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DATE DUE ' DATE DUE DATE DUE ii - fiL__ l | __JL_:|___I -@ MSU In An Mariam Mon/Emil Opportunlty Iii-tumor! Wanna-m EXISTENCE AND MULTIPLICITY OF POSITIVE SOLUTIONS OF NONLINEAR INTEGRAL AND DIFFERENTIAL EQUATIONS By H aiyan Wang A DISSERTATION Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1997 ABSTRACT EXISTENCE AND MULTIPLICITY OF POSITIVE SOLUTIONS OF NONLINEAR INTEGRAL AND DIFFERENTIAL EQUATIONS By H aiyan Wang In this dissertation, we use a fixed point theorem of cone expansion and com- pression combined with degree arguments to study the existence, nonexistence and multiplicity of positive solutions for systems of nonlinear integral equations. We obtain, in a unified manner, existence, nonexistence and multiplicity results for positive radial solutions of elliptic systems on an annulus which generalize and extend earlier results. Using the same method we also study existence, nonexis- tence and multiplicity of a class of integral equations which arise from differential equations with nonlinear boundary conditions. To my parents and wife iii ACKNOWLEDGMENTS I would like to express my sincere gratitude to Professor D.R. Dunninger, my dissertation advisor, for his constant encouragement and kind guidance. I would also like to thank the committee members Professor Michael Frazier , Professor Patricia Lamm, Professor Jerry Schuur and Professor Zhengfang Zhou for their time. iv TABLE OF CONTENTS 1 Integral Equations 3 1.1 System Of Nonlinear Integral Equations ............... 3 1.1.1 Introduction ........................... 3 1.1.2 Main Results .......................... 4 1.1.3 Preliminaries .......................... 6 1.1.4 Proofs of Main Results ..................... 12 1.2 Case IV Continued ........................... 18 1.2.1 Introduction ........................... 18 1.2.2 Main Results .......................... 18 1.2.3 Upper-lower solutions ...................... 19 1.2.4 Preliminaries .......................... 22 1.2.5 Proof of Main Result ...................... 26 1.2.6 Elliptic Systems ......................... 28 1.2.7 Remarks ............................. 31 2 Nonlinear Boundary Conditions 33 2.1 Nonlinear Integral Equations ...................... 33 2.1.1 Introduction ........................... 33 2.1.2 Main Results .......................... 34 2.1.3 Preliminaries .......................... 36 2.1.4 Proofs of main results ..................... 42 2.1.5 Elliptic Systems ......................... 47 2.1.6 Remark ............................. 50 2.2 Case IV Continued ........................... 54 2.2.1 Introduction ........................... 54 2.2.2 Main Result ........................... 54 2.2.3 Upper-lower solutions ...................... 55 2.2.4 Preliminaries .......................... 59 2.2.5 Proof of Main Result ...................... 63 2.2.6 Elliptic Systems ......................... 63 BIBLIOGRAPHY 64 Bibliography 64 vi Introduction In recent years, existence of positive radial solutions for semilinear elliptic equations in an annulus have received considerable attention . For single equations, Arcoya[1], Bandle, Coffman and Marcus [2], Dang and Schmitt [5], Dang, Schmitt and Shivaji [6], Wang [19], Erbe, Hu and Wang [10, 11], Henderson and Wang [13] examine various conditions on the nonlinearity which guarantee the existence of one or more positive solutions for the problem. It has been shown in these papers that various combinations of superlinear and sublinear conditions on the nonlinearity guarantee existence of positive radial solutions. Numerous methods were used in proving these results. In particular, shooting methods were used in [2], fixed point theorem of cone expansion and compression in [19] and the Mountain Pass Theorem in [1]. Dunninger and Wang [8], Ma [17] showed that these results can be extended to systems of equations. Related results for fourth order problems have been obtained by Dunninger [7], Eloe and Henderson [9] and Ma and Wang [18]. In view of recent advances in this area, it is natural to ask whether analo- gous results are valid for systems of integral equations and elliptic equations with nonlinear boundary conditions. In the first chapter we use a fixed point theorem of cone expansion and com- pression combined with degree arguments to study the existence, nonexistence and multiplicity of positive solutions for systems of nonlinear integral equations. One of the primary motivations for studying systems of nonlinear integral equations is to obtain, in a unified manner, existence, nonexistence and multiplicity results for positive radial solutions of elliptic systems on an annulus which generalize and extend earlier results in [2, 5, 8, 11, 12, 14, 16, 17, 19]. In the second chapter these same methods are used to study the existence, nonexistence and multiplicity of positive solutions for a class of nonlinear integral equations which arise from differential equations with nonlinear boundary condi- tions CHAPTER 1 Integral Equations 1.1 System Of Nonlinear Integral Equations 1.1.1 Introduction In this section we use a fixed point theorem of cone expansion and compres- sion to study the existence, nonexistence and multiplicity of positive solutions for systems of nonlinear integral equations of the form u = ,\ [G ki($,y)hi(y)f(v(y))dy (1.1) vs) = 11/0 k2h2g>dy where G is a bounded closed domain in Euclidean space R", and the nonlinear- ities f,g are subject to various combinations of sublinear and superlinear growth conditions at O and 00. We do not require any monotonicity assumptions on f and g. To be more precise, we introduce the notation f0 = lim ——f(”), 90 = lim —9(“) v—+0 v u—+0 u f00 = lim __f(v), 900 = lim __g(u) v—ioo v u—ioo u and distinguish the following four cases: Case I. f0 = go = 0 and foo = goo = 00 (superlinear). Case II. f0 = go 2 00 and fc,o = 900 z 0 (sublinear). Case III. f0 = go = f00 = 900 = 0 (super-sublinear). Case IV. f0 = go = f00 = goo = 00 (sub-superlinear). By a positive solution of (1.1) we understand a solution (u, v) E C (G) x C (G) with u 2 0,12 2 0 and either u i 0 or 1) $ 0. In what follows we are only interested in positive solutions. Our main results are given in Section 1.1.2, with the proofs given in Section 1.1.4. 