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Reversed Sshaped connected component for a fourthorder boundary value problem
Advances in Difference Equations volume 2017, Article number: 113 (2017)
Abstract
In this paper, we investigate the existence of a reversed Sshaped component in the positive solutions set of the fourthorder boundary value problem
where \(\lambda>0\) is a parameter, \(h\in C[0,1]\) and \(f\in C[0,\infty )\), \(f(0)=0\), \(f(s)>0\) for all \(s>0\). By figuring the shape of unbounded continua of solutions, we show the existence and multiplicity of positive solutions with respect to parameter λ, and especially, we obtain the existence of three distinct positive solutions for λ being in a certain interval.
1 Introduction
The fourthorder boundary value problem
describes the deformations of an elastic beam both of whose ends are simply supported at 0 and 1, see Gupta [1]. The existence and multiplicity of positive solutions for (1.1) with \(\lambda\equiv1\) have been extensively studied by many authors using topological degree theory, fixed point theorems, lower and upper solution methods, and critical point theory (see, for example, [2–17] and the references therein).
However, to the best of our knowledge, when parameter λ varies in \(\mathbb{R}^{+}\), there are few papers concerned with the global behavior of positive solutions of (1.1), see, for example, [18–20]. By using Rabinowitz’s or Dancer’s global bifurcation theorem, [18–20] investigated the global structure of the solutions set of (1.1), and accordingly, obtained the existence and multiplicity of positive solutions and nodal solutions. Notice that these results give no information on direction turns of the connected component.
Very recently, Kim and Tanaka [21] considered the global structure of positive solutions set of the pLaplacian problem
where the nonlinearity f is asymptotic linear near 0, superlinear at some point, and sublinear near ∞. Based upon Rabinowitz’s global bifurcation theorem, they proved that an unbounded subcontinuum of positive solutions of (1.2) bifurcates from the trivial solution, grows to the right from the initial point, to the left at some point, and to the right near \(\lambda=\infty\). Roughly speaking, they concluded that there exists an Sshaped connected component in the positive solutions set of problem (1.2) (see Figure 1). Motivated by the above work, in a later paper [22], the present authors have established the existence result of an Sshaped connected component in the positive solutions set of the fourthorder boundary value problem
Now it is natural to ask whether we can get a reversed Sshaped connected component in the positive solutions set (see Figure 2) if the conditions on the nonlinearity are in contrast to these in [21]. In this paper, we will deal with this topic for problem (1.1) with \(f(x,u)=h(x)f(u)\). More precisely, we will establish the existence result of a reversed Sshaped connected component in the positive solutions set of the problem
Throughout the paper, we assume that

(H1)
\(h(x) \geq0\) in \([0,1]\) and \(h\not\equiv0\) in any subinterval of \([0,1]\);

(H2)
there exist \(\alpha>0\), \(f_{0}>0\), and \(f_{1}>0\) such that \(\lim_{s\rightarrow0^{+}}\frac{f(s)f_{0}s}{s^{1+\alpha}}=f_{1}\);

(H3)
\(f_{\infty}:=\lim_{s\rightarrow\infty}\frac{f(s)}{s}=\infty\);

(H4)
there exists \(s_{0}>0\) such that \(0\leq s\leq s_{0}\) implies that
$$f(s)\leq\frac{60f_{0}}{\mu_{1}\widehat{h}}s_{0}, $$where \(\widehat{h}=\max_{x\in[0,1]}h(x)\) and \(\mu_{1}>0\) is the first eigenvalue of the linear problem corresponding to (1.3) defined in Lemma 2.1.
It is easy to find that if (H2) holds, then
that is, f is asymptotic linear near 0. Contrary to the condition in [21], f is superlinear near ∞ according to (H3), then we cannot find a constant \(f^{\ast}>0\) such that \(f(s)\leq f^{\ast}s\) for all \(s\geq0\). This will bring great difficulty to the study of the global structure of the positive solutions set of (1.3). On the other hand, the concavity and convexity of the solutions of secondorder BVPs can be deduced directly from the nonlinearity in the equation, but for fourthorder BVPs, this becomes complicated, especially when the nonlinearity changes sign. So in this paper we only consider the case that \(h\geq0\) and \(f(s)>0\) for all \(s>0\).
Arguing the shape of a component in the positive solutions set of problem (1.3), we have the following result.
Theorem 1.1
Assume that (H1), (H2), (H3), and (H4) hold. Then there exist \(\lambda_{\ast}\in(0,\frac{\mu_{1}}{f_{0}})\) and \(\lambda^{\ast}>\frac{\mu_{1}}{f_{0}}\) such that

