Exact multiplicity results for a p-laplacian problem with concave convex concave nonlinearities

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1 Nonlinear Analysis 53 (23) Exact multiplicity results for a p-laplacian problem with concave convex concave nonlinearities Idris Addou a, Shin-Hwa Wang b; ;1 a Ecole Polytechnique de Montreal, Departement de Mathematiques et de Genie Industriel, C.P. 679, Succursale Centre-Ville, Montreal, Quebec, Canada H3C 3A7 b Department of Mathematics, National Tsing Hua University, Hsinchu 3, Taiwan, ROC Received 18 October 21; accepted 11 March 22 Abstract We study the exact number ofpositive solutions ofa two-point Dirichlet boundary-value problem involving the p-laplacian operator. We consider the case p = 2 as well as the case p 1, when the nonlinearity f satises f() = and has two distinct simple positive zeros and such that f changes sign exactly twice on (; ). Note that we may allow that f changes sign more than twice on (; ). Some interesting examples ofquartic polynomials are given. In particular, for f(u)= u 2 (u 1)(u 2), we study the evolution ofthe bifurcation curves of the p-laplacian problem as p increases from 1 to innity, and hence are able to determine the exact multiplicity ofpositive solutions for each p 1.? 23 Elsevier Science Ltd. All rights reserved. Keywords: Exact multiplicity result; p-laplacian; Concave convex concave nonlinearity; Positive solution; Dead-core solution; Bifurcation; Time-map 1. Introduction In this paper we present exact multiplicity results ofpositive solutions for the nonlinear two-point Dirichlet boundary-value problem ( p (u (x))) = f(u(x)); 1 x 1; u( 1) = u(1)=; (1.1) Corresponding author. Tel.: ; fax: addresses: idris.addou@polymtl.ca (I. Addou), shwang@math.nthu.edu.tw (S.-H. Wang). 1 Work partially supported by the National Science Council, Republic ofchina X/3/$ - see front matter? 23 Elsevier Science Ltd. All rights reserved. PII: S X(2)298-5

2 112 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) where p 1; p (y)= y p 2 y and ( p (u )) is the one-dimensional p-laplacian, and f C 2 [; ) satises f() = and has two distinct simple positive zeros b c. For p =2; ( p (u )) = u, and problem (1.1) reduces to u (x)=f(u(x)); 1 x 1; u( 1) = u(1)=: (1.2) When the nonlinearity f(u) = u(u b)(u c) is a cubic polynomial satisfying b c=2, Smoller and Wasserman [14] rst gave exact multiplicity results of (classical) positive solutions ofproblem (1.2) by applying the time-map analysis. For problem (1.2), they showed the existence ofa critical such that for problem (1.2) has no positive solution, it has exactly one positive solution for =, and exactly two positive solutions for : Consequently, Wang and Kazarino [16] and Wang [15] studied problem (1.2) when f is a cubic-like nonlinearity. Let F(u)= u f(t)dt: Theorem 1.1 (Wang and Kazarino [16, Theorem 1 and Remark 2]). Suppose f C 2 [; ) and there exist b c such that the following conditions are satised: f() = f(b) = f(c) = ; (1.3) f(u) for u (;b) (c; ) and f(u) for u (b; c); (1.4) c f(u)du ; (1.5) there exists r (;c) such that f (u) for u r and f (u) for r u : (1.6) Then there exists such that (a) for, problem (1.2) has no positive solution, (b) for =, problem (1.2) has exactly one positive solution u 1 satisfying u 1 c, (c) for, problem (1.2) has exactly two positive solutions u 1 u 2 satisfying u 1 u 2 c. Remark 1. If f C[; ) satises (1.3) (1.5), then it can be shown that (i) By the maximum principle, every classical positive solution u of(1.2) satises u c:

3 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) (ii) Any two distinct positive solutions of(1.2) are strictly ordered. That is, let u and û be any two distinct positive solutions of(1.2) with u û ; then u û; see e.g. [16, Lemma 1]. For f a cubic-like nonlinearity and for problem (1.2), the same exact multiplicity results when f changes sign exactly once were proved by Korman et al. [6, Theorem 2.3]. Recently, Korman and Shi [1, Theorem 5] proved an exact multiplicity result which generalizes theorem [6, Theorem 2.3] by slightly weakening the convexity assumptions on f. We also note that Korman et al. [7] and Ouyang and Shi [12,13] have extended similar results on a n-dimensional ball (n 2), see [7,12,13] for details. The main tool used in [6,7,1,12,13] is a bifurcation theorem of Crandall and Rabinowitz [3]. We note that the case where f changes sign exactly twice has not been studied yet. Other related results are available in the literature; see for instance, Korman [5], Korman and Ouyang [8,9], and Maya and Shivaji [11]. For problem (1.1) with p 2, when f is a cubic-like nonlinearity satisfying (1.3) (1.5), little is known. We refer to Addou [2] in which f is assumed to changes sign exactly once. 2. Classication of positive solutions and the time-map method To state the main results, we rst give a classication ofpositive solutions ofproblem (1.1) inc 1 [ 1; 1]. By a positive solution to problem (1.1) we mean a positive function u C 1 [ 1; 1] with p (u ) C 1 [ 1; 1] satisfying (1.1). Let Z = {x [ 1; 1] : u (x)=}: We note that it is easy to show that, if f C and u is a positive solution ofproblem (1.1), then u C 2 [ 1; 1] if1 p62 and u C 2 ([ 1; 1] Z) ifp 2: For the proofwe refer to Addou [1, Lemma 6]. Let A + be the subset of C1 [ 1; 1] consisting ofthe functions u satisfying (i) u(x) for all x ( 1; 1); u( 1) = u(1)= u ( 1): (ii) u is symmetrical with respect to : (iii) The derivative of u vanishes once and only once. Let A + 1 be the subset of C1 [ 1; 1] consisting ofthe functions u satisfying (i) u(x) for all x ( 1; 1); u( 1) = u(1)= u ( 1): (ii) u is symmetrical with respect to : (iii) There exists k (; 1) such that for all x ( 1; 1); u (x) = ifand only if k 6 x 6 k: Note that ifa solution u A + 1, then it is usually called a dead-core solution ofproblem (1.1).

