..(14 +1)! 5.. flosfilhw .QY , 3,11“. .3 fin , JAAMWa :( FH« .. :. hi: 9.. xx V ‘ I, .2 4.7 .2, . c 2.5. c . .v v. 4,! 4.1.1. 2.. i 3;... , Laura.» .113: 3;... “DAMN“ . 7 , iii-«3.3.. ..I 5.2.13} :8 an? 1‘ a... ‘ . n... . 0.», fir 3?: u at: (5? 3 “fix .1: . ,t . w. 3.) Lst... r . urn: HG‘. ‘ .36. . $7?! .::7 ’I?. ., . u figfixfié {tinting 21.- . Lfirfwwymgwflm? , - \- _w®%§wm A?” 25:7 . ‘ 4006 LIBRARY Michigan State University This is to certify that the dissertation entitled ON A CLASS OF NONLOCAL EVOLUTION EQUATIONS Ph.D presented by GUANGYU ZHAO has been accepted towards fulfillment of the requirements for the degree in Mathematics Major Professor’s Signature A4AG&&Y7' lC)‘ 2cmis Date MSU is an Affirmative Action/Equal Opportunity Institution PLACE IN RETURN BOX to remove this checkout from your record. TO AVOID FINES return on or before date due. MAY BE RECALLED with earlier due date if requested. DATE DUE DATE DUE DATE DUE 2/05"c:/Cl"RC/Da—"_5ieom.mm ‘ ON A CLASS OF NONLOCAL EVOLUTION EQUATIONS By Guangyu Zhao A DISSERTATION Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 2005 ABSTRACT ON A CLASS OF NONLOCAL EVOLUTION EQUATIONS By Guangyu Zhao The thesis includes three parts. In the first part, we study a nonlocal evolution equation which describes the seed dispersal of single species and prove the existence, uniqueness and stability of positive steady state solution to this equation. In the second part, we study the principal eigenvalue problem and given a sufficient condition that ensures the existence of coexistence state to a nonlocal evolution system. Then we consider a competition model which involves two similar species. The existence of coexistence states and their stability are investigated. In the third part, we establish the existence, uniqueness and continuous dependence on initial values for the solutions to a nonlocal phase field system. We also discuss the asymptotic behavior of the solution and prove the global boundedness of the solutions. To my parents iii ACKNOWLEDGMENTS I would like to thank Professor Peter W. Bates, my dissertation advisor, for his invaluable guidance, expert advice, constant encouragement and support during my graduate study at Michigan State University and Brigham Young University. I also want to thank Professor Peter W.Bates for his patience of correcting numerous gram- mar mistakes in this dissertation. I would also like to thank Professor Kening Lu and Professor Tiancheng Ouyang for their helpful advice when I studied at Brigham Young University. Many thanks to my dissertation committee members Professor Guang Bao, Pro- fessor Milan Miklavcic, Professor Monxun Tang, and Professor Zhengfang Zhou for their time and advice. This work cannot be done without the support and understanding from my familyl would like to thank my wife for her love and support, and also my parents for their forever caring and love of me. iv TABLE OF CONTENTS 1 Introduction 1 2 Existence,Uniqueness and Stability of Steady Solution for a Nonlocal Evolution Equation 7 2.1 Existence and uniqueness of positive steady solution ............ 7 2.2 The Existence and asymptotic behavior ................... 21 3 Existence and Stability of Coexistence States for a Nonlocal Evolu- tion System 27 3.1 The existence of coexistence states ..................... 28 3.2 Two similar competing species ........................ 34 3.3 Appendix ................................... 48 4 The Cauchy Problem for a Nonlocal Phase-Field Equation 57 4.1 Existence and uniqueness .......................... 57 4.2 Asymptotic behavior of the solutions .................... 68 4.3 Global boundedness of the solutions ..................... 75 BIBLIOGRAPHY 93 CHAPTER 1 Introduction The spatial dispersal of cell or organisms is central to biology, it has important effects both from ecological and genetic point of view in a variety of situations. Obviously the mechanism of dispersal is of great importance in this context and has received much attention recently. It is now a major focus of theoretical interest. The models for this can be roughly placed into two categories. The first consists of models continuous in both time and space. Most continuous models related to dispersal are based upon reaction-diffusion equations, which have been extensively studied, (see [19, 30, 34, 38] and [39]). The exact nature of dispersal and its theoretical treatment has became an important question and in particular there is a variety of reasons that suggest that the class of reaction-diffusion models, with its fundamental assumption that motion is governed by a random walk, is too restrictive to model the seed dispersal. These consideration led to the derivation [33], based on a variation of a position-jump process, to the evolution equations of the form '2'? = “ii/n HI - y)u(y)dy - u(.r)] + f(u,.r)u, Q C IR" (1.1) under the constraint that k > 0. To gain a insight of the derivation of the model (1.1). We quote the section 2.1 of [33]. Consider a single species in an 72,-dimensional habitat where the population can be modelled by a single function u(x, t), which is the density at position x and at time t. To establish the continuous model, we shall focus on the case n = 1 for clarity of exposition ( It is straightforward to generalized to arbitrary dimension). Divide the habitat 1R into contiguous sites, each of length Ax. Discretize time into steps of size At. Let u(i, I.) be the density of individuals in site 2' at time 1.. With the assumption that the rate at which individuals are leaving site 2' and going to site j is constant, the total number should be proportional to: the population in the interval 2', which is 12(i,t)A.’r; the size of the target site, which is Arc; and the amount of time during which the transit is being measured, At. Let a(j, i) be the proportionality constant. Then, the number of individuals leaving site 2' during the interval [t,t + At] is 00 Z a( j, i)u(i, t)(A:z:)2At. (1.2) j=-ooj#i It is biologically reasonable to believe that the mean and variance of the distances moved are finite. Hence 2 a(j,z')Ar, Z IJ-ila(j.i)(Ar)2 and Z Ij—il2a(j,i)(Ax)3 j=-ooj#i j=-ooj¢i j=—ooj;éi will all be assumed finite. During the same time interval, the number of arrivals to site i from elsewhere is 00 Z a(j, i)u(i, amalgam. (1.3) j=—ooj¢z‘ Finally, with each site we allow for the birth and death of individuals. Let f (u(z', t), 2') denote the per capita net reproduction rate at site 2' at the given population density. We assume that this rate is constant over the time interval. Then the number of new individuals at site i is f(u(z’, t), i)u(i, t)A:z:At. With regard to f, the following assumption are made: f(0,:r)>0, %<0 With (1.2) and (1.3), we deduce that the population density at location 2' and time t + At is given by u(i,t+At)=u(i,t)+ Z a(i,j)u(j,t)A:rAt j=—ooj¢i — Z a(j,i)u(i,t)Aa:At+f(u(i,t),i)u(i,t)A;rAt. j=-ooj¢i Let both Ax —) 0 and At —+ 0, we obtain %(;r, i.) = / [(r(:1:,y)u(y, I.) — a(y,:r)u(:1:, 1.)](1y + f(’lt(.’l.‘,l),.’l})’ll.(.’L‘, 1). Assume that the rate of transition between the various patches, a(:c, y), is homoge- neous and only depends on the distance between patches i.e. upon [1: - y]. Above equation can be written as gg=wjfke—ymeny—uarhnmau (La where k is an even function with and Iw=[wanws (X) Notice that the dispersal rate 7, which represents the total number of the dispersing organism per unit time, play an important role in the model. It is worth to notice that Bates, Fife, Ren and Wang [14] studied a dissipative model with nonlocal interaction which is derived from the point of view of certain continuum limits in dynamic Ising models. It is similar to (1.4) but with the bistable nonlinearity. So far the habitat has been considered to be infinite in the extent. Somehow, it is biologically unrealistic. With this in mind (1.4) may be modified to apply to a habitat in R of length h by two natural ways which lead to the equations all h E 2" 7]] ‘70" - 1/)“(illdi/ - "(Ill + fill-’1’)" (1'5) 0 3 with either [Oh k(s)ds g 1(# 1) or I: k(s)ds E 1. (see [33] for details) Although the approach used to obtain (1.5) has similarities with the classical derivation of the Laplacian via random walk, it is not assumed that individuals move from a given patch with a binomial distribution. In contrast to the Laplacian, the integral operator J in equation (1.5) is not a local operator, thus (1.5) should be considered as a model with long-range interaction. With the assumption that (1.5) has a positive steady solution ii and the nonlinearity f (x, u) = u(a(.r) -- u), where a(x) is C1 continuous, in [33] it is shown that ii is the global attractors for all solutions whose initial data is non-trivial and nonnegative. Based on (1.5), The authors also proposed a competitive system of the form a h 1% = 71]!) k(:I: —— y)1i.(y)dy — ri(;1:)] + f(u + 'u, (1:)u, (1.6) ('91) h 5t- 2 e,2[/ k(:17 - y)v(y)(ly - v(.1:)] + [(11, + 1),:17)?’ (1.7) 0 to try to understand how competition drives selection of 7. Here it and v stand for the densities of two different species. They discovered an interesting fact that the slowest disperser as measured by dispersal rate is always selected. Their study also reveals that both (1.5) and (1.8) can display very rich dynamics and hence gives rise to many interesting mathematics issues. As pointed out in [33], the mathematical analysis of (1.1) appears to be difficult even though the dispersal is represented by a bounded operator. Unlike the reaction-diffusion equations, (1.1) no longer has a regularizing effect. In the study of classical dispersal models which are often based upon reaction- diffusion equations, many useful results on the global dynamics of diffusive equations were established in terms of principal eigenvalues of scalar elliptic eigenvalue prob- lems. In [23], these linear elliptic eigenvalue problems are carefully explored. The authors obtain several important properties of the principal eigenvalue which were then used to study the global dynamics of logistic models. A trichotomy of the global 4 asymptotics was also established. Meanwhile, by using monotone dynamical systems theory, in [24] the authors derived similar results for some quasimonotone reaction- diffusion systems with delays. Even though their approaches are not immediately applicable to ( 1.1) due to the lack of compactness, the importance of the principal eigenvalue is evident. Those approaches strongly suggest that an analogous idea us- ing the principal eigenvalue should be developed, particularly for the case where the reaction term is sublinear. In [33] the authors prove the existence of a principal eigen- value for the integral operator Ln := J :1: u + b(:r)u under certain conditions, where J =1: 11 = In J (x,y)u(y)dy. We shall use those ideas combined with a comparison argument to study the steady solutions of (1.1) and their stability. In 1997, Bates [9] proposed the study of a nonlocal phase-field equation in which motility of phase boundaries is temperature-dependent and temperature varies ac- cording to a heat equation with phase change becoming a heat source or sink through the latent heat of fusion. This system, when considered on a finite region, has the form g}; = LJQ: — y)u(y)dy — [fix/(33 — Sikh/“(13) _ flu) +161 (1'8) 8(9 +lu) _ T _ A6 (19) in (0, T) x Q, with initial and boundary conditions u(0,x) = uo(:t:), 6(O,:r) = 00(1), (1.10) 32],,” = 0, (1.11) where T > 0, S) C R" is a bounded domain. Here 6 represents temperature, u is an order parameter, I is a latent heat coefficient, the interaction kernel satisfies J(—Lr) = J(3:), and f is bistable. For this system, one expect to see spinodal decomposition or the spontaneous creation of a fine—scaled patterned structure from initial data that is close to homo— geneous but with in a certain range. The dissertation is organized as follows: In chapter 2, we adopt the sub-super solution methods and local bifurcation theory to study the positive steady solution to ( 1.1) and its uniqueness. We investigate the stability of this positive steady state and the long time behavior of solutions to (1.1). In chapter 3, we are more interested in the coexistence states of the competition system (if=dll/y(x,y))u(y)dy—b(sc)x21(1Ml+ulxl(z)+f 0,d,- > 0(2' 2 1,2) and An 6 1R. By studying the principal eigenvalue problem, we are able to apply abstract results from the bifurcation theory to obtain the existence of coexistence states. We also consider a special form of (1.12) which involves two similar species. The existence of coexistence states and their stability are investigated. In chapter 4, we focus our attention to the system (1.8)—( 1.11) and prove the existence, uniqueness and continuous dependence on initial data of the solution, in this case, we require initial data no 6 L°°(Q), and 00 E L°°(Q) fl W1’2(Q). We also discuss the asymptotic behavior of the solution. CHAPTER 2 Existence,Uniqueness and Stability of Steady Solution for a Nonlocal Evolution Equation In this chapter we study the existence,uniqueness and stability of positive steady state solutions of nonlocal evolutionary problem of the form 91‘- : / JCT, Blue/Ml} + b(:17)'u.