J. Math. Anal. Appl. 35 (26) 725 739 www.elsevier.com/locate/jmaa Layer structures for the solutions to the perturbed simple pendulum problems Tetsutaro Shibata Applied Mathematics Research Group, Graduate School of Engineering, Hiroshima University, Higashi-Hiroshima, 739-8527, Japan Received 28 July 24 Available online 2 August 25 Submitted by R.E. O Malley, Jr. Abstract We consider the perturbed simple pendulum equation u (t) = µf ( u(t) ) + λ sin u(t), t := ( T,T), u(t) >, t, u(±t)=, where λ>andµ R are parameters. The typical example of f is f(u)= u p u(p>). The purpose of this paper is to study the shape of the solutions when λ. More precisely, by using a variational approach, we show that there exist two types of solutions: one is almost flat inside and another is like a step function with two steps. 25 Elsevier nc. All rights reserved. Keywords: nterior layer; Boundary layer; Simple pendulum problems. ntroduction We consider the perturbed simple pendulum equation u (t) = µf ( u(t) ) + λ sin u(t), t := ( T,T), (.) u(t) >, t, (.2) E-mail address: shibata@amath.hiroshima-u.ac.jp. 22-247X/$ see front matter 25 Elsevier nc. All rights reserved. doi:.6/j.jmaa.25.6.49
726 T. Shibata / J. Math. Anal. Appl. 35 (26) 725 739 u(±t)=, (.3) where T>isaconstant and λ>, µ R are parameters. We assume that f satisfies the following conditions: (A) f C (R), f ( u) = f(u)for u R and f(u)>foru>. (A2) f () =. (A3) f(u)/uis increasing for <u<π. The typical example of f(u)is f(u)= u p u(p>). The purpose of this paper is to study the shape of the solutions of (.) (.3) when λ. More precisely, by using a variational approach, we show that (.) (.3) has two types of solutions: one is almost flat inside and another is like a step function with two steps. Therefore, it is shown that the structure of the solutions (.) (.3) is rich. Linear and nonlinear multiparameter problems have been investigated intensively by many authors. We refer to [ 4,6 9] and the references therein. n particular, one of the main topics for the nonlinear problems is to study the equations which develop layer type solutions. ndeed, concerning the layer structure of the solutions, a possible layer structure was brought out in [6,7] for one-parameter singular perturbation problems; and it is known that the solutions with layers appear for two-parameter problems, which are different from (.) (.3) (cf. [ 2,4]). Recently, Shibata [3] considered (.) (.3) for the case µ< by means of the following constrained minimization method. Let } U β := {u H (): ( ) cos u(t) dt = β, where <β<4t is a fixed constant and H () is the usual real Sobolev space. Regarding µ< as a given parameter, consider the minimizing problem, which depends on µ: Minimize 2 u 2 2 µ F ( u(t) ) dt under the constraint u U β, (.4) where F(u) := u f(s)ds. Then by Lagrange multiplier theorem, for a given µ<, a unique solution triple (µ, λ(µ), u(µ)) R 2 + U β was obtained, where λ(µ) is the Lagrange multiplier. Then the following result was obtained in [3]: Theorem. [3]. Let <θ β <π satisfies cos θ β = β/(2t). Then u(µ) θ β uniformly on any compact subset in and λ(µ) as µ. We see from Theorem. that u(µ) is almost flat inside and develops boundary layer as µ. We emphasize that this asymptotic behavior of u(µ) is the most characteristic feature of the solution of two-parameter problem (.) (.3) in the following sense. Let µ = µ < be fixed in (.) (.3) and consider a one-parameter problem
