# S-Shaped Connected Component for Nonlinear Fourth-Order Problem of Elastic Beam Equation

Authors: Jinxiang Wang; Ruyun Ma; Jin Wen
Journal: Journal of Function Spaces
 (2017)

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## Abstract

We investigate the existence ofS-shaped connected component in the set of positive solutions of the fourth-order boundary value problem: u′′′′x=λhxfux, x∈(0,1),u(0)=u(1)=u′′0=u′′1=0, where λ>0 is a parameter, h∈C[0,1], and f∈C[0,∞) with f0≔lims→0⁡(f(s)/s)=∞. We develop a bifurcation approach to deal with this extreme situation by constructing a sequence of functions f[n] satisfying f[n]→f and (f[n])0→∞. By studying the auxiliary problems, we get a sequence of unbounded connected components C[n], and, then, we find an unbounded connected component C in the set of positive solutions of the fourth-order boundary value problem which satisfies 0,0∈C⊂lim⁡sup⁡C[n] and is S-shaped.


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## Body

## 1. Introduction

The fourth-order boundary value problem(1)u′′′′x=λfx,ux,x∈0,1,u0=u1=u′′0=u′′1=0describes the deformations of an elastic beam with simple support at the end; see Gupta [1]. This kind of problems has been extensively studied by using topological degree theory, fixed point theorems, lower and upper solutions method, and critical point theory; see [2–17] and the references therein.Let(2)f0≔lims→0fx,ss.In this case, f0∈(0,∞), the global structure of solution set of (1) has been studied extensively by several authors via the well-known Rabinowitz global bifurcation theorem since studying an unbounded connected component bifurcating from the trivial solution at (π4,0); see Rynne [18], Ma et al. [19], and Dai and Han [20]. Very recently, Wang and Ma [21] considered the more general problem (3)u′′′′x=λfx,ux,u′′x,x∈0,1,u0=u1=u′′0=u′′1=0under the following assumption.(A) There exist constantsa,b∈0,∞ with a+b>0 and c>0, d>0 such that (4)lims2+p2→0fx,s,p-as-bps2+p21+c=-duniformlyforx∈0,1.Since (a+b)∈(0,∞), the Rabinowitz global bifurcation theorem can be used to guarantee an unbounded connected component bifurcating from the trivial solution at (λ1(a,b),0). However, in the extreme situation f0=∞, the Rabinowitz global bifurcation theorem cannot be directly used to get a connected component bifurcating from the trivial solution anymore.To overcome this difficulty, we have to consider a special problem(5)u′′′′x=λhxfux,x∈0,1,u0=u1=u′′0=u′′1=0,where f0=∞. We will develop a bifurcation approach to deal with this extreme situation by constructing a sequence of functions fn satisfying (6)fn⟶f,fn0⟶∞.By studying the auxiliary problems 1.6nu′′′′x=λhxfnux,x∈0,1,u0=u1=u′′0=u′′1=0,we get a sequence of unbounded connected components C[n] of the set of positive solutions of 1.6n, and, then, we find an unbounded component C in the set of positive solutions of (5) which satisfies (0,0)∈C⊂lim⁡sup⁡C[n].More precisely, we will prove the following.Theorem 1.
Assume that(H1)
h∈C[0,1], h(x)≥0 in [0,1], and h(x)>0 in [1/4,3/4];(H2)
f∈C0,∞, f(s)>0 for all s>0, f(0)=0, and f0≔lims→0⁡(fs/s)=∞;(H3)
f∞≔lims→∞⁡fs/s=0;(H4)
there existss0>0 such that 0≤s≤s0 implies that (7)fs≤60h0s0,h0=maxx∈0,1⁡hx;(H5)
there existss0>s0 such that (8)mins∈s0,4s0⁡fss≥16π4h0,h0=minx∈1/4,3/4⁡hx.Then there exist 0<λ∗<1 and λ∗>1 such that(i)
(5) has at least one positive solution if 0<λ<λ∗;(ii)
(5) has at least two positive solutions if λ=λ∗;(iii)
(5) has at least three positive solutions if λ∗<λ<λ∗;(iv)
(5) has at least two positive solutions if λ=λ∗;(v)
(5) has at least one positive solution if λ>λ∗.For other results on the shape of the connected component of solution set, see Wang and Ma [21] and Sim and Tanaka [22].The rest of this paper is arranged as follows. In Section2, we give some preliminaries. In Section 3, we consider the corresponding auxiliary problems and obtain a sequence of unbounded components with rightward direction near the initial point. Section 4 is devoted to showing the direction turns of the component, and we complete the proof of Theorem 1.