(:r) + f(;1:,'u.) at Q ”(3, 0) = (Mil?) where J, b and f are sufficiently smooth functions and J is positive. 2.1 Existence and uniqueness of positive steady solution Let Q be a bounded domain of IR" of class C'" for some '7 6 (0,1). Let H be the Hilibert space L2(Q) with inner product (~, ). Let X := C (5) be the Banach space of real continuous functions on LI. Throughout this paper, X is considered as an ordered Banach space with a positive cone X +, where X + = {u E X In 2 0}. It is well known that X + is generating, normal and has nonempty interior, which we denote by intX+, (see [1] for more details). For ¢,cp E X, we write at 3 cp if 50 -— <23 6 X+, (,0 >> 43 if :p — (b E intX+ and 4p << q) if gt — go 6 intX+. An operator T : X —-+ X is called positive if TX + g X+. Definition 2.1.1 An operator A is said to be resolvent positive if the resolvent set p(A) of A contains an interval (a, 00) and (Al — A)‘1 is positive for sufliciently large A E p(A) H R. We also denote the spectral bound of an operator A by 5(A) = sup {Re/\ : /\ E o(A)} where 0(A) is the spectrum of A. Consider the linear operator L on X defined by Lu(3) :22 [$2 J(3,y)u(y)dy + b(3)u(3) (2.2) where we assume that (H1) .](:z:, y) 6 C(5 x 5, HP) is symmetric. (H2) J(3, y) > 0, for any 3, y 6 fl. Lemma 2.1.2 Let L be given by (2.2). Assume that (H1) and (H2) are satisfied and b E X. Then L is a nesolvent positive operator on X and 5(L) E o(L). If there ezist A 6 IR and a continuous function (i) E X+\ {0} such that Lei 2 Act. Then A = 5(L). Moreover, 5(L) is an isolated eigenvalue and ker(L — 5(L)I) = span {qt}. Proof. First, we prove that L is a resolvent positive operator on X. In fact, L is a bounded, linear operator on X. Thus, p(L) contains (||L|], 00). Choose a) 6 EV such that w > supIE-g—2 | L, J (3,y)dy| + supxefi ]b(3)|. Obviously, A E p(L) whenever 8 A Z w. To prove that (Al — L)‘1 is positive for all /\ 2 w, it is sufficient to show that (L— A1)v g 0 implies v 2 0 for all A 2 w. Let U = v+ — v”, where v“ = max{—v,0} and v+ = max {v, 0}. If v‘ 75 0, then straight calculation yields that 0 s (J * v+.v‘) s ((L - A)v‘.v‘) and so 0 g ((L — A)v",v“) S (supl J(a:,y)dyl +sup ]b] — A)(v—,v‘) < 0. :66 n are?) The contradiction shows that v‘ E 0 in Q, and so u 2 O in 5. Since 5(L) > —00, by ([45],Theorem 3.5), 5(L) E 0(L). Now, assume that there exist 43 E X +\ {0} and A E R such that J * a5 = (A — b(:17))qb. It is easy to see that J a: q) >> 0. Consequently, we have that (b >> O and A - b(;z:) >> 0. If 5(L) > A, then the linear operator K, defined by K u = (5(L) —b(:r))u, is continuous and bijective on X because that 5(L) —b(:z:) is bounded and 5(L) —b(;r) >> 0. Note that (L — 5(L)I)u = J *u - Ku. We infer that the linear operator L — 5(L)I is Fredholm of index zero because that J : u —-> J a: u is compact on X (see [46], Theoren15.C, p295). By ([32], Proposition 2.3 and 2.4 p.151), (see also [36]), 5(1)) is an eigenvalue with finite algebraic multiplicity and there exists a positive eigenfunction (p E X+\ {0} associated with 5(L). Since L is self-adjoint when considered as an Operator on H, its eigenfunctions corresponding to distinct eigenvalues are orthogonal, hence we have (¢,;p) = 0. But this contradicts the fact that both g6 and «,9 are strictly positive. Thus, 5(L) = A. We now show that ker(L — s(L)) = span {45}. Suppose this is not true, then there exists an eigenfunction 21; associated with 5(L) such that 111 74 to for all t 6 IR. Since (b >> 0, there exist I. such that id) +11) 2 0. Let t = inf {1. E tho +1]; 2 0}. Note that we have J =1: (to + 1.0) = (5(L) — b)(t¢ + 1b) and to + 2b # 0. Again, J * (to + w) >> 0 implies M + 1/2 >> 0, which violates the definition of t. The contradiction yields the desired conclusion. To prove that 5(L) is isolated, we assume to the contrary that there exists a sequence {an}:l C 0(L) such that linian/in = 5(L). By ([46], Proposition1,p300), it is evident that L — unl is a Hedholm operator of index zero and hence an is an eigenvalue of L on X if n is sufficiently large. On the other hand, thanks to 5(L) - b(:r) >> 0, we have [tn - b(:r) _>_ 6 for some 6 > 0 provided n sufficiently large. Let 0,, be the corresponding eigenfunction with ][6nllx = 1. Due to the compactness of J and the fact that the sequence {6"} is bounded in X, along some subsequence, still labelled n, limn—wOHOJn - b(~r))’1J * 6n - t9||x = 0 for some 19 E X. From (an — b(:r))‘1J * 6,, = 6", it follows that (L —- 5(L)I)t9 z 0 a11d||19||x =1. Thus 19 is an eigenfunction associated with 5(L). Since 19 does not change sign over fl— and is bounded away from zero, the convergence of (un — b(:1:))'1J * 0,, implies (6", 19) > 0 and we arrive at a contradiction again. Therefore, 5(L) is isolated and the proof is completed. Lemma 2.1.3 Let all assumptions in Lemma 2.1.2 be satisfied. Then the following three statements are equivalent. (i) There exists a H E X+\ {0} such that —LTi E X+\ {0}. (ii) 5(L) < 0. (iii) For each f E X, Lu 2 f has exactly one solution in X. Moreover, if to is a solution to Ln = f and f g 0, then m 2 0. Proof. (i)=>(ii). Suppose 5(L) 2 0. Let g 2 Lu — 5(L)u. Obviously, g S 0 and g 75 0. Let (b be the positive eigenfunction associated with 5(L), then (o, g) < 0. On 10 the other hand, we have (<25. 9) = (e. (L - 5(L)1)fi) = ((L - 5(L)1)¢fli) = 0. which is a contradiction. Thus, 5(L) < 0. (ii)=> (i). This is trivial since the eigenfunction 4) >> 0 and satisfies 11¢ << 0. (iii)=>(ii). Clearly, 0 E p(L), thus 5(L) < 0. (ii)=>(iii). The existence of a unique solution is ensured by the fact that 0 E p(L). Suppose w is the solution to Ln = f and f S 0 with w 9E X +. Let (b be the positive eigenfunction associated with 5(L). There exists t > 0 such that t4) + w 2 0. Once again, we let t = inf {t E lR+It + w 2 0}. Obviously, tgb+ w aé 0 and Lad +w) s 0. Now, let x0 6 5 be a point such that t¢(.r0) + w(a:0) = 0, then we have 0 s A Jan, are + My = Leas + mm) s 0. Since J (1:0, 3;) > 0 and id + w is nonnegative, we have tcb + w E O. The contradiction leads to t¢+ w >> 0, which of course violates the definition of t. Consequently, w 2 0 and the proof is completed. Lemma 2.1.4 Assume (H1) and (H2). Suppose that b1,b2 E X with b1 2 b2 and b1 74 b2. IfL,-¢, = u,gb,-,i = 1,2, where L,u :2 J * u + b,-u and 45,- E X+\ {0} ,i = 1,2. Then #1 = 5(L1) > [12 = 5(L2). Proof. From Lemma 2.1.2, it follows that #1 = 5(L1) and m = 5(L2). Furthermore, J* (b, = ([1, — bi)¢i,i=1,2 and J*¢,‘ >> 0 indicate fibi >> 0,1 = 1 2. NOW, If [1.1 S [12, then (L2 — [121)451 = L1451 + (1)2 — b1)¢1 — [12491 = (HI — lt2)¢1 + ([12 - b1)¢1 S 0- According to Lemma 2.1.3, 5(L2 — m) < 0. This is impossible, because 5(L2 — u2) = 0. Therefore, 5(L1) > 5(L2). 11 Next we consider the existence of solutions to [a J(:, y)u(y)dy + b(x)u(:1:)+ f(x, u) = o (2.3) For the remainder of this paper , we assume that (H3) f(re,s) E C(fi x lR+,R). t33f(:r,s) E C(fi x R+,IR) and f(rr,0) E 0. (H4) f(x, ) is strictly sublinear,i.e., for any a E (0, 1), f(x, as) > af(x, s), where s > 0. (H5) f (-, s) and are continuous, uniformly for s in bounded sets. We also let M for u > O 9(rc. 2!) = u (2.4) Buf(:r,0) foru = 0. It is clear that g E C(O x 5, 1R) and f(:1:,u) = g(:r, u)u. Definition 2.1.5 A function in X is said to be a subsolution of (2.3) if / Jo, y>u(y>dy + bu + f(x. u) 2 o. (2.5) n A supersolution is defined similarly by reversing the inequality. Theorem 2.1.6 suppose that (H1)—(H5) are satisfied. If ( 2. 3) has a supersolution fl and subsolution u in X +\ {0} such that y g E. Then (2. 3) has a unique solution in intX+. Proof. We define V = {u E C (S—l)|g S u _ a}. By condition (H3), we see that 7 _ {—I-(x, u) is uniformly bounded on Q x V and flu 0f 53(22, u) + b(:r) +13 >> 0 (2.6) for (:r, u) 6 fl— x V provided )8 is sufficiently large. We define the mapping for 6 > 0 large as follows: v = Tu if / J(:r., y)v(y)(ly + b(:1:)v(:z:) - fiv(.'1:) = —[f(.*1:,u) + [in]. (2.7) n 12 We also define Jae) := f 1e. many + bu - ave). Since .7 is invertible on X if 6 is sufficiently large and the right hand side of (2.7) belongs to X for each u E V, T : V —> X is well defined. Next, we show that T is monotone in the sense that w S wg implies Twl S ng, provided both ml and U12 belonging to V. In fact, if wl < 102 then le E f(r,w1) +fiw1 S ng E f(x,w2) +flw2, thanks to (2.6). Notice that \7(Twi) = ‘Fwi: thus, we have n.7(T1U2 — Twl) S 0. By virtue of Lemma 2.1.2, (—,7)‘1 is a positive operator as long as 6 is sufficiently large. Hence we obtain Twl S ng From this, we deduce that the sequence defined inductively by u1 2 TH and u,, = Tun_1 is monotone decreasing. Similarly, v1 2 Tu and v" = Tv,,_1 define a monotone increasing sequence. Furthermore, we can show by induction that 2301 S'“ Svn'“ 0 such that b(:£) + g(:c, u‘) S -—6, for all :r E O. Furthermore, for any 1:1, '12 E O, we find that J * ”1171)“ J * ”'(32) + [b(a:1) — b(332)lu‘(1‘1)+lffl‘liu‘fflhl)‘ f(fzau’CT/Illl (2-8) = be.) + axe-2.621%» + (1 — 0>u‘>](u'(:r1)— we»), 14 where 0 S 6 S 1. Without loss generality, we may assume u‘ (1:1) 2 u“(:r2). Since (H4) gives that g(:c, -) is decreasing and 8,, f (1:, s) S g(:1:, s) for all s > 0, the following inequalities are true Buf($2,0U‘(I1)+(1 “ 9)u‘($2)) S 9(172,9u‘($1)+(1 - 0)U*(1‘2)) S 9($2,U‘($2)) Hence, we have —[b(a:2) + Buffing, 0u*(:1:1) + (1 — 6)u"(:c2))] Z 6. (2.9) From (2.8) and (2.9), we conclude that u‘ is continuous. We now show the uniqueness of positive solutions in X +. We shall argue by contradiction. Let (p1 7e p2 be two positive continuous solutions of (2.3). Then it is easy to see that legal is a supersolution of (2.3) which is greater than both cpl and p2 provided I: is sufficiently large. Actually k can be chosen so that legal > cpl + 502. Hence we may assume without loss of generality that (,02 << (p1. On the other hand, the fact that g(:r,',c1) < g(:1:,<,92) and Lemmas 2.1.2 and 2.1.4 yield 0 = 5(L + 9(r.s01)) < 5(L + 9(r. s02» = 0 This contradiction completes the proof. Next we assume that (2.3) can be written as / Jo, y)udy + f(sr. w) = o, (2.10) n a as is the case when it arises as the Euler-Lagrange equation. The existence of a positive solution will be established by Crandall and Rabinowitz’s bifurcation theorem which allows less restriction on f than is considered in [33]. In the following, we define 00w) = [2J(:r.y)u(y)dy - 21/ J(~r.y)dy + f(auait) (2-11) S2 15 and 2‘: {(Wt) ERX XIGUW) =0} 2+ = {H E 2] for some up 6 X+\ {0} , (MUM) E 2}. Clearly, (2.10) has a positive solution if 2+ is not empty. We also assume that (H6) f(x, -, ) E C'2 uniformly for all :1: 6 fl. f(x, O,u) E 0 and sum, 0, 0) E 0. (H7) f0 Buaflflx, 0, (Dd): §£ 0. Theorem 2.1.7 Suppose that (H1),(H2),(H6} and (H7) are satisfied. Then (0,0) is a bifurcation point of G (u, u) = 0 and 2+ is not empty. Proof. First, we see that G E C2(R x X,X) and C(u,0) :- 0 for all u 6 1R. Moreover, DuG(O, 0)u. = f (I J(:r,y)u(y)dy — u/ J(:r,y)dy Q and D/tDuG“); 0) = 31L3#f($,0,0). Since DuG(0, 0)1 = 0, Lemma 2.1.2 implies that 5(D1,,G'(0, 0)) = 0, kerDr,,,G(0, 0) = span {1}, and DuG(O, 0) is a Hedholm operator on X whose index is zero. Here 1 stands for constant function whose value is 1. According to Crandall-Rabinowitz’s theorem (cf [28]), to show that (0,0) is a bifurcation point of C(p, u) = 0, we only need to verify that D11, 011,0(0, 0)]. ¢ rangeD'u,G(0, 0). In fact, if this is not true, then there exists in E X such that DuG(0, 0)1U = auauf('13, O). 