4 114 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) Let B + be the subset of C1 [ 1; 1] consisting ofthe functions u satisfying (i) u(x) for all x ( 1; 1); u( 1) = u(1)==u ( 1): (ii) u is symmetrical with respect to : (iii) The derivative of u vanishes once and only once. It is easy to derive an energy relation ofsolutions u ofproblem (1.1), see e.g. [4, p. 421]. Denote by p = p=(p 1) the conjugate exponent of p: Lemma 2.1 (Energy relation). Let p 1 and assume that u is a positive solution of problem (1.1), then ( u (x) p + p F(u(x))) = for all x [ 1; 1]: Remark 2. By Lemma 2.1, it can be shown that if u is a positive solution ofproblem (1.1), then u is symmetrical with respect to : In addition, u A + A+ 1 B+ if f satises (1.3) (1.5) for some numbers b c: We omit the proof, see e.g. [2, Lemma 6]. To study problem (1.1), we make use ofthe time-map method (quadrature method). Suppose f C[; ) satises (1.3) (1.5) for some numbers b c. For any E and s, let G(E; s) :=E p p F(s) and X (E)={s :s dom G(E; ) and G(E; u) for all u (;s)}; Let r(e)= { if X (E)= ; sup(x (E)) otherwise: D = {E :r(e) dom G(E; );G(E; r(e))= } r(e) and (E p p F(t)) 1=p dt : Dene the time-map T(E)= r(e) (E p p F(t)) 1=p dt; E D: By Lemma 2.1 and arguments in [4], we have the following theorem. Note that in this paper, by Remark 2, we restrict ourselfon positive solutions u A + A+ 1 B+ : Theorem 2.2. Consider problem (1.1). Suppose f C[; ) satises (1.3) (1.5) for some numbers b c. Let E : Then T is a continuous function of E D:

5 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) Moreover, (i) Problem (1.1) has a solution u A + satisfying u ( 1) = E if and only if E D {} and T (E)=1; and in this case the solution is unique. (ii) Problem (1.1) has a solution u A + 1 satisfying u ( 1) = E if and only if E D {} and T (E) 1; and in this case the solution is unique. (iii) Problem (1.1) has a solution u B + satisfying u ( 1)= if and only if D and T()=1; and in this case the solution is unique. Remark 3. In practice, we rst study the variations ofthe real-valued function G(E; ), then compute X (E) and deduce r(e): Next, we compute D: For this, we rst compute the set D = {E :r(e) dom G(E; );G(E; r(e)) = and f(r(e)) } and then we deduce D by observing that D D {} D; we omit the proof. (Note that D is the closure of D.) After that, we dene the time-map T on D and then compute its limits at the boundary points of D: Next, we study the variations of T on D: We achieve our study by discussing the number ofsolutions to (i) Equation T(E)=1 for E D {} in case oflooking for solutions in A + : (ii) Inequality T (E) 1forE D {} in case oflooking for solutions in A + 1 : (iii) Equation T() = 1 in case where D and when we are looking for solutions in B + : 3. Main results We determine the exact multiplicity ofpositive solutions ofproblem (1.1) for each : We assume that f satises condition (f4) below, which implies for the particular case p =2; that f changes sign exactly twice on (; ), i.e., f is concave convex concave on (; ). Note that we may allow that f changes sign more than twice on (; ), i.e., we may allow that f is concave convex concave convex on (; ); see Remark 6. For f; recalling that F(u)= u f(t)dt; we let p (u) :=pf(u) uf(u); p(u) :=u p(u) p (u)=puf(u) u 2 f (u) pf(u); (3.1) { c } p / p := (F(c) F(u)) 1=p du p (; ]; (3.2) { } p p := (F() F(u)) /p 1=p (; ]; (3.3)

6 116 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) p := inf 66c { } p / (F() F(u)) 1=p du p ; (3.4) where c are dened behind. We shall show that p for p 1: For all ; we denote S the positive solution set of(1.1). For xed p 1; suppose f C 2 [; ) and there exist b c such that the following conditions are satised: (f1) f()=; (f2) f(u) f(u) for u b; f or b u c; f(u) f or u c; (f3) c f(u)du (then there exists a unique (b; c) satisfying f(u)du = ), (f4) there exist r p s p c such that (p 2)f (u) uf (u) for u r p ; (p 2)f (u) uf (u) f or r p u s p ; (p 2)f (u) uf (u) for s p u c (it can be weakened later; see Remark 6); (f5) there exists a unique p (s p ;c) satisfying (p 1)f( p ) p f ( p )= and such that p( p ) p(r p ): Remark 4. The geometric meaning of (f5) is as follows: On the xy-plane, consider the graph of y = p (x); by (3.1), the y-intercept ofthe tangent line through the point (r p ; p (r p )) is greater than or equal to the y-intercept ofthe tangent line through the point ( p ; p ( p )). Remark 5 (cf. Remark 1). If f C[; ) satises (f1) (f3), then by applying Lemma 2.1, it can be shown that (i) Every positive solution u ofproblem (1.1) satises 6 u 6 c. (ii) Any two distinct positive solutions ofproblem (1.1) are strictly ordered. That is, let u and û be any two distinct positive solutions ofproblem (1.1) with u û ; then u û. The next theorem gives complete description ofthe set S for p and f satisfying certain conditions.