## 2. Preliminaries

In this section, we state some preliminary results.Definition 2 (see [23]).
LetX be a Banach space and Cn∣n=1,2,… be a family of subsets of X. Then the superior limit D of {Cn} is defined by (9)D≔lim⁡supn→∞⁡Cn=x∈X∣∃ni⊂N,xni∈Cni,such  thatxni⟶x.Definition 3 (see [23]).
A component of a setQ means a maximal connected subset of Q.Lemma 4 (see [24]).
LetX be a Banach space, and let {Cn} be a family of connected subsets of X. Assume that(a)
there existzn∈Cn, n=1,2,…, and z∗∈X, such that zn→z∗;(b)
limn→∞⁡rn=∞, where rn=sup⁡x∣x∈Cn;(c)
for everyR>0,(⋃n=1∞Cn)∩BR is a relatively compact set of X, where BR=x∈X∣x≤R.
Then there exists an unbounded componentC in D and z∗∈C.Lemma 5 (see [19, Theorem  2.1]).
Assume that (H1) holds. Then the linear problem(10)u′′′′x=λhxux,x∈0,1,u0=u1=u′′0=u′′1=0has a positive simple principal eigenvalue (11)μ1=inf⁡∫01u′′x2dx∫01hxu2xdx∣u∈C40,1,u≢0,u0=u1=u′′0=u′′1=0.Moreover, the corresponding eigenfunction ϕ is positive in (0,1).
Letg∈C[0,1]; it is well known that the fourth-order linear problem (12)v′′′′=gt,t∈0,1,v0=v1=v′′0=v′′1=0has a unique solution (13)vt=∫01∫01Gt,sGs,τgτdτds,where (14)Gt,s=s1-t,0≤s≤t≤1,t1-s,0≤t≤s≤1.Moreover, if g≥0 and g≢0, then (15)v′′t=-∫01Gt,sgsds≤0;that is, v≥0 is concave.
Since Green’s functionG(t,s) has the properties (i)
0≤G(t,t)G(s,s)≤G(t,s)≤G(s,s), ∀t,s∈(0,1),(ii)
G(t,s)≥(1/4)G(s,s), ∀t∈[1/4,3/4], s∈(0,1),then, for any t∈[1/4,3/4], we have (16)vt=∫01Gt,s∫01Gs,τgτdτds≥14∫01Gs,s∫01Gs,τgτdτds≥14v∞.