16 Since DuG(0, 0) is a bounded, self—adjoint operator on H, the Fredholm alternative yields (1, auapflx, 0)) 2/‘18ur9yf(x, 0)d:z: = 0. This contradicts the given condition. Therefore, (0, 0) is a bifurcation point. By the Crandall-Rabinowitz theorem, there is a nontrivial continuously differentiable curve (p(s),u(-,s)) through (0,0) such that (p(s),u(-,s)) 6 2, where s 6 (-6,6) for 6 > 0 and (p(O), u(-, 0)) = (0, 0). Birthermore, u(', s) 2 s1 + 0(5) and so 2+ is not empty. Corollary 2.1.8 Assume (H1) and (H2). In addition, assume f(x,u) = ua(:z:)u + h(:r,u)u, where a 6 X and h 6 02(5 x O, R), h(:c,0) E 0, auh(x,0) < 0. Then 2+ C IR“L for s 6 (0,6) provided fa a+(:r)d:1: > In a’(a:)da:, where a’ = max{—a,0} and a+ = max {a, 0}. Moreover, if a(a:) >> O in 5, then (0, 0) is the unique bifurcation point for positive solutions. Proof. Let (p(s), u(-, s)) be the continuously differentiable curve ensured by The- orem 2.1.7. Notice that, supposing the dependence on s, LU * u(:z:) — LJ(x,y)dyu(z)]dx = — L[ua(x) + h(.r,u)]ud:c. The symmetry of J implies /n[ua(;r) + h(:r, u)]udx = 0. (2.12) If s is sufficiently small, we have fn[pa(:r) + h(:1:, u)]ud:1: = s /Q[u(s)(a+ -— a”) + 81,h(17, O)s]d.r + 0(32), (2.13) and from (2.12) and (2.13), we deduce that p(s) > 0 as s 6 (0, 6). Now, assume a(:z:) >> 0, suppose that M is another bifurcation point. Then there is a sequence (ltn,’ll.n) E R“ x X+ such that (Hm’U-n) -—» ([t1,0). If [11 = 0, nothing needs to be done. Therefore, we assume u] > 0. Let un(:r") = minfi u,,(:c). Note that J * un(.1:") — / J(:L'", y)dyu,,(.1:”) 2 0. Q 17 Consequently, u,,a(1:") +h(:r", un(:I:")) S 0. On the other hand, we may choose N > 0 such that una(a:") _>_ gm mfiina(:c) and |h(:c",u,,(x"))| S 3‘“ mfiina(x) whenever n > N. The contradiction shows that (0, 0) is the unique bifurcation point. Throughout the rest of this section, We shall focus on the case that Q C IR and establish the necessary and sufficient condition for the existence of a steady state solution to (2.3). Besides the condition (H1)-(H5), we assume that (H8) 0 = (0,1),0 < l < 00. (H9) 8“ f (:r, 0) is Lipschitz continuous on 5. (H10) f (-, s) is Lipschitz continuous, uniformly for s in bounded sets. First, we need the following lemma which can be found in [33]. Lemma 2.1.9 Assume that {H1 ),(H2} and {H8} hold and that b E X is Lipschitz continuous. Then L is a bounded, self-adjoint operator on H and has a simple eigen- value /\0 given by A0 = max (LU,’U.). IIUIIL2 = 1 The maximum is attained by a strictly positive eigenfunction 9b 6 X. Also 0(L) C (—00, A0] . Lemma 2.1.10 Suppose that (H1)-{H4) and (H8)-(H9) hold. In addition, b is Lip- schitz continuous on fl. If5(L+g(:I:. 0)) S 0. Then (2.3) does not possess any solution in X+\ {0}. Proof. Suppose this is not true. Let w be a solution in X +\ {0}. By Lemma 2.1.2, we infer that 5(L + g(:z:, w)) = 0. On the other hand, Lemma 2.1.2 and 2.1.9 ensure that 5(L + g(:1:, 0)) is a simple eigenvalue having an eigenfunction in intX+, by (H 4) and Lemma 2.4, we have 0 = 5(L + g(.r, w)) < 5(L + g(.1:,0))S 0. The contradiction gives the desired conclusion. 18 Lemma 2.1.11 Suppose that (H1)-(H4) and (H8)-(H9) hold and that b E X is Lip- schitz continuous. The following statements are equivalent. (i) (2.3) admits a subsolution u E X+\ {0} such that the inequality (2.4) is not identical to zero. (ii) (2.3) admits an arbitrarily small subsolution in X +\ {0}. (iii) 5(L + g(:1:, 0)) > 0. Proof. (i)=> (ii). It follows from the fact that Leg + f(r, (y) _>_ Leg + ef(.r,1_i) 2 0 forany0<6<1. (ii)=> (i). This is trivial. (i)=> (iii). Because of (i) and (H4), (L + KICK. 0))u 2 Lil. + 11(417. ulu = Lu + f(r. u). that is, (L + g($,0))(—y) S 0. As a consequence of Lemma 2.1.3, we have 5(L + g(2:,0)) Z 0. We next show 5(L + g(:r,0)) 75 0. Suppose this is not the case. Let 1]; E intX+ be an eigenfunction corresponding to the eigenvalue 0, then 0 = ((L + {/(:1:,O))'1/’, —y) = ('1/1, (L +g(;1:,0))(—15)) < 0. The contradiction shows 5(L + g(:1:, 0)) > 0. (iii)=> (i). Again, let 112 E intX+ be an eigenfunction associated with the eigenvalue 5(L + g(:r,0)), then LE'l/J + g(:1:, 0)a/) >> 0 for any 6 > 0. By the continuity of g(;r, -), we have Lei + f(z, at) = Lei + go. giant 2 0 for sufficiently small 6 > 0. Hence, at: is a subsolution of (2.3). Theorem 2.1.12 Suppose that (H1)-(H5) and (H8)-(H9) hold and that b E X is Lipschitz continuous. If (2. 3) has a positive supersolution ii 6 X, then ( 2. 3) has a unique positive continuous solution if and only if 5(L + g(:r., 0)) > 0 19 Proof. We first prove the necessity. Let go 6 X +\ {0} be a positive solution to (2.3). Then we have, in fact, cp >> 0 and 5(L + g(x, w)) = 0. Consequently, 5(L + MI, 0)) > 5(L + g(.r, w)) = 0. Now, suppose that 5(L + g(x, 0)) > 0. Let 1,!) be the positive eigenfunction asso— ciated with 5(L + g(a:, 0)), whose existence is guaranteed by the condition (H9) and Lemma 2.1.9. In addition, the proof of Lemma 2.1.11 shows that 51/) is a subsolution of (2.3) for arbitrarily small 5. Also, i E X+ and -(b(:1:) +g(:1:, i017 2 J :1: 17 >> 0 force 5 >> 0, and hence 52/) << ii for some 8 > 0. It follows that (2.3) has a solution it in X+ with 5d) S u S ii. Corollary 2.1.13 Assume (Hlj-(H4) and (H8)—(H10) and that b E X is Lipschitz continuous. Suppose 5(L + g(r, 0)) > 0. Then the following statements are equivalent (i) Problem (2.3) has a positive solution in X. (ii) Problem {2.3) has a positive supersolution in X. (iii) Problem ( 2. 3) has an arbitrarily large positive supersolution in X. (iv) There exists v E intX+, which is Lipschitz, such that 5(L + g(.r, v)) S 0 Proof. The equivalence of (i) and (ii) is an immediate consequence of Theorem 2.1.12 and the fact that Law +f(.1:,ow) S 0(Lw + f(:1:,w)) = 0, where a > 1 and w E X + is a solution of (2.3). The equivalence of (ii) and (iii) comes from the fact that if is is a supersolution of (2.3), then Lk-z/J + f(me/i) S Help + kf(.r,i/1) S O, for all k > 1. If (2.3) has a positive solution w in X, then 5(L + g(:1r, 111)) = 0. Let v E intX+ be Lipschitz continuous. Due to (H10), b(r) + g(:r, v) is Lipschitz continuous. Lemmas 20 2.1.2 and 2.1.9 imply that 5(L + g(:1:, v)!) is an eigenvalue of the linear and bounded operator L + g(:1:, v)I on X and there is a strictly positive eigenfunction associated with 5(L + g(:c,v)l). Choose U such that v >> w, By condition (H4), g(:r,-) is nonincreasing, it follows from Lemma 2.1.4 that 5(L + g(a:, v)I) < 5(L + g(r, w)). Therefore, (i) implies (iv). We now complete the proof by showing that (iv) implies (ii). Let ()3 be the strictly positive eigenfunction corresponding to 5(L + _q(:r, v)). Then we have Limb + f(r. W) = Lk¢ + kg(:r. k¢)¢ S Lk¢ + 1690:. v)¢ S 0 for sufficiently large k. Thus, kq’) is the desired supersolution of (2.3). 2.2 The Existence and asymptotic behavior In this section, we establish the basic existence and uniqueness results for (2.1) and study the long time behavior of the solution to (2.1). We shall first establish local existence and uniqueness in X. For t1 > 0, define X = C([0, t1], X) with norm Heb”? = maXtelo.t1] llpllx. Theorem 2.2.1 Assume that (H1}-(H3) hold and b E X. For each (to E X, there exists t1 > 0 such that (21) has a unique solution in X . Proof. We take the semigroup approach used in [9] to show the existence and uniqueness. As usual, we define the linear operator L on X by L = J at u + b(2:)u. For each at 6 X , we define mapping qu = u where t u = eLt¢(0) + / e“t “ S)f($a¢(5))d3 (2.14) 0 and eLt is the uniformly continuous semigroup on X generated by L because L is bounded on X. Now the equation (2.1) is reduced to (2.14). Since f (1:, -) is locally 21 Lipschitz, with an argument similar to that in [9], one can show that S is a contraction mapping on X for the suitable t1. Therefore, the existence and uniqueness of the solution to (2.1) follows from Banach’s fixed point Theorem. Definition 2.2.2 Let ST 2 H x (0, T) forO < T S 00. A function u E C1([0, 00), X) is said to be a subsolution of (2.1) in ST if nag/Jaimewy+maua+4umi (2w) (1 A supersolution is defined similarly by reversing the inequality. Proposition 2.2.3 Assume (H1), (H2),(H3) and (H4) are satisfied and that b E X. Then (2.1) has a global solution u(-,x,r,/)) for each ll’ 6 X+. Proof. Let ii be the solution to u, = f“ J(:1.‘, y)u(y)dy + b(:17)u(:17) +g(:1:, 0)u u(r. 0) = v. where g(:r, 0) is given by (2.4). Since L + g(2:, 0) is a bounded linear operator on X, we have IneiimuanL+flemMmuuSewdveflmmuu. This indicates that ii is a global solution. Furthermore, ii 6 X ,1 according to the comparison principle, see [33]. Due to (H5) f(;1:,u) = g(:1:, u)u S g(:17,0)n, whenever u 2 0. Thus, ii is supersolution of (2.1). By the comparison principle, 0 S WC. 3 I”) S 5(11. Mb). where u(:r, -, it) is the solution of (2.1) with u(0) = if). the desired conclusion follows immediately. Now, we are ready to give the main result in this section. 22 Theorem 2.2.4 Assume that (H1)-(H4) and (H8)-(H10) are satisfied and that b E X is Lipschitz continuous. Then one of following statement holds. (i) If 5(L + g(:1:,0)) < 0, then the zero solution of (2.1) is globally asymptotically stable in X +. (ii) If 5(L + .(l(:li,0)) = 0, then the solution of (2.1) u(;r:,l.,z/)) with '1/1 E X+\ {0} satisfies tlim u(-,t,i,b) = 0 ac. (iii) If 5(L + g(:r,0)) > 0, then (2.1) admits at most one stationary solution in intX+. If (2.1) has a stationary solution H E intX+, then if is globally asymptotically stable in X+. (iv) if 5(L + g(:1:, 0)) > 0 and (2.1) has no positive stationary solution, then gig; l|u(-. 1, MIX = oo. for 011% e X.\ {0} Proof. (1) Let «,0 be the eigenfunction associated with 5(L + g(:1:, 0)). According to Proposition 2.2.3, 0 S u(x, t, gt) S e(L + 9(I1 Olltcp, where u(r, t, (b) is solution to (2.1) with gt 6 X+\ {0}. Since 5(L + g(:r, 0)) < 0, for some M,(1 > 0 ||8L(L+g($,0))HX g ATE—at (see [31] Theorem 1.3.4). Thus, we find lim,_.oo ||u(-,t,q‘>)]|x = 0. (ii) Because of 5(L + g(x, 0)) = 0, lap is a supersolution of (2.1) if k > 0. By the comparison principle given in [33], we have u(-, t + h, kzp) = u(-,t, u(-, h, kcp)) S u(-,t, kcp) for each h > 0,that is, the function t H u(~, t, lap) is nonincreasing. This ensure that the pointwise limit 12 = £11111 u(-, t, Imp) (2.16) 23 exists. By The monotone convergence theorem and the continuity of f, L1? = f (:1: ii) and 0 S ii S kw. In fact, we have ii = 0 a.e. Otherwise, with J :1: ii >> 0, we may argue as in Theorem 2.1.6 to deduce that E E intX+ which contradicts the fact that zero is the only non-negative steady state of (2.1) which is ensured by Lemma 2.1.10 Let do 6 X +\ {0}. Then there exists k > 0 such that 0 S do << Imp. Again, the comparison principle gives 0 S u(o,t, do) S u(-, t, kcp), for allt 2 0. (ii) follows from this fact together with (2.16). (iii) By Lemma 2.1.11 and Corollary 2.1.13, (2.1) has an arbitrarily small subso- lution 6d and an arbitrarily large supersolution kit, where «p is the strictly positive eigenfunction corresponding to 5(L + g(:r,0)) and 0 < 6 < 1,k > 1. With the rea- soning similar to that for (ii), we find that the function u(cc, -,6 0 and 6 (k) > 0 such that u(-,t, 6(0) 2 6(6) and u(-, t, kit) 2 6(k) for all t Z 0. The uniqueness of the positive stationary solution of (2.1) together with previous argument yields that the pointwise convergence lim u(., t,6o both hold true. Because 6 is continuous, Dini’s Theorem gives lim ||u(-,t,6 0 such that u(o,h‘,d) >> 0 . Moreover, we have that u(~,t,14p) S u(-,t + h‘,d) for some 0 < 1 < 1 and u(-,t,d) S u(',l.fi'ii) for some 7 > 1. Therefore, (iii) follows from (2.17). (iv) Corollary 2.1.13 suggests 5(1. + g(:1:,v)) > 0 24 for each v E intX+ which is Lipschitz. It is well known that L + g(1:, v)! generates a uniformly continuous semigroup eta” +g(:r,v))I on X. Let '1/1‘ E intX+ be an eigenfunction corresponding to 5(L + g(:r, v)I). By the spectral mapping theorem, g'(et(L + g(:1:,v))) : et0(L + g($1v))1) and et(L + 9(131 v)I)d)‘ : et5(L + 9(32 v))w‘, The comparison principle gives ecu. + 9(x, v)1)¢0 2 et(L + g(:1:,v)1)pw. = pets(L + g(:1:,v)1)w., where do 6 intX+ and p is a positive number such that pd“ S do. Since 5(L + g(2:,v)I) > 0 ,1qu lie-”(L + ‘(l($’v)l)¢ollx = oo (2.18) for any do 6 intX+. Note that (2.18) also holds for eta” +g(:1:,v)I),p0 with do 6 X+\ {0} because that e(t + h)(L +g(a:,v)1) >> 0 for some h > 0 (see [33]). In the following, we shall adopt an idea in [23] to complete the proof. Suppose there is wo E X+\ {0} such that HU(-,t.wo)I|x S c < 00 for all t 2 0, where c > 0 is constant. Therefore, we may choose positive constant m > e such that O _<_u('it1w0) S m for all t Z 0. Consequently, Ut(',',1U0) : LU(-,',’LU())+f($,u.