Ω (cid:12)(cid:12)(cid:12) ∇ v ( γ ) (cid:12)(cid:12)(cid:12) dx = 1 γ + 2 Z Ω (cid:16) n ( γ ) (cid:17) γ +2 dx + t Z Ω R ( γ ) v ( γ ) dx ≤ e G T (cid:16) k n (0) k ∞ , Ω + γ (cid:17) γ + 2 Z Ω v ( γ ) dx + G e G T (cid:18) k n (0) k ∞ , Ω + 1 γ (cid:19) t Z Ω v ( γ ) dx. (3.25)Since (cid:12)(cid:12)(cid:12)n n ( γ ) ( x, t ) ≥ o(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)n n ( γ ) ( x, t ) < o(cid:12)(cid:12)(cid:12) = | Ω | , the inequality (3.22) implies (cid:12)(cid:12)(cid:12)n n ( γ ) ( x, t ) < o(cid:12)(cid:12)(cid:12) > (1 − σ ) | Ω | . TISSUE GROWTH MODEL WITH AUTOPHAGY 19
Evidently, (cid:16) v ( γ ) − (cid:17) + = 0 on (cid:8) n ( γ ) ( x, t ) < (cid:9) .This puts us in a position to apply Lemma 2.3. Upon doing so, we arrive at(3.26) Z Ω (cid:16) v ( γ ) − (cid:17) + dx ≤ c Z Ω (cid:12)(cid:12)(cid:12)(cid:12) ∇ (cid:16) v ( γ ) − (cid:17) + (cid:12)(cid:12)(cid:12)(cid:12) dx = c Z { n ( γ ) ( x,t ) ≥ } (cid:12)(cid:12)(cid:12) ∇ v ( γ ) (cid:12)(cid:12)(cid:12) dx. To estimate the first term on the right-hand side of (3.25), we use ( n ( γ ) − + as a test function in(3.1) to get(3.27) sup ≤ t ≤ T Z Ω h ( n ( γ ) − + i dx + γ Z Ω T (cid:16) n ( γ ) (cid:17) γ (cid:12)(cid:12)(cid:12) ∇ ( n ( γ ) − + (cid:12)(cid:12)(cid:12) dxdt ≤ c. For each ε > Z Ω v ( γ ) dx = Z { n ( γ ) ( x,t ) ≥ } v ( γ ) dx + Z { n ( γ ) ( x,t ) < } v ( γ ) dx ≤ Z Ω (cid:16) v ( γ ) − (cid:17) + dx + c ≤ c Z { n ( γ ) ( x,t ) ≥ } (cid:12)(cid:12)(cid:12) ∇ v ( γ ) (cid:12)(cid:12)(cid:12) dx + c = c ( γ + 1) Z Ω (cid:16) n ( γ ) (cid:17) γ (cid:12)(cid:12)(cid:12) ∇ ( n ( γ ) − + (cid:12)(cid:12)(cid:12) dx + c ≤ ε Z Ω (cid:16) n ( γ ) (cid:17) γ dx + c ( ε )( γ + 1) Z Ω (cid:16) n ( γ ) (cid:17) γ (cid:12)(cid:12)(cid:12) ∇ ( n ( γ ) − + (cid:12)(cid:12)(cid:12) dx + c ≤ ε k n ( γ ) k ∞ , Ω T Z Ω v ( γ ) dx + c ( ε )( γ + 1) Z Ω (cid:16) n ( γ ) (cid:17) γ (cid:12)(cid:12)(cid:12) ∇ ( n ( γ ) − + (cid:12)(cid:12)(cid:12) dx + c. By choose ε suitably small, we immediately get(3.28) Z Ω v ( γ ) dx ≤ c ( γ + 1) Z Ω (cid:16) n ( γ ) (cid:17) γ (cid:12)(cid:12)(cid:12) ∇ ( n ( γ ) − + (cid:12)(cid:12)(cid:12) dx + c. Use this in (3.25), then integrate, and apply (3.27) to obtain1 γ + 2 sup ≤ t ≤ T Z Ω t (cid:16) n ( γ ) (cid:17) γ +2 dx + γγ + 1 Z Ω T t (cid:12)(cid:12)(cid:12) ∇ v ( γ ) (cid:12)(cid:12)(cid:12) dxdt ≤ c ( γ + 1) Z Ω T (cid:16) n ( γ ) (cid:17) γ (cid:12)(cid:12)(cid:12) ∇ ( n ( γ ) − + (cid:12)(cid:12)(cid:12) dxdt + c Z Ω T tv ( γ ) dxdt + c ≤ c Z Ω T t |∇ v ( γ ) | dxdt + c ≤ γ γ + 1) Z Ω T t (cid:12)(cid:12)(cid:12) ∇ v ( γ ) (cid:12)(cid:12)(cid:12) dxdt + c. Subsequently, Z Ω T t (cid:12)(cid:12)(cid:12) ∇ v ( γ ) (cid:12)(cid:12)(cid:12) dxdt ≤ c. By a calculation similar to (3.26), Z Ω T t (cid:16) v ( γ ) (cid:17) dxdt ≤ Z Ω T t (cid:20)(cid:16) v ( γ ) − (cid:17) + (cid:21) dxdt + c ≤ c Z Ω T t (cid:12)(cid:12)(cid:12)(cid:12) ∇ (cid:16) v ( γ ) − (cid:17) + (cid:12)(cid:12)(cid:12)(cid:12) dxdt + c ≤ c. This completes the proof of Lemma 3.2. (cid:3)
We see that the sequence { v ( γ ) } is bounded in L ( τ, T ; W , (Ω)) for each τ ∈ (0 , T ). Thus wemay assume that (1.22) holds. Proof of (1.28) and (1.29) . We shall employ an argument from [10]. For each δ > ( γ ) δ = n ( x, t ) ∈ Ω T : n ( γ ) ( x, t ) ≥ δ o . We argue by contradiction. Suppose that (1.28) is not true. Then there is a δ > (cid:12)(cid:12)(cid:12) Ω ( ∞ )2 δ (cid:12)(cid:12)(cid:12) > . We claim(3.31) lim inf γ →∞ (cid:12)(cid:12)(cid:12) Ω ( γ ) δ (cid:12)(cid:12)(cid:12) ≡ c > . To see this, we estimate from (3.9) that Z Ω T n ( γ ) n ( ∞ ) χ Ω ( ∞ )2 δ dxdt = Z Ω ( ∞ )2 δ ∩ Ω ( γ ) δ n ( γ ) n ( ∞ ) dxdt + Z Ω ( ∞ )2 δ \ Ω ( γ ) δ n ( γ ) n ( ∞ ) dxdt ≤ e G T (cid:18) k n (0) k ∞ , Ω + 1 γ (cid:19) (cid:12)(cid:12)(cid:12) Ω ( γ ) δ (cid:12)(cid:12)(cid:12) + (1 + δ ) Z Ω ( ∞ )2 δ n ( ∞ ) dxdt. If c in (3.31) is 0, we take γ → ∞ in the above inequality to derive(3.32) Z Ω ( ∞ )2 δ n ( ∞ ) n ( ∞ ) dxdt ≤ (1 + δ ) Z Ω ( ∞ )2 δ n ( ∞ ) dxdt. This is possible only if (cid:12)(cid:12)(cid:12) Ω ( ∞ )2 δ (cid:12)(cid:12)(cid:12) = 0. But this contradicts (3.30). Thus (3.31) holds. On the otherhand, for each τ ∈ (0 , T ) we have(3.33) c ≥ Z Ω ( γ ) δ tv ( γ ) dxdt ≥ Z Ω ( γ ) δ ∩ (Ω × ( τ,T )) tv ( γ ) dxdt ≥ τ (1 + δ ) γ +1 (cid:12)(cid:12)(cid:12) Ω ( γ ) δ ∩ (Ω × ( τ, T )) (cid:12)(cid:12)(cid:12) . That is, lim sup γ →∞ (cid:12)(cid:12)(cid:12) Ω ( γ ) δ ∩ (Ω × ( τ, T )) (cid:12)(cid:12)(cid:12) ≤ τ ∈ (0 , T ).Obviously, this contradicts (3.31). This completes the proof of (1.28).Fix τ ∈ (0 , T ). First, we claim(3.34) lim γ →∞ Z Tτ Z Ω (cid:12)(cid:12)(cid:12) − n ( γ ) (cid:12)(cid:12)(cid:12) v ( γ ) dxdt = 0 . To see this, let ε ∈ (0 ,
