The existence of positive solutions for multi-point boundary value problem at resonance on the half-line

We establish new results on the existence of positive solutions for the multi-point boundary value problem at resonance on the half-line. Our results are based on the Leggett-Williams norm-type theorem due to O’Regan and Zima, which requires appropriate Banach spaces and proper operators. An example is given to illustrate the main results.

In this paper, we will discuss the existence of positive solutions for the multi-point boundary value problem

where \(f \in C([0,+\infty)\times\mathbf{R}\rightarrow\mathbf {R})\), \(f(t,0)\) is not always equal to 0, \(t\in[0,+\infty)\), \(\alpha _{i}>0\), \(\sum_{i=1}^{m-1}\alpha_{i}=1\), \(i=1,2,\ldots,m-1\), \(0= \xi _{1}<\cdots<\xi_{m-1}<\infty\).

Boundary value problems of differential equations are applied to more and more disciplines, and the existence of one or multiple positive solutions for multi-point BVPs has been attracting more and more authors, for details see [1–16]. Generally speaking, the boundary value problems of differential equations can be roughly divided into two parts. One is boundary value problems on the finite interval; Infante and Zima [17] obtained the existence of positive solutions for the problem

with \(0< \eta_{1}<\eta_{2}<\cdots<\eta_{m-2}<1\), \(\alpha_{i}>0\), \(\sum_{i=1}^{m-2}\alpha_{i}=1\). The other is boundary value problems on the infinite interval, for details see [18–22]; [20] obtained the existence of positive solutions for the problem

and

where \(f:[0,+\infty)\times\mathbf{R}^{2}\rightarrow\mathbf {R}\), \(e:[0,+\infty)\rightarrow\mathbf{R}\) are continuous and \(\eta\in (0,+\infty)\).

To the best of our knowledge, only few authors studied the existence of positive solutions for boundary value problems at resonance on the half-line. In [21], the authors dealt with the second order boundary value problem with integral boundary conditions on a half-line

In [22], the authors investigated the existence of positive solutions for the two-point problem at resonance on the half line,

where \(D_{0^{+}}^{\alpha}\) is the standard Riemann-Liouville fractional derivative.

Inspired by the works above, we will study the existence of positive solutions for the problem (1.1).

Define 1.1

We call u is a positive solution of the boundary value problem (1.1), if \(u\geq0\), \(u\neq0\), and satisfies the problem (1.1).

Let us recall some standard facts and the Leggett-Williams norm-type theorem due to O’Regan and Zima.

Let X, Y be real Banach spaces. A linear mapping \(L:\operatorname {dom}L\subset X\rightarrow Y\) is called a Fredholm operator of index zero if ImL is closed and \(\dim \operatorname {Ker}L=\operatorname {codim}\operatorname {Im}L< \infty\), which implies that there exist continuous projectors \(P:X\rightarrow X\) and \(Q:Y\rightarrow Y\) such that \(\operatorname {Im}P= \operatorname {Ker}L\) and \(\operatorname {Ker}Q= \operatorname {Im}L\). Moreover, since \(\dim \operatorname {Im}Q=\operatorname {codim}\operatorname {Im}L\), there exists an isomorphism \(J:\operatorname {Im}Q\rightarrow \operatorname {Ker}L\). Denote by \(L_{P}\) the restriction of L to \(\operatorname {Ker}P \cap \operatorname {dom}L\rightarrow \operatorname {Im}L\) and its inverse by \(K_{P}\). So \(K_{P}:\operatorname {Im}L\rightarrow \operatorname {Ker}P\cap \operatorname {dom}L\) and the coincidence equation \(Lx=Nx \) is equivalent to \(x=(P+JQN)x+K_{P}(I-Q)Nx\).

A nonempty convex closed set \(C\subset X\) is said to be a cone provided that

1. (i)

   \(\lambda x \in C\), for \(x\in C\), \(\lambda\geq0\);

2. (ii)

   \(x, -x\in C\) implies \(x=0\).

Note that every cone \(C\subset X\) induces a partial order in X by \(x\preceq y\) if and only if \(y-x\in C\).

Let \(\gamma:X\rightarrow C\) be a retraction, i.e. γ is a continuous mapping such that \(\gamma(x)=x\), \(x\in C\).