1.1.2 Main Results The following conditions will be needed throughout: (A1): A and p are positive parameters. (A2): h1,h2 : G —> [0, 00) are continuous and do not vanish identically on any subset (of positive measure) of G'. (A3): [€1,162 : G x G —-) [0, 00) are continuous. (A4): There exists closed sets G1,G2 C G, with positive measures, and con- stants 0 < 81,52 < 1 such that for i = 1, 2 Isa-(any) > 0 for all (3:, y) E G,- X G.- Isa-(2:, y) 2 e.k,~(z,y) for all x 6 Ci, y E G, 2 E G. (A5): f,g : [0, 00) -—> [0,00) are continuous. (A5): f(v) > O for v > 0 and g(u) > O for u > 0. The main results of this section can be stated as follows: Theorem 1.1.1 (i) Assume (A1) — (A5) and Case I hold. Then system (1.1) has at least one positive solution. (ii) Assume (A1) — (A5) and Case II hold. Then system (1.1) has at least one positive solution. Theorem 1.1.2 (i) Assume (A1)—(A6) and Case III hold. Then there are positive constants 01, 02 such that system (1.1) has at least one positive solution if A, p = 01, at least two positive solutions if A, u > 01, and no positive solutions if A, u S 02. (ii) Assume (A1) - (A6) and Case IV hold. Then there are positive constants 03, 04 such that system (1.1) has at least one positive solution if A, u = 03, at least two positive solutions if A, p < 03 and no positive solutions if A, p Z 04. Corollary 1.1.3 Assume (A1) — (A6) hold. (i) If f0 = go = 0 or foo = goo = 0, then there is a positive constant 05 such that system (1.1) has at least one positive solution for A, [1. Z 05. (ii) If f0 = go = 00 or foo = goo = 00, then there is a positive constant 06 such that system (1.1) has at least one positive solution for A, u S 06. 1.1.3 Preliminaries Let Ava) = A/0k1(x.y)hi(y)f)dy Bum) = u/0k2(z.y)h2dy F(u,v)($) = (Av($),BU($))- Then (1.1) is equivalent to the fixed point equation F(u, v) = (u,v) in the usual Banach space X = C(G) xC(G) with ||(u, v)“ = max(||u||, ||v||), where IIUII = sup IU($)|- xEG The following fixed-point theorem of cone expansion and compression will be crucial in the arguments that follow. Theorem 1.1.4 ([15]) Let X be a Banach space and let K C X be a cone in X. Assume $21, 522 are open subsets ofX with 0 E 91, S21 C 92 and let F:Kfl(Qg\Ql)—+K be a completely continuous operator such that either (i) “Full 3 Hull, u e K0891 and “Full 2 “u“, u E Kflaflz 07' (ii) HFUII 2 Hall, u E K0391 and ||Fu|| 3 Nu”, u E K0892. Then F has a fixed point in K (1 (5-22 \ (21). In order to apply Theorem 1.1.4, let K be the cone defined by K = {(u, v) e X: u,v 2 0, min u(:r) 2 egllull, min ice) _>_ 51||v||} $602 $601 where 51, 52 are defined in (A4). We now establish several lemmas that will be used throughout. Let U={(u,v)€X: u,v20}. ThenK CU. Lemma 1.1.5 Assume (A1) — (A5) hold. Then F : U —> U is completely contin- uous and F(U) C K. PROOF To prove F(U) C K, choose (u,v) E U. Then for :r 6 G1, Av 2 an [G k1(z.y)hi(y)f(v(y))dy = est/122(2) for all z E G, and so min Av(x) 2 elllAvH. :rEGl Similarly, ' > gig; Butt) _ 62IIBUI| i.e., F(u, v) E K; hence F (U) C K. The complete continuity of F is obvious. [I In the following we set M = k , , k , >0 max{(z,1I/I)1€aC)¥(XG 1(3: y) (@3163ng 2(1): 31)} = ' m'n k as, , min k 51:, >0. m mln{($,y)€élel 1( y) (z,y)€szGz 2( 31)} Lemma 1.1.6 Assume (A1) — (A5) hold. Let 01 = {(u,v) E X: “(u,v)” < H1}. (a) If f0 = go = 0, then for H1 sufficiently small ||F(u,v)|| S “(u,v)“, (u,v) E K0091, (b) If f0 = go = 00, then for H1 sufliciently small ”F(u’v)” Z “(u,v)”, (11,0)6 K0301. PROOF (a) Since f0 = go 2 0, we can choose H1 > 0 so that f (v) g 611 and g(u) S eu for 0 S u, v 3 H1, where the constant e > 0 satisfies (1.2) We [G mm s 1, uMe [G h.(y)dy s 1. If (u, v) E K (1 691, we have Av (x) |/\ M [a hi(y)f(v(y))dy AMe [G h. 0 so that f(v) 2 17v, g(u) Z nu for 0 S u, v 3 H1, where n > 0 is chosen so that (L3) Amnei [G h1(y)dy Z 1, #7717782 [G h2(y)dy Z 1. l 2 If (u, v) E K n 691, then either min v(:i:) Z elllvll = 51|l(uav)ll 2:601 01‘ min M58) 2 €2IIUI| = €2|l(u. v)||- $602 In the first case, we have for any :5 6 G1 IV Av(x) Am (a. hie/wands Amn (a. hi(y)v(y)dy Amneinwm [G1 mom ||('u, v)“ IV IV IV and therefore, ||F(u.v)ll Z Av($) Z “(u,v)” for (u, v) E K D 801. In the second case, we similarly find that ||F(u,v)|| Z ”(u,v)ll 11 for (u, v) 6 KfldQl. CI Lemma 1.1.7 Assume (A1) — (A5) hold. Let Q2 = {(u,v) E X: ”(u,v)“ < H2}. (a) If foo = 900 = 0, then for H1 sufi‘iciently large ||F(u,v)|| S “(u,v)“, (WI) 6 K0392- (b) If f00 = goo = 00, then for H2 sufi‘iciently large ||F(u,’v)|| Z “(u,v)“. (WI) 6 K0392- PROOF (a) Define two new functions f’(v) = max f (s) and g‘(u) = max 9(3). 0535:: OSSSU Note that f ‘(v) and g*(u) are nondecreasing in their respective arguments. More- over, it easily follows from foo = goo = 0 that fgo = 9;, = 0 . Therefore, there is an H2 such that f*(v) s so, g‘(u) 3 ea for u,v 2 H2, where the constant e > 0 is chosen as in (1.2). If (u, v) E K n 692, we have Av(:c) |/\ AM [monotony AM Anni/mummy AM Limits/Milan AM5H2 [0111mm H2. |/\ |/\ |/\ |/\ 12 Similarly, Bu(:c) 3 H2, and so ||F(uiv)|| S H2 = “(u,v)“ for (u, v) E K r1892. (b) Since foo = goo = 00, there is an H > 0 such that f (v) 2 17v, g(u) 2 nu for u,v 2 H, where n is chosen as in (1.3). Let H2 2 max{II/el,H/eg}. If (u, v) E K n 602, then either min ”(117) Z 51W)”: 51W”: ’0)” 2 H $601 01' min u(a:) 2 Quit“ = 52||(u,v)|| 2 PI. 2:602 Now by a computation similar to the one done in Lemma 1.1.6(b), we find that ||F(uav)|| Z “(u,v)“ for (u, v) E K0802. Cl 1.1.4 Proofs of Main Results Proof of Theorem 1.1.1(i) According to Lemma (1.1.6) (a) and Lemma (1.1.7) (b), we can choose numbers 13 H1 < H2 such that “F(u’v)“ S “(“7”)”: (”’7”) E K0601 and ||F(u,v)|| Z “(u,v)”, (u,v) 6 K0392 where 9,- = {(u, v) e X: ”(u,v)“ < 11,} (i = 1,2). Hence, by Theorem (1.1.4), F has a fixed point (u, v) 6 K 0 (5-22 \ (21) such that H1 5 ||(u,v)|| 3 H2, and so (1.1) has a positive solution. Proof of Theorem 1.1.1(ii) This part follows similarly since, by Lemma 1.1.6 (b) and Lemma 1.1.7 (a), we can choose numbers H1 < H2 such that ||F(u,v)|| Z ”(u,v)“, (u,v) 6 K0391 and “F(uw)“ S “(u,v)”. (WI) 6 K0392 where Q,- = {(u, v) E X: “(u,v)“ < H,} (i = 1,2). Proof of Theorem 1.1.2(i) u... as... .2. .4 .. . . 