(',',ul())) = Ln(-, -, wo) + g(.I.', u(-, -, wo))u(-, -, mo) 2 Lu(-, -, uro) + g(:1:, m)u(-, -, mo) in H x (0,00). This fact and the comparison principle imply that et(L + g(.r, 711)) 11.10 S u(', t,wo) 25 for t 2 0. Thus, (2.18) yields 31210 IIu(°1 tiwolllx : 00' The contradiction completes the proof. 26 CHAPTER 3 Existence and Stability of Coexistence States for a Nonlocal Evolution System In this chapter we study the existence of component-wise positive steady state solu- tions of nonlocal evolutionary problem of the form 3,6 — :1? E [0,h] (31) 52- : dQLU + U['Ym(-T) + 90311)) + C(I, u" v)u] { 9-1: — dlLu + u[/\l(1:) + f(x,u) + F(.r,u,v)v], h . where Lu := / J(:1:,y)u(;1/)(ly — b(:r)u(.-1:) and h > 0. (l,- > 0(i = 1, 2) and A, 7 6 IR, We also assoume (H1) J(-, ) E C([0, h]>< [0, h], IR“) is symmetric and J(1:, y) > 0 for any 1:, y E [0, h]. (112)th mm + 2,1,) — J(.r,y)[dy g 512.] for some 6 > 0. (H3) b, l, m E X+\{0} are Lipschitz. (H4) f (x, w) and g(.r, w) are defined on [0, h] x [0, 00), and 01 continuous in both :12 and w, such that f(r,0) = g(x,0) = 0, 8wf(.7:, ) < 0, 0wg(;1:, ) < 0. 27 (H5) F(x,u, v) and G(x,u,v) are defined on [0, h] x [0, 00)2 and Cl continuous in x and (u, v). The system (3.1) admits three types of componentwise nonnegative solution cou- ples in X x X that are independent of time. Namely, the trivial one (0,0), the semitrivial positive solutions (11,0) and (0,11),where u, v are positive solutions of dlLu + u[Al(x) + f(x, u)] = 0, x E [0, h], (3.2) (62141) + v[7m(x) + g(x, v)] = 0, x E [0, h] (3.3) respectively, and the coexistence states, which are the solution couples (u, v) with both components positive. (0, 0) is always a solution of (3.1). Moreover, (3.1) has a semi- trivial coexistence state of the form (a, 0) (resp.(O, v)) if and only if (3.2) (resp.(3.3)) has a positive solution in X. 3.1 The existence of coexistence states We now consider the one-parameter family of eigenvalue problems (L + AK)u = uu, A E IR, 11 E X (3.4) where the operator L is given by (3.1), b E X is Lipschitz. K E £(X) is the multipli- cation operator by k E X. L(A) = L + AK. we shall denote the principal eigenvalue of L(A) by u(A) , that is, an eigenvalue associated with an eigenfunction u E X +. Lemma 3.1.1 Suppose that k E X is Lipschitz, then the linear operator L(A) has a unique principal eigenvalue equal to 5(L(A)) and the mapping A 1—> 5(L(A)) is analytic. Moreover, it is convex. Proof. The existence and uniqueness of an principal eigenvalue for L(A) is an con- sequence of Lemma 2.1.2 and 2.1.9. Moreover, the principal eigenvalue is equal to 28 s(L(A)) for fixed A. Therefore we only need to show that A 1—> s(L(A)) is analytic. Let A E IR be an arbitrarily fixed number. Note that A H L(A) is analytic from (A — 1, A + 1) into C(X) and 5(L(A)) is a simple eigenvalue of L(A) with eigenvector d(A) E intX+. By the Proposition 4.5.8 in [15] , there exists 6 > 0 and analytic curve {noon/4..) ; .- e (X — (,X + a} c 11 x x such that(7r(0),1/)(0)) = (5(1.(X)),./.(X)), (Les-Ms) = nos-we) and 111(8) = we) + no). where 77(3) E range(L(A) - 5(L(A))I) and 7r(s) is a simple eigenvalue of L(s). For sufliciently small 6, 211(3) E intX+ if |s| < 6. Lemma 2.1.2 implies 5(L(s)) E 7r(s). Since A is arbitrary, the desired result follows. Finally, u is convex with respect to A because u(A) = SUPIIUIIH=1(L + AKu, u) is affine for fixed 11. Proposition 3.1.2 Suppose that k E X is Lipschitz. Let R. = {x E [0,h]] k(:lt) > 0}, F- = [O,h]\I‘+. Then u(A) —-> 00 as A ——> oo ifF+ 75 0 and u(A) —. 00 as A -—r ~00 if IL 75 (ll. Proof. We only give a proof for K. 96 Ill. The case that I‘- 75 0 can be treated similarly. Choose x’ E F+ and 6 > 0 such that (:r’ - 6, .r’ + 6) C I‘+ and min[,1_5,,,r+51 k(x) > c for some positive constant e. Let 6 E Co([0, h]) H X + such that suppfi C (x’ — 6, x’ + 6) and 6 E 1 in (x’ — $611:I + $6). A Straightforward calculation gives ~ ~ h ~ ((L+/\K)0.9)2 — max / J(x.y)dy-llbllLllllwllim+Ac5- x E [0, h] 0 Consequently, u(A) = max ((L + AK)u,u) —+ 00 as A -—> 00. Hull}! =1 Next, we consider the eigenvalue problem (L + A1065 = 0, (3.5) 29 denoting a principal eigenvalue by A(k) and distinguish two cases h h (D) / J(x.y)dy s (#)b(x) (N) / J(x,y)dy s be) 0 0 Remark 3.1.3 By the definition, A(k) is a root of u(-). Lemma 3.1.4 Assume (case D) (i) (3.5) has two principal eigenvalues A1(k) < 0 < A2(k) if 1“,. ¢ 0 and 1‘. ¢ 0. (ii) (3.5) has a unique principal eigenvalue A(k) > 0 if F- = 0 and A(k) < 0 if I} = 0 . Proof. (i) Note that L(0)1 S 0. Lemma 2.1.3 implies [1(0) < 0. On other hand, Proposition 3.1.2 shows that u(M) > 0 and u(—M) > 0 for some M > 0. By the continuity of u, the result follows. (ii) Suppose F- = (b. By (i), We see that (3.5) has a principal eigenvalue A‘(k) > 0. Since k is nonnegative, by Lemma 2.1.4, u(A) is monotone with respect to A. Therefore, uniqueness follows. Lemma 3.1.5 Assume {case N). Suppose 1",. 51$ 0 and F+ 3f 0, then (3.5) has a principal eigenvalue A(k) > 0 if fob k(x)dx < 0; A(k) < 0 if fob k(x) > 0 and 0 is the unique principal eigenvalue if f: k(x) = 0 and k(.1:) 75 0. Proof. Let (u(A),d(A)) be the eigenvalue-eigenfunction pair for (3.4). Differentiating (3.4) with respect to A and noting u(O) = 0 and d(0) = 1, we find Ld'(0) + K1 = #'(0)¢(0). (l where ’ 2: FA" Integrating both side , we have 11 / k(:I:)(l:1r = Ill/(0). 0 30 If fob k(x)d:1: < 0 then u’(0) < 0 and by the convexity of u, u(A) > 0 for any A < 0 . Furthermore, by Proposition 3.1.2, u has a positive zero. The case that foh k(x)dx > 0 follows similarly. Next, we assume that I: k(x) = 0 and [C(55) 74 0. Clearly, u’(0) = 0 and Ld’(0) = —k(x). Taking the derivative of (3.4) twice gives L¢"(0) + K¢'(0) = u"(0)- Consequently, we have h h Wm) = /0 k(x)d’(0)d:r = — ]0 Ld’(0)d’(0)d:1: > o. (3.6) The last inequality is true because supHuHH=1(Lu, u) = 0 and the supreme is attained only by constant. The convexity of u and (3.6) implies p(A) > 0 for any A 75 0. Throughout the remainder of this section, o[dL, K] will stand for the principal eigenvalue of (dL + AK)d = 0 Proposition 3.1.6 Assume (H1)-{115), then dle + w(Al(x) + f(x, w)) 2 0 in [0, h] (3.7) has a unique positive continuous solution defined on [0, h] if and only if A > o[d1 L, Al]. Furthermore, if 6,\ is a positive continuous solution to {3.7), then 6* E int X+ is Lipschitz and the operator L1,, defined by LLA = (11]; + Al(.’1‘)+ QADWII-(JS, 61) + f(LlI, 6A) is invertible. Proof. The existence of a unique positive Lipschitz continuous solution is an consequence of Theorem 2.1.6. To prove the other part, we observe that 5(d1L + (Al(x) + f(x,0A)I)) = O and GADwf(x,0A) < 0, which implies 5(L1,,\) < 0. Then Lemma 2.1.3 gives the desired conclusion. 31 Theorem 3.1.7 Assume (H1)-(H5). Let 9,1 E X+ be the solution to (3 7). Consider 7A 2: 0[d2L, 7771 + C(IE, 9A: 0)9,\]. Then a continuum (3+ C IR x intX+ x intX+ of coexistence states of (3.1} emanates from the point (71, 91, 0). Proof. We fix A E IR and treat '7 as the bifurcation parameter. We first show that (71, 9x. 0) is a bifurcation point to a branch of coexistence states. To do that, we use the theorem by Crandall and Rabinowitz in [28]. Consider the Operator .7: : leXxX—>Xdeefinedby dlLu + u[Al(x) + f(x, u) + F(x, u, v)v] f(y, u, v) =2 dgL'U + v['7m (x) + g(:r, u) + C(x, u, 1.1)11] f is an operator of class C2 in (u,v) and analytic in 'y and the zeros of .7: are the solutions to (3.1). Since 91 solves (3.1), f(7,6,\,0) = 0 for all 7 E IR and hence (7, 61, 0) can be regarded as the known branch of solution to (3.1). The linearization of f at (‘11, 6A, 0) with respect to (u, v) is given by , [41,1 FLT. 91.0)91 17(7) 3: D(u,v)f(716A10) = 1 O (12L + 7111(1) + g(:r:, 91) + C(x, 01, 0)6A where L1,), is the operator defined in Proposition 3.1.6. Let go >> 0 denote the prin- cipal eigenfunction of d2L + 7rn(x) + g(;1:,6,\) + C(x, (A, 0)()A,which is unique up to multiplicative constants. Then the null space of JO“) is Nero.» = span {(Lii(F(I.91.0)91r). - 0 and a curve 3 -—r (7(s),u(s), 11(5)), [8] < E, of class C'1 such that (7(5). 11(8). 11(8)) = (71 + 0(8).91 + 0(3), 3,0 + C(82)). 8 r 0. 33 and f(“/(8).U(8),v(3)) = 0, Is] < a. As (91%?) E intX+ x intX+, (u(s),v(s)) is component-wise positive if s > 0 is sufficiently small and hence (3.1) has a coexistence state for some 7 near 7,. 3.2 Two similar competing species In this section, we consider a special case of (3.1) where u and v stand for the densities of two competing species. We assume that two species are very similar. Namely, in what follows, we shall focus our attentions on the system (“)11 gt)— — uLNu + u(o(x) + 76(23) — u — v), in [0, h] (3.11) St = uLNv + v(a(x) — u - v) where LNw 2: foh J(x, y)w(y)dy—f0h J(x, y)dyw, a, 6 E X are Lipschitz in [0, h], and fob a(x)dx > 0. Assume that 7' is positive but very small. Clearly, when r = 0, u and v play identical roles in this system. For each fixed u, there is a set of nonnegative equilibria {(317, (1 — s)i.7)|0 S s S 1}. Here 6 is the unique positive continuous solution to the equation uLNv + v(a(x) -— v) = 0. (3.12) 17 depends smoothly on u. (see the appendix for more details) Following the approach employed in [34], we let M : (0,00) —1 IR be defined by h M(/i) :2/ fi(;zr)i7(x,/i)dx, n > 0, (3.13) 0 For r z 0, we look for triples (u, v, u) that satisfy 34 iuwu+uovi+u%v-u-”*=Q inmii (in) anvv + v(01(33) - U - v) = 0 and that are close to the curve Ffi x {)7} for some [I > 0, where r¢={eaiiu-sncfim03331r Note that for any n, 1“,, x {u} is a curve of solutions of (3.14) for r = 0. Also, for any small r, (3.14) has the semitrivial solutions (0, fi(-, u), u) (independent of r) (17(1/1'1 T): 01 “)7 where 17(-, u, T) is the positive solution of (3.12) with a(x) being replaced by 7(x) = a(x) + rfi(x). For our purpose, we introduce the following function space. Y=X> 0. Then for any ii there exists a neighbor- hood U of the curve FIE x {II} in Y x (0,00) and 6 > 0 with the following properties (i) If M (fl) 75 0, then for r E (0,6) there are no solutions of (3.14) in U other than the semitrivial solutions of (3.14). (ii) If MOI) = 0 and M’(fi) 75 0, then for T E (0,6) the set of solutions of (3.14) in U consists of the semitrivial solutions and of the set 2 D U, where Z is a smooth curve given by 35 2 = {((u(r,s),v(r,s),p(r,s)) : —6 S s S 1+ 6}. (3.15) Here (7', s) H (u(r, s), u(r, s)) E Y and (r, s) H u(r, s) E (0,00) are smooth functions on [0, 6) x (—6, 1 + 6) satisfying the following relations (u(r,0),v(r,0)) = (0,fi(-,u(r,0))), (3.16) (u(r,1),v(r,1)) = (fi(-,u(r,1),r),0), (3.17) (”(013)1v(018)1#(013)) : (35('1fi)1(1 - S)fi(°1fi)1fi) (318) Remark 3.2.2 In other words, a branch of coexistence states bifurcates from the branch of semitrivial equilibria (8,0) at p = u(r, 1) and meets the other branch of semitrivial equilibria (0, ii) at u = u(r, 0). For 1' = 0 the branch coincides with 1‘3. Note that from (3.18), it follows that the function u(r, s), v(r,s) and u(r, s) have the following expansion for —6 S s S 1 + 6 and [r] << 1 : u(r, s) 2 s6(-, II) + 7111(3) + 0(72), (3.19) v(r, s) = (1 — s)17(-, II) + rv1(s) + 0(r2), (3.20) ,u.