7 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) Fig. 1. Bifurcation diagrams of problem (1.1) with f()=. (a) 1 p6 2; (b) f ()= =f (m 1) ()= and f (m) () for some positive integer m 2, and 2 p6 m +1: Theorem 3.1 (S, see Fig. 1). Assume that p 1 and f C 2 [; ) satises (f1) (f5). (a) If1 p6 2; then: (i) For p ; S =. (ii) For = p, there exists u A + such that S = {u }. Moreover, u c. (iii) For p, there exist u ;v A + such that u v and S ={u ;v }. Moreover, u v c. (b) If f () = = f (m 1) () = and f (m) () for some integer m 2 (we allow m = ); and 2 p6 m + 1 then p p. Moreover: (iv) For p ; S =. (v) For = p, there exists u A + such that S = {u }. Moreover, u c. (vi) For p 6 p, there exist u ; v A + such that u v and S = {u ;v }. Moreover, u v 6 c; and v = c if and only if = p. (vii) For p, there exist u A + and v A + 1 such that u v and S = {u ;v }. Moreover, u v = c. In what follows, we consider the remaining cases: (C1) f () and p 2: (C2) f ()= =f (m 1) ()= and f (m) () for some integer m 2, and p m+1: By Remark 2, S A + A+ 1 B+, then S =(S A + ) (S A + 1 ) (S B + ): Theorem 3.2 (resp. Theorems 3.3 and 3.4) gives complete description ofthe set S A + (resp. S A + 1 ;S B + ). We shall show that, for p 2; p p and p p ; see Lemmas 4.7 and 4.5.

8 118 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) Fig. 2. Bifurcation diagrams of problem (1.1) with f() = ; p 2; and f satisfying either (C1) or (C2). (a) p p p; (b) p p = p; (c) p p p. Theorem 3.2 (S A +, see Fig. 2). Assume that p 2. In addition to (f1) (f5), suppose f C 2 [; ) satises either (C1) or (C2) above. Then p p and p p, and each eventual solution u of (1.1) in A + satises u 6 c: Moreover, (a) If p p p ; then: (i) For p ; S A + = : (ii) For = p ; there exists u A + such that S A + ={u }: Moreover, u c: (iii) For p 6 p ; there exist u ;v A + such that u v and S A + ={u ;v }: Moreover, u v 6 c; and v = c if and only if = p. (iv) For p p ; there exists u A + such that S A + = {u }: Moreover, u c: (v) For p ; S A + = : (b) If p p = p ; then: (vi) For p ; S A + = : (vii) For = p ; there exists u A + such that S A + = {u }: Moreover, u c: (viii) For p p = p ; there exist u ;v A + such that u v and S A + = {u ;v }: Moreover, u v c: (ix) For = p = p ; there exists u A + such that S A + = {u }: Moreover, u = c: (x) For p = p ; S A + = : (c) If p p p ; then: (xi) For p ; S A + = : (xii) For = p ; there exists u A + such that S A + = {u }: Moreover, u c: (xiii) For p p ; there exist u ;v A + such that u v and S A + ={u ;v }: Moreover, u v c:

9 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) (xiv) For p 6 6 p, there exists u A + such that S A + = {u }: Moreover, u 6 c; and u = c if and only if = p. (xv) For p ; S A + = : Theorem 3.3 (S A + 1 ; see Fig. 2). Assume that p 2. In addition to (f1) (f5), suppose f C 2 [; ) satises either (C1) or (C2) above. Then each eventual solution u of (1.1) in A + 1 satises u = c: Moreover, (i) For 6 p ; S A + 1 = : (ii) For p, there exists u A + 1 such that S A + 1 = {u }: Theorem 3.4 (S B +, see Fig. 2). Assume that p 2. In addition to (f1) (f5), suppose f C 2 [; ) satises either (C1) or (C2) above. Then each eventual solution u of (1.1) in B + satises u =. Moreover, (i) For = p, there exists u B + (ii) For p ; S B + = : such that S B + = {u }: We note that, for p 2 and for each p, it can be proved that there do not exist dead-core solutions u ofproblem (1.1) satisfying u = since f() and by applying Lemma 2.1. Remark 6. As one can check the proofoftheorems , the convexity assumption of p (u)=pf(u) uf(u) on(;c) in (f4) in Theorems can actually be weakened as follows: (f4 ) There exist 6 r p s p c such that f f (p 2)f (u) uf (u) for u r p (it is not necessary if r p =); (p 2)f (u) uf (u) or r p u s p ; (p 2)f (u) uf (u) or s p u c (it can be weakened below): See Section 3.1 for examples. We note that in (f 4 )ifr p = then (f5) is automatically satised since it can be easily shown that, for p 1, there exists a unique p (s p ;c) satisfying (p 1)f( p ) p f ( p ) = and such that p( p ) = p(r p ): We also note that in (f4 ) the assumption (p 2)f (u) uf (u) f or s p u c can actually be weakened as p(u)=(p 1)f(u) uf (u) { 6 f or sp u p ; f or d 6 u 6 c; (3.5) p(u)=(p 2)f (u) uf (u) f or p 6 u d; (3.6)