4 4 teams fr‘mu: 1L4}:— ,:z:=JJ 33L 3 14 If q > 0, then it follows from (A1) — (A6) that 01(q)E m n mad/0, h1(y)f(v(y))dg,,) 2 0 mi (u,v)eK, “(u,v h2(y)f(U(y))dy) Z 0 012(0) m min (/ (u,v)EK, “(u,v)”:q G2 with (011(q))2 + (02(Q))2 > 0. For any number 0 < H3, let on = max{H3/al(H3),H3/a2(H3)} with the un- derstanding that H3/a,(H3) = 0 if 01,—(H3) = 0, and set 523 = {(u,v) E X: “(u,v)“ < H3}. Suppose 0:1(H3) > 0. Then for A 2 01 and (u,v) E K (1 603, we have for any .1} 6 G1 A’le) Z (hm/G h1(y)f(v(y))dy = 0101(H3) 2 H3- 1 Similarly, if a2(H3) > 0, we find that Bu(a:) 2 H3 holds if ,u 2 01. Hence ||F(u.v)|| 2 H3 = “(u,v)“ for (u, v) E K n 693. Note that if A,u > 01, then it is easy to see that F(u, v) # (u, v) for (u, v) E K D 693. Since f0 2 go 2 0 and foo = goo = 0, it follows from Lemma 1.1.6 (a) and Lemma 1.1.7 (a), that we can choose H1, H2, such that H1 < H3 < H2 and “F(uw)“ S “(u,v)“ 15 for (u, v) 6 Km 69,- (i = 1,2), where Q,- = {(u, v) E X: ”(u,v)” < H,} (i = 1,2). Applying Theorem 1.1.4 to 91, Q3 and {23, S22, and assuming A, ,u > 01, we get a positive solution (u1,v1) of (1.1) such that H1 _<_ ||(u1,v1)|| < H3 and another positive solution (u2,v2) of (1.1) such that H3 < [[(u2,v2)|| 3 H2. If A = ,u = 01, then we can only assert the existence of a single positive solution of (1.1). To prove the nonexistence part, we first note that from the given limit condi- tions, there is a constant c1 such that f (v) _<_ clv for v 2 0 and g(u) S clu for u 2 0. Now let (u, v) E X be a positive solution of (1.1), where A, u are small enough so that AMC1/Gh1(y)dll<1a uMci [Gh2(y)dy< 1. Then for a: E G, we have |/\ AMci/Gh1v(y)dy _ AMenw)” [Chin/My < ”(Hi vlll /\ with a similar computation for v(:i:), which is a contradiction. Proof of Theorem 1.1.2(ii) 16 If q > 0, then it follows from (A1) — (A5) that mg) a Mametttinlh.h1(y)f‘”‘y))dy) > o no) a M (u,v)epfiWJ/Gz monotony) > 0 3(4) 5 maX(fi1(CI), fi2(4)) For any number 0 < H4, let 03 = H4/fl(H4) and set 94 = {(u, v) E X: ”(u,v)“ < H4}. Then for A S 03 and (u, v) E K (1 604, we have Ave) s cm [G mummy 5 0.301.) = H. and similarly, for u S 03, Bu(:c) 3 H4, which implies ”F(Uiv)” S H4 S ”(u,v)” for (u, v) E K0894. Again, it is easy to see that if A, u < 03, then F(u, v) 79 (u, v) for (u, v) 6 6524. Since f0 = go = 00 and fc,o = goo = 00, it follows from Lemma 1.1.6 (b) and Lemma 1.1.7 (b), that we can choose two numbers H1, H2, such that H1 < H4 < H2 and ||F(u,v)ll 2 “(u,v)“ 17 for (u, v) E K n 652,- (i = 1,2), where Q,- = {(u,v) E X: ”(u,v)“ < H,} (i = 1,2). As above, applying Theorem 1.1.4 to (21, Q4 and Q4, 92, and assuming A, u < 03, we get two positive solutions of (1.1), whereas if A = a = 03, then we can only assert the existence of a single positive solution of (1.1). To prove the nonexistence part, we again note that from the given limit con- ditions, there is a constant c2 such that f (v) 2 cm for v _>_ 0 and g(u) 2 cm for u 2 0. Let (u, v) E X be a positive solution of (1.1). By Lemma 1.1.5, (u, v) E K. Now choose A, a large enough so that Amc281/G h1(y)dy > 1, [1.7726282]; h2(y)dy > 1. l 2 If “(u,v)“ = ||v||, then for a: 6 Cl, we have u(:r) 2 Amcz [G h1(y)v(y)dy l 2 Amazeiuw)” [G h1(y)dy l > “(u,v)“- Similarly, if ||(u,v)|| = ||u||, then for a: 6 G2, we arrive at v(II?) > “(u,v)”- In either case, we have an obvious contradiction. 18 Proof of Corollary (1.1.3)(i) and (ii) Referring to the proof of Theorem 1.1.2(i), we apply Theorem 1.1.4 to 01 and $23 if f0 = go = 0, and to (22 and (23 if foo = goo = 0. Similarly, referring to the proof of Theorem 1.1.2(ii), we apply Theorem 1.1.4 to (21 and 94 if f0 = yo = 00, and to {22 and 9.; if foo = goo = 00. 1.2 Case IV Continued 1.2.1 Introduction In this part we consider system (1.1) with A = u : nos) = A [G ki(x.y)hi(y)f(v(y))dy (1.4) Ax) = A [G k2(x.y)h2(y)g(u(y))dy We will prove a more precise existence theorem for case IV under some added positivity and monotonicity conditions on f and g. 1 .2.2 Main Results The following conditions will be assumed throughout: (A’l): A is positive parameter. (All): hl, I12 : G —> [0, 00) are continuous and do not vanish identically on any subset (of positive measure) of G. (Ag): k1, k2 : G x G —> [0, 00) are continuous. (Ag): There exists closed sets G1,G2 C G, with positive measures, and con- l9 stants 0 < 51,52 < 1 such that for i = 1, 2 k,(:1:,y) > 0 for all (113,31) 6 G; X G,‘ k,(a:,y) 2 e,k,~(z,y) for all :1: E G,, y E G, 2 E G. (A’s): f, g : [0, 00) —> (0, 00) are continuous and nondecreasing. (Ali): foo = 900 = 00 Our main theorem follows. Theorem 1.2.1 Assume (A’l) — (Ag) Then there exists a positive number A" such that system (1.4) has at least two positive solutions for 0 < A < A“ , at least one positive for A = A" and no positive solution for A > A“. In proving this result, we shall employ upper and lower solution methods to- gether with degree theory arguments. 1.2.3 Upper-lower solutions We shall say that the pair (a, o) E C(G) x C (G) is an upper solution for the system (1.4) if am _>_ A [G k.(x.y>hi(y)f)dy (1.5) 17(3) _>_ A [G k2(z,y)h2(y)g(fi(y))dy. and the pair (u, y) E C(G) x C(G) is a lower solution if 20 u(x) S A [G k1(:r, y)hi(y)f (vb/Dds (1.6) 21(17) 3 A [G k.