Let \(\Psi:=P+JQN+K_{P}(I-Q)N\) and \(\Psi_{\gamma}:=\Psi\circ\gamma\).

Theorem 2.1

([12])

Let C be a cone in X and \(\Omega _{1}\), \(\Omega_{2}\) be open bounded subsets of X with \(\overline{\Omega}_{1}\subset\Omega_{2}\) and \(C\cap(\overline{\Omega}_{2}\setminus\Omega_{1})\neq\emptyset\). Assume that \(L:\operatorname {dom}L\subset X\rightarrow Y\) is a Fredholm operator of index zero and the following conditions are satisfied.

1. (C1)

   \(QN:X\rightarrow Y\) is continuous and bounded and \(K_{P}(I-Q)N:X\rightarrow X\) is compact on every bounded subset of X;

2. (C2)

   \(Lx\neq\lambda Nx \) for all \(x\in C\cap\partial\Omega _{2}\cap \operatorname {dom}L\) and \(\lambda\in(0,1)\);

3. (C3)

   γ maps subsets of \(\overline{\Omega}_{2}\) into bounded subsets of C;

4. (C4)

   \(d_{B}([I-(P+JQN)\gamma]\mid_{\operatorname {Ker}L},\operatorname {Ker}L\cap\Omega _{2},0)\neq0\), where \(d_{B}\) stands for the Brouwer degree;

5. (C5)

   there exists \(u_{0}\in C\backslash\{0\}\) such that \(\Vert x\Vert \leq\sigma(u_{0})\Vert \Psi x\Vert \) for \(x \in C(u_{0})\cap\partial\Omega _{1}\), where \(C(u_{0})=\{ x\in C:\mu u_{0}\preceq x\textit{ for some }\mu> 0\} \) and \(\sigma(u_{0})\) is such that \(\Vert x+u_{0}\Vert \geq\sigma(u_{0})\Vert x\Vert \) for every \(x\in C\);

6. (C6)

   \((P+JQN)\gamma(\partial\Omega_{2})\subset C\);

7. (C7)

   \(\Psi_{\gamma}(\overline{\Omega}_{2}\backslash\Omega _{1})\subset C\).

Then the equation \(Lx=Nx\) has a solution in the set \(C\cap(\overline {\Omega}_{2}\backslash\Omega_{1})\).

Lemma 2.1

([23])

Assume that \(V\subset X\) is bounded. V is compact if \(\{ \frac{u(t)}{1+t}:u\in V \}\) is equicontinuous on \([0,T]\), \(\forall T<\infty\), and equiconvergent at infinity.

In this paper, we will always suppose that the following condition holds.

(A₁) \(f:[0,+\infty)\times\textbf{R}\rightarrow\textbf{R}\) is continuous and \(f(t,0)\) is not always equal to 0. For any \(r>0\), there exists \(h_{r}(t)\in L[0,+\infty)\), \(h_{r}(t)>0 \) satisfying \(\vert f(t,(1+t)u)\vert \leq h_{r}(t)\), \(t\in[0,+\infty)\), \(\vert u\vert \leq r\), \(\alpha_{i}>0\), \(\sum_{i=1}^{m-1}\alpha_{i}=1\).

Let

with the norm \(\Vert u\Vert =\sup_{t\in[0,+\infty)}\frac{\vert u(t)\vert }{1+t}\), and

with the norm \(\Vert y\Vert _{1}=\int_{0}^{+\infty} \vert y(t)\vert \,dt+\sup_{t\in [0,+\infty)}\vert y(t)\vert \).

It is easy to prove that \((X, \Vert \cdot \Vert )\) and \((Y, \Vert \cdot \Vert _{1})\) are Banach spaces.

Define \(L:\operatorname {dom}L\subset X\rightarrow Y\) and \(N:X\rightarrow Y\) as follows:

where

Then the boundary value problem (1.1) can be written

For convenience, denote the function \(G(t,s)\) as follows:

Clearly, \(G(t,s)\leq\frac{3}{2\sum_{i=1}^{m-1}\alpha _{i}e^{-\xi_{i}}}+1\), for \(t,s\in[0,+\infty)\).

Lemma 3.1

L is a Fredholm operator of index zero.