10 12 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) where d ( p ;c] is dened by { c; if p (c) 6 p (t p ); d := inf {u ( p ;c]: p () p (t p ) for all (u; c]}; otherwise; (3.7) where t p is the unique zero of p(u) on(r p ;s p ): We summarize that (f4 ) can be weakened as follows: (f4 ) There exist 6 r p s p p d c such that f f f (p 2)f (u) uf (u) for u r p (it is not necessary if r p =); (p 2)f (u) uf (u) or r p u s p ; { 6 or sp u p ; p(u)=(p 1)f(u) uf (u) f or d 6 u 6 c; p(u)=(p 2)f (u) uf (u) or p 6 u d; where d is dened in (3.7). Note that if(3.5) and (3.6) are satised instead ofthe third condition in (f4) and (f4 ) then the result in Lemma 4.2 stated behind remains valid; that is, f (c). Also note that if d c then S p() ford c and Lemma 4.6 stated behind should read as follows: S p()+ p S p() for all (max{ p ;};d) and p 1: An example to Remark 6. Let p = 2 and f = f 1 (u)=u 2 (u 1)(u 3) 2. The function f 1 satises f 1 () = and f 1 (u)=2u3 84u 2 +9u 18 changes sign exactly thrice on (; ) with its zeros at r :259; s 1 1:336; s 2 2:65: It can be easily checked that f 1 satises (f1) (f5) in Theorem 3.1 except (f4). Nevertheless, it can be easily checked that f 1 satises (f4 ). Thus, for p=2, the bifurcation diagram ofproblem (1.1) is depicted in Fig. 1(a) Some examples For the sake ofsimplicity, we mainly restrict ourselves to the case p = 2 and rst give two examples ofquartic polynomials to Theorems 3.1 and 1.1. In Theorem 3.5, f(u)= u 2 (u b)(u c) satises f()=f ()=; f ()= 2bc and f changes sign exactly twice on (; ). In Theorem 3.6, f(u)= u(u â)(u b)(u c) satises f()=; f (), and f changes sign exactly twice on (; ) ifâb + bc + câ = 1 2 f () ; f changes sign exactly once on (; ) ifâb + bc + câ = 1 2 f () 6 :

11 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) Theorem 3.5. Let p =2: Let f(u)= u 2 (u b)(u c) satisfy c 5 c ) ( 3 b f(u)du : (3.8) Then f satises (f1) (f5) in Theorem 3.1. Thus the bifurcation diagram of problem (1.1) is depicted in Fig. 1(a). An example to Theorems 3.5, 3.1, 3.2 and Remark 6. We study the evolution ofthe bifurcation curves as p increases from 1 to innity for f = f 2 (u)= u 2 (u 1)(u 2) = 2u 2 +3u 3 u 4 satisfying f 2 () = f 2 () = and f 2 () = 4: First, for p =2; f 2 satises condition (3.8) in Theorem 3.5 and hence (f1) (f5) in Theorem 3.1. While for all p 1, it can be easily checked that (i) For 1 p 3; f 2 satises (f1) (f5) with r p = 36 9p p +17p 2 4 8p s p = 36 9p p +17p 2 4 8p p = 12 3p p + p 2 1 2p c=2; we omit the proof. Thus the bifurcation diagram of problem (1.1) is depicted in Fig. 1(a) for 1 p62; and in Fig. 1(b) for 2 p 3. (ii) For p =3; f 2 satises (f1) (f3), (f5) and (f 4 ) with the denition of r p = (since the function f 2 (u) uf 2 (u) has exactly one positive zero on (; 2)) and s p= 9 8 p = 3 2 c = 2. Thus the bifurcation diagram of problem (1.1) is depicted in Fig. 1(b) for p =3. (iii) For p 3; f 2 satises (f1) (f3). Moreover, it can be easily shown that: (A) For 3 p61; f 2 satises (f4 ) with the denition of r p = (since the function f 2 (u) uf 2 (u) has exactly one positive zero on (; 2)); we omit the proof. In this case (f5) is automatically satised, see Remark 6. (B) For p 1; f 2 satises (f4 ) with the denition of r p = (since the function f 2 (u) uf 2 (u) has exactly one positive zero on (; 2)); we omit the proof. In this case (f5) is automatically satised, see Remark 6. When p 3; we compute that s p = 36 9p p +17p 2 4 8p p = 12 3p p + p 2 1 2p c=2:

12 122 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) Fig. 3. Graphs of p and p:p 7:649: Numerical simulation (see Fig. 3) shows that there exists a number p 7:649 such that (a) for 3 p p ; p p : So the bifurcation diagram of problem (1.1) is depicted in Fig. 2(a). (b) for p = p ; p = p : So the bifurcation diagram of problem (1.1) is depicted in Fig. 2(b). (c) for p p6 1; p p : So the bifurcation diagram of problem (1.1) is depicted in Fig. 2(c). It follows immediately by the above analysis that, for all p 1, problem (1.1) has at most two positive solutions for each : Theorem 3.6. Let p =2: Let f(u)= u(u â)(u b)(u c) satisfy â b c (3.9) and Then c 1 6 (5â +5b + ( (5â +5b) 2 12âb) c ) f(u)du : (3.1) (i) If âb + bc + câ = 1 2 f () 6 ; then f satises all conditions (1.3) (1.6) in Theorem 1.1. (ii) If âb + bc + câ = 1 2 f () ; then f satises all conditions (f1) (f5) in Theorem 3.1. Therefore, in either case (i) or (ii), the bifurcation diagram of problem (1.1) is depicted in Fig. 1(a).

13 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) An example to Theorem 3.6. Let p = 2. The function f = f 3 (u)= u(u +1=2)(u 1) (u 3) satises conditions (3.9) and (3.1) in Theorem 3.6 with f 3 () = 2 : 4. Proofs of main results First, we have the next lemma which follows easily for function f C[; ) satisfying (f1) (f3). We omit the proof. Lemma 4.1. Assume that f C[; ) satises (f1) (f3). Consider the function dened on (; ) by s G(; E; s) :=E p p F(s); where p 1; E and are real parameters. Then (i) If E E c := (p F(c)) 1=p ; the function G(; E; ) is strictly positive on (; ): (ii) If E = E c, the function G(; E c ; ) is strictly positive on (;c) and vanishes at c. (iii) If E E c, the function G(; E; ) has on the open interval (; c) a unique zero ŝ(; E) and is strictly positive on the open interval (; ŝ(; E)): Moreover, (a) The function E ŝ(; E) is C 1 on (;E c ) 1)Ep 1 (; f(ŝ(; E)) for all E (;E c): (b) lim E + ŝ(; E)= and lim E E c ŝ(; E)=c: for s ; (iv) If E =, the function G(; ;s) = for s = ; for s6c: Now, for p 1; and E, we let X (; E):={s dom G(; E; )=(; ) :G(; E; u) for all u (;s)} (; ) if E E c ; (;c] if E = E c ; = (; ŝ(; E)] if E E c ; (;] if E =;