(x,y)h2dy In the following, we shall write (u1,v1) g (u2,v2) if u1(:2:) g u2(:z:), v1(:z:) g v2(:c) holds for all a: 6 G. We now establish several lemmas that will be used throughout. Lemma 1.2.2 Assume (A'l) - (Ag) . If there exists an upper solution (11, o) and lower solution (:4, y) of (1.4) such that (u, y) S (a, '17), then there is a solution (u, v) of system (1.4) such that (use) S (u,v) S (11,17)- PROOF We construct a sequence {am on}, by iteration, using the following equa- tions With {to = 17., 170 = 17 me) = A [knewhnywwnondy (1.7) 17n+1(37)= A [G k2($,y)h2(y)g(fin(y))dy- First we have a. — a. s A [G kt(x,y)hi(y)(f(fi(y)) — f(?7(y)))dy = o 271- v. _<. A [G to, y)h2(y)(9(fi(y)) — g>dy = o 21 which implies 111 S 2'10 and 271 S 170. Therefore, by the monotonicity of f and g, — a. = A [G kitz.y)h.(y) — f(v"o(y)))dy s o a. -— 271 = A [G kitz,y)h2(y)))dy s o and so 212 5 711 and 172 g 171. By induction and monotonicity of f and g we can show that and ...gt7,,g...goggolgo. Starting from yo = y, yo = y we construct another sequence {um an}, by iteration, using the following equations e...(s) = A [G ki(z.y)hi(y)f 0 such that g(u) Z nu, where n is chosen so that Aincmzewrs/ hi(y)dy/ h2(y)dy>1. G1 02 Choosing n large enough so that €2llunll 2 q, and taking 1:1 6 G1 and 222 6 G2, we have llunll Z un(ar1) Z Mme/G h1(y)’vn(y)dy l _>_ Amman/G windy and llvnll Z vn($2) 2 Anmn/G h2(y)un(y)dy Z A11777’5277llunll/G (It’ll/My- Therefore nu..u _>_ Amman/G handy I 2 Aincmzeieelc hi(y)dy/G h2(y)dyllun|| l 2 > llunll 24 which is a contradiction. C] Now let I‘ denote the set of A > 0 such that a positive solution of system (1.4) exists. Let A" = sup{I‘}. By Theorem 1.1.2(ii), I‘ is nonempty and bounded, and thus 0 < A“ < 00. We claim that A“ E I‘. To see this, let An —> A“ where An 6 I‘. Since the An are bounded, Lemma 1.2.3 implies that the corresponding solutions (un, v”) are bounded. By the compactness of the integral operators of system ( 1.4), it follows that A‘ 6 I‘. Now let (u*, v*) be a solution of system (1.4) corresponding to A“. Lemma 1.2.4 Assume (A’l) — (Ag). Then (u* + e, v* + e), for 8 small enough, is a positive upper solution of system (1.4) for any A < A“. PROOF Since (11*, v*) 2 0 there is a constant c1 that f(v*(t)) > c1 > 0,g(u*(t)) > c1 > 0 for all a: E G. By uniform continuity, there is an 5* such that [f(v*($) + E) - f(v*($))| < 610* - A)//\ |9(v*($) + 8) - 9(v*($))| < 610* - A)//\ for alle G,0 Se 35*. Now u* +5 2 A* [0de y)h1(y)f(v*(y))dy = A [G ki(x,y)ht(y)f(v* + e)(y)>1dy and let SI = {:13 E Q : v(:i:) g v*(at) +8} 52 = {x E Q : v(a:) > v*(at) + E}. In 81 we see, by monotonicity, that ~ f(v) = f(v) S f(v*+€) whereas in 32 ~ f(v) = f(v* +5). Consequently, u g (u* + e), and similarly, v S (u* + 5). Since u, v 2 0, we have proved the lemma. C] 1.2.5 Proof of Main Result Since (u*, v*) is an upper solution and (0,0) is a lower solution (f (0) > 0, g(0) > 0) for any A < A*, by Lemma 1.2.2 there is a solution (u,\, v,\) of system (1.4) such that (0, 0) g (u), v,\) S (u‘, v*). Thus for 0 < A S A“ a positive solution exists, whereas for A > A‘ a positive solution does not exist. Moreover, the pair w,\ = (u A, v1) 6 B. 27 In order to show that a second positive solution exists for A < A“, we will use degree theory arguments. Since F,\ is bounded for A in compact intervals, deg(I — FA, B(w,\, R), 0) :1 if R is large enough, where B (w)(, R) E C (G) is the ball centered at w,\ with radius R. If there exists a w = (u, v) E 8B such that w = Fi(w), then (f, g) = (f,g) and so a second solution exists for A. Now suppose w 7e F,\(w) for all w E 63. Then deg(I — FA, B, 0) is well defined. It follows from Lemma (1.2.5) that F ,\ has no fixed point in B (w;, R)) / B. We then have, by the excision property of degree, deg(I — FAsBiO) + deg(I - FAaB(w2\)R)/B’O) = deg(I — FA,B(wA,R)i0)=1- Since the second term on the left is 0 and deg(I — F,\, B, 0) = deg(I - FA, B, 0), it follows that deg(I - FA, B, 0) = I. On the other hand, by Lemma 1.2.3 all solutions are bounded for A in compact sets, and thus deg(I — FA, B(0, L), 0) = const for L large enough, where B (0, L) E C (G) is the ball centered at 0 with radius L. The latter degree must equal 0, since for A > A* no solutions exist. Finally, by the 28 excision property, 0 = deg(I — FAiB(01 L)10) : deg(I — FA1B(01L)/Ba0) + deg(I _ FAiBiO) which implies deg(I — FA,B(0,L)/B,0) = —1 and so a second positive solution of system (1.4) exists for 0 < A < A“. 1.2.6 Elliptic Systems Consider now the problem of finding positive radial solutions for an elliptic system of the form Au+ Ah1(|$|)f(v) = 0 in R1 < [ml < R2 Av + uh2(|:r|)g(u) = 0 in R1 < [col < R2 (1'9) Bu 8v alu+fll—T—l = 0, a2v+flzg = 0 on [ml 2 R1 u 0v 71u+615£ = 0, ’72?) +6251: = 0 on [11] = R2 where cu, fi,,7,~, 6,- 2 0 and P15 Vifli+01i7i+ CH5.- > 0 i=1i2- Here G = {:t E R" : R1 3 [ml 3 R2, R1, R2 > 0} is an annulus with boundary 8G and 6/6n denotes the outer normal derivative on 6G. By a positive solution of (1.9) we understand a solution (u, v) E C2(G) X C2(G), with u, v 2 0 and not both identically zero. Note that by the maximum principle, each nontrivial nonnegative 29 component of a solution (u, v) is clearly positive in Q. We shall seek criteria for the existence of positive radial solutions 21 = u(r) of (1.9) which then satisfy n—l u"(r) + r u'(r) + Ah1(r)f(v(r)) = 0 in R1 < r < R2 n —1 r 011.1.(R1) — ,61U’(R1) = 0, ’71U(R2) + 61UI(R2) = 0 u”(r) + v'(r) + uh2(r)g(u(r)) = 0 in R1 < r < Hz (1.10) (121}(R1) — flit/(R1) = 0, ’YQ'U(R2) + 62U,(R2) =7 0. By applying the change of variables 3 = — LR” (1 /t"’1)dt, followed by the change of variables t = (m — s)/m, where m = — £2(1/t""1)dt, system (1.10) can be brought into the form u"(r) + Ah1*(:r)f(v(:v)) = 0, 0 < :c < 1 (1.11) u"(z) + uh2*(:r)g(u($)) = 0, 0 < :1: < 1 alu(0) — Blu'(0) = 0, 'ylu(1) + 61u’(1) = 0 0211(0) — flgv'(0) = 0, 72v(1) + 62v'(1) = 0. where h1*(t) = m2r2<"-1)(m(1—t))hi(r(m(1—t))) h2*(t) = m2r2("“1)(m(1 — t))h2(r(m(1 — t))). It is easy to check that h1*, h2* satisfy (A2) or (A’Z). On the other hand, system (1.11) is equivalent to the system of integral equa- 30 tions use) = A [0 ki(x.y)hi*(y)f(v(y))dy vo) = 11/01k2(z,y)h2*(y)f(U(y))dy where [9,-(23, y) (i = 1, 2) is the Green’s function (1.12) 1 (71+ 51— 7i$)(16i+ Qty), 3/ S 17 ki(z’y) = __ p‘ (fl.