(r, s) = fl + 7121(3) + 0(9), (3.21) where (111, vl) E Y and II E IR are smooth function of s. Proof. We look for any triple (u,v,u) near PI]. x {)7} such that (u,v) can be written as 36 (111v) : (83(3/‘01 (1 _ 8)8(1f‘l')) + (w) 2), (3'22) where s E IR and (w, z) E X2, and they are in or near [0, 1] and {(0, 0)}, respectively. Rewrite (3.22) as ('11.,v — 17(-, [1)) = .9(1I(°,[1.),—17(',[l)) + (111,2). We shall find 3 and (w, z) from (111,2) = Q(1l)(U,’U—6(°,/J)), 86611). 47611)) = (I - Q(u))(u. v - 5611)). where I is the identity on Y and Q(u) is the projection of Y onto X2 along the subspace X101) I: Span {(8(1H')1 —1)\(,ll))} ' (Note that X1(/i) is a complement of X2 in Y for u z ii.) In particular, we find the following values of s and (w, z) for semitrivial equilibria: (075611)) = (0,5610) + (0.0). (3-23) (80,11,710) : (38(‘1fila (1 — 5)6(= fill + ("(7-1”): C(vafllv WltIlS = “(Ta/1): (3'24) where (r), C) and o are smooth functions of (r, /1.) taking values in X2 and IR, respec- tively. Clearly, 0(0:u) = 1.» (77(0. xtl.C(01#)) = (0.0): (3-25) since fi(-,,u,0) = v(-,u). For a small 6 > 0 let F be the map on Y x (—6,6) x (—6, 1 + 6) x ([2 — 6.;2 + 6) defined by 37 F(w, z,r,s,u) = I‘LNU’ - (w + $30.11) + (01 — 17(3 It))1/ - (10+ 2)?!) + 7/3830. It) + T/tw uLNz - (w + z)(1—— s)v‘(.,p) + (a -— fi(-,u))z — (w + z)z It is clear that F is well defined and smooth (in fact polynomial) as a Y-valued map. By plugging (3.22) into (3.14), we see that, in order to finds solution to (3.14), we need to solve the equation F(w, z, r, s,u) = 0, (3.26) with (w, z) E X2. By examining the properties of F(w, z, r, s,u) for (w, z) E X2. we find F(O, 0,0, s,u) E 0 (s E (—6, 1 + 6),,u E (,1? — 6,fi+ 6)), (3.27) F(O, O, r, 0,;1) E 0 (r E (—6,6),,u E (ii — 6,fi + 6)), (3.28) F(72(T.#),C(T.#),Ta 00.10.70 5 0 (T E (-5. (5)111 E (I7 - 547+ 5»- (3-29) Define E(s,p) :2 D(w, z)F(0,0,s,,u) E £(Y, Y), that is, uL — 36 + a — i7 —sfi E(s, ,1) = N ( ) . (3.30) —(l — s)i7 ”LN — (l — s)ii+ ((1 — ii) We may rewrite (3.30) as Eta/t) = K01) - V6311). where J :1: u 0 mom)?” 2: ” 0 uJ*v 38 and foh J(x, y)dy + 36 + (i7 — a) 517 V(s,;i) 2: (1 - 3).? f0“ J(x, y)dy + (1 — 5).? + (.7 — 61) Since 6 E intX+ is a solution of (3.12), The fact that J =1: 15 = (th J(x, y)dy + ii- (1)6 immediately implies foh J (x, y)dy + if — a > 0 for any x E [0, h]. Therefore, V(s, p.) is an invertible operator on Y and it is a standard consequence of the compactness of K (u) that E (s,u) is a Fredholm operator of index zero. By the definition of F, the vector (fi(-,u), —1’7(-,u))T is in the kernel of E(s,/,1), or equivalently it is an eigenfunction corresponding to the eigenvalue 0 of E (s, u). By the positivity of ’6, zero must be a simple eigenvalue of E(s, u) and KerE(8.u) = Span {(3611). -5(-.u))T} = X101)- Indeed, let (a, E) be an eigenfunction associated with 0. Plugging it into the equation E(s, )1)(u, v)T = 0, we find uLN(U + v) + (a — 2am +6) = o. By virtue of Lemma 2.1.4, s(,uLN + (a — 2fi)I) < 0. Consequently, uLN + (a — 26)I is invertible. It follows that 6+6 :— 0 and (uLN+a—fi)fi = 0. By Lemma 2.1.2, one sees that u = of? for some constant c. As will become clear late, ker E (s, )1) (I) R(E(s, )1.)) = Y. This will confirm that zero is a simple eigenvalue. Now, let P(s, p) be the continuous linear projection of Y onto X 1(11) along the range of E(s, )1) (the range R(E (s, [1)) is a closed subspace of Y of codimension one). P(s, u) can be explicitly written as follows: ‘ hfi- iw—s hfiz P(s,11)(w.z)T= (1 ”fofhgz)” f" (r(-.u),—v(-,u))T. (3.31) 39 In fact, we have R(P(s.u)) = X101). (14841))2 = P641). and P(8.1L)E(s.1u) = 0. These properties can be verified by a straightforward computation. Formula (3.31) in particular implies Y2(s.#) == R(E(Sa#)) = [(10.2) E Y: (1- «(fl/oh 5(-.#)w - 8 Aha-.102 = 0]- Also note that (s,u) H P(s,;i) is smooth (in the operator norm). Following the Lyapunov—Schmidt scenario, we now consider the system P(s,u)F(w,z,r,s,u) = 0, (3.32) (1*P(8.#))F(w.2.r,8.u) = 0. (333) where (w, z) E X2 and I is the identity on Y. If u is sufficiently close to II (and we make 6 small enough for that to hold for all u E (II — 6, )3 + 6), then kerE(s,[1) (I X2 = {0}. It follows that E(s,)1) is an isomorphism of X 2 onto Y2(s, )1). By the implicit function theorem, we can thus solve (3.32), which leads to the following conclusion. There exists 61 > 0, a neighborhood U of (0, 0) E X2, and a smooth function (7.131“) H (w(7131#)12(7,31#)) : (_61161) X (—6111 + 61) X (I: _ 611;; + 61) _* X2 such that (u.1(0,s,)1),z(0,s,/i))= (0,0) and (w, Z,T,S,/l) E U x (—-61,61) x (—61,1+ 61) x (,6 — 61,)? + 61) satisfies (3.26) if and only if (w, z) = (w(r,s,u), z(r,s,,u)) and (r, s, u) solves the bifurcation equation P(s,u)F(w(r, s,u), z(r,s,1i)),r, 3,;1) = 0. 40 By (3.28) and (3.29), w and z satisfy (w(r, 0,11), 2(r, 0,u)) = (0, 0), (3.34) (W(T,0(T.H).u). 26.06.110.11)» = (77(TaulaC(T:#))- (335) Now, defining ((r, s, u) by €(T381#)(a(.1#)1 —6('1H))T = P(31#)F(w(7131H)1Z(T137l~f))17131l‘)1 the bifurcation equation is equivalent to {(r, s,u) = 0. (3.36) We immediately have the following solution of (3.25): {(0,841) E 5(7) 0.11) 566.06.70.11) 2 0. (337) These identities hold because for each of the indicated values of (1', s, u) E (—61, 61) x (—61,1+61)x(fi—61,fi+61), there is a solution (w, z) E U of (3.26); see (3.27)-(3.29). Recall that from these solutions, the triples (r, 0,;1) and (r, o(r, n), )1) correspond to semitrivial equilibria of (4.10); see (3.23),(3.24) and (3.35). It follows from (3.37) that €(7’, 51H) : .S(0’(T, ff) — 3)T61(T131f‘) for some smooth function {1(r,s,u). Solutions of (3.36) different from (3.37) are found by solving 61(7. 8. it) = 0. (3.38) Observe that 8T5“)? 31/1) E8(1— 3)€l(0151#)a 41 as 0(0, p) E 1. The derivative 011 the left-hand side is computed from (601“):—fi('1/‘))87'§(0132l1) : 67(P(81H)F(w(7-1 8,11),Z(T, 81/5))1T131f‘))IT :0 = P(s, u)FT(0, 0,0,3, )1) + P(s, u)E(s, [1.)(1U7', zT) P(si LOFT“), 010131 H) (recall that R(E(s, u)) = ker P(s,u)). Using (3.31), we find s(1 - s) foh 662 h *2 IO” P(S,/1)F7-(O, 0101 8,11) : P(31#)(S:88(1#)10)T = (8(1#)1_a('1l1))T' Thus ('9 __ s(1 — s)M(u) 1 TE“), 8,11) _- foh 6‘2 , i.e., AI A (1(0, .S‘,[l)[# = I; = —fl—— (3.39) item) To complete the proof, consider first the case M (ii) 79 0. Making 61 smaller, if necessary, we infer from (3.39) that (3.38) has no solution in (—61,61) x (—61, 1 + 61) x (II — 61,17 + 61). This implies statement (1) of Proposition 3.2.1. Now assume M (II) = 0. Then a I M’(fi) 0, s, l = A 2 h A ' #51( ll) / I f0 52“,”) If M ’(fi) 7S 0, the implicit function theorem implies that for some 62 > 0, all solutions of (3.38) in (—61,61) x (—61, 1 + 61) x (i2 — 61,ZZ+ 61) are given by u : m(r,s), T E (—52.52), 3 E (‘62211' 62), where m(T, s) is a smooth function satisfying m(0, s) _=. II. Thus, in addition to the solutions given by (3.37), the bifurcation equation (3.36) has the family of solutions {(7131m(7—19)) : T E (_62162la S E (—6211+ 62)} ' (340) In this family, the point (r, 0, m(r, 0)) is also contained in the set of solutions found in (3.37), and it corresponds to the semitrivial solution (0,6(-,p)) of (4.10) with 42 u = m(r, 0). Next we look for points (7, s, m(r, s)) in the family corresponding to the semitrivial equilibria (ii(-, p, r), 0). We see that s and r are found from the equation 3 = 0(7', m(r, s)). (3.41) Since 0(0,/1) E 1, for r z 0 there is a unique solution s = .‘s-(r) of (3.41), and it depends smoothly on r. Hence for each fixed r as 0, (r,‘s’(r),m(r, 3(1)» is a point contained in the family (3.40) which corresponds to the semitrivial solution (ii(-,;i,r),0) of (4.10) with [l = m(r,.‘s’(r)). Using the scaled variable 3' = 53(7), we define “(Tel = 337(wm(7.§l)+w(T.3.m(T,§)). v(T»S) = (1 - 318(-.m(r.3)>+ 2(713,m(r.3‘)). u(r, s) = m(r, 3). These are smooth functions of (r, s) E (—6, 6) x (-6, 1 +6) if 6 is sufficiently small, and (u(r, s), 12(r, s)) is a solution of (3.14) for u = m(r, 3). These solutions together with the semitrivial equilibria contain all solutions of (3.14) in a small neighborhood of I}, x {17} for r E (—6, 6). The relations (w(0,s,p),z(0,s,u)) = (0,0), m(0,s) = ii, and 's‘(0) = 1 imply (3.18). The correspondences between the solution (7', s(r), m(r, s(r))), (r, 0, u(O, r)) of (3.36) and the semitrivial equilibria, as discussed above, imply (3.16), (3.17). This completes the proof. Next we consider the stability of coexistence states on the curve 2 given in Propo- sition 3.2.1. A crucial step is to analyze the corresponding linear eigenvalue problem [iLNcp + ((1 + r73 — 2n - v)(0 + (—u)u’) = Acp uLNd + (—v):p + (a — u — 2v)d = Ad where (u,v) is the coexistence state of (4.10). When r = 0, we have (u,v) = in [0, h], (3.42) (36, (1 — s)i7) and (3.42) has an eigenvalue A = 0, the corresponding eigenfunction being (6, —fi). Let N(r, 3,;1) be the linear, bounded operator defined by 43 N(7' s “X90 d)T -: uLNcp + (a + Tl? — 2v — v)Ip ad I , I up uLNi/i + (a — u — 2v)i/2 , it is easy to see that 0 is an eigenvalue of N(0, 5,11) with corresponding eigenfunction being (6,6). Moreover, 0 is (algebraically) simple and equal to s(N(0,s,u)). By spectral perturbation theory, for [7] << 1, N (r, s, )1) has a simple eigenvalue, denoted by A(1', 3), such that lim,_,o A(r, s) = 0, and A(7', 3) corresponds to an eigenfunction in the interior of 17+. Fhrthermore, we can show A(r, s) = 5(N(r, s, [1)) Hence the sign of A(r, 3) determines the stability of coexistence states 011 E. (For more details, see Appendix). Note that A(r, 0) = A(7', 1) = 0 (3.43) for u(r,0) and u(r, 1) are bifurcation points (points of intersections of )3 with the branch of semitrivial solutions). By straightforward but tedious computation, we can obtain a formula for A(r, s) (the proof is given in the appendix). For the formulation we introduce some nota— tion. Take the linear subspace spanned by 6 to be 6) and let Oibe its orthogonal complement in H. We have the self-adjoint operators defined on H .C—ll = fiLN+(Y-26. Notice that .C is a Fredholm operators of index zero 011 H and 0 = SUP||w||H=1(£w1 w) is the principal eigenvalue of [I (the eigenfunction is 6). By Lemma 2.3 and 2.4, [I — 6 has the bound inverse (L — 6)”lon H. Now we define £‘lon Oi by setting .C'ld = d if and only £1!) = d and d, d E Oi. Then we have the following results about A(r, .3). Proposition 3.2.3 For 0 S s S land 0 < 7' << 1, the following statement hold: 44 (i) 2 foh ‘32 where C (r, s) is some constant uniformly bounded for s E [0, 1] and r << 1. h A(T, s) = s(s — 1)12{ ]0 136[((£ — 6)'1— £'1]f’36+ C(T, s)r}, (ii) With 61 as in (3.10), one has 1 h m.) = m [0 /i6[2s(£ — (s))-1W) + (1 — age-1(a)]. Note that, by our assumption, A107) = f” 662 = 0, so that £‘1(fi6) is well defined. Proof. see the appendix. Lemma 3.2.4 The following holds for any nontrivial (,0 E 9*: h / ,o[(£ — 17)“ - £“],: > 0. (3.44) 0 Proof. For t > 0 let h W) =/ 99(13 - “6‘30- 0 Due to the fact that [I — t6 is invertible for t > 0, h is well defined. We claim that h is strictly increasing. To show this, we set (I) = (l: — r6)‘1(p . With resolvent , it is easy to see that a 70% = (c —t6)'1(6<1>). (3.45) Then dh h 64> " M1- 3,- — fora—fowfi-ti) (W) h = f ((L — 161-161(64) h = /62>0. 0 The last inequality is strict since # 0. In the following we Show that h limt _, 0+ h(t) = / (all—Id (3.46) o for every (,0 E Oi, from which (3.44) follows. To prove (3.46), let Su := 6u for u E H. We always assume that r > 0 is small and C,- are strictly positive constants independent of 7. Then we have IT’Si C 9* and “fl” S Cl. (3.47) (-£g.g) 2 Czllgllii. ((5 E 9L. (348) IISII S Ga. (3.49) (Sin/2) 2 Cillwll’i. d e H. (3.50) Consider the equation (C - 69%)? = g. .