T. Shibata / J. Math. Anal. Appl. 35 (26) 725 739 727 v (t) = µ f ( v(t) ) + λ sin v(t), t, (.5) v(t) >, t, (.6) v(±t)=. (.7) Then for a given λ, there exists a unique solution (λ, v λ ) R + C 2 ( ). Moreover, if λ, then v λ π locally uniformly in (cf. [5]). Therefore, we do not have any solutions {v λ } of a one-parameter problem (.5) (.7) such that v λ θ β (< π) as λ. t should be pointed out that only a flat solution has been obtained in [3], since only the case where µ< has been considered. ndeed, if µ< is assumed, then we see from [5] that the maximum norm of the solution is less than π. Therefore, by the variational method (.4), it is impossible to treat the solution of (.) (.3) with maximum norm larger than π. To treat both cases mentioned above at the same time, we adopt here another sort of variational approach. Namely, we regard λ> as a given parameter here and using different type of variational approach from (.4), we show that (.) (.3) has both flat and step function type solutions. t is shown that the maximum norm of step function type solution is bigger than π. We now explain the variational framework used here. Let M α := {v H (): Q(v) := F ( v(t) ) } dt = 2TF(α), (.8) where α> is a constant. Then consider the minimization problem, which depends on λ>: Minimize K λ (v) := 2 v 2 2 λ ( ) cos v(t) dt Let under the constraint v M α. (.9) β(λ,α) := min K λ (v). v M α Then by Lagrange multiplier theorem, for a given λ>, there exists (λ, µ(λ), u λ ) R 2 M α which satisfies (.) (.3) with K λ (u λ ) = β(λ,α), where µ(λ), which is called the variational eigenvalue, is the Lagrange multiplier. Now we state our results. Theorem.. Let <α<π be fixed. Then (a) µ(λ) < for λ. (b) u λ α locally uniformly on as λ. (c) µ(λ) = C λ + o(λ) as λ, where C = sin α/f (α). The following Theorem.2 is our main result in this paper.
728 T. Shibata / J. Math. Anal. Appl. 35 (26) 725 739 Theorem.2. Let π<α<3π be fixed. Then (a) µ(λ) >. (b) µ(λ) as λ. More precisely, as λ, λ exp ( ( ) ) ( ( ) ) t α + o() λ <µ(λ)<λexp tα o() λ, where t α := (F (α) F(π))T/(F(3π) F(π)), which is positive by the condition π<α<3π and (A). (c) Assume 3F(α)<F(3π)+ 2F(π). (.) Then, as λ, u λ 3π locally uniformly on ( t α,t α ), u λ π locally uniformly on ( T, t α ) (t α,t). Remark.3. (i) Theorem.(b) implies that for <α<π, u λ is almost flat inside and develops boundary layer as λ. On the other hand, Theorem.2(c) implies that for π<α<3π satisfying (.), u λ has both boundary layers and interior layers. Moreover, u λ is almost flat in ( t α,t α ) and ( T, t α ) (t α,t). n other words, u λ is almost a step function with two steps in this case. Therefore, the structures of u λ for <α<π and π<α<3π are totally different. (ii) The rough idea of the proof of Theorem.2(c) is as follows. We first show that u λ has boundary layers at t =±T. Secondly, we show that u λ has a interior layer in (,T) and is almost equal to π and 3π. nequality (.) is a technical condition to obtain the estimate of u λ from above. Then the position of the interior layer is established automatically. f f(u)= u p u (p >), then (.) implies π<α<((3 p+ + 2)/3) /(p+) π. For instance, if p = 7, then (.) is equivalent to π<α<(6563/3) /8 π = 2.65... π. (iii) t is certainly important to consider the asymptotic shape of u λ as λ for the case α = π. Clearly, u λ is almost equal to π in ( T,) (,T). By Theorem.