(i)
(1.3) has at least one positive solution if \(0<\lambda <\lambda_{\ast}\);

(ii)
(1.3) has at least two positive solutions if \(\lambda =\lambda_{\ast}\);

(iii)
(1.3) has at least three positive solutions if \(\lambda_{\ast}<\lambda<\frac{\mu_{1}}{f_{0}}\);

(iv)
(1.3) has at least two positive solutions if \(\frac {\mu_{1}}{f_{0}}\leq\lambda<\lambda^{\ast}\);

(v)
(1.3) has at least one positive solution if \(\lambda =\lambda^{\ast}\);

(vi)
(1.3) has no positive solution if \(\lambda>\lambda ^{\ast}\).
Remark 1.1
Indeed, condition (H2) pushes the direction of bifurcation to the left near \(u=0\), while conditions (H4) and (H3) guarantee that the bifurcation curve grows to the right at some point and grows to the left near \(\lambda=0\), respectively.
Remark 1.2
Let us consider the functions
and
Obviously, h satisfies (H1), \(\widehat{h}=1\) and the first eigenvalue of the linear problem corresponding to (1.3) is \(\mu _{1}=\pi^{4}\). It is easy to check that f satisfies (H3) and (H2) with
Denote
By using Mathematica 9.0, we deduce that f is increasing on \([0,\infty)\), and the equation
has exactly three roots: \(s_{1}=0\), \(s_{2}\doteq0.356\), \(s_{3}\doteq 1.513\). Combining this with \(f_{0}=1>\frac{60}{\pi^{4}}=w'(0)\) and \(f_{\infty}=\infty\), we conclude that for each fixed \(s_{0}\in[s_{2},s_{3}]\),
that is, f satisfies (H4).
Notice that the Conditions (H1)(H4) are fulfilled, then Theorem 1.1 guarantees that there exist \(\lambda _{\ast}\in(0,\pi^{4})\) and \(\lambda^{\ast}>\pi^{4}\) such that the conclusions (i)(vi) are correct for problem
The rest of this paper is arranged as follows. In Section 2, we show global bifurcation phenomena from the trivial branch with the leftward direction. Section 3 is devoted to showing that there are at least two direction turns of the component and to completing the proof of Theorem 1.1.
2 Leftward bifurcation
In this section, we state some preliminary results and show global bifurcation phenomena from the trivial branch with the leftward direction.
Let \(g\in C[0,1]\), then the solution of the fourthorder linear boundary value problem
can be expressed by
where
Moreover, if \(g(t)\geq0\), \(g\not\equiv0\), then
and \(v(t)\geq0\) is concave.
Since the Green’s function \(G(t,s)\) has the properties:

(i)
\(0\leq G(t,s)\leq G(s,s)\), \(\forall t,s\in[0,1]\);