## 3. Auxiliary Problems and Rightward Bifurcation

For eachn∈N, we define fn(s):0,∞→0,∞ by (17)fns=fs,ifs∈1n,∞,s2nf1n+2n-α-2nf12n+2f12n-22n-1-α-f1n,ifs∈12n,1n,2nf12ns-s1+α,ifs∈0,12n. Then f[n]∈C0,∞,0,∞ with f[n](s)>0 for all s∈(0,∞) and (18)fn0≔lims→0⁡fnss=2nf12n>0. By (H2), it follows that limn→∞⁡fn0=∞.We extendf to an odd function g:R→R by (19)gs=fs,ifs≥0,-f-s,ifs<0. Similarly we may extend f[n] to an odd function g[n]:R→R for each n∈N.Now let us consider the auxiliary family of the equationsPnu′′′′x=λhxgnux,x∈0,1,u0=u1=u′′0=u′′1=0. We rewrite Pn by (20)u′′′′x=λhxgn0ux+λhxgnux-gn0ux,x∈0,1,u0=u1=u′′0=u′′1=0.Since (18) implies (g[n])0=2nf(1/2n)>0, then, using Rabinowitz’s global bifurcation theorem and following the similar arguments in the proof of Theorem1.1 in [19] or Theorem2.2 in [20], we have the following.Lemma 6.
Assume that (H1) and (H2) hold; then, for each fixedn∈N, from (μ1/(g[n])0,0) there emanates an unbounded subcontinuum C[n] of positive solutions of Pn in the set R×E, where E=u∈C30,1∣u0=u1=u′′0=u′′1=0 with the norm u=u∞+u′∞+u′′∞+u′′′∞.Lemma 7.
Assume that (H1) and (H2) hold. For each fixedn∈N, let λk,uk∣k=1,2,… be a sequence of positive solutions to Pn which satisfies λk→μ1/(g[n])0 and uk→0 as k→∞. Let ϕ be the first eigenfunction of (10) which satisfies ϕ∞=1. Then there exists a subsequence of {uk}, again denoted by {uk}, such that uk/uk∞ converges uniformly to ϕ on [0,1].Proof.
Setvk≔uk/uk∞. Then vk∞=1. For every (λk,uk), we have (21)uk′′x=-λk∫01Gx,shsgnuksds.From the boundary condition uk(0)=uk(1)=0, there exists x^k∈(0,1) such that uk′(x^k)=0. Integrating (21) on [x,x^k], we obtain (22)uk′x=λk∫xx^k∫01Gt,shsgnuksdsdt,x∈0,1.Dividing both sides of (22) by uk∞, we get (23)vk′x=λk∫xx^k∫01Gt,shsgnuksuksvksdsdt,x∈0,1.Since uk→0 implies uk∞→0, then, by (18), there exists a constant m1>2nf(1/2n) such that (24)gnuksuks<m1,∀k∈N,s∈0,1.From λk→μ1/(g[n])0, it follows that there exists a constant m2>0 such that (25)λk≤m2,∀k∈N.Then, for x∈[0,1], (23) implies that (26)λk∫xx^k∫01Gt,shsgnuksuksvksdsdt≤m1m2vk∞∫xx^k∫01Gt,shsdsdt≤m1m2vk∞∫01∫01Gt,shsdsdt=Mvk∞=M,M=m1m2∫01∫01Gt,shsdsdt;that is, (27)vk′∞≤M,∀k∈N.Since vk′∞ is bounded, by the Ascoli-Arzela theorem, a subsequence of {vk} uniformly converges to a limit v∈C[0,1] with v∞=1, and we again denote by {vk} the subsequence.
For every(λk,uk), we have (28)ukx=λk∫01Gx,s∫01Gs,τhτgnukτdτds.Dividing both sides of (28) by uk∞, we get (29)vkx=λk∫01Gx,s∫01Gs,τhτgnukτukτvkτdτds.Since uk∞→0, we conclude that g[n](uk(τ))/uk(τ)→(g[n])0 for each fixed τ∈[0,1]. Then Lebesgue’s dominated convergence theorem shows that (30)vx=μ1gn0∫01Gx,s∫01Gs,τhτgn0vτdτds=μ1∫01Gx,s∫01Gs,τhτvτdτds, which means that v is a nontrivial solution of (10) with λ=μ1, and hence v≡ϕ.Lemma 8.
Assume that (H1) and (H2) hold. For each fixedn∈N, let C[n] be as in Lemma 6. Then there exists δ>0 such that (λ,u)∈C[n] and λ-μ1/gn0+u≤δ imply λ>μ1/(g[n])0.Proof.
Assume to the contrary that there exists a sequenceλk,uk∣k=1,2,…⊂C[n] such that λk→μ1/(g[n])0, uk→0, and λk≤μ1/(g[n])0. By Lemma 7, there exists a subsequence of {uk}, again denoted by {uk}, such that uk/uk∞ converges uniformly to ϕ on [0,1]. Multiplying the equation of Pn with (λ,u)=(λk,uk) by ϕ and integrating it over [0,1], we have (31)∫01ϕxuk′′′′xdx=λk∫01hxgnukxϕxdx.By simple computation, one has that (32)∫01ϕxuk′′′′xdx=∫01ϕ′′′′xukxdx=μ1∫01hxϕxukxdx. Combining (31) with (32), we obtain (33)∫01hxgnukxϕxdx=μ1λk∫01hxϕxukxdx; that is (34)∫01hxϕxgnukx-gn0ukxdxuk∞1+α=∫01hxϕxμ1/λkukx-gn0ukxdxuk∞1+α.Since uk→0 implies uk∞→0, then, from the definition of g[n] and (17), we have (35)limk→∞⁡gnukx-gn0ukxukx1+α=-1,∀x∈0,1.Then Lebesgue’s dominated convergence theorem, Lemma 7, and (35) imply that (36)∫01hxϕxgnukx-gn0ukxdxuk∞1+α=∫01hxϕxgnukx-gn0ukxukx1+αukxuk∞1+αdx⟶-∫01hxϕ2+αxdx<0,ask⟶∞.Similarly, (37)∫01hxϕxμ1/λkukx-gn0ukxdxuk∞1+α=μ1-gn0λkλkuk∞α∫01hxϕxukxuk∞dx; by Lebesgue’s dominated convergence theorem and Lemma 7 again, we conclude that (38)∫01hxϕxukxuk∞dx⟶∫01hxϕ2xdx>0,ask⟶∞;this contradicts (34).