- + a.$)(7.- + 6.- - 713/). x S y. We now verify that k,(a:, y) satisfies condition (A4) or (A;) . Set G = [0,1],G, = [1/4, 3/4]. Clearly k,(:c,y) Z %(5.- + %)(,Bi + %) > 0 for all :v,y E G,. Moreover, for a: E G,, we have 1 (51+ 1.1;")(161'1' oat-y). 0 S y S :1- kiW-‘D 2 ; (6.+%*)(B.-+%i). i y. Then the given problem is equivalent to the fixed point equation Au = u in the usual Banach space X = C([0,1]) with “u“ = sup |u(:r)|. Note that (1.14) zE[0,l] is a special case of (1.13). In Section 1.2.6, we showed that k(:t:, y) 2 i%k(z,y) for all :2: E [1/4,3/4], y,z E [0,1], and therefore, g(x, y) 2 1i6g(z,y) for all :1: E [1/4,3/4], y,z E [0,1]. Consequently, the previous results are clearly seen to apply to the above problem. CHAPTER 2 Nonlinear Boundary Conditions 2.1 Nonlinear Integral Equations 2.1.1 Introduction In this section we study the existence, nonexistence and multiplicity of positive solutions for systems of nonlinear integral equations of the form 1 (21) MI) = uP(U(0), U(1))I + uQ(U(0). u(1)) + A/o M23, y)h(y)f(U(1/))dy- The above integral equation arises from problems in differential equations with nonlinear boundary conditions as we will see in Section 2.2.6. The nonlinearities f, P and Q are subject to various combinations of sublinear and superlinear growth conditions at 0 and 00. We do not require any monotonicity assumptions on f, P and Q. 33 34 We introduce the notation f0 = lim f—(T—Q, f00 = lim M u—+0 u u-*00 11 P0 _ hm F(uw), Q0 _ Q(U.v) _ (u,v)—>0 ’U. + ’U — (u,v)-+0 u + ’U P..— ,m F(uw), Qm_ out» — 1 — lim ||(uiv)||-+°° U + v ||(u,v)||-+co u + ’U and distinguish the following four cases: Case I. f0 = P0 = Q0 = 0 and foo = 00 (superlinear). Case 11. f0 = 00 and ft,0 = Pco = Q00 = 0 (sublinear). Case III. f0 = foo = P0 = Q0 = 0 (super-sublinear). Case IV. f0 = foo = 00 (sub-superlinear). By a positive solution of (2.1) we understand a solution u E C([0,1]) with u 2 0 and u i 0. In what follows we are only interested in positive solutions. Our main results are given in Section 2.1.2, with the proofs given in Section 2.1.4. 2.1.2 Main Results The following conditions will be needed throughout: (A1): A and u are positive parameters. 35 (A2): h : [0, 1] —+ [0, 00) is continuous and does not vanish identically on any subset (of positive measure) of [0,1]. (A3): k(:c, y) is nonnegative and continuous on [0,1] x [0,1]. (A4): There exists a closed interval [0, B] C [0, 1], 0 < a < B < 1, and constant 0 < 81 < 1 such that May) > 0 for all (WI) 6 [CW] >< [CW] k(:c,y) Z elk(z,y) for all a: E [01,6], y E [0,1], 2 E [0,1]. (A5): f : [0, 00) —> [0,00) is continuous. (A6): P and Q: [0,oo) X [0,00) to (—00, 00) are continuous; moreover F(u, v):1: + Q(u,v) 2 0 for all a: E [0,1] and u,v 2 0. (A7): f(u) > 0 for u > 0. The main results of this section follow: Theorem 2.1.1 (1) Assume (A1) — (A6) and Case I hold. Then (2.1) has at least one positive solution. (ii) Assume (A1) — (A6) and Case II hold. Then (2.1) has at least one positive solution. Theorem 2.1.2 (i) Assume (A1)—(A7) and Case III hold. Then there are positive constants 01, 02 such that (2.1) has at least one positive solution if A = 01 ,a > 0, at 36 least two positive solutions if A > 01 u > 0, and no positive solutions if A, u S 02. (ii) Assume (A1) — (A7) and Case IV hold. Then there are positive constants 03,04 such that (2.1) has at least one positive solution if A, u = 03, at least two positive solutions if A, u < 03 and no positive solutions if A 2 04, u > 0. Using arguments as in the proof of Theorem 2.1.2, we get the following results. Corollary 2.1.3 Assume (A1) — (A7) hold. (1) If f0 = P0 = Q0 = 0 or foo = PC,o = Q00 2 0, then there is a positive constant 05 such that (2.1) has at least one positive solution for all A Z 05, p > 0. (ii) If f0 = 00 or foo = 00, then there is a positive constant 05 such that (2.1) has at least one positive solution for all A, u g 05. 2.1.3 Preliminaries Let AWE) = uP(U(0), U(1))$ + uQ(u(0), u(1)) + /\ [01 Mar, y)h(y)f(U(y))dy. Then (2.1) is equivalent to the fixed point equation Au = u in the usual Banach space X = C([0, 1]) with “all = sup IU($)| xE[0,1] 37 In order to apply Theorem (1.1.4), let K be the cone defined by K = {u E X: 21. _>_ 0, reninmuh) 2 52[[u[|} where 52 = min{a, 1 — 6,51}. We now establish several lemmas that will be used throughout. Let U={uEX: uZO}. ThenK CU. Lemma 2.1.4 Assume (A1) — (A6) hold. Then A : U —) X is completely contin- uous and A(U) C K. PROOF To prove A(U) C K, choose a E U. Since P(U(0),U(1))$ + Q(U(0),U(1)) is a linear function of :r, it follows that P(u(0), u(1))a: + Q(u(0), u(1) 2 min{Pa + Q, PB + Q} for a: E [(1, 6]. Thus Aucc) = uP(U(0).u(1))$+uQ(U(0),U(1))+/\/01k(x.y)h(y)f(U(z/))dy 38 IV umin{a, 1 — B}(P(u(0), 11(1))2 + Q(u(0)1u(1))+ +Ae. [01 Hz. nut/Wendy lemme). u(1))z + o)+ As. [01 k(z. y)h(y>f(u>dy €2A’U.(Z) IV IV for all z E [0,1], and so ' A > A , $33,] u(-'L‘)_€2l| u|| i.e., A(u) E K; hence A(U) C K. The complete continuity of A is obvious. D In the following we set M— — max km, >0 (:1:,y)E[0,l]x[0,1] ( y) = min k 11:, >0. (I.y)€[a.fll> 0 so that f (u) S eu and umax(Q(u, v), P(u, v) +Q(u,v)) S %(u+v) for 0 S u, v S H1, where the constant e > 0 satisfies 1 (2.2) AM: [0 h(y)dy g [\DIH Ifu E K0891, we have Au(:1:) |/\ umax(Q(u(0), 11(1)), P(u(0), “(Ill + Q(u(0). 