EXISTENCE OF SOLUTIONS FOR QUASILINEAR ELLIPTIC DEGENERATE SYSTEMS WITH L 1 DATA AND NONLINEARITY IN THE GRADIENT

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1 Electronic Journal of Differential Equations, Vol (2013), No. 142, pp ISSN: URL: or ftp ejde.math.txstate.edu EXISTENCE OF SOLUTIONS FOR QUASILINEAR ELLIPTIC DEGENERATE SYSTEMS WITH L 1 DATA AND NONLINEARITY IN THE GRADIENT ABDELHAQ MOUIDA, NOUREDDINE ALAA, SALIM MESBAHI, WALID BOUARIFI Abstract. In this article we show the existence of weak solutions for some quasilinear degenerate elliptic systems arising in modeling chemotaxis and angiogenesis. The nonlinearity we consider has critical growth with respect to the gradient and the data are in L Introduction Reaction-diffusion systems are important for a wide range of applied areas such as cell processes, drug release, ecology, spread of diseases, industrial catalytic processes, transport of contaminants in the environment, chemistry in interstellar media, to mention a few. Some of these applications, especially in chemistry and biology, are explained in books by Murray [26, 27] and Baker [10]. While a general theory of reaction-diffusion systems is detailed in the books of Rothe [34] and Grzybowski [21]. Various forms of this problems have been proposed in the literature. Most discussions in the current literature are for linear or nonlinear systems and different methods for the existence problem have been used, see Alaa et al [1] [9], Baras [11, 12], Boccardo et al [15], Boudiba [16] and Pierre et al [29]-[32]. This is a relatively recent subject of mathematical and applied research. Most of the work that has been done so far is concerned with the exploration of particular aspects of very specific systems and equations. This is because these systems are usually very complex and display a wide range of phenomena remain poorly understood. Consequently, there is no established program for solving a large class of systems. For example a system of Chemotaxis, which is a biological phenomenon describing the change of motion of a population densities or of single particles (such as amoebae, bacteria, endothelial cells, any cell, animals, etc.) in response (taxis) to an external chemical stimulus spread in the environment where they reside see for example [28]. The simple mathematical model which describes such a phenomenon 2000 Mathematics Subject Classification. 35J55, 35J60, 35J70. Key words and phrases. Degenerate elliptic systems; auasilinear; chemotaxis; angiogenesis; weak solutions. c 2013 Texas State University - San Marcos. Submitted December 7, Published June 22,

2 2 A. MOUIDA, N. ALAA, S. MESBAHI, W. BOUARIFI EJDE-2013/142 reads as follows u t D u u + (κ(u) u + χ(v) v) = 0 in (0, T ) v t D v v + (ζ(u) u + η(v) v) = 0 in (0, T ) u(0) = u 0, v 0 (0) = v 0 here u and v are the population densities. For a simple expansion, we include a(x) = κ(u) χ(v) ζ(u) η(v), b(x) =, c(x) =, d(x) = u v u v f = (κ(u) u + χ(v) )v, g = (ζ(u) u + η(v) v). Then the system can be written as u t D u u + a(x) u 2 + b(x) v 2 = f in (0, T ) v t D v v + c(x) u 2 + d(x) v 2 = g in (0, T ) u(0) = u 0, v 0 (0) = v 0. In this work we are interested in the quasilinear elliplic degenerate problem u D 1 u + a(x) u 2 + b(x) v α = f(x) v D 2 v + c(x) u β + d(x) v 2 = g(x) u = v = 0 on in in (1.1) (1.2) (1.3) where is an open bounded set of R N, N 1, with smooth boundary, the diffusion coefficients D 1 and D 2 are positive constants, a, b, c, d, f, g : [0, + ) are a non-negative integrable functions and 1 α, β 2. We are interested in the case where the data are non-regular and where the growth of the nonlinear terms is arbitrary with respect to the gradient. To help understanding the situation, let us mention some previous works concerning the problem when a, b, c, d L (). if f, g are regular enough (f, g W 1, ()) and for all α, β 1, the method of sub- and super-solution can be used to prove the existence of solutions to (1.3). For instance (0, 0) is a subsolution and a solution, w = (w 1, w 2 ), of the linear problem w 1 D 1 w 2 = f(x) w 1 D 1 w 2 = g(x) w 1 = w 2 = 0 on, in in (1.4) is a supersolution. Then (1.3) has a solution (u, v) W 1, 0 () W 2,p (); see Lions [23]. If f, g L 2 () and 1 α, β 2, then u α, v β L 1 (). Many authors have studied this problem and showed that (1.3) has a solution (u, v) H0 1 () H0 1 (), see Bensoussan et al [14], Boccardo et al [15] and the references there in. If f, g L 1 () and 1 α, β < 2, Alaa and Mesbahi [1] proved that (1.3) has a non negative solution (u, v) W 1,1 0 () W 1,1 0 (). The case where f, g M + B () (f, g are a finite non negative measures on ) has treated by Alaa and Pierre [9]. They proved that if 1 α, β 2 and the supersolution w = (w 1, w 2 ) H0 1 () H0 1 (), then the problem (1.3) has a non negative solution (u, v) H0 1 () H0 1 ().