Proof

It is easy to get

and

Define \(Q:Y\rightarrow Y\) by

Clearly, \(\operatorname {Ker}Q=\operatorname {Im}L\), \(\operatorname {Im}Q=\{y\mid y=ce^{-t},t\geq0, c\in\textbf{R}\}\), and \(Q:Y\rightarrow Y\) is a linear projector. In fact, for \(y(t)\in Y\), we have

For \(y\in Y\), we have \(y=(y-Qy)+Qy\), \(Qy\in \operatorname {Im}Q\), \((I-Q)y\in \operatorname {Ker}Q=\operatorname {Im}L\). So we obtain \(Y=\operatorname {Im}Q+\operatorname {Im}L\). Take \(y_{0}\in \operatorname {Im}Q\cap \operatorname {Im}L\). \(y_{0}\in \operatorname {Im}Q\) means that \(y_{0}\) can be written \(y_{0}=ce^{-t}\), \(c\in\textbf{R}\). At the same time, by \(y_{0}\in \operatorname {Im}L\) and (3.2), we get

i.e. \(c=0\). This implies that \(y_{0}=0\). Thus, \(Y=\operatorname {Im}Q\oplus \operatorname {Im}L\) and \(\dim \operatorname {Ker}L=\operatorname {codim}\operatorname {Im}L=1<+\infty\). Observing that ImL is closed in Y, L is a Fredholm operator of index zero. □

Define \(P:X\rightarrow X\) as

Clearly, \(P:X\rightarrow X\) is a linear continuous projector and

Thus, \(X=\operatorname {Im}P\oplus \operatorname {Ker}P=\operatorname {Ker}L\oplus \operatorname {Ker}P\).

Define \(K_{P}:\operatorname {Im}L\rightarrow \operatorname {Ker}P\cap \operatorname {dom}L\) by

By simple calculations, we have \((K_{P}L_{P})u=u\), \(\forall u\in \operatorname {dom}L\cap \operatorname {Ker}P\), and \((L_{P}K_{P})y=y\), \(\forall y\in \operatorname {Im}L\). So \(K_{P}=(L_{P})^{-1}\), where \(L_{P}=L\mid_{\operatorname {dom}L\cap \operatorname {Ker}P}:\operatorname {dom}L\cap \operatorname {Ker}P\rightarrow \operatorname {Im}L\).

Define the linear isomorphism \(J:\operatorname {Im}Q\rightarrow \operatorname {Ker}L\) as

Thus, \(JQN+K_{P}(I-Q)N:X\rightarrow X\) is given by

Lemma 3.2

\(QN:X\rightarrow Y\) is continuous and bounded and \(K_{P}(I-Q)N:\overline{\Omega}\rightarrow X\) is compact, where \(\Omega\subset X\) is bounded.

Proof

For convenience, denote \(M_{r}:=\frac{\int_{0}^{+\infty}h_{r}(\tau)\,d\tau}{\sum_{i=1}^{m-1}\alpha_{i}e^{-\xi_{i}}}\).

We will prove that \(QN:X\rightarrow Y\) is continuous and bounded.

Since \(\Omega\subset X\) is bounded, for \(u\in\overline{\Omega}\), there exists a constant \(r>0\), such that \(\Vert u\Vert < r\). By the condition (A₁), we have

So \(QN:X\rightarrow Y\) is bounded. By (A₁) and the Lebesgue dominated convergence theorem, we see that \(QN:X\rightarrow Y\) is continuous.

Now, we will prove that \(K_{P}(I-Q)N:\overline{\Omega}\rightarrow X\) is compact.

First of all, by the condition (A₁) and \(u\in\overline{\Omega}\), we have

i.e. \(K_{P}(I-Q)N:\overline{\Omega}\rightarrow X\) is bounded.

Second, for \(u\in\overline{\Omega}\), \(0< t_{1}< t_{2}< T<\infty\),

By the uniform continuity of \(\frac{t-s}{1+t}\) in \([0,T]\times[0,T]\) and \(\frac{t}{1+t}\) in \([0,T]\), and the absolute continuity of the integral, we see that \(K_{P}(I-Q)N:\overline {\Omega}\rightarrow X\) is equicontinuous on \([0,T]\), \(\forall T>0\).