71(1)» + +AM/01 h(y)f(U(y))dy gnu“ + AM»: / h(y)u(y)dy |/\ l 0 1 1 gllull + AMeuuu f0 h(y)dy IIUII- |/\ |/\ Hence HAUII S [IUII for u E K (1 691. (b) If f0 = 00, we can choose H1 > 0 so that f(u) 2 nu, for 0 S u S H1, where n > 0 is chosen so that l3 (2.3) Amn52 / h(y)dy 2 1. a If u E Imam, then . > . e’e’llflnum _ €2IIUI| 40 We have for any x E [0, H] 19 Am) 2 Am l. hf>ds l3 2 Amn/a h(y)U(y)dy l3 _>_ Amn€2IIUI| / (new > IIUII and therefore, “A(U)“ 2 ”U“ foruEKflBQl. [:1 Lemma 2.1.6 Assume (A1) — (A6) hold. Let {22 = {u E X: [lull < H2}. (a) If foo 2 P00 = Q00 = 0, then for H2 sufi‘lciently large. IIAUII S “all, u 6 K0 392- (b) If f00 = 00, then for H2 sufliciently large [[Aull 2 [[u[[, u e K n 312,. PROOF (a) Define new functlons f*(u) 2 £133. f(s) , Q (11) = ogféuQ(s’ t) and 41 (P + Q)‘(u) = 0<19:1.t12u(P(s,t)+ Q(s, t)). Note that f“(u), Q‘(u) and (P + Q)*(u) are nondecreasing in. their respective arguments. Moreover, it easily follows from f00 = PC,o = Q00 = 0 that jg, = Q30 2 (P + Q):o = 0 . Therefore, there is an H2 such that f*(u) S eu, pmax(Q"(u), (P + Q)‘(u)) S in for u 2 H2, where the constant e > 0 is chosen as in (2.2). If u E K n 692, we have Au(:1:) l/\ pmax(Q(u(0), 21(1)), P(u(0), u(1)) + Q(U(0), u(1))) + +AM [01 h(y)f(U(y))al:t/ AmaX(Q’(2H2), (P + @1211.» + AM [01 h(y)f*(n(y))dy 1 1 5H2 + AME/0 h(y)U(y)dy 1 $172 + AMeH2 [0 h(y)dy H2. |/\ l |/\ |/\ Hence ”A(U)” S H2 = “UN for u E K n 692. (b) Since f00 = 00, there is an II > 0 such that f (u) 2 nu for u 2 II, where n is chosen as in (2.3). Let H2 2 3/52. If u E K 0 892, then min 2: >5 >H. as)“ )_ nun- 42 We have for any :15 E la, 6] 3 Ana) 2 Am l. hf>ds fl 2 Amn/a hudy B Z Amn€2HUII/a h1(y)dy 2 ”UN and therefore, “A(u)“ Z IIUII foruEKflBflg. D 2.1.4 Proofs of main results Proof of Theorem 2.1.1(i) According to Lemma 2.1.5 (a) and Lemma 2.1.6 (b), we can choose numbers H1 < H2 such that “A(u)” S Ilull. u e K o 891 and ”A(U)” Z HUII. u 6 K0002 where Q,- = {u E X: [lull < H,-} (i = 1,2). Hence, by Theorem 1.1.4, A has a fixed point 11 E K (1 (92 \ 01) such that H1 S 43 [lull _<_ H2, and so (2.1) has a positive solution. Proof of Theorem 2.1.1(ii) This part follows similarly since, by Lemma 2.1.5 (b) and Lemma 2.1.6 (a), we can choose numbers H1 < H2 such that “A(U)“ 2 “all. u E K0391 and “A(U)” S IIUII, u 6 K F7002 where {Ii 2 {’U. E X I [lull < Hi} (121,2). Proof of Theorem 2.1.2(i) For q > 0 it follows from (A1) — (A7) that [3 a a min (m / h(v)f(U(y))dy) > o. “6K1 IIuII=q ‘1 For any number 0 < H3, let 01 = H3/a(H3) and set 93 = {u E X: [lull < H3}. Then for A Z 01 and u E K 0 6523, we have for any a: E la, 6] [6 Aux) 2 mm / h(y)f(U(y))dy 2 claw.) = H. a 44 Hence “A(U)“ 2 H3 = IIUII for u E K n 603. Since f0 = P0 = Q0 = 0 and foo = Poo = Q00 2 0, it follows from Lemma 2.1.5 (a) and Lemma 2.1.6 (a), that we can choose H1, H2, such that H1 < H3 < H; and ”A(U)“ S “U“ for u E K n 052,- (i = 1,2), where 0i = {’U. E X I “It“ < H,} (i=1,2). Applying Theorem 1.1.4 to (21,523 and 03,92, and assuming A > 01, we get a positive solution 111 of (2.1) such that H1 S [lulll < H3 and another positive solution ug of (2.1) such that H3 < ”“2“ S H2. If A = 01, then we can only assert the existence of a single positive solution of (2.1). To prove the nonexistence part, we first note that from the given limit condi- tions, there is a constant c1 such that f (u) S clu for u 2 0 and max(Q(u. 12). P(u. v) + Q(u. v)) s ci(u + v) for u, v 2 0. Now let u E X be a positive solution of (2.1), where A, u are small enough so that 1 1 l AM01/ h(y)dy < —, #01 < —. o 2 4 45 Then for a: E [0, 1], we have |/\ uc1(u(0) + 11(1)) + AMcl /01h(y)u(y)dy gnu“ + AMclnun [0‘ man < IIUII u(1r) l/\ which is a contradiction. Proof of Theorem 2.1.2(ii) For q > 0, let 1 = M ax / h d > 0. A(q) uEKrlllull=q( 0 (y)f(U(y)) 3/) For any number 0 < H4, let {24 = {u E X: [lull < H4}. We then choose 03 > 0 such that 1 03 S 5114/5014) and 03 maX(Q(U(0),U(1))iP(U(0),U(1)) + Q(U(0),U(1))) S gm, u E K 0 094. 46 Then for A, u S 03 and u E K (1 694, we have Aux) 3 $11. + 03M [01 h(y)f(U(y))dy s $114+ 0.501.) s H... which implies ”A(U)“ S H4 S IIUII for u E K n 604. Since f0 = go 2 00, it follows from Lemma 2.1.5 (b) and Lemma 2.1.6 (b), that we can choose two numbers H1, H2, such that H1 < H4 < H2 and ”A(U)“ 2 IIUH for u E K (1 852,- (i=1, 2), where Q,- : {u E X: [lull < H,} (i=1,2). As above, applying Theorem 1.1.4 to S21, S24 and Q4, 92, and assuming A, u < 03, we get two positive solutions of (2.1), whereas if A = u = 03, then we can only assert the existence of a single positive solution of (2.1). To prove the nonexistence part, we again note that from the given limit condi- tions, there is a constant c2 such that f (u) 2 C; for u 2 0. Let u E X be a positive solution of (2.1). By Lemma 2.1.4, u E K. Now choose A large enough so that 5 ATHCgEQ/ h(y)dy > 1 a 47 Then for :1: E la, 6], we have no) 2 Ame/fuming [3 2 Am0252ll’ull/a hdy >|WH which is an obvious contradiction. Proof of Corollary 2.1.3(i) and (ii) Referring to the proof of Theorem 2.1.2(i), we apply Theorem 1.1.4 to I21 and $23 if f0 = P0 = Q0 = 0, and to $22 and 93 if foo = P00 2 Qco = 0. Similarly, referring to the proof of Theorem 2.1.2(ii), we apply Theorem 1.1.4 to {21 and $24 if f0 = 00, and to {22 and 94 if f.>0 = 00. 2.1.5 Elliptic Systems Consider now the problem of finding positive radial solutions for an elliptic system of the form All + Ah(l$l)f('l.l.) = 0 III R1< l$l< R2 Bu (2-4) au + 6—1; = ugl(u(l:1:l)), on [:1:l = R1 ’7“ + 65:- = moan», on Ix) = R. where a, 6, 7, 6 Z 0 and pEah+®+afi¢0 and 91,92 : [0,00) —) [0,00) are continuous. By a positive solution of (2.4) we understand a solution u E 02([0, 1]), with u 2 0 and not identically zero. 48 As in Section 1.2.6, the existence of positive radial solutions of (2.4) leads us to the system u"(x) + Ah*(a:)f(u(:t)) = 0, 0 < :1: < 1 0M0) - AUTO) = 1191040)). 