3 EJDE-2013/142 EXISTENCE OF SOLUTIONS 3 We are particularly interested in the case of a system (1.3) when a, b, c, d, f, g are not regular, more precisely, a, b, c, d, f, g are in L 1 (). Let us make some specifications on the model problem u D 1 u + b(r) v α = f v D 2 v + c(r) u β = g u = v = 0 on B in B in B (1.5) where B is the unit ball in R N, r = x and { ln r if N = 2 b(r) = c(r) = r 2 N if N 3. (1.6) In this case, b(r), c(r) are in L 1 loc (B) but not in L (B). As a consequence the techniques usually used to prove existence and based on a priori L -estimates on u and u fail. To overcome this difficulty, we will develop a new method which differ completely of the previous approach. We have organized this article as follows. In section 2 we give the precise setting of the problem and state the main result. In section 3 we present an approximate problem and we give suitable estimates to prove that (1.3) has a solution in the case where the growth of the nonlinearity with respect to the gradient is arbitrary. 2. Assumptions and statement of main results Let f, g, a, b, c, d are functions that satisfies the following assumptions f, g L 1 (), f, g 0 (2.1) a, b, c, d L 1 loc(), a, b, c, d 0 (2.2) First, we have to clarify in which sense we want to solved problem (1.3). Definition 2.1. We say that (u, v) is a weak solution of (1.3) if u, v W 1,1 0 () a(x) u 2, b(x) v α, c(x) u β, d(x) v 2 L 1 loc() u D 1 u + a(x) u 2 + b(x) v α = f(x) v D 2 v + c(x) u β + d(x) v 2 = g(x) (2.3) We are interested to proving the existence of weak positive solutions of the problem (1.3). For this, we define the truncation function T k C 2, such that T k (r) = r T k (r) k + 1 if 0 r k if r k 0 T k(r) 1 if r 0 T k(r) = 0 if r k T k (r) C(k). (2.4)

4 4 A. MOUIDA, N. ALAA, S. MESBAHI, W. BOUARIFI EJDE-2013/142 For example, the function T k can be defined as T k (r) = r in [0, k] T k (r) = 1 2 (r k)4 (r k) 3 + r in [k, k + 1] (2.5) T k (r) = 1 (k + 1) for r > k Then we define the the space τ 1,2 () = { w : R measurable, such that T k (w) H 1 () for all k > 0 } This enables us to state the main result of this paper. Theorem 2.2. Assume that (2.1) and (2.2) hold, and 1 α, β < 2. If there exists a function θ τ 1,2 () and a sequence θ n L () such that 0 a, b, c, d θ in θ n θ a.e. T k (θ n ) T k (θ) strongly in L 2 () ( 1 lim sup T k (θ n ) 2) = 0 k n k Then the problem (1.3) has a non negative weak solution. (2.6) Remark 2.3. (i) If a, b, c, d L (), then (2.6) is satisfied. Indeed, θ can take the value of any non negative constant C, such that C max { } a L, b L, c L, d L (2.7) (ii) Hypothesis (2.6) holds for the functions ξ = b or c given in (1.6). Indeed ξ = λ is in this case the measure of Dirac which is a finite non negative measure on. By consequent, we take θ = ξ and θ n solution of θ n = λ n in (2.8) θ n = 0 on, where λ n C 0 (), λ n λ in L 1 () and λ n λ. Then, we can applied Theorem 2.2 and conclude the existence of the non negative weak solution for our model problem (2.1). 3. Proof of theorem An approximation scheme. In this paragraph, we define an approximated system of (1.3). For this, we truncate the functions a, b, c, d, f, g by introducing the sequence a n, b n, c n, d n, f n, g n defined as follows and a n = min{a, θ n }, b n = min{b, θ n }, c n = min{c, θ n }, d n = min{d, θ n } f n C 0 (), f n f in L 1 (), f n f g n C0 (), g n g in L 1 (), g n g Then the approximate problem is u n, v n W 1, 0 () u n D 1 u n + a n (x) u n 2 + b n (x) v n α = f n (x) v n D 2 v n + c n (x) u n β + d n (x) v n 2 = g n (x). (3.1) (3.2)