14 124 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) by Lemma 4.1. Therefore, if E E c ; c if E = E c ; r(; E) :=supx(; E)= ŝ(; E) if E E c ; if E =; D p ():={E :r(; E) dom G(; E; )=(; ); G(; E; r(; E)) = and f(r(; E)) } =(;E c ): Notice that the denition domain ofthe time-map T p is given by r(; E) dom G(; E; )=(; );G(; E; r(; E))=; D p () := E : r(; E) and (G(; E; u)) 1=p du : In the present case, (;E c ) D p () [;E c ]. We dene, for E D p (), the time-map T p (; E):= = r(; E) r(;e) (G(; E; u)) 1=p du (E p p F(u)) 1=p du r(; E) =(p ) 1=p (F(r(; E)) F(u)) 1=p du; since G(; E; r(; E)) = E p p F(r(; E))=: For p 1; we dene S p () := (F() F(u)) 1=p du for all [; c]: (4.1) Thus our time-map may be written as T p (; E)=(p ) 1=p S p (r(; E)): We observe that (f2) implies that f (c) 6 : Actually, it can be proved that f (c) by using (f2) and (f4). We omit the proof. Lemma 4.2. Suppose f C 2 [; ) satises (f1), (f2) and (f4). Then f (c) : Lemma 4.3. S p (c)= if and only if 1 p6 2:

15 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) Proof. By Lemma 4.2, f (c) : Then ( ) 1=p (F(c) F(u)) 1=p f (c) (c u) 2=p near c : 2 Hence, by (4.1), easy computation shows that S p (c)= ifand only if1 p6 2: Lemma 4.4. (i) If f (), then S p ()= if and only if 1 p6 2: (ii) If f ()= and f (), then S p ()= if and only if 1 p6 3: (iii) If f () = = f (m 1) ()= and f (m) () for some integer m 2, then S p ()= if and only if 1 p6 m +1: (iii) If f (k) ()= for any integer k =1; 2;:::; then S p ()= for all p 1: Proof. By (4.1), the integral which represents S p () presents two singularities, at and at. By the fact that f(), it follows (F() F(u)) 1=p ( f()) 1=p ( u) 1=p near + : Since p 1, it follows that ( u) 1=p du for any, small enough. Therefore (F() F(u)) 1=p du : We study the singularity at in the following: (i) Assume that f () : Since F()=F()=; by using l Hopital s rule twice, we obtain ( ) 1=p F(u) (F() F(u)) 1=p =( F(u)) 1=p = u 2=p u 2 ( f ) 1=p () u 2=p near + : 2 Hence, by (4.1), easy computation shows that S p ()= ifand only if1 p62: (ii) Assume that f () = and f (). Similarly, using l Hopital s rule three times, we obtain ( ) 1=p F(u) (F() F(u)) 1=p =( F(u)) 1=p = u 3=p u 3 ( f ) 1=p () u 3=p near + : 6 Hence, by (4.1), easy computation shows that S p ()= ifand only if1 p63. (iii) Suppose f () = = f (m 1) () = and f (m) () for some integer m 2: By applying similar arguments in above, we can show that S p ()= ifand only if 1 p6m +1: (iii) Suppose f (k) () = for any integer k =1; 2;::: : Let p 1. By applying l Hopital s rule p times it follows: F(u) f(u) lim u + u p = lim = lim u + pup 1 u + f (u) f (p 1) (u) = = lim = p(p 1)up 2 u + p!

16 126 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) then there exists = p such that F(u) u p for u p : Since F()= F()=, (F() F(u)) 1=p u 1 for u p : Hence, by (4.1), easy computation shows that S p ()= for all p 1. The proofoflemma 4.4 is complete. Now, after computing the limits of the time-map T p (; ), we study its exact variations in the interior ofits denition domain; that is, (;E c ): Notice that, to study the variations of T p (; ) in(;e c ), it suces to study those of S p ( ) in the interior ofthe range of r(; ); that is, (; c): One has S p()= 1 p where p () p (u) du for (; c); (4.2) (F() F(u)) 1=p p (u) :=pf(u) uf(u) (4.3) for u [;c]; p 1: By (4.3), it follows that p(u)=(p 1)f(u) uf (u); (4.4) p(u)=(p 2)f (u) uf (u): (4.5) Thus by (f1) and (f4), p ()=; (4.6) p()=(p 1)f()=; for u r p ; p(u) f or r p u s p ; f or s p u c: In addition, by (f3), (f2) and Lemma 4.2, p ()= f() ; p (c)=pf(c) cf(c)=pf(c) ; p(c)=(p 1)f(c) cf (c)= cf (c) : Hence there exist t p (r p ;s p ) and p (s p ;c) such that p is strictly increasing on (;t p ); (4.7) p is strictly decreasing on (t p ; p ); (4.8) p is strictly increasing on ( p ;c): (4.9)