7U(1)+ 511(1) = 1192010)) (2.5) where h*(t) = m2r2("‘1)(m(1—t))h(r(m(l -t))). It is easy to check that h* satisfies (A2)- On the other hand, (2.5) is equivalent to the integral equation (2.6) use) = wow). mm + uQ(u(0).u(1)) + A [0‘ to, y)h*(y)f(U(y))dy where P(u, v) = 0192(1)) ; 791(1”), Q(u, v) = (7 + (5)91 (1;) + 592(1)) and k(:1:, y) is the Green’s function (7+5-723)(l3+ay). ySsc Hz 11):}- , ’0 (fi+a$)(7+5-'ry), xSy. As in Section 1.2.6, k(:1:, y) satisfies condition (A4). Moreover, it is easy to see that (a + mum) + 69100) 2 0. P(u, v) + Q(u, v) = p 49 Therefore P(u, v):t + Q(u, v) 2 min(Q(u, v), P(u, v) + Q(u, v)) 2 0 for all :1: in [0,1]. It follows from (g1)0 = (92)o = 0 that P0 = Q0 = 0, and from (g1)oo = (g2)co = 0 that Poo = Q00 = 0. Thus, conditions (A4), (A6) are satisfied and all the previous results apply to (2.4). Theorem 11.7 (i) Assume (A1).(A2).(As).(A7), fo = (900 = (92)o = 0 and foo = 00. Then (2.4) has at least one positive radial solution. (ii) Assume (A1):(A2)1(A5),(A7), f0 = 00, (91)o = (92)o = 0 and foo = (91).,o = (92).» = 0. Then (2.4) has at least one positive radial solution. Theorem 2.1.8 (i) Assume (A1),(A2),(A5) and (A7) and f0 = foo = (g1)°o = (g2)°° = 0. Then there are positive constants 01,02 such that (2.4) has at least one positive radial solution if A = 01 u > 0, at least two positive radial solutions if A > 01 ,u > 0, and no positive radial solutions if A, u S 02. (ii) Assume (A1),(A2),(A5) and (A7) and f0 = fc,o = 00. Then there are positive constants 03, 04 such that (2.4) has at least one positive radial solution if A, u = 03, at least two positive radial solutions if A, u < 03 and no positive radial solutions if A, u Z 04. 50 2.1.6 Remark In the previous sections we have assumed that f is not identically zero. Let’s consider the problem (2.4), or equivalently the problem (2.5) when f (u) = 0 for all u E [0, 00). The equivalent integral equation is (2-7) u(:17) = ”P(u(0), 11(1)):10 + uQ(U(0),U(1)) P(u, v) = a92(v) ; 791W), Q(u, v) = (7 + 6)91(:l + .592“). Let (gi)o=lgf(l) u . (gi)oo=ul_l_{§o u for i=1,2. Then all previous theorems are true. For example, we can prove the following theorem Theorem 2.1.9 (i) Assume (g1)o = 0 and (92)“, = 00. Then (2.4) has at least one positive solution for 0 < u < 00 (ii) Assume (g1)o = 00 and (g2),o = 0. Then (2.4) has at least one positive solution for 0 < [1 < oo. 51 In order to prove this theorem, let K be the cone defined by K={uEX:u=a:1:+b20a,bElR, :1:E[0,1]}. If u(:1:) = a2: + b E K, it follows that “till = maX(u(0),U(1)) and min(a, a + b) S u(:1:) S max(a, a + b) for all :1: E [0,1]. The proof of the theorem is based on the two lemmas Lemma 2.1.10 Let 01 = {'U. E X I ”U” < H1}. (a) If (gl)o == (g2)o = 0, then for H1 sufl‘iciently small [[Aull g [lu[[, u 6 1m on, (b) If (g1)o = (92)0 = 00, then for H1 sufl‘lciently small [[Au[[ 2 llull, u E Kfl 8521. Lemma 2.1.11 Let $22 = {u E X: [lull < H2}. 52 (a) If (g1)oo = (g2)oo = 0, then for H2 sufiiciently large [[Aull 3 “all, u e K n 09,. (b) If (g1)°o = (g2)<,o = 00, then for H2 sufiiciently large [[Aull 2 [[u[[, u e K n 39,. Let 1411(2) = uP(U(0). U(1))$ + uQ(U(0). 11(1)). 01‘ age(U(1)) - 791(U(0))$ + ”(7 + 5)91(U(0)) + 592(U(1))_ Aux: () u p p It is obvious that A maps K —+ K and is compact from C[0,1] to C[0,1]. Proof of Lemma 2.1.10 If (g1)o = (g2)o = 0, we can choose H1 > 0 so that 91(11) S EU for u S H1, where the constant e > 0 satisfies M(7+6)e+fle <1, and 53 If u E KG 8521, we have Au(:1:) S umax{Au(0),Au(1)} (7 + 6)gi(u(0)) + (392041)) = #max{ p S Mllullmax{(y+6)e+fle,(7+6)05+fie} 5 Hull- Hence IIAUII S IIUII for u E Kflafll. If (g1)o = (g2)0 = 00, we can choose H1 > 0 so that g,(u) Z Ku for u S H1, where K satisfies (7+6):{+6K Z 1. If u E Kn Bill, we have ”(’7 + 5)91(U(0)) + (392041)) p (7+6)K+6K p Au(0) lV #ll’ull IIUII. IV 54 Thus for H1 sufficiently small [[Aull 2 Hull, u E K0891. By the same arguments, one can prove that the lemma (2.1.11) is true. It then follows that Theorem(2.l.9) is true. 2.2 Case IV Continued 2.2.1 Introduction We now consider problem (2.5) with A = u : u"(23) + Ah($)f (“(10) = 0 (2-8) 13001) (0) = 0111(0) - MTG) = A91 (11(0)) Bl (10(1) = 711(1) + 5U'(1) = A92(U(1)) where a, 6,7, 6 Z 0 and psyfl+a7+ad > 0. We will prove a more precise existence theorem for the case IV under some added positivity and monotonicity conditions on f, g1, g2. 2.2.2 Main Result The following conditions will be assumed throughout: (A’l): A is a positive parameter. 55 (A’z): h : [0, 1] —+ [0, 00) is continuous and does not vanish identically on any subset (of positive measure) of [0, 1]. (Ag): f : [0, 00) —> (0,00) is continuous and nondecreasing. (Ag): g1, g2 : [0, 00) —> (0,00) are continuous and nondecreasing. (Ag): f0, = limum L92 = 00. The main result follows. Theorem 2.2.1 Assume (All — (Ag). Then there exists a positive number A“ such that (2.8) has at least two positive solutions for 0 < A < A‘ , at least one positive solution for A = A“ and no positive solutions for A > A’. Our approach to this problem is similar to the approach we used in Section 1.2. 2.2.3 Upper-lower solutions For (2.8), 11 E C2[0, 1] is an upper solution if fl"($13) + Ah($)f(fi($)) S 0 3003(0) 2 Audi-1(0)). Bi(fi)(1) Z Agni-1(1)) and u E C2[0, 1] is a lower solution if 9"(513) + Ah($)f(u($)) Z 0 Bo(u)(0) S Ag1(y_(0)), 31(9)“) S Agzflllll- 56 We now establish several lemmas that will be used throughout. The first result is a simple consequence of the classical maximum priciple. In particular it shows that all solutions of (2.8) are nonegative. Lemma 2.2.2 Suppose a,fi,7,6 _>_ 0 and 76 + a”) + (16 > 0. If u E C2[0,1] satisfies u" S 0, x E (0,1), au(0) — flu’(0) 2 0, and 'yu(1) + 6u’(1) Z 0, then u 2 0 for allx E [0,1]. Let 11, u be upper and lower solutions for ( 2.8) and define f(fi(x)). u(x) _>. 11(1) f*(u(a=)) = f(u(:v)). 2(2) 3 u(x) s u(x) f(u(2=)). u(36)Su(:r) 9107(3)), u(113) 2 WC) 91*(u(x)) = g1(u(x)), u(x) S u(x) S u(x) 9101127)), u(at) S u(x) 92(fi(x)). u(102 fl(2:) 92*(u($)) = 9201(2)). u(x) _<_ u(x) s so) 92(y.