(b), we see that if α<π and α is very close to π, then u λ is almost flat and equal to π inside when λ. On the other hand, by Theorem.2(c), if α>π and α is very close to π, then u λ is almost flat and equal to π in ( T, t α ) (t α,t), and u λ is almost equal to 3π in ( t α,t α ). Since α>π and α is nearly equal to π, we see that t α is very small by definition of t α and t α asα π. Therefore, if α = π, then the asymptotic shape of u λ when λ is expected to be a box with spike at t =. Therefore, it is quite interesting to determine whether the asymptotic shape of u λ is like a box with spike at t = asλ when α = π. However, it is difficult to treat this problem by our methods here. The reason whyisasfollows.weregardα as a parameter and denote the minimizer by u λ = u λ,α if u λ M α. Then it is quite natural to consider a sequence of minimizer {u λ,α } for a fixed λ, and observe a limit function u λ,π = lim α π u λ,α. Then it is not so difficult to show that u λ,π is also a minimizer of (.9) for α = π, and satisfies (.) (.3). Therefore, it is expected that u λ,π 3π as λ. However, to show this, the uniform estimate
T. Shibata / J. Math. Anal. Appl. 35 (26) 725 739 729 u λ,α 3π δ for all λ>λ and π<α<π+ δ for some <δ is necessary. Since this estimate is quite difficult to show, it is so hard to show whether u λ 3π or π as λ. From these points of view, it is not easy to study the case where α = π by the simple calculation. The future direction of this study is certainly to extend our investigation to the case where α = π,3π,... The remainder of this paper is organized as follows. n Section 2, we prove Theorem.2. Section 3 is devoted to the proof of Theorem.. For the sake of completeness, we show the existence of (λ, µ(λ), u λ ) R 2 M α in Appendix A. 2. Proof of Theorem.2 n what follows, we denote by C the various constants which are independent of λ. n particular, the several characters C, which appear in an equality or an inequality repeatedly, may imply the different constants each other. Proof of Theorem.2(a). Assume that µ(λ). Then u λ > satisfies u (t) ( ) µ(λ) f u(t) = λ sin u(t), t, u(±t)=. Then it follows from [5] that u λ <π. ndeed, let <m λ <π satisfy µ(λ) f(m λ ) = λ sin m λ. Then we know from [5] that u λ <m λ. This is impossible, since u λ M α and α>π. We next prove Theorem.2(c). To do this, we need some lemmas. For a given γ>, let t γ,λ [,T] satisfy u λ (t γ,λ ) = γ, which is unique if it exists, since u λ (t) <, <t T. (2.) Lemma 2.. Let <δ be fixed. Then t π δ,λ T as λ. Proof. Since u λ M α (π<α<3π), we see that u λ >π. Therefore, there exists unique t π,λ. By (.), we have { u λ (t) + µ(λ)f ( u λ (t) ) + λ sin u λ (t) } u λ (t) =. This implies that for t [,T] 2 u λ (t)2 + µ(λ)f ( u λ (t) ) + λ ( cos u λ (t) ) constant = µ(λ)f ( ) ( ) u λ + λ cos uλ (put t = ) = 2 u λ (t π,λ) 2 + µ(λ)f (π) + 2λ (put t = t π,λ ). (2.2) By this and (2.), we see that for t [,T]