17 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) Fig. 4. Graph of p. In addition, there exist p (t p ; p ) and p ( p ;c) such that p ( p )= p ( p )=: (4.1) The typical graph of p (u) on[;c] is depicted in Fig. 4. Lemma 4.5. Assume that p 1: Then (i) If p 2 then S p(c)= : (ii) If p then S p() for all (; p ]: (iii) If = p and S p () then 6 S p() : (iv) If p and S p () then S p()= : Proof. Recall (4.2), S p()= 1 p () p (u) du for [; c]: p (F() F(u)) 1=p (i) Assume that p 2. By Lemma 4.2, f (c). This implies that p (c) p (u) (F(c) F(u)) 1+1=p 21+1=p c( f (c)) 1=p (c u) 1+2=p near c : Thus easy computation shows that S p(c)=. (ii) Assume that ( p ) p.by(4.6) (4.8) and (4.1), it follows that p () p (u) for all (; p ] and for all u (;): So S p() for all (; p ]: (iii) If = p, then p () p (u) = p ( p ) p (u) for all u (;): So 6 S p() : (iv) Since F()=F() =, it follows that the integral representing S p() has two singularities; one at and the other at : So we write S p()= 1 p (I + I );

18 128 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) where I := p p () p (u) (F() F(u)) 1+1=p du and I p () p (u) := du: p (F() F(u)) 1+1=p We next show I and I = : First we show I : By (c ) p ; it follows that p() and therefore, p () p (u) (F() F(u)) 1+1=p p() 1 (f()) 1+1=p ( u) 1=p near : Since 1=p 1 and p()(f()) 1 1=p, easy computation shows that I. Secondly we show I = : We distinguish two cases: (I) Assume that f () : Note that p () p (u) (F() F(u)) 1+1=p p () 1 ( f ()=2) 1+1=p u 2(1+1=p) near + : Since 2(1 + 1=p) 1 and p ()( f ()=2) 1 1=p ; easy computation shows that I =. (II) Assume that f ()=: It is easy to see that f () 6 by (f2). Using l Hopital s rule three times, it follows that lim u + F(u)=u3 = f ()=6 6 : (a) Assume that f (). Note that p () p (u) (F() F(u)) 1+1=p p () 1 ( f ()=6) 1+1=p u 3(1+1=p) near + : Since 3(1 + 1=p) 1 and p ()( f ()=6) 1 1=p ; easy computation shows that I = : (b) Assume that f ()=. Note that lim u + ( F(u)=u 3 ) 1 1=p = ; then there exist K and p such that ( F(u) ) 1 1=p 1 u 3 u 3(1+1=p) 1 K u 3(1+1=p) for u : On the other hand, p () p (u) for u p, then ( p () p (u) (F() F(u)) 1+1=p = F(u) ) 1 1=p p () p (u) u 3 u 3(1+1=p) 6 K p() p (u) u 3(1+1=p) for u

19 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) and p () p (u) u 3(1+1=p) p() u 3(1+1=p) near + : Since 3(1 + 1=p) 1 and p () ; it follows that ( p () p (u)) u 3(1+1=p) du = ; which implies that I = : The proofoflemma 4.5 is complete. Lemma 4.6. For p 1; S p()+(p=)s p() for all (max{ p ;};c): Proof. Easy computations show that for all (; c); where S p()= p +1 (p) 2 ( p () p (u)) 2 1 du + (F() F(u)) 2+1=p p 2 p (u) := p(p +1)F(u)+2puf(u) u 2 f (u) for u (;c): Then S p()+ p S p()= 1 p 2 for all (; c); where + p +1 (p) 2 p() p(u) (F() F(u)) 1+1=p du ( p () p (u)) 2 du (F() F(u)) 2+1=p p(u) := p (u)+p p (u)=u p(u) p (u) for all u (;c): Hence S p()+ p S p() 1 p 2 p () p (u) du; (F() F(u)) 1+1=p p() p(u) du: (4.11) (F() F(u)) 1+1=p Note that p (u)=u p(u): Therefore, by (f4), it follows that p is strictly increasing on [;r p ); is strictly decreasing on (r p ;s p ) and is strictly increasing on (s p ;c): On the other hand, p()= p() p ()=; p()= p()=; p( p )= p p( p ) p ( p )= p ( p ) :

20 13 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) Fig. 5. Graph of p: Therefore, p(r p ) and by (f5) it follows that p() p(u) for all u (;) and all (max{ p ;};c): The typical graph of p(u) on[;c] is depicted in Fig. 5. Therefore, the integral in (4.11) is strictly positive for all (max{ p ;};c); which proves Lemma 4.6. Lemma 4.7. For p 1; S p () has exactly one critical point, a minimum, on (; c): More precisely, there exists a unique m p (; c) such that S p () is strictly decreasing on (; m p ) and is strictly increasing on (m p ;c). Proof. We prove existence and uniqueness of m p separately. Uniqueness of m p : We distinguish two cases: (i) p : By Lemma 4.6, the uniqueness of m p in ( p ;c) follows, and by Lemma 4.5(ii), the non-existence of m p in (; p ] follows. So uniqueness of m p in (; c) follows. (ii) p : The uniqueness of m p in (; c) follows by Lemma 4.6 immediately. Existence of m p : We distinguish two cases: (i) 1 p6 2: By Lemmas 4.3 and 4.4, it follows that S p ()=S p (c)= : So the existence of m p follows. (ii) p 2: By Lemmas 4.3 and 4.5(i), it follows that S p (c) and S p(c) =. On the other hand, there are many cases to be considered for S p (): However, by Lemmas 4.4 and 4.5, it follows that in all these cases, either S p () = or (S p () and S p () is strictly decreasing on a right neighborhood of ). So the existence of m p follows in all cases. The proofoflemma 4.7 is complete. Let u be a positive solution ofproblem (1.1), by Lemma 2.1, (i) if u 6 c then u ( 1)=(p F( u )) 1=p, and hence u A + A+ 1 ;