(1 6 9L. (3-51) We first show that ”(NIH S CsIIQIIH- (3-52) To prove this, set (,9 = (:6 + d, where c E IR and d E 96. Substituting in (3.51), we have 132/; — tcS6 — tSd = g. (3.53) Take the inner product with 6 and use (3.49) and (3.50) (sum/(86,6): s c.1141)” (3.54) ICI = 46 Take the inner product of (3.53) with ——d and use (3.48), (3.49) and (3.54), we get C2||wllii s (-£(/2.w)=-(g,w)-tC(SE¢)-t(Sv,¢) (3.55) IIQIIHIWJIIH + “Cs +1lcslll/Jlli1» (3-56) l/\ which implies that Hill!” S C7II9IIH (3-57) if t is small enough. Estimate (3.54) and (3.57) prove (3.52). For (p given by (3.51), we set _1 -1 (S£_lg,;f)A r 6 — t(,c [I g + (56,6) v). (3.58) It is easy to see that (L — TS)6 = h (3.59) where S£“lq 6) h=S—1——————( " S". L 9 (sat) ” Clearly, ||h|[H S CSIIQIIH- Applying (3.52) to (3.59), we also find ”6“” S CSIIhIIH S CQIIQIIH- Finally, for any 9 E Of, from (3.58) h(T) = (9.90) = (9.519) + “9.0). Since [IQHH is bounded, we find that lim,_,o+ h(t) = (g, D’lg). This proves Lemma. Theorem 3.2.5 Assume foh a(:1r)(l.1: > 0. For any )7 > 0, the following statements hold true: 47 (i) If MOI) ¢ 0, then there exists 6 > 0 such that for )1 E (Ii — 6,,6 + 6) and r E (0,6) problem (4.10) has no coexistence states. (ii) If M (II) = 0 and M’ (II) 75 0, then for any sufficiently small 6 > 0, there exists 6 = 6(6) > 0 with the following property. For every 7' E (0, 6), there exist E < fl with EJI E ()7 - 6, 17+ 6) such that for any )1 E [ii —- 6,;’Z+ 6], (4.10) has a coexistence state if 11 E (E1 H); moreover, any coexistence state, if it exists, is asymptotically stable. Proof. We have p(r, s) = fl + 7771(3) + 0(7'2). By Proposition 3.2.3 and lemma 3.2.4, for )1 z fl and r z 0,/1(r, -) is strictly monotone. It follows that the first statement of theorem hold with g = inf u(m). H = sup u(m) SE[0,1] 36[0,1] and that the coexistence state on the branch 2 is unique for each fixed 11 E (g, '11). By Proposition 3.2.3 and Lemma 3.2.4, A(7', 5) > 0 for small 7', and thus the coexistence state is stable. 3.3 Appendix Proof of Proposition 3.2.3. The proof of proposition 3.2.3 is given here after some computational results, throughout, (u, v) will be a coexistence state of (3.14), u1,vl, fi are given by (3.19), (3.20) and (3.21) respectively, and (C, c) is the solution of (3.42). Throughout the appendix, we assume that 111(6) 2 0, where M is given by (3.13). Note that C = 6, q = —6 for r =0. Lemma 3.3.1 The following statements hold ( 1‘) h / fiuv = 0. (3.60) 0 48 (W h h 1(T,s)-_—T /0 B(Cv+(u)//0 (Cv+(u). (3.61) Proof. (i) Multiply (3.14a) by v, (3.14b) by u and subtracting, we obtain u(LNuv — LNvu) + Tfiuv = 0. (3.62) The result follows from integrating (3.62) over (0, h). (ii) Multiplying (3.42a) by v, (3.42b) by u and subtracting, we find A(r, s)(Cv —— (u) = v[11LNC + (a + T6 —- u - v)] — u[uLN< + (a — u — u)]. (3.63) Integrating (3.63) over (0, h) and using (3.14) we deduce that A(T,s)/0 (Cv-qu) = f0{([uLNv+(a+r6—u—v)]—§[,11LNv+(a—u—v)]} h = r/ 6(Cv + Cu). 0 Set A = (c - imam). (3.64) B = (L — 6)‘1(,B6), (3.65) C = £'1({36), (3.66) and expand the eigenvalue C, c in the form C = 1.7+ TCI('15) + 726201735)! (367) 49 g : -6+ TC1(-, s) + 72C2(°, r, s). (3.68) Lemma 3.3.2 For some 7,- E R, we have m 111 = —s[A + 38 +(1 — s)C] + 716, (3.69) v1 = —(1-— s)[A + 33 — sC] — 716, (3.70) 00 C1 = —-A — 253 + (23 —1)C' + 726, (3.71) (1 = A + (23 — 1)B — (25 — 1)C — 7212.03.72) Proof. (i) From direct calculation, the following hold in [0, h] uLNul + ((3 — 6)ul + s6(,L'3 — u) — v1) + sflLN6 = 0, (3.73) uLNv1+(/3 — 6)vl + (1 — s)6(—u1 — vl)+(1— s)[ZLN6 = 0. (3.74) Multiplying (3.73),(3.74) by (1 — s) and s respectively, and subtracting, we find that £[(1 - s)". — 5m] + ..(1 -s)/317 = 0. from which it follows on taking the inverse and using definition (3.66) that (1 — s)u1 — svl = —s(1 — s)C' + 736. (3.75) Adding (3.73) and (3.74), we have in a similar manner 50 111+ v1 = —A — 38. (3.76) Then (3.69) and (3.70) follows from (3.75) and (3.76) by straightforward manipula- tion. (ii) Since A,(0, s) = 0, it is easy to check that C1 and 81 satisfy the following 1341— s6(C1+ q) + fllLN8+ 6([3 — u, — v)) = 0, (3.77) £C1_(1‘- S)U(Cl + C1) — filLNll‘i" U(U1+ ‘01) = 0. (3.78) Adding (3.77) and (3.78), by an argument similar to that used previously, we find (1 + <1 = —B. (3.79) By (3.76), 111 + v1 = —A — 38. Substituting this and (3.79) in (3.77), we obtain the equation which determines C1 up to an additive term 746. Using definition (3.64), (3.65) and (3.66), it is easy to see that Cl given by (3.71) satisfies that equation, which verifies (3.71). this and (3.79) yield (3.72). Lemma 3.3.3 The following holds: h / ,B6[2A + 233 + (1 — 2s)C] = 0. (3.80) 0 Proof. By (3.19)—(3.21) and (3.60), h h 0 2/ Bur = T/ (36].va + (1 — s)ul] + 0(7'2) 0 o . h ’A smce f0 (31.12 = 0. Therefore I: / fiflsm +(1 — s)ul] = 0. (3,81) 0 51 The result follows from (3.81) together with (3.69)—(3.70). Proof of Proposition 3.2.3 In the following, C,(r,s) denote quantities that are uniformly bounded for s E [0,1] and small 1'. Piom (3.67) and (3.68), h h / (Cv —— ('11) = / 62 + TC, (T, s). (3.82) 0 0 We now use (3.69)-(3.72) to obtain )1 h h / [3(Cv + g‘u) = (1 - 2.5)] [16+ T/ B6[vl — 111 + (1 —- s)C] +sg1] + 7202(r, s) o o 0 h = 7/ [36[(4s — 2)A + (632 — 4.5)]? + (6s — 1 — 6.92)(,‘] + 7203(7, s). 0 From this and (3.80), h h / B(Cv + cu) = 2s(1— s)r/ 66(0 — B) + 7204033). (3.83) 0 0 As a consequence of (3.61), (3.82) and (3.83), we have 2s(1 — s) [0" new — C) foh '72 Since A(r,0) = A(r, 1) E 0 for all 1 (see (3.43)), we have 05(0,r) = 05(1,r) E 0. A(r, s) = 7’2[ + TC5(7', 3)]. (3.84) This implies that we can write (.75 as (15(7, s) : s(1 — s)( 15(r, s). This proves part (i) of Proposition 3.2.3. Part (ii) follows directly from Lemma 3.3.3 and the relation A = 171(13 - 6)‘1(LN6) = ‘21er The latter equality is obtained by differentiating the equation (3.12) for 6 with respect to 11. In the remainder of the appendix, we simplify the notion by writing N(T) = N(T,s,u), A(r,s) = A(r). 52 Lemma 3.3.4 N(r) is a resolvent positive operator on Y and 5(N(r)) = A(r). Proof. We first prove that N (r) is a resolvent positive operator on Y. Choose 17 E p(N(r)) such that h > max Jx, (1+4 max ox +61: . 77 {093,0 (1)1 quuml I()l} It is sufficient to show that (N(r) — AI)(w, z)T S 0 implies w 2 0 and z 2 0 if A 2 7). Let w+ = max(0,w) and w“ = max(O, -w). 2+,z" are defined similarly. Then we have uLNww- + (a + TB — 2u — v)ww_ + uzw‘ — Aww' S 0 (3.85) and [lLNZZ_ + ((1 — 11 — 2v)zz" + vurz‘ — Azz" S 0 (3.86) Integrating (3.85) and (3.86), we find —(uLNw_,w_)H—((a+r—2u—v)w',w')H-(uz‘,w‘)H+(Aw",w—)H S 0 (3.87) and ——(/1LNz’, 2')” — ((01 — 2n — v)::_, 2‘)” — (vur’, 2')” + (A2”, 2")” S 0 (3.88) Adding (3.87) to (3.88) and Multiplying the both side of resulting inequality by —1, we have (uLN(w- + z‘), (w' + z"))H + ((01 + 7/3 — 2u — v)w", w’)H +(((1 — 2u — v)z“, 2’)” + ((u + v)(w— + z‘), (w- + z-))H —-A((w' +2‘),(w— +z—))H 20 With the positivity of u and v, we get (uL~(w' + 2‘), (w’ + 2’))H + 2((lorl + lfll + u +v)(w' + Z‘). (w" + 2’))H — /\(('w' + z"), (w- + 27))” 2 0 and so h Osmaaxsh 0 J(Sl?,y)dy+40SmaaxS h(|a(r)|+|fl(:c)|—A}((w +2: ),(w +2 ))H20, This would lead to a contradiction if 20‘ + z‘ 76 0 on (2. Thus we have w” E 0 and z“ E 0 on Q and hence w 2 0 and z 2 0 in Q by the continuity of w and 2. We observe that N(T) = KM) + W(T) where i.] * w 0 K(/1.)('m, z)T :2 I 0 pJ*z and h a + 713 — 2n —‘1) — / J(z1:,y)dy u W(T) = 0 h v a—u-2v—/ J(zr,y)dy 0 When T = 0, as we proved in Proposition 3.2.1, W(0) is invertible and N (0) is a Fred— holm operator of index zero. Moreover, O is an eigenvalue of N (0) with corresponding eigenvalue (53,8). By the spectral perturbation theory, for |T| << 1, N (7') has a simple eigenvalue, denoted by /\(T) such that lim MT) = 0 T —-) and lim (90(7), (fi(‘r))T 2 (ii, 8)T in Y (3.89) T—>0 54 where (cp(T),¢(T)) is an eigenfunction associated with /\(T). Furthermore, due to (3.19)-(3.21), we find h h lim0[(a+Tfi—u—v—‘/o‘ J(:1:,y)dy)-—)\(T)]=0r—i7—f0 J(x,y)dy<0 T —) and so W(T) — wl is invertible for any w 2 A(T) if T is sufficiently small. On the other hand, with the same reasoning, we see that W(T) is also a resolvent positive operator and hence 5(W(T)) 6 0'(W(T)). Consequently, we obtain that 5(W(T)) < /\(T) S S(fl/(T)) provided T is sufficiently small. Since K01) is a positive and compact operator on Y, Theorem 4.7 in [45] shows that s(N(T)) is an eigenvalue of N (T) associated with positive eigenfunction of N (T). Next we show that Suppose this is not true. Let (@(T), \Il(T)) E intX+ x intX+ be an eigenfunction corresponding to s(N(T)). For any t E R", define got = (p(T) - t(T)) and ¢t = (¢(T) — t\II(T)). By virtue of (3.89), for sufficiently small T, ¢(T) >> 0, d)(T) >> 0 and then there exists 2 such that 90; >> 0, (15,, >> 0 for t < i, so; 2 0, (I)? Z 0 and either (p; or as; has a zero in [0, h]. We may assume without loss of generality that there exists :51 E [0, h] such that $2031) 2 0. Then (Tamera + (a + w — 2n -v)(s02)(-'r1) + mam) — s(N(T))¢z(-'Ir1) s o and so h u [0 ,,(,,,, yMO/My s o. (3.90) Since ’1th J(x,y)go;(y)dy Z 0, (3.90) implies «p; E 0 and hence a; E O, which is impossible. Therefore, A(T) = s( N (T)) and the proof is completed. 55 Proposition 3.3.5 Assume that 01(33) is Lipschitz continuous in [0, h] with I: a(x)da: > 0 and u > 0. Then for each u uLNu + u(a(:1:) — u) = 0 (3.91) has a unique solution in intX+ which continuously depends upon u. Proof. According to Lemma 3.1.5, 5(uLN + 01]) > 0. Notice that 012;); |a(:t:)| + 1 is a supsolution to (3.91). Thus it follows from Theorem 2.1.12 that (3.91) has a unique solution w(u) in intX+ for each u > 0. Furthermore, in light of the proof of Theorem 2.1.6, it is evident that w(u) is also Lipschitz continuous in [0, h]. To show the continuous dependence of w upon u, we define G : R x X —+ X by C(u, u) = uLNu + u(a(:z:) — u). Clearly, G is C l in both u and u and each zero of G is a solution to (3.91). Now let fl be an arbitrarily fixed positive number. A linearization of G with respect to )1 leads to the operator A : X ——> X Au 2: flLNu + u(a — 2u). By Lemma 2.1.3 , s(A) < 0. Consequently, A is invertible and so there exist a pair (u, u(u)) in a small neighborhood U of (fi,w(fl)) such that G(u,u(u)) = 0 and 3133; ||u(u) — w(fi)|| = 0 in term of the implicit function theorem. Here U C (fl - 6, [7 + 6) x X for some 6 > 0. By the positivity of w(fi), u(u) is also positive if u is sufficiently close to ,3. Then the uniqueness of w(u) implies w(u) E u(u). Thus the proof is completed. 56 CHAPTER 4 The Cauchy Problem for a Nonlocal Phase-Field Equation Consider the following problem u: = [a J(sc — y>u(y)dy — fn J(z — y)dyutr) — f(u) + w, (6 + lu)t = A0 in (0, T) x Q, with initial and boundary conditions u(0,x) : u0(2:), 6(0,1:) 2 (90(1), 36 film — 0) (4.3) (4.4) where T > 0, Q C IR” is a bounded domain. Here 6 represents temperature,u is an order parameter, 1 is a latent heat coefficient, the interaction kernel satisfies J (—:1:) = J (2:), and f is bistable. 4.1 Existence and uniqueness In order to prove the existence, we make the following assumptions (A1) M E supfQ |J(:r — y)|dy < 00 and f E C(R). 57 (A2) There exist c1 > 0, c2 > 0, c3 > 0, c4 > 0 and r > 2 such that f(u)u Z CII’UIr —' CQIUI, and |f(’U.)| S C3IUIr-l + C4. Note that (A2) implies F(u) = [Du f(s)ds 2 c5|ulr — c6|u| (4.5) for some positive constants c5 and C6- We prove the existence of a solution to (4.1)-(4.4) by the method of successive approximation. Define 6(°)(t, x) := 60(x) and for k 2 1 (u(k), 9W) iteratively to be solutions to the system at” = / J(r—wu‘kiywy— / J(x—y>dyu<*>(x)—f(u<*‘>)+w<*-l>, (4.6) (I n of“) — A6“) + 6““) = 424’“) + 6”“) (4.7) in (0, T) x Q, with initial and boundary conditions um(0.x) = 210(1), 9(“(0419 = 90(1), (4-8) 390:) (in Ian = 0- (49) Lemma 4.1.1 With h = 1, for anyT > 0, ifuo E L°°(Q), and 60 E HlflL°°(Q), then there exists a unique solution (u,0) to system (4.6) -(4.9). Furthermore, 71.“), u?) E L°°((0, T), L°°(Q)) and 0“) e L°°((0,T), mm» m L2((0,T), 112(9)). Proof. Since the right hand side of equation (4.6) is locally Lipschitz continuous in L°°((0,T), L°°(Q)), local existence follows from standard ODE theory. In order to prove the global existence, we prove global boundedness of the solutions. For any p > 1, multiplying equation (4.6) by |u(1)|”’1u and integrating over Q, we obtain +__Id_t dl/lu“|p+1d1‘+/f(u(1)) )|u(1)|” 1udrr: P 2: f/J(.’L‘ — ?