Third, for \(\varepsilon>0\), there exists a constant \(l>0\), such that

Since \(\lim_{t\rightarrow+\infty}\frac{t-l}{1+t}=1\), \(\lim_{t\rightarrow+\infty}\frac{t}{1+t}=1\), there exists a constant \(T>l\) such that

For \(u\in\overline{\Omega}\), \(T\leq t_{1}< t_{2}\), we have

Thus, \(K_{P}(I-Q)N:\overline{\Omega}\rightarrow X\) is equiconvergent at infinity.

By Lemma 2.1, we see that \(K_{P}(I-Q)N:\overline{\Omega}\rightarrow X\) is compact. □

Theorem 3.1

Assume that (A₁) and the following conditions hold.

(A₂) For \(u\geq0\), there exist three nonnegative functions \(\mu (t)\), \(\beta_{i}(t)\), \(i=1,2\), such that

where \(\mu(t)ue^{-t}, \beta_{2}(t), \beta_{1}(t)ue^{-t}\in L[0,+\infty )\), \(\inf_{t\in[0,+\infty)}\beta_{1}(t) :=\beta_{0}>0\) and \(\mu(t)\) satisfying

1. (i)

   \(\sup_{t\in[0,+\infty)}\mu(t):=\mu _{1}<\frac{2\sum_{i=1}^{m-1}\alpha_{i}e^{-\xi _{i}}}{3+2\sum_{i=1}^{m-1}\alpha_{i}e^{-\xi_{i}}}\),

2. (ii)

   there exists \(t_{0}\in[0,+\infty)\), such that \(d_{0}:=\frac {t_{0}}{1+t_{0}}\int_{0}^{+\infty}[1-G(t_{0},s)\mu (s)](1+s)e^{-s}\,ds>1\), \(G(t_{0},s)\geq0\).

(A₃) There exists \(R>\frac{\mu_{1}+\alpha_{1}\beta_{0}+1}{\alpha_{1}\beta _{0}}\int_{0}^{\infty}\beta_{2}(s)\,ds\), such that \(f(t,Rt)<0\), \(t\in[0,+\infty)\).

Then the problem (1.1) has at least one positive solution.

Proof

Take a cone

Set

where \(d_{0}\) is given by the condition (A₂) and \(R>\frac{\mu_{1}+\alpha_{1}\beta_{0}+1}{\alpha_{1}\beta _{0}}\int_{0}^{\infty}\beta_{2}(s)\,ds\). Clearly, \(\Omega_{1}\), \(\Omega_{2}\) are open and bounded sets of X, \(\overline{\Omega}_{1}= \{u\in X |\frac{1}{d_{0}}\Vert u\Vert \leq\frac{\vert u(t)\vert }{1+t}\leq r< R \}\subset\Omega_{2}\), and \({C\cap(\overline{\Omega}_{2}\setminus\Omega_{1})\neq\emptyset}\).

In view of Lemmas 3.1 and 3.2, L is a Fredholm operator of index zero and the condition (C1) of Theorem 2.1 is fulfilled.

Suppose that there exist \(u_{1}(t)\in C\cap\partial\Omega_{2}\cap \operatorname {dom}L\) and \(\lambda_{0}\in(0,1)\) such that \(Lu_{1}=\lambda_{0}Nu_{1}\), i.e. \(u_{1}''(t)+\lambda_{0} f(t,u_{1}(t))=0\). By \(u_{1}(t)\in \operatorname {dom}L\), we have

It follows from (A₂) that

So

Considering (A₂), (3.7), and

we obtain

These contradict \(u_{1}(t)\in C\cap\partial\Omega_{2}\cap \operatorname {dom}L\). So (C2) is satisfied.

Let \((\gamma u)(t)=\vert u(t)\vert \), \(u(t)\in X\). Then \(\gamma:X\rightarrow C\) is a retraction and maps subsets of \(\overline{\Omega}_{2}\) into bounded subsets of C, i.e. (C3) holds.

Let \(u(t)\in \operatorname {Ker}L\cap\partial\Omega_{2}\), then \(u(t)=ct\), \(t\geq0\). Define

where \(c\in\{-R,R\}\) and \(\lambda\in[0,1]\). Suppose \(H(ct,\lambda )=0\), by (A₂), we obtain

Hence \(H(ct,\lambda)= 0\) implies \(c\geq0\). Furthermore, if \(H(Rt,\lambda)=0\), we have

which is a contradiction to the condition (A₃).