73 T. Shibata / J. Math. Anal. Appl. 35 (26) 725 739 u λ (t) = 2λ ( + cos u λ (t) ) + 2µ(λ) ( F(π) F ( u λ (t) )) + u λ (t π,λ) 2. (2.3) By this and Theorem.2(a), for t π δ,λ t T, u λ (t) 2λ( cos δ). (2.4) By this, we obtain π δ = T t π δ,λ u λ This implies our conclusion. (t) dt 2λ( cos δ)(t tπ δ,λ ). Lemma 2.2. Assume that u λ 3π for λ. Let an arbitrary <δ be fixed. Then t π+δ,λ t 3π δ,λ as λ. Lemma 2.2 can be proved by the same argument as that in Lemma 2.. So we omit the proof. Lemma 2.3. Let an arbitrary <δ be fixed. Then for λ t π,λ t π+δ,λ >t π δ,λ t π,λ. (2.5) Proof. By (2.3) and putting θ := π u λ (t), we obtain t π δ,λ t π,λ = Similarly, by (2.3) t π δ,λ t π,λ u λ (t) dt 2λ( + cos u λ (t)) + 2µ(λ)(F (π) F(u λ (t))) + u λ (t π,λ) 2 δ dθ =. (2.6) 2λ( cos θ)+ 2µ(λ)(F (π) F(π θ))+ u λ (t π,λ) 2 δ dθ t π,λ t π+δ,λ =. (2.7) 2λ( cos θ) 2µ(λ)(F (θ + π) F(π))+ u λ (t π,λ) 2 By this and (2.6), we obtain (2.5). Lemma 2.4. Assume that there is a constant <δ satisfying lim sup λ u λ > 3π + δ. Then for <δ δ and λ t 3π,λ t 3π+δ,λ >t π,λ t π+δ,λ. (2.8) Proof. Let <δ δ be fixed. Put t = t 3π,λ in (2.2). Then we obtain 2 u λ (t π,λ) 2 + µ(λ)f (π) + 2λ = 2 u λ (t 3π,λ) 2 + µ(λ)f (3π)+ 2λ. (2.9)
T. Shibata / J. Math. Anal. Appl. 35 (26) 725 739 73 By this and (A), we see that u λ (t 3π,λ) 2 <u λ (t π,λ) 2. (2.) By this and (2.2), for t [t 3π+δ,λ,t 3π,λ ],wehave 2 u λ (t)2 = λ ( + cos u λ (t) ) µ(λ) ( F ( u λ (t) ) F(3π) ) + 2 u λ (t 3π,λ) 2 <λ ( + cos u λ (t) ) µ(λ) ( F ( u λ (t) ) F(3π) ) + 2 u λ (t π,λ) 2. (2.) This along with (2.) and the same argument as that to obtain (2.6) implies that t 3π,λ t 3π+δ,λ δ dθ >. (2.2) 2λ( cos θ) 2µ(λ)(F (θ + 3π) F(3π))+ u λ (t π,λ) 2 By (A3), it is easy to see that for θ δ, F(θ + 3π) F(3π)>F(θ + π) F(π). (2.3) Then (2.7), (2.2) and (2.3) imply (2.8). Thus the proof is complete. Proof of Theorem.2(c). We first show that lim sup u λ 3π. (2.4) λ To do this, we assume that there exists a constant <δ and a subsequence of { u λ }, which is denoted by { u λ } again, such that u λ > 3π + δ and derive a contradiction. For <δ<δ and λ, we put π,δ,λ := t π δ,λ t π+δ,λ, 3π,δ,λ := t 3π,λ t 3π+δ,λ. Then we see from Lemmas 2.3 and 2.4 that π,δ,λ < 2 3π,δ,λ, π,δ,λ + 3π,δ,λ <T. (2.5) By (2.5), for λ, we obtain π,δ,λ < 2T/3. Since u λ M α and u λ (t) = u λ ( t) for t [,T], by this and Lemmas 2. and 2.2, we obtain T TF(α)= F ( u λ (t) ) dt F(π δ) π,δ,λ + F(3π δ)(t π,δ,λ ) + o() F(3π)T ( F(3π) F(π) ) π,δ,λ + O(δ) + o() F(3π)T 2 3 T ( F(3π) F(π) ) + O(δ)+ o() = 3 TF(3π)+ 2 TF(π)+ O(δ) + o(). (2.6) 3
732 T. Shibata / J. Math. Anal. Appl. 35 (26) 725 739 Since <δ is arbitrary, this contradicts (.). Therefore, we obtain (2.4). Then there are two possibilities: (i) lim λ u λ = π or (ii) lim λ u λ = 3π. However, (i) is impossible, since u λ M α (π < α < 3π). Hence, we obtain (ii). Then we see from Lemmas 2. and 2.2 that for any t [,T), we have only two possibilities: (i) lim λ u λ (t) = π or (ii) lim λ u λ (t) = 3π. Then by (2.) and u λ M α, obviously, Theorem.2(c) holds, since u λ (t) = u λ ( t) for t [,T]. Thus the proof is complete. Proof of Theorem.2(b). The proof is divided into three steps. Step. We first show that for λ u λ 3π. (2.7) ndeed, if there exists a subsequence of { u λ }, which is denoted by { u λ } again, such that u λ > 3π, then by putting θ = u λ (t) 3π and δ λ := u λ 3π >, we see from the same calculation as that to obtain (2.2) that t 3π,λ > δ λ dθ. (2.8) 2λ( cos θ) 2µ(λ)(F (θ + 3π) F(3π))+ u λ (t π,λ) 2 Further, by putting δ = δ λ in (2.6) and (2.7), we obtain t π δλ,λ t π,λ = δ λ t π,λ t π+δλ,λ = δ λ dθ, (2.9) 2λ( cos θ)+ 2µ(λ)(F (π) F(π θ))+ u λ (t π,λ) 2 dθ. (2.2) 2λ( cos θ) 2µ(λ)(F (θ + π) F(π))+ u λ (t π,λ) 2 By (2.3) and (2.8) (2.2), we obtain t 3π,λ >t π,λ t π+δλ,λ >t π δλ,λ t π,λ. (2.2) By this and the same argument to obtain (2.6), we obtain a contradiction. Therefore, we obtain (2.7). Step 2. Assume that there exists a subsequence of {µ(λ)}, denoted by {µ(λ)} again, such that µ(λ) δ >. By (2.2), we have 2 u λ (t)2 = µ(λ) ( F ( ) ( u λ F uλ (t) )) + λ ( ) cos u λ (t) cos u λ. (2.22) Let <δ be fixed. By mean value theorem and (A.3), for t [,t 3π 2δ,λ ] and λ, we obtain
T. Shibata / J. Math. Anal. Appl. 35 (26) 725 739 733 F ( u λ ) F ( uλ (t) ) f(3π 2δ) ( u λ u λ (t) ) ( f(3π) Cδ )( u λ u λ (t) ). (2.23) By (2.7) and the fact that u λ 3π as λ,fort [,t 3π 2δ,λ ] and λ, cos u λ (t) cos u λ = sin u λ ( uλ (t) u λ ) 2 cos( θ u λ + ( θ)u λ (t) )( u λ u λ (t) ) 2 2( Cδ o() )( uλ u λ (t) ) 2, (2.24) where <θ<. By (2.22) (2.24), for λ u λ (t) 2 ( f(3π) Cδ ) µ(λ) ( u λ u λ (t) ) + λ ( Cδ o() )( u λ u λ (t) ) 2 ( := A λ uλ u λ (t) ) 2 ( + Bλ uλ u λ (t) ), (2.25) where A λ = λ ( Cδ o() ), By this, for λ, t 3π 2δ,λ = = t 3π 2δ,λ dt u λ 3π+2δ < Aλ 3δ t 3π 2δ,λ B λ = 2 ( f(3π) Cδ ) µ(λ). u λ (t) Aλ ( u λ u λ (t)) 2 + B λ ( u λ u λ (t)) dt Aλ θ 2 + B λ θ dθ dθ (θ + B λ /(2A λ )) 2 Bλ 2/(4A2 λ ) = [log ( Aλ 3δ + B λ + 3δ + B ) ] 2 λ B2 λ 2A λ 2A λ log B λ. (2.26) 2A λ Step 3. There are three cases to consider. Case (i). Assume that there exists a subsequence of {µ(λ)} satisfying µ(λ)/λ as λ. Then since µ(λ) δ,asλ, t 3π 2δ,λ C ( C + C log λ ) C ) λ (C + C log λδ. λ µ(λ) This along with Lemmas 2. and 2.2 implies that u λ (t) π (t \{}) asλ.this is a contradiction, since u λ M α and π<α. Case (ii). Assume that there is a subsequence of {µ(λ)} satisfying C µ(λ)/λ C. Then by this and (2.26), as λ, 4A 2 λ
734 T. Shibata / J. Math. Anal. Appl. 35 (26) 725 739 t 3π 2δ,λ C λ. (2.27) By the same reason as that of case (i), this is a contradiction. Case (iii). Assume that there exists a subsequence of {µ(λ)} satisfying µ(λ)/λ. Then by this and (2.26), as λ, ( ) t 3π 2δ,λ o λ. (2.28) By the same reason as that of case (i), this is a contradiction. Therefore, we see that µ(λ) asλ. Finally, we show the decay rate of µ(λ) as λ. Since t 3π 2δ,λ t α as λ,by (2.26), L := Aλ < Aλ 2δ 3δ Aλ θ 2 + B λ θ dθ < t 3π 2δ,λ Aλ θ 2 + B λ θ dθ := L 2. (2.29) Since µ(λ) asλ, L = ( ) ( + o() log ( 3δ + o() ) ) λ ( ) + log <t α + o(). (2.3) λ Cµ(λ) This implies that log λ µ(λ) <t ( ) α + o() λ. (2.3) By this, we obtain µ(λ) > λ exp ( ( ) ) t α + o() λ. (2.32) By the same argument as above, we also obtain µ(λ) < λ exp ( ( ) ) t α o() λ. (2.33) Thus the proof is complete. 3. Proof of Theorem. To prove Theorem., we follows the idea of the proof of [3, Theorem 2]. Proof of Theorem.(a). We assume that µ(λ) and derive a contradiction. There are three cases to consider. Case. Assume that there exists a subsequence of { u λ }, denoted by { u λ } again, such that u λ π.let<δ be fixed. Then by (2.2), for t [t π δ,λ,t]