/)u(l)(y)]u(1)Ip"lii(1)d:lrdy (4°10) —//J(:c-—y)u(1)(:c)|u(1)|”’1u(1)dxdy+l/0(0)|u(1)|”_ludr. Using Holder’s and Young’s inequalities and conditions (A 1) and (A2), we have —|u(l)|1"+ldf£+(/—"/‘|u(1)|p+1r 1uclr _<_ C(p)Clp+1, l where Cl is a constant independent of p and limp _. ooC(p)p +1 S C2 with 02 independent of p. We need the following version of Gronwall’s lemma (see Ternam [44]): Lemma 4.1.2 ( Uniform Gronwall’s inequality) Let y be a positive absolutely contin— uous function on (0,00) which satisfies with m > 1,11 > 0,6 2 0. Then, fort Z 0, we have i —1 m +(1/(m —1)t)m - 1. (4.12) y(t)S( ) VIOn Using this and (4.11), we have 59 P+1 —(p+1) WNHKi_«XmCT9p+T—l+lCh-2fllT’Q - M13 Therefore, 1 p + 1 —1 mflmsiscow+lanp+T-1+«ar-a0r-1 (4M) Letting p —+ 00, we have llu(1)lloo S 0 (4-15) for some constant C. Also from condition (A2) and equation (4.6), we have Wu... _<. C. (4.16) Since equation (4.7) is a linear parabolic equation, by inequality (4.16) and stan- dard parabolic theory, we have 0“) E L°°((0, T), L°°(S1)) n L2((0, T), H2(Q)). By induction, there exist unique solution (u(k), 6"”) of system (4.6)-(4.8). Further- more, ”W, 11;") e L°°((0, T), L°°(Q)) and 0“" e L°°((O, T), L°°(a))nL2((0, T) mm» for every k. Now we prove that there exists a uniform bound for u(k), utk) and 6‘“. Multiplying equation (4.7) by I6‘kllp‘16(k)(a:) for p > 5’ and integrating over $2, we have /|6(k)|p—16(k)6(k)::+/V(|9(k)lp—16(k))'V6(k)d$+/|6(k)|p+ldx = -1 I/J(r- U‘W y)|6"°’|" l9"’dyd:z:+z [f(u (’0 )|6(")|"’16(")dx (4.17) +1 / / J(x -y)u(k)(:1:)|6(k)lp‘16(k)dyd:r+(1—12)]|6(k)|p_16(k)6(k‘1)d;r. 60 Since p+1 WW 2 l2_ (P+1)2 2 ————=|6|P"‘|v0|2 L§IJ—1—)—V(|0)P-10)-vo, (4.18) using Holder’s and Young’s inequalities, we obtain p +1 1 a“) P+ldx +-——— 4”] v 60‘) 2 2dr + - f 6"” Wax 3 cap) —1/ Iur+ldx+c2(z,p) / |6<*-1>IP+‘dr+ / |f(u"‘))|"“dx for some positive constants c1(l, p) and c2(l, p) which depend only on p and l. Multiplying equation (4.6) by |u(k)|(r ‘- I)? ‘" 1um, and integrating over 0, we obtain -1)P+1 (k) (k)(7-1)P-1u( (k) (1-—_——11)p+1d%/|u“‘)|(dx+/f(u)|u| dx =//.]( :c -— y) u“ )(y)|u(k )r-|( 1p) -111.(k)(l;1:dy +l/0(k“l)|u(k)|(r _ 1),) — 1u(kldat ‘//J(4—y>u<*>(4)lu“l( ‘ 1”) ‘lu‘k’dxdy (4.20) Condition (Ag) implies f (u)luI(’ ‘ 1)? ‘ 1,, 2 c1|11|("1)(”+1)— QM” -1)P (4.21) and |f(u)|”+1 S C7Iu|(’" ‘1“? +1) + c8 (4.22) for some positive constants c-, and c8. F rom equation (4.20), inequality (4.21), Holder’s and Young’s inequalities, we have 61 _i__i (u(k)|(7‘ — 1)P + 1‘13: + 2 (u(k)|(7” -1l(P + lldx (1' — 1)p +1dt 2 (4.23) 3 cap) + clap, 1) / Ifi‘k‘1’l”“dx for some positive constants c(r, p) and c1(r, p, l). Integrating (4.23) from 0 to t, we obtain 1 c t . __ (k) (r — 1)p+ 1 _1 (k) (1 —1)(p + 1) 4 (r—1)p+1/|u | dx+2/(;/|u I (11' t s c(r.p>t + clap, 1)] f0IP+144 + / luol(" — ”P + 1 (4.24) o t s C(u0.T,T,P) + c.(r,p,z) / / 0"‘“’|"+‘drr o for some positive constants c(uo, T, r, p) and c1(r, p, 1). Integrating inequality (4.19) from 0 to t, using (4.22) and (4.24), we have t / |6|P+1dxds) (4.25) (I 0 for some positive constant (:(uo, 90, p, r, l, T) which does not depend on k. By induction, we have / [6(k)|p+1d3: S cect (4.26) a for some positive constant c which does not depend on k. Similarly from inequalities (4.23) and (4.26), we also have / |u|P+‘dx _<_ C, (4.27) (I and f If (u(“WH‘dI S C (4.28) 12 62 for some positive constant C which does not depend on 1:. Equation (4.6), inequalities (4.26)-(4.28), and Young’s inequality imply / |u]k’|P+ldx g C (4.29) {2 for some positive constant C which does not depend on k. This implies 411]“ + 60"” E Lp+l((0, T), LP+1(Q)) and H — ta)“ + WWI... s C (4.30) for some positive constant C which does not depend on k. The following lemma may be found in [29]. Lemma 4.1.3 Consider the following linear parabolic equation: 6,,—A6+0=g in (0,T)xQ, “0.35) = 90(13), (4-31) (96 files: = 0. Up > 3, g E L"((0,T),LP(Q)) and 60 E L°°(Q) flWl’2(Q), then 6 E L°°((0,T), L°°(Q ), and we have T _ sup IIOIImSCmax{||90|loo,(/ Ilgniw}, (432) 0 — a|24x+ / |V(6"‘+" — WW”;- /(9 — am)? (4.39) g [2 / W“) —u§")|2dx+ / IOU" —6”“1)|2dx. Since Hum”00 S C, from equation (4.6), and condition (A2), we have / |u - u(k)|2dx g C(T) / lam — 6<'°-1>|24x, (4.40) and f WW”) — f(u‘k’)|2dx = / WW” + (1 — A)u)(u<*+“ — Wards: (4.41) g C(T) / wk“) — u|2dx. Therefore, equation (4.6), and inequalities (4.40)-(4.41) imply llnng) — irék)|2d:r s 4 f I / J(x — yxum” — u>dylzdx + 4 f(/ J(x - y)dy)2 — :4“)de + 4 /(f(u(k+1)) — f(u("))2d;z: + 4 [(9%) — 0)'~’dx s (31(7) / W) — 0|2dx (4.42) for some positive constant C1(T) which does not depend on k. Inequalities (4.39)—(4.42) yield 21% / W“) — 0(k)|2d:r g C(T) / )0“) — 6(k—“de: (4.43) for some positive constant C (T) which does not depend on k. By induction, this implies 65 / |6— 6(kllzdx < at): 3/0 / [91 -6°|d:rds (4.44) So 6"“) is a Cauchy sequence in C([0, T], 142(0)). Therefore, there exists 0 E C([O,T], L2(Q)) such that 0““) —-+ 0 in C([O,T], L2(Q)). From (4.33)-(4.35), we have HWWSQ an) /T/ |A6|2dzcdt g (1', (4.46) 0 T0 2 [a [n W dxdt g C. (4.47) Also from (4.36), (4.40)-(4.42), we have u(k) —» u in C([0, T], L2(Q)), (4.48) u"; l —» u, in C([O,T], L2(S2)), (4.49) f(um) _. f(u) m C([O,T], 112(0)). (4.50) Therefore, letting k —+ 00 in equation (4.6), we have ”t = / J(ZL‘ — y)'u(y)(ly — / J(JZ —- y)cly11.(;lr) — f(u) +19 (4.51) Q Q for t > 0 and a.e. :1: E 9. Since a)“ .4 at, a)“ _. 6,, AW) —» A0 in mm, T), L2(o)), letting k —+ 00 in the weak form of equation (4.7), we have T T / f(lut + 9t)€(t,:1:)d:rdt = / A65“, :r)d.r.dt (4.52) 0 n 0 s2 66 for {(t,2:) E L2((0, T), L2(Q)). Since it is true of 6““), we also have [T / n(t)(A0

9 in C([O,T], 52(9)), and since ()(kl(t,m) and 0(t,:r) are con- (4.54) tinuous with respect to t in L262), by taking k arbitrarily large we can see that 6(0,a:) = 00 a.e. in 0. Similarly, u(0,:r) = no a.e. in 0. Equations (4.51)-(4.54) imply that n and 0 are solutions of system (4.1‘)-(4.4) in a weak sense. To prove uniqueness and continuous dependence on initial data, let 6,0 E L°°(Q) fl W1'2(Q), um E L°°(Q), and for R > 0, ll9i0llL°° S R, llui0llL°° g R, where i = 1,2. Let u, and 6,- be solutions corresponding to initial data a“) and 6,0, then we have llgillL‘” S C(T,R), and lluille S C(T,R). Denote v = ul — 112,11) = 61— 62. We have 1;, = / J(:z: — y)'v(y)(ly — f .](:I: — y)(ly2r(:1t) —- f’()\n1 + (1 — A)'ug)v + lm, (4.55) Q 9 (w + lv)t = Aw (4.56) in (0,T) x Q, for some Mas, t) E [0,1]. We also have initial and boundary conditions v(0,.1:) = v0(.i:), w(0,:c) = 1110(1), (4.57) 32....-. (4.8) 67 Multiplying equation (4.55) by of, integrating over 0, multiplying equation (4.55) by v, integrating over Q, multiplying equation (4.56) by to, integrating over 9, we flvzl =//J( ((Ir-y)vy )dyvtd1-/J(I-y)dyv($)vt (4.59) —/(f'(/\u1+(1 - /\)u2)vvt + lwvt)d:r:, / -_—— f [a J(x — y)v(y)dyvd$ - [n J O we have /|;;%dlt|2dr+ it/lulzda +(:1/|71,| Ida: +é %/|0|2d;r +—- d/(F(u)+c6|u|)dx+ / |V0|2dz (4.69) dt 0 n 3 C(5) / mum} / lut|2dx+e/ (912mm. 9 2 (I Q 69 From Poincare’s inequality f(6 — 6)2d:r g C f |V6|2dz, where 6 = '71,] f 6dr, and since f(6 + lu)d:1: = 10 E f(60 + lu0)da:, we have [62dzr g C(Io)/u2 + C(10) + C/ |V6|2dx. (4.70) Denote Y(t) = In |u|2d$ + f” |6|2dx + fn(F(u) + €6|u|)d:r. Using Holder’s and Young’s inequalities again, (4.68), (4.69) and (4.70) yield dY :1? + CI(IO)Y S 02(10) (4‘71) for some positive constants C1(Io) and C2(Io) which do not depend on initial data. Gronwall’s inequality implies Y s 02(10) + Yoe-CIW. (4.72) We have: Theorem 4.2.1 There exists a constant C(10) and a time t0(10) which does not oth- erwise depend on initial data such that Hull. S C(10). (4.73) lle S C(10) (4.74) fort Z t0([0). Next we estimate ||V6||2. The following lemma may be found in [44] 7O Lemma 4.2.2 Let g, h, y be positive locally integrable functions on (to, 00) such that y’ is locally integrable on (to, 00), and which satisfy l (cl—3t, S gy + It for I, Z to t+l t+l 1+1 / 9(-9)d3 S 01. / h(8)d8 S 02, / y(s)ds g a3 t t t fort Z to, where a1, a2, a3, are positive constants. Then y(t + 1) 3 (a3 + a2) exp(a1) for any t Z to. Multiplying equation (4.2) by 0t, and integrating over Q, we obtain [(90 2+§%/|V0|2dr=/( —lut0t)d:r:. Holder’s and Young’s inequalities and (4.77) imply i IVO 2dr 3 c 'u. )2dx. dt ’ By inequality (4.69), we also have t+l / / |V6|2dr S c t n for t Z t0(10). (4.75) (4.76) (4.77) (4.78) (4.79) Multiplying (4.1) by at, integrating over 9, and using Holder’s and Young’s in- equalities, we have 1 d 2/(ut)2d$+— d; (F(u)+66|11|)d113 S C/u2dx+/62d:r+C. 71 (4.80) Integrating from t to t + 1, and using inequality (4.69), we obtain [HI f(ut)2dx S c (4.81) Applying Lemma 4.2.2 to (4.78), using (4.79) and(4.81), we have for t Z t0(10). / |V6|2dx g c (4.82) for t 2 50(10) + 1. Theorem 4.2.3 There exists constants C(10) and t1(10) E t0(IO) + 1 independent of initial data such that ||V<9||2 S C(Io) (4-83) for t Z t1(10). Corollary 4.2.4 If n 2 1, there exists an absorbing set in an afline subspace of L°° x W1’2(Q) Proof. It follows from Theorem 4.2.1 and Theorem 4.2.3 that there exists an absorbing set in W1'2(Q) for 6. We need to prove there exists an absorbing set for u in L°°. In fact, since n=l, W1'2(Q) L—> CHI-l) is compact, there exist constants C3([0) and t0(10) which do not otherwise depend on initial data such that Hmmsam) MM) for t Z t0(10). 72 Using a similar argument to that in the proof of Lemma 4.1.1, we also have ”“1100 S C(10) (4-85) for some constant C(10) which does not depend on initial data and for t _>_ t0(Io) Since —E( (,u e) =—/flu idx—LIV0I2dx, (4.86) integrating equation (4.86) from O to l, we have E(u0,e0) -E(u(t),e(t)) = I [a ugdxds+ [0. [a |V0|2dxds. (4.87) From (4.5 ,r > 2 and the Cauchy-Schwartz inequality, we have E=—)//J( x—y) (u(x)—u(y) )zdxdy+/(F((xu ))+ $07112) Z f/J(x — y)(u(x) — u(y))2dxdy + c5/Iulrdx — /:,|u|dx (4.88) 2 c7 / Inlrdx — 7:8 for some constants c7 and c8. Therefore E(u,e) is bounded below, we have E (170,730) — E (-u(l.),e(l.)) g C for some positive constant C which does not depend on t. This implies / [ugdxds+/ /|V0|2dxds S C. (4.89) 0 n 0 n Multiplying equation (4.2) by A0, and integrating over (2, we obtain 2dt‘/Q’V6|2dx+/IA6’2dx: —l/utA0d..x (4.90) 73 It follows from Holder’s and Young’s inequalities that 1] |V0|2dx+C1/ |A0|2dx g 02/ |utl2dx (4.91) dt 0 71 n for some constant C1 and C2. Since / |V(0 - 60% = [(0 - 60476 — 604x .<. ”0 — 60020476 — 6002. Q Q from Poincare’s inequality, we also have ||V0||2 g C ||A0||2. The inequality (4.91) implies d _/1V0lzdrr+C3/lV612d-T S 02/ |ut|2dx (4'92) dt Q :2 0 for some constant C3 and C2. Claim: ffl |V0|2dx —> 0 as t ——+ 00. Multiplying eC3‘ on both sides of inequality (4.92), and integrating from t to 2t, we have .2tC’3117472t. )113 — 97:3va, ->|I3 2t (4.93) 3C1/ eC3s/Iut(s,x)|2dxds t n This implies ||V9(2t. -)H§ - €_’C’||V9(t= -)||§ (4.94) 2t S (11/ 7203(3—2’)/ |?tt(b‘,fII)|2d.’IIdS. t 0 Since ||V0(t, )||§ S C and 1000152 |ut(s,x)|2dxds S C, from (4.89), letting t —> oo in inequality (4.94) gives 74 tlim ||V0(2t, -)||§ = 0. (4.95) Since suptll0|lz S C, by inequality (4.89), we have: Theorem 4.2.5 If (u, 0) is a solution of system (4.1) {4.4), there exists a sequence 1;. -» 00 such that [III/”J(x - y)uk(y)dy - [2J(x - y)dyuk(x) — f(uk) + l0k|2dx —> 0, (4.96) 0;, —-> c in W"2 (4.97) for some constant c, where uk(x) = 71(1),, .17) and 0k(:r) = 0(1),, x). 4.3 Global boundedness of the solutions In section 3, we proved that u and 0 are uniformly bounded with respect to t for dimension n = 1. In this section, we shall study the global boundedness of solutions for higher space dimensions if the initial data 00 has better regularity. We prove that the solution (71(1), 0(1)) is uniformly bounded in suitable function space on the entire interval [0, 00). We are also concerned with the regularity of solutions. Within this framework, we are able to prove a sharper result on the asymptotic behavior of (4.1)-(4.2). Through out this section, we fix p such that p = 2, if n = 1, 2, 3 and p > g, if n > 3. Let A be the unbounded linear operator from LP(§2) into itself defined by the following: Au 2 ~Av + 11,11 E D(A) 81! (4.98) D(A) = {v e LP(Q) : Au 6 MW), 87’" =0}. It is well known that A generates an analytic semigroup on the space LP(Q). Also let X a be the space D(A°) endowed with the graph norms ”-1171 of A“ for 27—1 < a < 1. P Define a linear operator on L" for q > 1: Lu IE / J(x — y)udy — / J(x — y)dyu. (4.99) n a Denote QT = [0, T] x Q. We have the following comparison principle. Lemma 4.3.1 Let u E C’([0, T], L") for (q > 1) and “t S Lu + 7:(t,:17)u, (4.100) a.e on Q. Assume c(t, x) S M1 on QT for some positive constant M1, then u(t, x) S 0 a.e on 9 fort E [0,T]. Proof. Let p = A11 + M + 1 and v = ire—p", where M = sup f” J(x — y)dy, we have v1: vie—pt — pile—pt S e—pt(Lu + cu) — pu = Lv + cv — pv. (4.101) Multiplying inequality (4.101) by (11+)“1 and integrating over Q, we obtain (l/(11+)qdzr)t S / I.1.7(v+)q7l;1;+/7:(v+)q71.r —p/(v+)q7lx. (4.102) (I n n n 72 Since LLv(v+)"—1dx = [I/QJQ —yll’+dy(v+)q‘ldI _ [a [a J(x-y)1’_dy(v+)q_’dx— f” A J(I_y)dy(v+)q_ldx (4.103) Sf/J(l"ylt'WdI/(U‘LV—ldlfi (I Q 76 It follows from Holder’s and Young’s inequalities that LLJQ — y)v+dy(v+)q—’dx 1 q - 1 (4-104) 3 7/‘17/n J(x — mung/14057 / (Wan—q— s M. f (49144. Inequalities (4.102)-(4.104) yield (% Lanna), + (p — M1 — M2) ((59qu g 0 (4.105) Integrating (4.105) from 0 to t, we have ||v+(t)||3 S Hv+(0)|lg- (4-105) v+(0) = 0 implies the conclusion. Lemma 4.3.2 Instead of condition (A2), we assume (Ag) There exist a,- > 0, b, > 0, c, > 0 and r > 2 such that S a1|u|"2u + blu + c], if u _<_ 0 f (u) : (4.107) > aglu|'—2u — 0211, — C2, if U > 0 Suppose that the condition (A1) and (A2) are satisfied, then there are positive con- stants n1 and n2 such that 1 ||lt(8)l|oo S (m + “Halli;l +712 sup 116(Tllloolr — 1. (4-108) 0 S r S s where constants nl and 71.2 only depend on 71,-, 11,-, 75 and r. 77 Proof. Choose any T > 0. Let M(s) = sup ||0(r)||30 for each s E [0, T) and 0 S T S s set 1 11(T)"=‘(n1+||u0|l;l+ "241(30’" _ 1, (4.109) 1 35(7) E —(n1 +H"0H;1+ 12.2M(3))7‘ — 1 (4.110) for each 7 E [0, s], where 111 and n2 are two constants which will be determined later. Notice that both 11 and y are constant on the interval [0, s] for fixed 3. Thus 1‘1),—fnJ(x-y)1‘1dy+/flJ(x—y)dyu+f(u)=f(u), (4.111) and (9)1 - [2J0 - 109719 + f” J(x - y)dya + f(9) = f(a). (4-112) We claim that f(fi) Z [V(s) and f(y) S —l\«[(s) on [0,3] for suitably chosen n, and n2. In fact, from condition (A2) and the definitions of 11 and y, we have 1 M) — M7s) 292n1- an: — 1 + 42117411: — 1.71.07)... 1 (4.113) 1 _ _ _ . 7-1 . ~—-1 C2+ (712112 1)M(s) 02712 M(s)’ , and 1 474) — M7412 - blur ‘ 1 + «1.1777712: — b.1174”... (4.114) 1 1 _ ‘ _ _ 7-1 , . ~—1 c1 + ((11112 1)M(s) bln2 M(s)7 . 78 b r —1 + (4)” “ 2 + 1} then ai r—l F“ ‘t 1 ' = ~— _— irs , c ioosmg 122 max, _ 112{a,-(r _ 2) 1 (1.2712 ‘— 1 — 0271;. — 1 2 0 and (1171.2 — 1 —1)17‘l; _ 1 2 0. Inequalities (4.113)-(4. 114) yield 1 1 1 —_/- _ T—1_ 93r_2_ _ T—l f(u) M(s) 2 (1271.1 11271,l b2( ) c2 11271.2 , 02 and 1 1 1 r — 1 1’1 — r — 1 “f(ul-M(5)Zain1—bin1 _b’(71—)r-2 ‘Ci—bln2 - 1 Choosing 711 such that 1 1 1 0277.1 —bgnir _1 —bg(§'2-)T - 2 — C2 —b2n2r -1 2 0, (1.2 and 1 1 1 b1 )T_2—C1—bln2r-1 20. 01 0.1711 — 01711,. -1 — 01( It follows from (4.111)-(4.118) that a. — / J74 — 41277417721 + / J74 — 704177 + 7722) 2 26, (2 and at - / .I(:1: - 309719 + / J (:17 - 107199 + [(1.7) S 19 n n on [0, s]. Denote 171 = u -— 1'1 and w = y - u, then we have 1227- - / J(x -— y)w(y)dy + / J(x - y)dyw + CID S 0, 11 $2 79 (4.115) (4.116) (4.117) (4.118) (4.119) (4.120) (4.121) ‘fi rn'kw w and a. - [9 J74 ~— 0474144 + [9 J74 — 10444 + 44 s 0, 74.122) 1 1 where E =/ -f'(/\fi + (1 — A)u)d)\ and g =/ —f'()\y_ +(1— A)u)d)\. 0 0 Lemma 4.3.1 and condition (A2) imply It: |/\ C |/\ {21 (4.123) on [0, 3]. Therefore, 1 11474111.. s (n1+lluoll;;‘+ 420344 1147011..)7 - 1. 74.124) Theorem 4.3.3 Suppose that conditions (A1) and (A2) are satisfied and (u0,00) E L°° XX“ with 225 < 07 < 1. Then there exists a unique solution (77, 0) to (1.1) and (1.2) which possesses the property described in Theorem 2.2. Moreover, (u, 0) E L°° x X a sup “0(1)”ch S C1(a,-, b,, c,, ]|00]|Xa, ||u0]]°o) (4.125) S t < oo SUP IIU(t)||oo S 020147194164. ||90||xm ll’ltolloo) (4-126) 0 S t < 00 Proof. It follows from Theorem 4.1.4 that system (4.1)-(4.4) has a unique solution (11,0) because 00 E X“ C WI'2 0 L°°. Since at E L°°((0,T), LP), a basic regularity result implies that 0 E C ([0, T], X 0). Without loss of generality, we may assume that p>r—1+1and |f(U)| S (041+ a2)|u|"“ +021 + b2)|U| + (Cl + 62). (4-127) 80 Note that 1 (/ |u,|p(l:lt)p = (/ IIm. + f(u) +10|pd;L')p Q Q 1 1 s (f | / J(x — y)'u.dy|”dr)” + (f I / J(x — y)dy/MIMI)? Q Q Q Q 1 l + (/ If(u)|pd:c)p + (/ |0|p(1;1:)p Q Q _<_ (QAII + ()1 + b2)(/ lulp(l:r)p + ((11 + (1.2)(/ lulrpd:1:)p n n l 1 + u] 1mm +(c1+ calm?) Q p-(r) r—1(p—1)—1 r-1+1 g (2M1+ ()1 +112 + (11+ ”mu”... 7’ + ”an... 7’ mun. P p - 2 3 Nuance? 11611.? +PIs21 (4.128) p -— 1 1 . . Let A = — — —. Due to inequality (4.128) and Lemma 4.3.2, we see that P 7'1) HMS. ')llp 5 Cl( sup |19(T)l|oo)A + C2l|9(8)||§o + 03- (4-129) 0 S T S .9 Therefore, sup Huds, )H. s cm sup 116(8)”...1‘ + cs, (4.130) SsSt OSSSt where 0,, (i = 1, 2, ...5) are positive constants which only depend on (1,, b,, (‘1', MI, Huolloo, “60”.» and 9. By using the operator A, ( 1.2) can be formulated in terms of the abstract Cauchy problem in the Banach space LP(Q) as follows: 013+ A9 2 6 -— [at (4.131) 6(0) = 90. (4.132) Therefore t 110(1)”... 3 Ila—Atfiollxa + / Ila—A“ — ”(we -) - 6(8))llxads ‘ t s ra‘a‘llflollxa + 0S<111<)t(ll'ut(s.-)|lp + ll0(--)llp) / (t — srae-“l — 3M»- m g Hoonxa +( sup Hanna/‘06 / {ac-“8.1.9. 0 < < t o _ 3 _ (4.133) Using the embedding X a =—» L°°(Q), we obtain SUP ||9(t)||oo S Co: 811p ||9(t)|l)(a S C's( SUP ||6(t)||°o)* + Ca- OStum1ds t (4.142) S C(lonHom |l90||oo)h- Consequently, IIUt(t + h) - "illoo S 2Mill'U(t + h) — u(tlllao + ||f'(n)IIU(t + h) - u(Ulla. (4143) + “W + h) — 9(t)lloo S C1(l|'uvo||ao, |l0ill|oollt°‘l- 83 for n = (u(t + h) + (1 — §)u(t). A similar argument also shows that u), E C"([O, 00), LP). (4.144) This completes the proof. Proposition 4.3.5 If the condition (A2) is replaced by (A3) There exist a3 > 0, b3 > 0, d1 > 0, d2 > 0 and r > 2 such that a3lu|'_2u + b3u — (11 S f(u) S a3|u|r"2u + b3u + d2. (4.145) and (u(t),6(t)) are solution of {1.1-1.2} with (u0,60) E L°° x X“, then there is a constant K independent of uo and 60 such that timooHB—allwqu =0, (4.146) limsupll6||00 S K, (4.147) t —> oo lim sup IIuHoo S K, (4.148) t —1 00 where 6 = [317' f0 6dr, n < q < 71. +6,, and 6 S 2n(a — n/2p) Proof. By interpolation theorem, we have H9 — éllwm s cue — 9113,1316 41133.. (4.149) From inequality (4.134), and suptll6lloo < C, we have 84 sup||6 — 6||Xa < C. t From section 3, we have lim 9—6 x, =0. 14 OOH II”, 12 This implies (4.146). Since ”6|le _<_ ”‘9 “ éllwla ‘l‘ lléllwlm and HéIIWl. < 0116”? for some constants C and fl. Inequality (4.74) implies lim sup ||6||W1,q 3 C(10), t —» 00 where 10 = foam) + 60)d:r.. Embedding theorem implies (4.147). (4.150) (4.151) (4.152) (4.153) (4.154) Using a similar argument to that in the proof of Lemma 4.1.1 gives (4.148). Remark 4.3.6 Inequalities (4.133), (4.147) and (4.148) also imply: 111118111) llgllX" S C(10). t—1 00 Next, we prove that there exists a subsequence tk such that u(t,.) ——+ u‘, 6(tk) —~> c, and (u‘, c) is a steady state solution of system (4.1)-(4.4). We assume: (A3) M E sup I In J’(.r. — y)dyl < 00, and f E C1(R). (A4) There exists a constant c7 > 0, such that f’ (u) + 0(1) 2 c7 > 0, where a(:r) = In J(x — y)dy. We have: Theorem 4.3.7 Assume that conditions (A1), (A2), and (A3) are satisfied, let (11,6) be a solution of system (4.1)-(4.4), we have llutll2 —+ O as t —> 00, (4.155) ”van. —+ 0 as t —. 00. (4.156) In addition, if (A4) is also satisfied, then there eaists a subsequence {tk}, 11 E L262), and constant C such that ||u(tk,x) — u(x)||2 —+ 0 as k —> 00, (4.157) ||6—C||2—+Oas k—+oo, (4.158) l/udx + C|Q| = [0, (4.159) 12 where Io = fn(luo+60)dr. And (5,0) is a steady state solution to system (4.1)-(4.4). Proof.The conclusion (4.156) follows from (4.95). Taking derivative with respect to t in equation (4.1), multiplying by “t and i11te~ grating ovcr Q, we have 86 li/lutlzZ//J(:1:—y)ut(y)'ut(:l‘)d.77dy—f/J(.’l?-y)1l.:(:r)dy(lflt 2C“ 0 Q Q Q Q —/f’(u)u§(x)d$+l/6tutdx n a By (4.136) and Holder’s and Young’s inequalities, we have d a—t/|u¢|2deC1/IutI2dx+C2/I6t|2dx n n a (4.160) (4.161) Multiplying equation (4.2) by 6t, integrating over 9, and using Holder’s and Young’s inequalities, we have 1d 1 -— 62 — 62 <0 2 2(it/KIWI‘1'2/Qltldilc_fnlutl Inequality (4.89), (4.95), and (4.162) imply / / l0t(I)l2d-T S C, / / l“t(-T)l2d-T S (I 0 Q 0 Q for some constant C. Let c > 0, it follows from (4.163) that there exists {sn} with / |ut(s,,,:1:)|2d:1: —+ 0, n and there exists N such that and for any n > N and t > Sm we have / |ut(sn,:1:)|2d:1: < c, n 1 / lut(s, r)|2d:rds < c, "Q 87 (4.162) (4.163) (4.164) (4.165) (4.166) t f / |6t(s,x)|2d:1:ds <4. (4.167) s" 9 For any t > 3", from (4.162), (4.165)-(4.167), we have / |u,(s,a:)|2dx < (2 + c)e. (4.168) Q This implies (4.155). — — I l Denote w = 6 — 6, since 6 = T1,— [n6drr = l — — / udrr = (:1 — c2/udrc, we ' ' IQI IQI n a have 6 = w + Cl — c2 f0 udx. By (4.156) and Poincare inequality, we also have ||w||2 -> 0 as t —+ 00. (4.169) Using this and (4.1), we have a(:r)u(t, 51:) + f(u) = [(J(:1: — y) — lc2)u(t, y)dy + lw +lc1 — ut. (4.170) 12 Since operator K: L2(Q) —) L2(Q) defined by K(u) E L(lh — y) — l(.'2)u(t, y)dy is compact, there exists a sequence {tk} such that K (u(tk)) converges to h in L2(Q). This and (4.155), (4.169) imply that a(:r)u(t),.) + f(u(tk)) is a Cauchy sequence in 112(8)). From condition (A4), we have CYHUW) — u(tmlllz < ||a(.1‘)u(t1-)+ f(uUkll — a(l‘)lt(tm) + f(U(tm))ll-2- (4-171) 88 Therefore, u(tk) is a Cauchy sequence in L2(Q), there exists {1 such that ||u(tk) — The boundedness of ||6HW1,2(Q) and (4.156) imply (4.158). [(111 + 6) = I0, This, (4.157) and (4.158) imply (4.159). Since Since u(tk) and ii are globally bounded, we have ||f(u(tk) — f(filll2 —> 0. This, (4.157),(4.158), and (4.170) imply that (11,0) is a steady state solution to system (4.1)-(4.4). We complete the proof. If the initial data are smooth enough, we also have: Corollary 4.3.8 If 110(3) E W’"’(Q), 60(23) E W2"’(Q) for o > n, conditions (A1), (A2), (A3), (A4) are satisfied, then there exists a subsequence tk such that u(tk) —» u‘ in C71 (Q), (4.172) 6a,.) -> c m 072(5)) (4.173) where (u‘,c) is a steady state solution of system (4.1)-(4.4), 0 < ”)1, 72 < 1 are two constants. Proof. Since u0(;1:) E W1=U(Q), 60(23) E W2*0(Q) for o > n, by Theorem 4.3.3, there exist solutions 11 E L°°((O, oo), L°°(SZ)), 6 E C([0, oo), W2'°(Sl)). This implies we) 6 C([0, oo), w1v0(a)). (4.174) u(a: + hei) — u(zr) h From equation (4.1), we have Denote A3,“ = 89 Ahmt— -— ”Ahiflr — )yu(y)dy — [g AhiJ(1: - y)dyu(I) ’ (4.175) _ u/{;J(x - y)dyAhiu — f'(€)Ah.-u + lAhi6, where f = Au(:r + h) + (1 -— /\)u(:r). Since ||A26||La(nr) S ||V6||La(m for h < disf.(t)§2’,(')Q), and W” H L°°, we have SUP HAWlloo S C (4-176) t for some positive constant C which does not depend on 11. Multiplying equation (4.175) by [AiuIU‘zAZw integrating over 12’, using Holder’s and Young’s inequalities, (4.176), and condition (A4), we have 6616/" lA),u|” + C 11' [ALuIU S Cf (4.177) for some constants C, C1 which do not depend on h. Gronwall’s inequality implies sup ||A),u||0 S C2 (4.178) t for some constant C2 which does not depend on h. This implies sup lluxillg S C2. (4.179) t Since ”Ulloo < C, we have Sltlp Ilullwl, (r S C. (4.180) 90 Therefore there exists a subsequence tk such that u(t,.) _. 1p in w1,0(a), u(tk) —+ u" in C'“(Q) This and Theorem 4.2.5 yield fnJ(x — y)u‘(y)dy — [1M3 — y)dyu‘(:c) — f(u”) + lc = 0 (4.181) where c is a constant. Also (4.146) and Theorem 4.2.5 imply that there exists a sequence {tk} such that a 111nm ”6,. — CHWL q = 0. (4.132) Since q > n, by embedding theorem, we have lim “6,. — 4|ch = 0. (4.133) lk —> 00 This completes the proof. 91 BIBLIOGRAPHY 92 BIBLIOGRAPHY [1] H. Amann, Fixed point equations and nonlinear eigenvalue problems in ordered Banach spaces, SIAM Rev,18 (1976), 620-709. [2] H. Amann and J. Lopez-Gomez, A priori bounds and multiple solutions for super- linear indefinite elliptic problems, J Differential Equations 146 (1998),336—374. [3] N. D. Alikakos, L1D bounds of solutions of reaction-diffusion equaitons, Comm. P. D. E. 4(8) (1979), 827-868. [4] N. D. Alikakos, P. W. Bates, and X. 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