1) be given. We estimate from (3.16) that Z Tτ Z Ω (cid:12)(cid:12)(cid:12) − n ( γ ) (cid:12)(cid:12)(cid:12) v ( γ ) dxdt = Z { | − n ( γ ) | ≤ ε }∩ (Ω × ( τ,T )) (cid:12)(cid:12)(cid:12) − n ( γ ) (cid:12)(cid:12)(cid:12) v ( γ ) dxdt + Z { n ( γ ) > ε }∩ (Ω × ( τ,T )) (cid:12)(cid:12)(cid:12) − n ( γ ) (cid:12)(cid:12)(cid:12) v ( γ ) dxdt + Z { n ( γ ) < − ε }∩ (Ω × ( τ,T )) (cid:12)(cid:12)(cid:12) − n ( γ ) (cid:12)(cid:12)(cid:12) v ( γ ) dxdt ≤ cε + c (cid:12)(cid:12)(cid:12) { n ( γ ) > ε } ∩ (Ω × ( τ, T )) (cid:12)(cid:12)(cid:12) + c (1 − ε ) γ +1 . Subsequently,(3.35) lim sup γ →∞ Z Tτ Z Ω (cid:12)(cid:12)(cid:12) − n ( γ ) (cid:12)(cid:12)(cid:12) v ( γ ) dxdt ≤ cε. TISSUE GROWTH MODEL WITH AUTOPHAGY 21
Since ε can be arbitrarily small, we yield (3.34).Observe from (3.1) that the sequence { ∂ t n ( γ ) } is bounded in L (cid:0) τ, T ; (cid:0) W , (Ω) (cid:1) ∗ (cid:1) . We can inferfrom Lions-Aubin’s lemma that { n ( γ ) } is precompact in C (cid:0) [ τ, T ]; (cid:0) W , (Ω) (cid:1) ∗ (cid:1) . We may assumethat(3.36) n ( γ ) → n ( ∞ ) strongly in C (cid:0) [ τ, T ]; (cid:0) W , (Ω) (cid:1) ∗ (cid:1) .With this in mind, we can deduce from (1.22) that Z Tτ Z Ω (cid:16) − n ( γ ) (cid:17) v ( γ ) dxdt = Z Tτ h − n ( γ ) , v ( γ ) i dt = Z Tτ h − n ( ∞ ) , v ( ∞ ) i dt = Z Tτ Z Ω (cid:16) − n ( ∞ ) (cid:17) v ( ∞ ) dxdt. This together with (3.34) and (1.28) implies(3.37) (cid:16) − n ( ∞ ) (cid:17) v ( ∞ ) = 0 , from which (1.29) follows.The proof of (1.23) is similar to Lemma 2.8. We use t ∂ t v ( γ ) as a test function in (3.1) to get( γ + 1) t Z Ω (cid:16) n ( γ ) (cid:17) γ (cid:16) ∂ t n ( γ ) (cid:17) dx + γ γ + 1) ddt Z Ω t |∇ v ( γ ) | dx = γγ + 1 Z Ω t |∇ v ( γ ) | dx + t Z Ω R ( γ ) ∂ t v ( γ ) dx. (3.38)To estimate the last integral in the above equation, we compute from (3.3) that − Dt Z Ω n ( γ )2 ∂ t v ( γ ) dx = − D ddt Z Ω t n ( γ )2 v ( γ ) dx + 2 Dt Z Ω n ( γ )2 v ( γ ) dx + Dt Z Ω ∂ t n ( γ )2 v ( γ ) dx = − D ddt Z Ω t n ( γ )2 v ( γ ) dx + 2 Dt Z Ω n ( γ )2 v ( γ ) dx − γDt γ + 1 Z Ω n ( γ )2 n ( γ ) (cid:12)(cid:12)(cid:12) ∇ v ( γ ) (cid:12)(cid:12)(cid:12) dx + Dt Z Ω R ( γ )2 v ( γ ) dx. Integrate and then apply (3.16) to deduce(3.39) − D Z τ t Z Ω n ( γ )2 ∂ t v ( γ ) dxdt ≤ c. Similarly, t Z Ω G ( d ( γ ) ) n ( γ ) ∂ t v ( γ ) dx = γ + 1 γ + 2 ddt Z Ω t G ( d ( γ ) ) (cid:16) n ( γ ) (cid:17) γ +2 dx − γ + 1) tγ + 2 Z Ω G ( d ( γ ) ) (cid:16) n ( γ ) (cid:17) γ +2 dx − γ + 1 γ + 2 Z Ω t G ′ ( d ( γ ) ) ∂ t d ( γ ) (cid:16) n ( γ ) (cid:17) γ +2 dx. Integrate and then use (H5), (3.16) and (3.12) to derive(3.40) Z τ Z Ω t G ( d ( γ ) ) n ( γ ) ∂ t v ( γ ) dxdt ≤ γ + 1 γ + 2 Z Ω τ G ( d ( γ ) ) n ( γ ) v ( γ ) dx + c. Integrate (3.38) and then take into consideration of (3.39) and (3.40) to obtain( γ + 1) Z τ Z Ω t (cid:16) n ( γ ) (cid:17) γ (cid:16) ∂ t n ( γ ) (cid:17) dxdt + γ γ + 1) Z Ω τ |∇ v ( γ ) | dx ≤ γ + 1 γ + 2 Z Ω τ G ( d ( γ ) ) n ( γ ) v ( γ ) dx + c. (3.41)We easily infer from (3.27) that Z Ω v ( γ ) dx ≤ c Z Ω |∇ v ( γ ) | dx + c ≤ ε Z Ω |∇ v ( γ ) | dx + c ( ε ) , ε > . (3.42)Use this in (3.41) and choose ε suitably small in the resulting inequality to derive(3.43) ( γ + 1) Z Ω T t (cid:16) n ( γ ) (cid:17) γ (cid:16) ∂ t n ( γ ) (cid:17) dxdt + sup ≤ t ≤ T Z Ω t |∇ v ( γ ) | dx ≤ c. This combined with (3.42) yields(3.44) sup ≤ t ≤ T Z Ω t (cid:16) v ( γ ) (cid:17) dx ≤ c. Use t (cid:0) v ( γ ) − v ( ∞ ) (cid:1) as a test function in (3.1) to deduce Z Ω t ∂ t n ( γ ) (cid:16) v ( γ ) − v ( ∞ ) (cid:17) dx + t γγ + 1 Z Ω ∇ v ( γ ) · ∇ (cid:16) v ( γ ) − v ( ∞ ) (cid:17) dx = t Z Ω R ( γ ) (cid:16) v ( γ ) − v ( ∞ ) (cid:17) dx. (3.45)Note that(3.46) Z Ω t ∂ t n ( γ ) v ( γ ) dx = 1 γ + 2 ddt Z Ω t (cid:16) n ( γ ) (cid:17) γ +2 dx − tγ + 2 Z Ω (cid:16) n ( γ ) (cid:17) γ +2 dx. Integrate to get Z Ω T t ∂ t n ( γ ) v ( γ ) dxdt = 1 γ + 2 Z Ω T (cid:16) n ( γ ) (cid:17) γ +2 dx − γ + 2 Z Ω T t (cid:16) n ( γ ) (cid:17) γ +2 dxdt → γ → ∞ . The last step is due to (3.44). Keeping this and (3.45) in mind, we calculatelim sup γ →∞ Z Ω T t (cid:12)(cid:12)(cid:12) ∇ (cid:16) v ( γ ) − v ( ∞ ) (cid:17)(cid:12)(cid:12)(cid:12) dxdt ≤ lim sup γ →∞ Z Ω T t ∇ v ( γ ) · ∇ (cid:16) v ( γ ) − v ( ∞ ) (cid:17) dxdt ≤ Z T h t∂ t n ( ∞ ) , tv ( ∞ ) i dt + lim sup γ →∞ Z Ω T t R ( γ ) (cid:16) v ( γ ) − v ( ∞ ) (cid:17) dxdt. (3.47)Observe that R ( γ ) = (cid:16) G ( d ( γ ) ) − G ( d ( ∞ ) ) (cid:17) n ( γ ) + G ( d ( ∞ ) ) n ( γ ) − Dn ( γ )2 . TISSUE GROWTH MODEL WITH AUTOPHAGY 23
Remember that { tn ( γ ) } , { tn ( γ )2 } are precompact in C ([0 , T ]; (cid:0) W , (Ω) (cid:1) ∗ ). Furthermore, we have G ( d ( ∞ ) ) ∈ L ∞ (0 , T ; W , ∞ (Ω)) due to (H5) and (3.13). Hencelim γ →∞ Z Ω T t R ( γ ) (cid:16) v ( γ ) − v ( ∞ ) (cid:17) dxdt = lim γ →∞ Z T D tn ( γ ) , tG ( d ( ∞ ) ) (cid:16) v ( γ ) − v ( ∞ ) (cid:17)E dt − D lim γ →∞ Z T D tn ( γ )2 , t (cid:16) v ( γ ) − v ( ∞ ) (cid:17)E dt = 0 . (3.48)Use this in (3.47) to obtain(3.49) lim sup γ →∞ Z Ω T t (cid:12)(cid:12)(cid:12) ∇ (cid:16) v ( γ ) − v ( ∞ ) (cid:17)(cid:12)(cid:12)(cid:12) dxdt ≤ Z T h t∂ t n ( ∞ ) , tv ( ∞ ) i dt. Let Ψ ( ∞ ) ( s ) = (cid:26) s ≤ ∞ if s > ( ∞ ) ( s ) is convex and lower semicontinuous ([12], p.49). We compute the subgradient ∂ Ψ ( ∞ ) of Ψ ( ∞ ) ( s ) to get ∂ Ψ ( ∞ ) ( s ) = ϕ ∞ ( s ) , where ϕ ∞ ( s ) is given as in (1.11). Even though n ∞ / ∈ L ( τ, T ; W , (Ω)), we