5 EJDE-2013/142 EXISTENCE OF SOLUTIONS 5 One can see that a n, b n, c n, d n are in L (). On the other hand, (0, 0) is a subsolution of (3.2) and (U n, V n ) a solution of the linear problem U n D 1 U n = f n V n D 2 V n = g n U n, V n W 1, 0 () in in (3.3) is a supersolution, then by the classical results in Amann and Grandall [13] and Lions [23, 24], there exists (u n, v n ) solution of (3.2) such that 0 u n U n for all n 0 v n V n for all n 3.2. A priori estimates. To prove theorem 2.2, we propose to send n to infinity in (3.2). For this we will need some estimates passing to the limit. Lemma 3.1. Let u n, v n, a n, b n, c n, d n be sequences defined as above. Then (i) T k (u n ) 2 k f L 1 () T k (v n ) 2 k g L1 () and (ii) b n. T k (v n ) α k f L1 () c n. T k (u n ) β k g L1 () Proof. (i) By multiplying the first equation of (3.2) by T k (u n ) and the second equation by T k (v n ) and integrating over, we have T k (u n ) 2 + D 1 T k (u n ) 2 + a n T k (u n ) T k (u n ) 2 + b n T k (u n ) T k (v n ) α f n T k (u n ) and T k (v n ) 2 + D 2 T k (v n ) 2 + c n T k (v n ) T k (u n ) β + d n T k (v n ) T k (v n ) 2 g n T k (v n ) Thanks to the positivity of a n, b n, c n, d n, the assumptions on f n and g n, the definition of the function T k, we deduce the result. (ii) Integrating the first equation of (3.2) over, we obtain u n D 1 u n + a n (x) u n 2 + b n (x) v n α = f n (x) (3.4) On the other hand, it is well know that for every function y in W 1,1 0 () such that y = H, H L 1 () y 0

6 6 A. MOUIDA, N. ALAA, S. MESBAHI, W. BOUARIFI EJDE-2013/142 there exists a sequence y n in C 2 () C 0 () which satisfies y n y strongly in W 1,1 0 () y n y The regularity of y n allows us to write y n = strongly in L 1 () but y n 0 on and y n = 0 in. Then yn υ limit that y 0. Therefore The relation (3.4) yields u n + By (3.1); we conclude that u n + a n (x) u n 2 + u n 0 y n υ dσ, 0. We deduce by passing to the a n (x) u n 2 + b n (x) v n α f n (x). b n (x) v n α f L 1 (). In the same way, if we integrate the second equation of (3.2) over, we obtain v n + c n (x) u n β + d n (x) v n 2 g L1 (), hence the result follows. Remark 3.2. (1) Using the assertion (ii) of lemma 3.1, and the compactness of the operator where 1 q < L 1 () W 1,q 0 () G ϑ N N 1, and ϑ is the solution of the problem ϑ W 1,q 0 () αϑ ϑ = G we conclude the existence of u, up to a subsequence, still denoted by u n for simplicity, such that see Brezis [17] (2) Assertion (i) implies that u n u strongly in W 1,q 0 (), 1 q < N N 1, (u n, u n ) (u, u) a.e. in (T k (u n ), T k (v n )) (T k (u), T k (v)) weakly in H 1 0 () H 1 0 () Lemma 3.3. Let (u n, v n ) be a solution of (3.2), then ( 1 ) ( 1 lim sup T h (u n ) 2 dx = lim h + n h sup h + n h ) T h (v n ) 2 dx = 0

7 EJDE-2013/142 EXISTENCE OF SOLUTIONS 7 Proof. We first remark that u n satisfies u n f n If we multiply this inequality by T h (u n ) and integrate on, we obtain for every 0 < M < h, T h (u n ) 2 ft h (u n ) + ft h (u n ) {u n M} {u n>m} M f + h hence fχ {un>m} 1 T h (u n ) 2 M f + fχ {un>m} h h {u n > M} = dx 1 {u n>m} M u n L 1 C M ( Then lim M + supn {u n > M} ) = 0 On other hand, since f L 1 (), we have for each ε > 0 there exists δ such that for for all E, E < δ f ε E 2. Taking into account the above limit, we obtain that for each ε > 0, there exists M ε such that for all M M ε, ( ) sup fχ [un>m] ε n 2 Taking M = M ε and letting h tend to infinity, we obtain ( 1 lim sup T h (u n ) 2) = 0. h n h Lemma 3.4. Let η n be sequence such that η n η, a.e. in and η n 2 C then η n η in L α () for all 1 α < 2. Proof. We show that η n is equi-integrable in L α (). Let E be a measurable subset of ; we have η n α E (2 α)/2( η n 2) α/2 C E (2 α)/2 E E Since 1 α < 2 then 0 < 2 α 1. We choose E = ( ε C )2/(2 α), we obtain E η n α ε Convergence. The aim of this paragraph is to prove that (u, v) (obtained in the previous section) is in fact a solution of problem (1.3). According to definition 2.1, we have to show only that u D 1 u + a(x) u 2 + b(x) v α = f(x) v D 2 v + c(x) u β + d(x) v 2 = g(x)