T. Shibata / J. Math. Anal. Appl. 35 (26) 725 739 735 2 u λ (t)2 λ ( + cos u λ (t) ) λ( cos δ). By this and (2.), π δ = T π δ,λ u λ (t) dt 2λ( cos δ)(t t π δ,λ ). This implies that t π δ,λ T as λ. This is a contradiction, since u λ M α and < α<π. Case 2. Assume that there exists a subsequence of { u λ }, denoted by { u λ } again, such that u λ π as λ. Then we see that α + δ< u λ <π for <δ. Then it is clear that t α+δ,λ T as λ. ndeed, if t α+δ,λ T as λ, then 2TF(α)= lim sup Q(u λ ) 2TF(α+ δ). λ This is a contradiction. Therefore, there exists a constant <ɛ such that <t α+δ,λ T ɛ for λ. We choose φ C () satisfying supp φ (T ɛ,t). Since u λ (t) α + δ for t (T ɛ,t)by (2.), we see that for t (T ɛ,t)and λ sin u λ (t) δ >. (3.) u λ (t) Then for λ µ(λ) = inf v H () v v 2 2 λ sin u λ (t) u λ (t) f(u λ (t)) u λ (t) v 2 dt v 2 dt φ 2 2 λδ f(u λ (t)) u λ (t) φ2 dt φ 2 dt <. (3.2) Case 3. Assume that there exists a subsequence of { u λ }, denoted by { u λ } again, such that lim λ u λ <π. Since (3.) holds for any t [,T] in this case, we choose φ C () and obtain (3.2). Thus the proof is complete. For simplicity, we put µ(λ) := µ(λ) >. Then (.) is equivalent to u (t) + µ(λ)f ( u(t) ) = λ sin u(t), t. (3.3) Therefore, in what follows, we consider (3.3) with conditions (.2) and (.3). Lemma 3.. µ(λ) Cλ for λ. Proof. Let θ λ ={t (,π): µ(λ)f (θ) = λ sin θ}. Then by [5], we see that u λ <θ λ <π. Furthermore, since u λ M α, we see that for λ u λ >δ >. By the same calculation as that to obtain (2.2), we have (3.4) (3.5)
736 T. Shibata / J. Math. Anal. Appl. 35 (26) 725 739 2 u λ (t)2 λ cos u λ (t) µ(λ)f ( u λ (t) ) = λcos u λ µ(λ)f ( ) u λ By (A3), (3.5) and (3.6), we obtain = 2 u λ (T )2 λ. (3.6) µ(λ)f (δ ) µ(λ)f ( ) u λ = 2 u λ (T )2 + λ ( ) cos u λ 2λ. Thus the proof is complete. We put g λ (u) := Then by (3.3), sin u µ(λ)f (u)/λ. (3.7) u u λ (t) = λg( u λ (t) ) u λ (t), t. (3.8) Lemma 3.2. g λ (u λ (t)) locally uniformly on as λ. Proof. We assume that there exists a constant δ>, t [,T)and a subsequence of {λ}, denoted by {λ} again, such that g λ (u λ (t )) δ for λ. Since g λ (u) is decreasing for <u<π by (A3), we see from (2.) that for any t [t,t)and λ ( g λ uλ (t) ) ( g λ uλ (t ) ) δ. (3.9) We choose φ C () with supp φ (t,t). Then by (3.8), we obtain λ = inf v H () v v 2 2 g λ(u λ (t))v 2 dt φ 2 2 δ φ 2. 2 This is a contradiction. Thus the proof is complete. Lemma 3.3. λ C µ(λ) for λ. Proof. We assume that there exists a subsequence of {λ/ µ(λ)}, denoted by {λ/ µ(λ)} again, such that λ/ µ(λ) as λ, and derive a contradiction. Multiply (3.3) by u λ. Then integration by parts along with (3.4), (3.8) and Lemma 3.2 implies that, as λ, 2 2 = λ u λ (t) sin u λ (t) dt µ(λ) f ( u λ (t) ) u λ (t) dt u λ ( = λ g λ uλ (t) ) u λ (t) 2 dt = o(λ). (3.) By (3.4) and the assumption, µ(λ) f ( u λ (t) ) u λ (t) dt = o(λ).