Thus, \(H(u,\lambda)\neq0\), for \(u\in \operatorname {Ker}L\cap\partial\Omega_{2}\), and \(\lambda\in[0,1]\). Therefore

Thus, (C4) holds.

Let \(u_{0}(t)=t\), \(t \in[0,+\infty)\), then \(u_{0}\in C\setminus\{0\}\), \(C(u_{0})=\{u\in C| u(t)\geq\mu t \mbox{ for some } \mu>0, t \in[0,+\infty )\}\), and we take \(\sigma(u_{0})=1\). Let \(u \in C(u_{0})\cap\partial \Omega_{1}\), we have \(\frac{1}{d_{0}}\Vert u\Vert \leq\frac{\vert u(t)\vert }{1+t}\leq r\), \(t \in[0,+\infty)\).

For \(u\in C(u_{0})\cap\partial\Omega_{1}\), by (A₂), we get

Thus, \(\Vert u\Vert \leq\sigma(u_{0})\Vert \Psi u\Vert \), for \(u\in C(u_{0})\cap\partial \Omega_{1}\). So (C5) holds.

For \(u(t)\in\partial\Omega_{2}\), \(t\in[0,+\infty)\), by the condition (A₂), we have

which means that \((P+JQN)\gamma(\partial\Omega_{2})\subset C\). Hence, (C6) holds.

For \(u(t)\in\overline{\Omega}_{2}\setminus\Omega_{1}\), \(t\in[0,+\infty)\), by the condition (A₂), we have

So \(\Psi_{\gamma}(\overline{\Omega}_{2}\setminus\Omega_{1})\subset C\), i.e. (C7) is satisfied.

By Theorem 2.1, we confirm that the equation \(Lu=Nu\) has a positive solution u, i.e. the problem (1.1) has at least one positive solution. □

Let us consider the following boundary value problem:

Here, \(f(t,u(t))=te^{-t}-\frac{1}{90}u(t)e^{-t}\), \(\alpha _{1}=0.68\), \(\alpha_{2}=0.018\), \(\alpha_{3}=0.302\), \(\xi_{1}=0\), \(\xi _{2}=0.5\), \(\xi_{3}=0.95\). Take \(h_{r}(t)=te^{-t}+\frac{r}{90}(1+t)e^{-t}\), \(\mu (t)=\frac{1}{80}\), \(\beta_{1}(t)=\frac{1}{100}\), \(\beta_{2}(t)=te^{-t}\), \(t\in[0,+\infty)\), \(t_{0}=1.05\), \(R=160\), \(r=150\).

Obviously, \(\vert f(t,(1+t)u)\vert \leq h_{r}(t)\), \(t\in[0,+\infty)\), \(r>0\), \(\vert u\vert < r\). By our calculations, we can get \(0.0029\leq G(t,s)\leq2.8571\) and

\(\mu_{1}=\sup_{t\in[0,+\infty)}\mu(t)=\frac {1}{80}\), \(\beta_{0}=\inf_{t\in[0,+\infty)}\beta_{1}(t)=\frac {1}{100}\), \(\beta_{2}(t)\in L[0,+\infty)\); \(f(t,Rt)<0\), \(t\in[0,+\infty)\). By simple calculations, we can get that \(0.348< G(1.05,s)<1.478\), so

So the conditions (A₁)-(A₃) hold. By Theorem 3.1, we can conclude that the problem (4.1) has at least one positive solution.

The authors are grateful to anonymous referees for their constructive comments and suggestions which led to improvement of the original manuscript. This work is supported by the Natural Science Foundation of China (11171088) and the Natural Science Foundation of Hebei Province (A2013208108).

Affiliations

1. College of Sciences, Hebei University of Science and Technology, Shijiazhuang, Hebei, 050018, P.R. China

   • Weihua Jiang
   •  & Caixia Yang

Corresponding author

Correspondence to Weihua Jiang.

Competing interests

The authors declare that there is no conflict of interests regarding the publication of this article.

Authors’ contributions

The authors declare that the study was realized in collaboration with the same responsibility. All authors read and approved the final manuscript.

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