can easily de-rive from (3.43) and (3.44) that the conclusions of Lemma 2.1 still hold here. That is, t R Ω Ψ ( ∞ ) ( n ∞ ( x, t )) dx is an absolutely continuous function on (0 , T ) and ddt Z Ω Ψ ( ∞ ) ( n ∞ ( x, t )) dx = h ∂ t n ( ∞ ) , v ( ∞ ) i . Therefore, Z T h t∂ t n ( ∞ ) , tv ( ∞ ) i dt = Z T t h ∂ t n ( ∞ ) , v ( ∞ ) i dt = Z T t ddt Z Ω Ψ ( ∞ ) ( n ∞ ( x, t )) dxdt = Z T ddt Z Ω t Ψ ( ∞ ) ( n ∞ ( x, t )) dxdt − Z T Z Ω t Ψ ( ∞ ) ( n ∞ ( x, t )) dxdt = 0 . (3.50)The last step is due to the fact that Ψ ( ∞ ) ( n ∞ ( x, t )) ≡
0. Combing (3.50) with (3.49) yields (1.23).To complete the proof of Theorem 1.3, we still need to verify (1.26). To this end, we multiplythrough (3.1) by v ( γ ) to get1 γ + 2 ∂ t (cid:16) n ( γ ) (cid:17) γ +2 − γγ + 1 (cid:16) div( v ( γ ) ∇ v ( γ ) ) − |∇ v ( γ ) | (cid:17) = R ( γ ) v ( γ ) . Even though it is not clear if { tv ( γ ) } is precompact in L (Ω T ) because we do not have any estimateson ∂ t v ( γ ) , (1.23) and (3.48) are enough to justify passing to the limit in the above equation, therebyobtaining (1.26). This finishes the proof of Theorem 1.3. (cid:3) References [1] D. G. Aronson and P. B´enilan, R´egularit´e des solutions de l’´equation des milieux poreux dans R N , C. R. Acad.Sci. Paris S´er. A-B , (1979), A103-A105.[2] P. B´enilan, M.C. Crandall, and P. Sacks, Some L existence and dependence results for semilinear ellipticequations under nonlinear boundary conditions, Appl. Math. Optim. , (1988), 203-224.[3] P. B´enilan, M.C. Crandall, The continuous dependence on ϕ of solutions of u t − ∆ ϕ ( u ) = 0, Indiana Univ. Math.J. , (1981), 161-177.[4] F. Bubba, B. Perthame, C. Pouchol, and M. Schmidtchen, Hele-Shaw limit for a system of two reaction-(cross-)diffusion equations for living tissues, Arch. Rational. Mech. Anal. , (2020), 735-766.[5] L. Caffarelli and A. Friedman, Asymptotic behavior of solutions of u t − ∆ u m = as m → ∞ , Indiana Univ. Math.J. , (1987), 711-728.[6] N. David and B. Perthame, Free boundary limit of tumor growth model with nutrient, J. Math. Pures Appl. ,(2021), https://doi.org/10.1016/j.matpur.2021.01.007. arXiv:2003.10731v1 [math.AP], 2020.[7] P. Degond, S. Hecht, and N. Vauchelet, Incompressible limit of a continuum model of tissue growth for two cellpopulations,