(ii)
\(G(t,s)\geq\frac{1}{4}G(s,s)\), \(\forall t\in[\frac {1}{4},\frac{3}{4}]\), \(s\in[0,1]\),
then, for \(t\in[\frac{1}{4},\frac{3}{4}]\), we have
Let us consider the linear eigenvalue problem
Lemma 2.1
see [19], Theorem 2.1
Assume that (H1) holds. Then the linear problem (2.2) has a positive simple principal eigenvalue
Moreover, the corresponding eigenfunction ϕ is positive in \((0, 1)\).
Extend f to \(\mathbb{R}\) with the oddity and rewrite (1.3) by
Since condition (H2) implies (1.4), then following an argument similar to that in the proof of Theorem 1.1 in [19] or Theorem 2.2 in [20], we have the following.
Lemma 2.2
Assume that (H1) and (H2) hold, then from \((\frac{\mu_{1}}{f_{0}},0)\) there emanates an unbounded subcontinuum \(\mathcal{C}\) of positive solutions of (1.3) in the set \(\mathbb{R}\times E\), where \(E=\{u\in C^{3}[0,1]\mid u(0)=u(1)=u''(0)=u''(1)=0\}\) with the norm \(\Vert u \Vert = \Vert u \Vert _{\infty}+ \Vert u' \Vert _{\infty}+ \Vert u'' \Vert _{\infty }+ \Vert u''' \Vert _{\infty}\).
Lemma 2.3
Assume that (H1) and (H2) hold. Let \(\{ (\lambda_{n},u_{n})\}\) be a sequence of positive solutions to (1.3) which satisfies \(\lambda_{n}\rightarrow\frac{\mu_{1}}{f_{0}}\) and \(\Vert u_{n} \Vert \rightarrow0\). Let ϕ be the first eigenfunction of (2.2) which satisfies \(\Vert \phi \Vert _{\infty}=1\). Then there exists a subsequence of \(\{u_{n}\}\), again denoted by \(\{u_{n}\}\), such that \(\frac{u_{n}}{ \Vert u_{n} \Vert _{\infty}}\) converges uniformly to ϕ on \([0,1]\).
Proof
Set \(v_{n}:=\frac{u_{n}}{ \Vert u_{n} \Vert _{\infty}}\). Then \(\Vert v_{n} \Vert _{\infty}=1\). For every \((\lambda_{n},u_{n})\), we have
From the boundary condition \(u_{n}(0)=u_{n}(1)=0\), there exists \(\widehat{x}_{n}\in(0,1)\) such that \(u_{n}'(\widehat{x}_{n})=0\). Integrating (2.4) on \([x,\widehat{x}_{n}]\), we obtain
Dividing both sides of (2.5) by \(\Vert u_{n} \Vert _{\infty }\), we get
Since \(\Vert u_{n} \Vert \rightarrow0\) implies \(\Vert u_{n} \Vert _{\infty}\rightarrow0\), then by (1.4) there exists a constant \(m_{1}>0\) such that
From \(\lambda_{n}\rightarrow\frac{\mu_{1}}{f_{0}}\), it follows that there exists a constant \(m_{2}>0\) such that
Then, for \(x\in[0,1]\), (2.6) implies that
that is,
Since \(\Vert v_{n}' \Vert _{\infty}\) is bounded, by the AscoliArzela theorem, a subsequence of \(\{v_{n}\}\) uniformly converges to a limit \(v\in C[0,1]\) with \(\Vert v \Vert _{\infty}=1\), and we again denote by \(\{v_{n}\}\) the subsequence.
For every \((\lambda_{n},u_{n})\), we have
Dividing both sides of (2.11) by \(\Vert u_{n} \Vert _{\infty}\), we get
Since \(\Vert u_{n} \Vert _{\infty}\rightarrow0\), we conclude that \(\frac{f(u_{n}(\tau))}{u_{n}(\tau)}\rightarrow f_{0}\) for each fixed \(\tau\in[0,1]\). By recalling (2.7), Lebesgue’s dominated convergence theorem shows that
which means that v is a nontrivial solution of (2.2) with \(\lambda =\mu_{1}\), and hence \(v\equiv\phi\). □
Lemma 2.4
Assume that (H1) and (H2) hold. Let \(\mathcal{C}\) be as in Lemma 2.2. Then there exists \(\delta>0\) such that \((\lambda,u)\in\mathcal{C}\) and \(\vert \lambda\frac {\mu_{1}}{f_{0}} \vert + \Vert u \Vert \leq\delta\) imply \(\lambda<\frac{\mu_{1}}{f_{0}}\).
Proof
Assume to the contrary that there exists a sequence \(\{(\lambda_{n},u_{n})\}\subset\mathcal{C}\) such that \(\lambda_{n}\rightarrow\frac{\mu_{1}}{f_{0}}\), \(\Vert u_{n} \Vert \rightarrow0\) and \(\lambda_{n}\geq\frac{\mu_{1}}{f_{0}}\). By Lemma 2.3, there exists a subsequence of \(\{u_{n}\}\), again denoted by \(\{u_{n}\}\), such that \(\frac{u_{n}}{ \Vert u_{n} \Vert _{\infty}}\) converges uniformly to ϕ on \([0,1]\). Multiplying the equation of (1.3) with \((\lambda,u)=(\lambda_{n},u_{n})\) by ϕ and integrating it over \([0,1]\), we have
By a simple computation, one has that
Combining (2.14) with (2.15), we obtain