T. Shibata / J. Math. Anal. Appl. 35 (26) 725 739 737 By this and (3.), as λ, u λ (t) sin u λ (t) dt. (3.) Since u λ M α, by (3.4) and (3.), we see that u λ (t) π, t [,t ), (3.2) u λ (t), t (t,t], (3.3) where t := F(α)T/F(π). Then by (3.4), (3.6), (3.2) and (3.3), for t (t,t] and λ 2 u λ (t)2 = λ ( ) ( ( cos u λ (t) cos u λ + µ(λ) F uλ (t) ) F ( )) u λ = 2 ( + o() ) λ. (3.4) Let <δ be fixed. Then by (2.) and (3.4), for λ π>u λ (t + δ) u λ (t + 2δ) = t +2δ t +δ This is a contradiction. Thus the proof is complete. u λ (t) dt 2 λδ ( + o() ). Proof of Theorem.(b). Let an arbitrary <t 2 <T be fixed. We first prove that u λ (t 2 ) δ for λ. ndeed, if there exists a subsequence of {u λ (t 2 )} λ, denoted by {u λ (t 2 )} λ again, such that u λ (t 2 ) asλ, then by (A2) and Lemmas 3. 3.3, as λ, sin u λ (t 2 ) u λ (t 2 ) ( = g λ uλ (t 2 ) ) + µ(λ) f(u λ (t 2 )). (3.5) λ u λ (t 2 ) This is a contradiction, since the left-hand side of (3.5) tends to as λ. This implies that u λ (t) δ for any t t 2 and λ. Now let ξ>beanarbitrary accumulation point of { µ(λ)/λ}. We see that θ λ π as λ. ndeed, if θ λ π as λ, then µ(λ) λ = sin θ λ f(θ λ ). This contradicts Lemma 3.3. Therefore, we see from (3.4), (3.5) and the argument above that for t [,t 2 ] and λ δ u λ (t) π δ. By this and Lemma 3.2, for t t 2,asλ, ( µ(λ) sin ξ = lim λ λ = lim uλ (t) λ f(u λ (t)) g ) λ(u λ (t))u λ (t) sin u λ (t) = lim f(u λ (t)) λ f(u λ (t)). (3.6) Since sin u λ (t)/f (u λ (t)) is increasing for t [,T], and u λ M α, this implies that u λ α locally uniformly as λ and ξ = C. Now our assertion follows from a standard compactness argument. Thus the proof is complete.
738 T. Shibata / J. Math. Anal. Appl. 35 (26) 725 739 Acknowledgment The author thanks the referee for helpful comments that improved the manuscript. Appendix A n this section, we show the existence of (λ, µ(λ), u λ ) R 2 M α, where u λ is the minimizer of the problem (.9). Let λ> and α> be fixed. Since K λ (v) 4Tλfor any v M α, we can choose a minimizing sequence {u n } n= M α such that, as n, K λ (u n ) β(λ,α) 4Tλ. (A.) Since K λ (u n ) = K λ ( u n ) and u n M α by assumption (A), without loss of generality, we may assume that u n forn N. By(A.),forn N, u 2 n 2 2 K λ(u n ) + 4Tλ<C. Therefore, we can choose a subsequence of {u n } n=, denoted by {u n} n= again, such that, as n, u n u λ weakly in H (), (A.2) u n u λ in C( ). (A.3) By (A.3), we see that u λ M α. n particular, u λ in. Furthermore, by (A.2) and (A.3), K λ (u λ ) = u 2 λ 2 2 λ ( cos uλ (t) ) dt lim inf n lim inf n u 2 2 n 2 lim λ ( cos un (t) ) dt n ( u 2 2 n 2 λ ( cos un (t) ) dt This implies that u λ is a minimizer of (.9). Then Q (u λ )u λ = f ( u λ (t) ) u λ (t) dt >, ) = β(λ,α). where the prime denotes the Fréchet derivative of Q. Now we apply the Lagrange multiplier theorem to our situation and obtain (λ, µ(λ), u λ ) R 2 M α, which satisfies (.) and (.3) in a weak sense. Here µ(λ) is the Lagrange multiplier. Then by a standard regularity theorem, we see that u λ C 2 ( ) and it follows from the strong maximum principle that u λ > in. Thus the proof is complete.
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