Networks & Heterogeneous Media , (2020), 57-85.[8] X. Dou, Jian-Guo Liu, and Z. Zhou, Modeling the autophagic effect in tumor growth: a cross diffusion modeland its free boundary limit, Preprint, 2020.[9] A. Friedman, A hierarchy of cancer models and their mathematical challenges, Discrete Contin. Dyn. Syst. Ser.B , (2004),147-159. Mathematical models in cancer (Nashville, TN, 2002).[10] A. Friedman and Shao Yun Huang, Asymptotic behavior of solutions of u t = ∆ φ m ( u ) as m → ∞ with inconsis-tent initial values, Analyse math´ematique et applications , Gauthier-Villars, Montrouge, 1988, pp. 165–180. MR956958.[11] D. Gilbarg and N. S. Trudinger,
Elliptic Partial Differential Equations of Second Order , Springer-Verlag, Berlin,1983.[12] A. Haraux,
Nonlinear Evolution Equations-Global Behavior of Solutions , Lecture Notes in Mathematics, 841,Springer-Verlag, Berlin Heidelberg New York, 1981.[13] Q.A. Ladyzenskaja, V.A. Solonnikov, and N.N. Ural’ceva,
Linear and Quasi-linear Equations of Parabolic Type ,Tran. Math. Monographs, Vol. 23, AMS, Providence, RI, 1968.[14] B. Perthame, F. Quir´os, and J. L. V´azquez, The hele–shaw asymptotics for mechanical models of tumor growth,
Archive for Rational Mechanics and Analysis , (2014), 93–127.[15] B. Perthame, F. Quirs, M. Tang, and N. Vauchelet, Derivation of a Hele-Shaw type system from a cell modelwith active motion, Interfaces Free Bound. , (2014), 489-508.[16] B. Perthame, M. Tang, and N. Vauchelet, Traveling wave solution of the Hele-Shaw model of tumor growth withnutrient, Math. Models Methods Appl. Sci. , (2014), 2601-2626.[17] B. Perthame and N. Vauchelet, Incompressible limit of a mechanical model of tumour growth with viscosity, Philos. Trans. Roy. Soc. A , (2050):20140283, 16, 2015.[18] B.C. Price and X. Xu, Global existence theorem for model governing the motion of two cell populations, Kinet.Relat. Models , (2020), 1175-1191. arXiv:2004.05939 [math.AP].[19] J. Simon, Compact sets in the space L p (0 , T ; B ), Ann. Mat. Pura Appl. , (1987), 65-96.[20] R. Temam, Navier-Stokes Equations: Theory and Numerical Analysis , AMS Chelsea Publishing, Providence, RI,2001.[21] X. Xu, Asymptotic behavior of solutions of hyperbolic conservation laws u t + ( u m ) x = 0 as m → ∞ withinconsistent initial values, Proc. Roy. Soc. Edinburgh Sect. A , (1989), 61-71.[22] X. Xu, The Continuous Dependence of Solutions to the Cauchy problem ddt A ( u ) + B ( u ) ∋ f on A and B andApplications to Partial Differential Equations , Ph.D. Thesis, the University of Texas at Austin, 1988.[23] X. Xu, Nonlinear diffusion in the Keller-Segel model of parabolic-parabolic type, J. Differential Equations , (2021), 264-286. arXiv:1911.05863 [math.AP] Email address : [email protected] Email address ::