that is,
Since
then Lebesgue’s dominated convergence theorem, Lemma 2.3, and condition (H2) imply that
Similarly,
By Lebesgue’s dominated convergence theorem and Lemma 2.3 again, we conclude that
this contradicts (2.16). □
3 Direction turns of the component and the proof of Theorem 1.1
In this section, we show that there are at least two direction turns of the component under conditions (H4) and (H3), that is, the component is reversed Sshaped, and accordingly we finish the proof of Theorem 1.1.
Lemma 3.1
Assume that (H1) and (H4) hold. Let u be a solution of (1.3) with \(\Vert u \Vert _{\infty}=s_{0}\), then \(\lambda>\frac{\mu_{1}}{f_{0}}\).
Proof
Let u be a solution of (1.3) with \(\Vert u \Vert _{\infty}=s_{0}\), then by condition (H4) and the property of \(G(x,s)\), we have
then \(\lambda>\frac{\mu_{1}}{f_{0}}\). □
Lemma 3.2
Assume that (H1), (H2), and (H3) hold. Let \(\mathcal{C}\) be as in Lemma 2.2. Then \(\sup\{\lambda\mid (\lambda ,u)\in\mathcal{C}\}<\infty\).
Proof
Note that if (H2) and (H3) hold, there exists a positive constant k with \(f_{0}\geq k\) such that
For any \((\lambda,u)\in\mathcal{C}\), by (3.2) and (2.1) and the property of \(G(x,s)\), we have
where
Then (3.3) implies that
□
Lemma 3.3
Assume that (H1), (H2), and (H3) hold. Let \(\{(\lambda_{n},u_{n})\}\) be a sequence of positive solutions to (1.3), then \(\Vert u_{n} \Vert \rightarrow\infty\) implies \(\Vert u_{n} \Vert _{\infty}\rightarrow\infty\).
Proof
From Lemma 3.2, we conclude that \(\{\lambda _{n}\}\) is bounded. Assume on the contrary that \(\Vert u_{n} \Vert _{\infty}\) is bounded. By recalling (2.5) and (2.4), we have that \(\Vert u_{n}' \Vert _{\infty}\) and \(\Vert u_{n}'' \Vert _{\infty}\) are bounded too.
From the boundary condition \(u_{n}''(0)=u_{n}''(1)=0\), there exists \(x_{n}^{*}\in(0,1)\) such that \(u_{n}'''(x_{n}^{*})=0\). Integrating the equation of (1.3) on \([x_{n}^{*},x]\), we obtain
then \(\Vert u_{n}''' \Vert _{\infty}\) is bounded. Finally, we conclude that \(\Vert u_{n} \Vert = \Vert u_{n} \Vert _{\infty }+ \Vert u_{n}' \Vert _{\infty}+ \Vert u_{n}'' \Vert _{ \infty}+ \Vert u_{n}''' \Vert _{\infty} \) is bounded, this deduces a contradiction. □
Lemma 3.4
Assume that (H1), (H2), and (H3) hold. Let \(\{(\lambda_{n},u_{n})\}\) be a sequence of positive solutions to (1.3), then \(\Vert u_{n} \Vert \rightarrow\infty\) implies \(\lambda_{n}\rightarrow0\).
Proof
For every \((\lambda_{n},u_{n})\), we have
Lemma 3.3 and Lemma 3.2 imply that \(\Vert u_{n} \Vert _{\infty}\rightarrow\infty\). Dividing both sides of (3.5) by \(\Vert u_{n} \Vert _{\infty}\), we get
Then by (H3) and the boundedness of \(\frac{u_{n}(x)}{ \Vert u_{n} \Vert _{\infty}}\), we conclude that \(\lambda _{n}\rightarrow0\). □
Proof of Theorem 1.1
Let \(\mathcal{C}\) be as in Lemma 2.2. By Lemma 2.4, \(\mathcal{C}\) is bifurcating from \((\frac{\mu _{1}}{f_{0}},0)\) and goes leftward. Since \(\mathcal{C}\) is unbounded, there exists \(\{(\lambda_{n},u_{n})\}\) such that \((\lambda _{n},u_{n})\in\mathcal{C}\) and \(\vert \lambda_{n} \vert + \Vert u_{n} \Vert \rightarrow\infty\). By Lemmas 3.2 and 3.4, we have that \(\Vert u_{n} \Vert \rightarrow\infty\) and \(\lambda_{n}\rightarrow0\). Lemma 3.3 implies that \(\Vert u_{n} \Vert _{\infty}\rightarrow\infty\), then there exists \((\lambda_{0},u_{0})\in\mathcal{C}\) such that \(\Vert u_{0} \Vert _{\infty}=s_{0}\), and Lemma 3.1 shows that \(\lambda _{0}>\frac{\mu_{1}}{f_{0}}\).
By Lemmas 2.2, 2.4, and 3.1, \(\mathcal{C}\) passes through some points \((\frac{\mu_{1}}{f_{0}}, v_{1})\) and \((\frac{\mu_{1}}{f_{0}}, v_{2})\) with \(\Vert v_{1} \Vert _{\infty}< s_{0}< \Vert v_{2} \Vert _{\infty}\), and there exist \(\underline {\lambda}\) and λ̅ which satisfy \(0<\underline {\lambda}<\frac{\mu_{1}}{f_{0}}<\overline{\lambda}\) and both (i) and (ii):