## 4. Direction Turns of Component and Proof of Theorem1

Lemma 9.
Assume that (H1) and (H2) hold; then there exists an unbounded connected componentC⊂R×E with (0,0)∈C in the solutions set of (5). Moreover, the component C grows to the right near (0,0).Proof.
Let us verify thatCn∣n=1,2,… satisfy all of the conditions of Lemma 4.
Since(39)limn→∞⁡μ1gn0=limn→∞⁡μ12nf1/2n=0,Condition (a) in Lemma 4 is satisfied with z∗=(0,0). Obviously (40)rn=sup⁡λ+u∣λ,u∈Cn=∞,and, accordingly, (b) holds. (c) can be deduced directly from the Ascoli-Arzela theorem and the definition of g[n]. Therefore, the superior limit of {C[n]}, that is, D, contains an unbounded connected component C⊂R×E with (0,0)∈C. From Lemma 8, the component C grows to right near (0,0).Lemma 10.
Assume that (H1), (H2), and (H3) hold. Letλk,uk∣k=1,2,…⊂C⊂R×E be a sequence of positive solutions to (5); then uk→∞ implies uk∞→∞.Proof.
Assume on the contrary thatuk∞ is bounded; we divide the proof into two cases.
Case  1
λ
k
I
s
B
o
u
n
d
e
d. By recalling (22) and (21), we have that uk′∞ and uk′′∞ are bounded.
From the boundary conditionuk′′0=uk′′1=0, there exists xk∗∈(0,1) such that uk′′′xk∗=0. Integrating the equation of (5) on xk∗,x, we obtain (41)uk′′′x=∫xk∗xuk′′′′sds=λk∫xk∗xhsfuksds,x∈0,1;then uk′′′∞ is bounded too. Finally, we conclude that uk=uk∞+uk′∞+uk′′∞+uk′′′∞ is bounded; this deduces a contradiction.
Case  2
(
λ
k
→
∞
). Since uk∞ is bounded, then, by (H2) and (H3), there exists constant C>0 such that (42)fukx≥Cukx,∀k∈N,x∈0,1.Since (λk,uk)∈C, combining (42) with (16) we have (43)ukx=λk∫01Gx,s∫01Gs,τhτfukτdτds≥λk∫01Gx,s∫1/43/4Gs,τhτCukτdτds≥C14uk∞λk∫01Gx,s∫1/43/4Gs,τhτdτds,which yields that {λk} is bounded; this deduces a contradiction.Lemma 11.
Assume that (H1), (H2), and (H3) hold. Then,C joins (0,0) to (∞,∞) in [0,∞)×E.Proof.
We divide the proof into two steps.
Step  1. We show that supλ∣λ,u∈C=∞.
Assume on the contrary thatsupλ∣λ,u∈C=:c0<∞. Let λk,uk∣k=1,2,…⊂C be such that λk+uk→∞; then uk→∞, and from Lemma 10, uk∞→∞.
Since(λk,uk)∈C, we have that (44)uk′′′′x=λkhxfukx,x∈0,1,uk0=uk1=uk′′0=uk′′1=0.Set ωk:=uk/uk∞, and then ωk∞=1, and (45)ωk′′′′x=λkhxfukxuk∞,x∈0,1,ωk0=ωk1=ωk′′0=ωk′′1=0.From (H3), we have that λkh(x)f(uk(x))/uk∞ is bounded uniformly; then {ωk′′′′} is bounded. By Ascoli-Arzela theorem, choosing a subsequence and relabelling it if necessary, it follows that there exists (λ^,u^)∈[0,c0]×E with u^∞=1 such that (46)limk→∞⁡λk,ωk=λ^,u^.Let(47)f~r=max⁡fs∣0≤s≤r;then f~ is nondecreasing and (H3) implies that (48)limr→∞f~rr=0.Since (49)fukxuk∞≤f~uk∞uk∞,this together with (48) and un∞→∞ implies that(50)limk→∞fukxuk∞=0,uniformlyforx∈0,1.Notice that (45) is equivalent to (51)ωkx=λk∫01Gx,s∫01Gs,τhτfukτuk∞dτds,x∈0,1.Combining this with (50) and using (46) and Lebesgue dominated convergence theorem, it follows that (52)u^=λ^∫01Gx,s∫01Gs,τhτ0dτds=0,x∈0,1.This contradicts with u^∞=1. Therefore, supλ∣λ,u∈C=∞.
Step  2. We show that supu∣λ,u∈C=∞.
Assume on the contrary thatsupu∣λ,u∈C=:M0<∞. Let {(λk,uk)}⊂C be such that (53)λk⟶∞,uk≤M0.Since uk≤M0 implies uk∞≤M0, then, following the same arguments in the proof of Case2 in Lemma 10, we can get a contradiction.Lemma 12.