8 8 A. MOUIDA, N. ALAA, S. MESBAHI, W. BOUARIFI EJDE-2013/142 By lemma 3.1, we know that a n (x) u n 2, d n (x) v n 2, b n (x) v n α, c n (x) u n β are uniformly bounded in L 1 (). Moreover a n (x) u n 2 0, d n (x) u n 2 0, b n (x) u n α 0, c n (x) u n β 0 and for almost every x in, we have a n (x) u n (x) 2 a(x) u(x) 2 d n (x) v n (x) 2 d(x) v(x) 2 b n (x) v n (x) α b(x) v(x) α c n (x) u n (x) β c(x) u(x) β Then there exists µ 1, µ 2 non negative measures, see Schwartz [35], such that Consequently, lim (u n D 1 u n + a n (x) u n 2 + b n (x) v n α ) n + = u D 1 u + a(x) u 2 + b(x) v α + µ 1 lim (v n D 2 v n + c n (x) u n β + d n (x) v n 2 ) n + = v D 2 v + c(x) u β + d(x) v 2 + µ 2 u D 1 u + a(x) u 2 + b(x) v α f v D 2 v + c(x) u β + d(x) v 2 g Therefore, to conclude the proof of Theorem 2.2, we must establish the opposite inequality. For this, Let H be a function in C 1 (R), such that 0 H(s) 1 { 0 if s 1 H(s) = 1 if s 1 2 To this end, we introduce the test functions Φ 1 = ψ 1 exp[ θ n D 1 H(u n, Φ 2 = ψ 2 exp[ θ n D 2 v n ]H( θ n H(v n where H denotes the function defined above and ψ 1, ψ 2 0, ψ 1, ψ 2 H0 1 () L (). We multiply the first equation in (3.2) by Φ 1 and we integrate on, we obtain f n Φ 1 = I j, where I 2 = D 1 I 1 = 1 j 7 u n Φ 1, u n ψ 1 exp[ θ n D 1 H(u n I 3 = u n u n θ n Φ 1

9 EJDE-2013/142 EXISTENCE OF SOLUTIONS 9 I 4 = D 1 k I 6 = D 1 k u n θ n ψ 1 exp[ θ n u n ]H ( θ n D 1 H(u n I 5 = (a n θ n ) u n 2 Φ 1 u n 2 ψ 1 exp[ θ n D 1 H ( u n I 7 = b n. v n α Φ 1 Next we study each term. For the first term, we have lim I 1 = lim T k (u n )ψ 1 exp[ θ n n + n + D 1 H(u n = uψ 1 exp[ θ u]h( θ D 1 H(u since ψ 1 exp[ θ n D 1 H(u n converges strongly in L 2 () to ψ 1 exp[ θ D 1 u]h( θ H(u in L2 () and T k (u n ) converges weakly to T k (u) in L 2 (), (see [24, lemma 1.3, p 12]). Concerning the second term, we get lim I 2 = lim D 1 T k (u n ) ψ 1 exp[ θ n n + n + D 1 H(u n = D 1 u ψ 1 exp[ θ u]h( θ D 1 H(u since ψ 1 exp[ θ n D 1 H(u n converges strongly in L 2 () to ψ 1 exp[ θ D 1 u]h( θ H(u. For I 3, we first remark that lim I 3 = lim T k (u n ) T k (u n ) T k (θ n )ψ 1 exp[ θ n n + n + D 1 H(u n = T k (u) T k (u) T k (θ)ψ 1 exp[ θ u]h( θ D 1 H(u since T k (u n ) T k (u) T k (θ n ) T k (θ) weakly in H 1 0 () strongly in H 1 0 () To study I 4 and I 6 we use Lemma 3.3. For I 4, we have [ 1 I 4 D 1 T k (u n ) 2 ψ 1 exp[ θ n u n ]H ( θ ] 1/2 n k D 1 H(u n