(i)
if \(\lambda\in(\underline{\lambda},\frac{\mu_{1}}{f_{0}})\), then there exist u and v such that \((\lambda,u),(\lambda,v)\in \mathcal{C}\) and \(\Vert u \Vert _{\infty}< \Vert v \Vert _{\infty}<s_{0}\);

(ii)
if \(\lambda\in[\frac{\mu_{1}}{f_{0}},\overline{\lambda})\), then there exist u and v such that \((\lambda,u),(\lambda,v)\in \mathcal{C}\) and \(\Vert u \Vert _{\infty}< s_{0}< \Vert v \Vert _{\infty}\).
Define \(\lambda_{\ast}=\inf\{\underline{\lambda}: \underline{\lambda} \text{ satisfies (i)}\}\) and \(\lambda ^{\ast}=\sup\{\overline{\lambda}: \overline{\lambda} \text{ satisfies (ii)} \}\). Then (1.3) has a positive solution \(u_{\lambda_{\ast}}\) at \(\lambda=\lambda_{\ast}\) and \(u_{\lambda ^{\ast}}\) at \(\lambda=\lambda^{\ast}\), respectively. Clearly, the component curve turns to the right at \((\lambda_{\ast}, \Vert u_{\lambda_{\ast}} \Vert _{\infty})\) and to the left at \((\lambda^{\ast}, \Vert u_{\lambda^{\ast}} \Vert _{\infty})\) (see Figure 2). That is, \(\mathcal{C}\) is a reversed Sshaped component, this together with Lemma 3.4 completes the proof of Theorem 1.1. □
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Acknowledgements
This work was supported by the Natural Science Foundation of China (No. 11671322, No. 11626016).
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RM completed the main study, carried out the results of this article, and drafted the paper. JW checked the proofs and verified the calculation. The two authors read and approved the final manuscript.
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Wang, J., Ma, R. Reversed Sshaped connected component for a fourthorder boundary value problem. Adv Differ Equ 2017, 113 (2017). https://doi.org/10.1186/s1366201711675
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DOI: https://doi.org/10.1186/s1366201711675