21 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) (ii) if u = then u ( 1)=(p F( u )) 1=p =(p F()) 1=p =, and hence u B + : By Lemma 4.3, S p (c) for p 2. In this case, suppose for = p =(S p (c)) p =p ; u p is the corresponding solution ofproblem (1.1) satisfying u p = u p () = c; u p ( 1)=(p p F(c)) 1=p ; and u p A +. Then for each p, u p (( p ) 1=p ( x 1+( p )1=p )); if1 ( p )1=p x 6 1; u (x) := c; if x 6 1 ( p )1=p is a C 1 dead-core solution ofproblem (1.1) satisfying u = c; u ( 1)=(p F(c)) 1=p (p p F(c)) 1=p = u p ( 1) and u A + 1. Hence Theorems follow immediately by Theorem 2.2 and Lemmas and Proofs of the examples The proofs of Theorems 3.5 and 3.6 are quite similar. Proof of Theorem 3.5. It is easy to see that f(u)= u 2 (u b)(u c) forc 5 3 b satises (f1), (f2) and (f4) with its two inection points at r = 1 4 (b + c (b + c) bc) ; s= 1 4 (b (b + c + + c) 2 8 3bc) c; since f () = 2bc. Also f satises (f3) since c f(u)du = 1 2 c4 (c 5 3 b) : Finally, for (f5), it can be shown that there exists a unique number = 1 3 (b + c + b 2 + c 2 bc) satisfying f(u) uf (u)=u 2 (3u 2 2(b + c)u + bc)= on (b; c). We then compute that 2736( () (r)) =256(b + c + b 2 + c 2 bc) 3 (b 2 + c 2 2bc +(b + c) b 2 + c 2 bc) + 243(b + c b 2 + c bc)3 (b 2 + c bc (b + c) b 2 + c 2 2 3bc): (4.12)

22 132 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) To estimate the above expression we need the following lemma which follows easily by (3.8). We omit the proof. Lemma 4.8. Assume (3.8). Then (i) b 2 + c bc b2 + c 2 2bc 4 25 c2, (ii) 2 5 c b 2 + c 2 bc 34 5 c, (iii) 2 5 c b 2 + c bc 34 5 c: Now by Lemma 4.8(i) and (ii), b 2 + c 2 2bc +(b + c) b + c bc 4 25 c2 +(b + c)( 2 5 c) ; (4.13) 256(b + c + b 2 + c 2 bc) 3 256(b + c c)3 : (4.14) Also, by Lemma 4.8(iii), 243(b + c b 2 + c bc)3 243(b + c 2 5 c)3 : (4.15) Since b; c ; we obtain ( ) b2 + c b bc b 2 + c c 2 23 bc 3 bc b + c and hence (b 2 + c b bc) (b + c) 2 + c bc : Then by Lemma 4.8(i) and (iii), it follows that ( 4 25 c2 (b + c) ) 34 5 c ( b 2 + c 2 Let 1 x := b + c c 3 5 c + c = 8 c 5 : Then by (4.12) (4.16), we compute that 2736( () (r)) 149 bc) 256(b + c c)3 ( 4 25 c2 +(b + c)( 2 5 c)) + 243(b + c 2 5 c)3 ( 4 25 c2 (b + c) 34 5 c) (b + c) b 2 + c bc : (4.16)

23 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) =c 5 {256(x )3 ( x) + 243(x 2 5 )3 ( x)} =c 5 { (5x +2) (x 2 5 )3 (4 5 34x)} (4.17) for 1 x 8 5 ; we omit the proof of the last inequality. Therefore (f5) follows. This completes the proofoftheorem 3.5. Proof of Theorem 3.6. To prove Theorem 3.6, we need the following lemma. Lemma 4.9. If, in addition to (3.9) and (3.1), f(u)= u(u â)(u b)(u c) satises then: âb + bc + câ : (4.18) (i) â 1 5 c: (ii) b 3 5 c: (iii) â + b + c : (iv) â 2 + b 2 + c 2 2(âb + bc + câ) 4 25 c2 : (v) â + b + c â 2 + b 2 + c (âb + bc + câ) : (vi) [â 2 +b 2 +c â (âb+bc+câ)] [(â+b+c) 2 + b 2 + c (âb + bc + câ)] : (vii) â 2 + b 2 + c 2 (âb + bc + câ) â 2 + b 2 + c (viii) 2 5 â c 2 + b 2 + c (âb + bc + câ) 39 5 c: (âb + bc + câ) 4 25 c2 : Proof. (i) By (3.1), we obtain 3 5 c2 c(â + b) +2âb ; which implies 3 5 c âb âb + bc + câ, and by (4.18), it follows that 5 c2 âb: Also, by (4.18), it follows that â + b âb=c ; and hence b â : Then âb â 2 and therefore 1 5 c2 â 2 : This implies immediately that â 1 5 c. (ii) By (3.1) and the fact that â b; we obtain 3 5 c 1 1 (5â +5b + (5â +5b) 2 12âb) = 1 1 (5â +5b + (5b 5â) 2 2âb) 1 1 (5â +5b + (5b 5â) 2 ) = b : (iii) Condition (4.18) implies â + b âb=c ; and hence â + b + c : (iv) Since â 2 ; 2(âb +âc) ; and by assertion (ii), we obtain â 2 + b 2 + c 2 2(âb + bc + câ) b 2 + c 2 2bc =(c b) c2 :