10 10 A. MOUIDA, N. ALAA, S. MESBAHI, W. BOUARIFI EJDE-2013/142 [ 1 T k (θ n ) 2 ψ 1 exp[ θ n u n ]H ( θ n k D 1 H(u n 1 D 1 [ ψ 1 L () T k (u n ) 2 ] 1/2 1 [ ψ 1 L k () k since exp[ θn D 1 u n ] 1, thus I 4 D 1 [ ψ 1 L δ k ] 1/2 [ ψ 1 L ρ k ] 1/2 ]1/2 T k (θ n ) 2 ] 12 Where δ k = sup( 1 T k (u n ) 2 ) and ρ k = sup( 1 T k (θ n ) 2 ) n k n k By Lemma 3.3, we have lim δ k = 0, k lim ρ k = 0 k Then lim sup (I 4 ) = 0 k n Similarly, for I 6, we have I 6 D 1 ψ 1 L δ k Then lim sup (I 6 ) = 0 k n Now we investigate the remaining term I 5. Since a n θ n and ψ 1 0, we have (a n θ n ) u n 2 exp[ θ n u]h( θ n D 1 H(u n 0 in Therefore, by Fatou s lemma, we obtain lim I 5 (a θ) u 2 exp[ θ u]h( θ n + D 1 H(u For I 7, we obtain lim I 7 = n + lim n + T k (b n ) T k (v n ) α ψ 1 exp[ θ n D 1 H(u n By a direct application of Lemma 3.3, we have T k (v n ) α T k (v) α strongly in L 1 (), then lim I 7 = b v α ψ 1 exp[ θ u]h( θ n + D 1 H(u We have shown that ω( 1 + uψ 1 exp[ θ u]h( θ D 1 H(u + D 1 u ψ 1 exp[ θ u]h( θ D 1 H(u u u θψ 1 exp[ θ u]h( θ D 1 H(u + (a θ) u 2 exp[ θ u]h( θ D 1 H(u + b v α ψ 1 exp[ θ u]h( θ D 1 H(u fψ 1 exp[ θ u]h( θ D 1 H(u where ω(ε) denotes a quantity that tends to 0 when ε tends to 0. Now we choose ψ 1 = ϕ 1 exp[ θ D 1 u]h( θ H(u

11 EJDE-2013/142 EXISTENCE OF SOLUTIONS 11 where ϕ 1 0, ϕ 1 D() and we replace ψ 1 by this value in the previous inequality to obtain w( 1 uϕ 1 H 2 ( θ H2 ( u D 1 u ϕ 1 H 2 ( θ H2 ( u ϕ 1 u u θh 2 ( θ H2 ( u ϕ 1 u 2 θh 2 ( θ H2 ( u D 1 ϕ 1 u θh ( θ k H( θ H2 ( u D 1 ϕ 1 u 2 H 2 ( θ k H(u H ( u + u u θϕ 1 H 2 ( θ H2 ( u b v α ϕ 1 H 2 ( θ H2 ( u a u 2 ϕ 1 H 2 ( θ H2 ( u + θϕ 1 u 2 H 2 ( θ H2 ( u fϕ 1 H 2 ( θ H2 ( u By developing calculations and remarking that the sixth and seventh terms are equivalent to ω( 1 k ), we can write uϕ 1 H 2 ( θ H2 ( u D 1 u ϕ 1 H 2 ( θ H2 ( u [a u 2 + b v α ]ϕ 1 H 2 ( θ H2 ( u + ω( 1 fϕ 1 H 2 ( θ H2 ( u Finally passing to the limit as k tends to infinity, we use the fact that lim k H( θ = 1, lim H(u k = 1 to conclude that for every ϕ 1 0, ϕ 1 D(), [u D 1 u + a(x) u 2 + b(x) v α ]ϕ 1 In the same way, we multiply the second equation in (3.2) by Φ 2 and we integrate on. By studying separately each term as in the previous case still using Lemmas 3.1, 3.3 and 3.4, we choose ψ 2 = ϕ 2 exp[ θ D 2 u]h( θ H( v where ϕ 2 0, ϕ 2 D() and we replace ψ 2 by this value in the inequality obtained to conclude that for every ϕ 2 0, ϕ 2 D() that [v D 2 v + c(x) u β + d(x) v 2 ]ϕ 2 gϕ 2 This completes the proof of theorem 2.2. Acknowledgments. We are grateful to the anonymous referee for the corrections and useful suggestions that have improved this article. fϕ 1

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