(2=)). u(x)Su(=v)- Consider the following problem u"(56) + Ah($)f*(u($)) = 0 (2.9) Bo(U)(0) = C"1(0) - flU'(0) = A91*(U(0)) 3100(1) = 711(1) + 51/0) = Ag2*(U(1))- 57 Lemma 2.2.3 Assume (A’l) — (A1,). If there is a solution u(x) of {2.9), then u(x) S u(x) S u(x) for x in [0,1]. In other words, u(x) is a solution of {2.8). PROOF We first prove that u(x) S u(x). Suppose to the contrary that u(xo) > u(xo) for 230 E [0, 1]. Then there are four cases: (a). u(x) > u(x) for all x E [0,1]. In this case, we have f*(u($)) = “71(30): 91*(U(0)) = 9101(0)), 92*(u(0)) = 9201(0))- Therefore (a — u)" S O 30(11 — 10(0) 2 0, 31(17 - 10(1) 2 0. which by lemma 2.2.2, implies that 11 2 u, which is a contradiction. (b). u(x) > u(x) for all x E (a, 1], where 0 < a < 1, and u(a) = u(a). Therefore (a -— u)" S 0 for all x E [a, 1] u(a) — u(a) = 0, B2(1‘1 — u)(1) Z 0. By Lemma 2.2.2 we conclude that a 2 u for all x E [a, 1], which is a contradiction. (c). u(x) > u(x) for all x E [0, a) where 0 < a < 1, and u(a) = u(a). The proof of this case is same as case (b). (d). u(x) > u(x) for all x E (a, b), where 0 < a, b < 1, and u(a) = u(a), u(b) = u(b). Therefore (11 — u)" S 0 for all x E [a, b] 58 {1(a) — u(a) = 0, u(b) — u(b) = 0. By Lemma 2.2.2 we conclude that 11 2 u for all x E [a, b], which is a contradiction. By the same arguments we can prove that u(x) S u(x) for x E [0, 1]. Since u(x) S u(x) S u(x) for x E [0,1], the above definition implies that u(x) is also a solution to (2.8) E] Lemma 2.2.4 Assume (A’1)—(Af,). If there exist upper and lower solutions 1'1. and u of (2.8) with u(x) S u(x) for x E [0,1], then there is a solution u to {2.8) such that u(x) S u(x) S u(x), for all x E [0,1]. PROOF Let’s consider problem (2.9). The equivalent integral equation is (2.1%) = AP*>x + AQ*(u(o>,u(1)) + A [0’ use, y)h(y)f*(U(y))dy where P*(u,v) = 0192*(0) ; 791*(1‘), Q*(u, v) = (7 + 5)91*(:) + 592*(71) and k(x, y) is the corresponding Green’s function. Let A(uxx) = AP*(u(0),u(1>>x + AQ*,u<1)) + A [0‘ use. y)hf*)dy. 59 Then A : C°[0,1] —> C°[0,1] is completely continuous. Since f*,g1*,g2* are bounded, A is bounded. By Schauder’s fixed point theorem, A has a fixed point u, which is a solution to (2.9). By Lemma 2.2.3, u is also a solution to (2.8) Cl 2.2.4 Preliminaries We first note that (2.8) is equivalent to the integral equation (2.11) u(x) = AP(u(0),u<1>)z + Q(uw). um) + A [01k(x.y)h(y)f(U(y))dy where P(u, v) = ag2(v) - 791(11), Q(u, v) 2 ('y + d)gl(u) + (392(1)) p p and k(:1:, y) is the corresponding Green’s function. Since P, Q satisfiy (A6), we can apply the theorems from Section 2.1 to (2.11). Lemma 2.2.5 Assume (A’l) — (A’5). Then there is at least one positive solution to (2.8) for small A > 0. PROOF This lemma is a direct consequence of Corollary 2.1.3. [I] Lemma 2.2.6 Assume (A’l) — (A’s). Then there exists a positive number A such that (2.8) has no positive solution for A > A. 60 PROOF This lemma is a direct consequence of Theorem 2.1.2. E] Lemma 2.2.7 Assume (A’l) — (Ag). Then all solutions of (2.8) are bounded for A belonging to a compact subset of (0,00). PROOF Suppose that there is an unbounded sequence (an) of solutions of (2.8), and thus of (2.11), with {An} bounded away from 0. Then u,, satisfies (2.11). By (Ag) there is a p > 0 such that f (u) 2 nu for u 2 p, where n satisfies —A,,n/l::4 k,(1/2 y)h (y)dy Z 2 for all n. If un —> 00 as n —) 00, there exists an N > 0 such that min u.< )2-1-nunn2 )0 xE[1/4, 3/4] fornZN. FornZNwe have un(1/2) 2 A [7’ k(.)hl/Zy (yy)f(un( ))dy 2 Ann/1:: k(1/2.y)h(y)un(y)dy 1 3/4 2 ZAnnllunII/ k<1/2.y)h 0 such that a positive solution of (2.8) exists. Let A’ = sup{I‘}. By Lemmas 2.2.5 and 2.2.6, P is nonempty and bounded, and 61 thus 0 < A“ < 00. Moreover we can show as before that A“ E I‘. Let u* be a solution of (2.8) corresponding to A‘. Define f(u*(x) + e), u(x) Z u*(x) + 6 flux» = f(utx», -e s u(x) s u*(x) + e f(_€)1 u(HF) S -E. g1(u*(x) + 5), u(x) Z u*(x) + e 9"1 ("(75» = 9101(1)). -E S Ma?) S u*(SE) + E gt(-€), 11(3) S -6, g2(u*(x) + 6). u(x) 2 u*(x) + «6 93(u(z)) = gz(u(x)), —e _<_ u(x) g u*(x) + e gzl-Eli “(13) S -8, where e > o and with the understanding that f(u) = f(O), gl(u) = 91(0) and 92m) = 92(0) if u < 0. Let mute» = M3(u(0),u(1))x+ )+ A /k( m .)hy t)))dy where ~ _ 0193(0) - 7930‘) ~ u v _ (”Y + (5)9300 + 292(7)) P(U,’U) _ p a Q( 1 ) "" p ' and define B={uEX:—e 0, sufi‘iently small, such that if u(x) E C[0,1] satisfies FA(u(x)) = u(x) for some 0 < A < A‘, then u E B PROOF Is is clear that 0 S u. Let’s prove that u S u* + e. We first show that u* +5 is an upper solution of (2.8). Since u* 2 0 it follows that f (u*(x)) > 01 > 0 for all x E [0,1]. By uniform continuity, there is an 50 such that [f(u*($) + 5) - f(u*($))| < 010* - A)/* for all :1: E [0,1], 0 S e S 50. Then (u* + e)” = (u*)” = —A*hf(u*) = —Ahf(u* + e) + Alhf(u* + e) — h f(u*)] + (A — A*)hf(u*) < —Ahf(u* + e) + c1(A* — A) + c1(A — A*) |/\ —Ahf(u* + 5). Thus (u* + e)" + Ahf(u* + E) S 0. By the same arguments, we can show that Beta" + 6) Z A91(U*(0) + 8) B1(u* + e) Z Agg(u*(1) + e), 63 for a suffiently small 5 > 0. Therefore u* +5 is an upper solution of (2.8). It follows from Lemma 2.2.3 that u S u* + 5. [:1 2.2.5 Proof of Main Result The proof of the main result now follows as in Section 1.2.5. 2.2.6 Elliptic Systems Consider now the problem of finding positive radial solutions for an elliptic equation of the form Au+ Ah(lx|)f(u) = 0 in R1 < [xl < R2 3 (2.12) a. + 2% = Agllullxlllr on lzl = R. )u + 66—: = A92(U(lzl)). on lxl = R. where a, 6,7, 6 Z 0 and pEa(7+6)+76>0. We showed in Section 2.1.5 that (2.12) is equivalent to (2.8). Therefore the following theorem is true. Theorem 2.2.9 Assume (A’l) — (A’5). Then there exists a positive number A‘ such that (2.12) has at least two positive radial solutions for 0 < A < A“ , at least one positive radial solution for A = A“ and no positive radial solutions for A > A". BIBLIOGRAPHY BIBLIOGRAPHY [1] D. 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