Assume that (H1), (H2), (H3), and (H4) hold. Let(λ,u)∈C be a solution of (5) with u∞=s0; then λ>1.Proof.
Letu be a solution of (5) with u∞=s0; then, by Condition (H4) and the property of G(x,s), we have (54)s0=u∞=maxx∈0,1⁡λ∫01Gx,s∫01Gs,τhτfuτdτds<λh060h0s0∫01Gs,s∫01Gs,τdτds=λs0;then λ>1.Lemma 13.
Assume that (H1), (H2), (H3), (H4), and (H5) hold. Let(λ,u)∈C be a solution of (5) with u∞=4s0; then λ<1.Proof.
Letu be a solution of (5) with u∞=4s0; then (16) implies that (55)s0=14u∞≤ux≤u∞=4s0,x∈14,34.Suppose on the contrary that λ≥1; then, by (H5) we have (56)u′′′′x=λhxfux≥λhxfuxuxux≥16π4ux,x∈14,34.Multiplying inequality (56) by sin[2π(x-1/4)] and integrating it over [1/4,3/4], we have (57)∫1/43/4u′′′′sin⁡2πx-14dx≥∫1/43/416π4uxsin⁡2πx-14dx.On the other hand, by simple computation, one has that (58)∫1/43/4u′′′′sin2πx-14dx=2πu′′14+u′′34-8π3u14+u34+∫1/43/416π4uxsin⁡2πx-14dx;since u(t)≥0 is concave, then 2πu′′1/4+u′′3/4-8π3u1/4+u3/4<0; this deduces a contradiction.Proof of Theorem1.
From Lemma9, there exists an unbounded connected component C in the positive solutions set of (5); moreover, C⊂R×E with (0,0)∈C and it grows to the right near (0,0). From Lemma 11, there exists a sequence {(λk,uk)}⊂C such that λk→∞ and uk→∞. Lemma 10 implies that uk∞→∞; then there exist (λ0,u0) and (λ0,u0)∈C such that u0∞=s0 and u0∞=4s0; Lemmas 12 and 13 imply that λ0>1 and λ0<1, respectively.
By Lemmas11, 12, and 13, there exist (λ∗,u∗) and (λ∗,u∗)∈C which satisfy 0<λ∗<1<λ∗ and u∗<u∗, such that the component C turns to the left at (λ∗,u∗) and to the right at (λ∗,u∗); that is, C is an S-shaped component; this together with Lemma 9 completes the proof of Theorem 1.Remark 14.
Let us take(59)fx,s,p≡hxfs;then Condition (A) in [21] (see (4)) implies (60)f0=a∈0,∞,which means f is in linear growth at zero. Equation (60) guarantees that Rabinowitz global bifurcation theorem can be directly used to bifurcate a connected component from (λ1/f0,0). However, in this paper, we deal with (5) under the superlinear growth condition at zero; that is,(H2)f0≔lims→0⁡(fs/s)=∞.In the situation, f0=∞, the Rabinowitz global bifurcation theorem cannot be directly used to get a connected component joining (0,0) with infinity anymore. To overcome this difficulty, we have to construct a sequence of functions fn which is in linear growth at zero and satisfies (61)limn→∞⁡fn⟶f,limn→∞⁡fn0=∞.By studying the corresponding auxiliary problems 1.6n, we obtain a sequence of unbounded connected components {C[n]} via Rabinowitz global bifurcation theorem. Now, by using the fact that the superior limit of certain infinity collection of connected components contains an unbounded connected component (see Ma and An in [24]), we get a connected component C: (62)0,0∈C⊂lim⁡supn→∞⁡Cnwhich joins (0,0) with infinity.Therefore, the key conditions, the conclusion, and the proofs of the main results in this paper and in [21] are very different.



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