24 134 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) (v) It is easy to prove that â 2 + b 2 + c (âb + bc + câ) by making use of assertion (iv) and (4.18). In fact, â 2 + b 2 + c (âb + bc + câ) =â2 + b 2 + c 2 2(âb + bc + câ) (âb + bc + câ) 4 25 c (âb + bc + câ) : Next, the inequality in assertion (v) follows easily by (4.18) and assertion (iii). (vi) Since b; c ; we obtain ( ) â2 + b 2 + c (âb + bc + câ) â â 2 + b 2 + c b 2 + c 2 23 (âb + bc + câ) 3 (âb + bc + câ) and hence â + b + c [â 2 + b 2 + c (âb + bc + câ)] [(â + b + c) â 2 + b 2 + c (âb + bc + câ)] : (vii) By (4.18), â 2 + b 2 + c 2 (âb + bc + câ) â 2 + b 2 + c (âb + bc + câ): In addition, since â 2 ; (âb +âc) and bc ; by assertion (ii), â 2 + b 2 + c (âb + bc + câ) b2 + c 2 2bc =(c b) c2 : (viii) By (4.18) and assertions (i) and (ii), â 2 +b 2 +c (âb + bc + câ) â 2 +b 2 +c c c2 +c 2 = 39 5 c: On the other hand, since â 2 ; â(b + c) ; and bc ; by assertion (ii), â 2 + b 2 + c (âb + bc + câ) b 2 + c 2 2bc = c b 2 5 c: The proofoflemma 4.9 is complete. We are now ready to prove Theorem 3.6. Case 1: âb + bc + câ 6 : In this case, it is easy to see that f(u)= u(u â)(u b)(u c) satisfying (3.9) satises conditions (1.3) and (1.4) in Theorem 1.1. Also, f satises (1.5) since c f(u)du = 1 6 c3 (3c 2 5c(â + b)+1âb)

25 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) by (3.1). Finally, since f () = 2(âb + bc + câ) and by Lemma 4.9(v), the quadratic polynomial f (u)= 12u 2 +6(â + b + c)u 2(âb +âc + bc) has exactly one positive zero r given by r = 1 4 â (â + b + c b 2 + c (âb + bc + câ)) c: So (1.6) holds. We summarize that f satises (1.3) (1.6) in Theorem 1.1. Case 2: âb + bc + câ : In this case, for f(u)= u(u â)(u b)(u c) satisfying (3.9), it is easy to see that f satises conditions (f1) and (f2) in Theorem 3.1. Also, f satises (f3) since c f(u)du = 1 6 c3 (3c 2 5c(â + b)+1âb) by (3.1). In addition, since f () = 2(âb + bc + câ) ; the quadratic polynomial f (u)= 12u 2 +6(â + b + c)u 2(âb +âc + bc) has exactly two positive zeros r s given by r = 1 4 â (â + b + c 2 + b 2 + c (âb + bc + câ)) ; s = 1 4 â (â + b + c b 2 + c (âb + bc + câ)) c: So f satises (f4). Finally, for (f5), it can be shown that there exists a unique number = 1 3 (â + b + c + â 2 + b 2 + c 2 (âb + bc + câ)) satisfying f(u) uf (u)=u 2 (3u 2 2(â+b+c)u+âb+bc +câ)= on (b; c). Similarly as before, let 1 x := â + b + c c b + c c Then by Lemma 4.9, we compute that 2736( () (r)) 3 5 c + c = 8 c 5 : =256(â + b + c + â 2 + b 2 + c 2 (âb + bc + câ)) 3 (â 2 + b 2 + c 2 2(âb + bc + câ)+(â + b + c) â2 + b 2 + c 2 (âb + bc + câ)) + 243(â + b + c â 2 + b 2 + c (âb + bc + câ))3 (â 2 + b 2 + c (âb + bc + câ) (â + b + c) â 2 + b 2 + c (âb + bc + câ))

26 136 I. Addou, S.-H. Wang / Nonlinear Analysis 53 (23) ( 256 â + b + c + 2 ) 3 ( ( )) c 25 c2 +(â + b + c) 5 c (â + b + c 25 ) ( 3 c 4 25 c2 (â + b + c) ( 39 5 c )) ( =c {256 5 x + 2 ) 3 ( ) ( 5 x x 2 ) ( 3 )} x { =c (5x +2) ( x 2 3 (4 5 5) } 39x) for 1 x 8 5 (cf. (4.17)); we omit the proofofthe last inequality. So (f5) holds. We summarize that f satises (f1) (f5) in Theorem 3.1. The proofoftheorem 3.6 is complete. Acknowledgements The authors thank Professors P. Korman and J. Shi for sending many of their recent papers. Much ofthe computation in this paper has been checked using the symbolic manipulator Mathcad 7 Professional for I. Addou and Mathematica 4. for S.-H. Wang. References [1] I. Addou, Multiplicity ofsolutions for quasilinear elliptic boundary-value problems, Electron. J. Dierential Equations 21 (1999) [2] I. Addou, Exact multiplicity results for quasilinear boundary-value problems with cubic-like nonlinearities, Electron. J. Dierential Equations 1 (2) 1 26 (addendum, 27 29). [3] M.G. Crandall, P.H. Rabinowitz, Bifurcation, perturbation of simple eigenvalues and linearized stability, Arch. Rational Mech. Anal. 52 (1973) [4] M. Guedda, L. Veron, Bifurcation phenomena associated to the p-laplace operator, Trans. Amer. Math. Soc. 31 (1988) [5] P. Korman, The global solution set for a class of semilinear problems, J. Math. Anal. Appl. 26 (1998) [6] P. Korman, Y. Li, T. Ouyang, Exact multiplicity results for boundary value problems with nonlinearities generalizing cubic, Proc. Royal Soc. Edinburgh 126A (1996) [7] P. Korman, Y. Li, T. Ouyang, An exact multiplicity result for a class of semilinear equations, Commun. Partial Dierential Equations 22 (1997) [8] P. Korman, T. Ouyang, Multiplicity results for two classes of boundary-value problems, SIAM J. Math. Anal. 26 (1995) [9] P. Korman, T. Ouyang, Exact multiplicity results for a class of boundary-value problems with cubic nonlinearities, J. Math. Anal. Appl. 194 (1995) [1] P. Korman, J. Shi, Instability and exact multiplicity ofsolutions ofsemilinear equations, Electron. J. Dierential Equations Conf. 5 (2)

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