Solvability of a third-order differential equation with functional boundary conditions at resonance

By using the coincidence degree theory due to Mawhin and constructing suitable operators, we study the existence of solutions for a third-order functional boundary value problem at resonance with $\operatorname{dim } \operatorname{Ker}L=1$.

A boundary value problem is said to be at resonance if the corresponding homogeneous boundary value problem has a non-trivial solution. Boundary value problems at resonance have been studied by many authors. We refer the readers to [1–9] and the references cited therein. In [10], the authors discussed the second-order differential equation

with functional boundary conditions

where $\Gamma_{1}, \Gamma_{2}$ are linear functionals on $C^{1}[0,1]$ satisfying the general resonance condition $\Gamma_{1}(t)\Gamma_{2}(1) = \Gamma_{1}(1)\Gamma_{2}(t)$. (The authors also studied the non-resonant scenario under condition ($A_{1}$): $\Gamma_{1}(t)\Gamma_{2}(1) \neq\Gamma _{1}(1)\Gamma_{2}(t)$.) To be specific, the following resonant cases received attention:

($A_{2}$):

$\Gamma_{1}(t), \Gamma_{1}(1), \Gamma_{2}(1) = 0$, $\Gamma _{2}(t) \neq0$;

($A_{3}$):

$\Gamma_{1}(t), \Gamma_{1}(1), \Gamma_{2}(t) = 0$, $\Gamma _{2}(1) \neq0$;

($A_{4}$):

$\Gamma_{1}(1), \Gamma_{2}(t), \Gamma_{2}(1) = 0$, $\Gamma _{1}(t) \neq0$;

($A_{5}$):

$\Gamma_{1}(t), \Gamma_{2}(1), \Gamma_{2}(t) = 0$, $\Gamma _{1}(1) \neq0$;

($A_{6}$):

$\Gamma_{1}(1), \Gamma_{1}(t), \Gamma_{2}(1), \Gamma_{2}(t) = 0$.

This paper is a study of third-order functional boundary value problems (FBVPs) at resonance. It improves and generalizes the results of [1, 7] and the results of [2] applicable to third-order problems. We consider

where $\varphi_{i}:C^{2}[0,1]\rightarrow\mathbb{R}$, $i=1,2,3$, are bounded linear functionals. To the best of our knowledge, this is the first paper devoted to a third-order FBVP at resonance. We present several generalizations to the existing results and improvements to the method based on Mawhin’s coincidence degree theory.

The framework of this paper is as follows. In Section 2, we present some notations and the fundamentals of coincidence degree theory. In Section 3, we study problem (1.1) under the conditions

respectively. In Section 4, we show the existence of a solution for problem (1.1) under the condition

(Here, if $\varphi_{2}(t^{j}) =0$ for some $j\in\{0,1,2\}$, then also $\varphi_{1}(t^{j}) =0$.)

For convenience, we denote

From the last three determinants we can define and derive the following three relations:

and $\Delta_{3}(Lx)= -2x(0)\triangle$. Also, $\Delta_{ij}$, $i,j=1,2,3$, $\Delta_{k}(y)_{ij}$, $i,k=1,2,3$, $j \in\{1,2,3\} \setminus\{k\}$, are the cofactors of $\varphi_{i}(t^{3-j})$ in Δ, $\Delta_{k}(y)$, $k =1,2,3$, respectively.

We introduce some notations and a theorem. For more details, see [11].

Let X and Y be real Banach spaces and $L: \operatorname{dom} L\subset X\rightarrow Y$ be a Fredholm operator of index zero, $P: X\rightarrow X$, $Q: Y\rightarrow Y$ be projectors such that

It follows that

is invertible. We denote the inverse by $K_{P}$.

If Ω is an open bounded subset of X, $\operatorname{dom} L \cap \overline{\Omega}\neq\emptyset$, the map $N:X\rightarrow Y$ is called L-compact on Ω̅ if $QN(\overline{\Omega})$ is bounded and $K_{P}(I-Q)N:\overline{\Omega}\rightarrow X$ is compact. We rely on Mawhin’s theorem for coincidences [8].

Theorem 2.1

Let $L: \operatorname{dom} L\subset X \rightarrow Y$ be a Fredholm operator of index zero and $N:X\rightarrow Y$ be L-compact on Ω̅. Assume that the following conditions are satisfied:

1. (1)

   $Lx \neq\lambda Nx$ for every $(x,\lambda)\in[(\operatorname {dom} L\setminus\operatorname{Ker} L) \cap\partial\Omega]\times(0,1)$;

2. (2)

   $Nx \notin\operatorname{Im} L$ for every $x\in\operatorname{Ker} L \cap\partial\Omega$;

3. (3)

   $\operatorname{deg}(JQN| _{\operatorname{Ker} L}, \Omega\cap\operatorname {Ker} L, 0)\neq0$, where $Q: Y\rightarrow Y$ is a projection such that $\operatorname{Im} L= \operatorname{Ker} Q$, and $J: \operatorname{Im} Q \to\operatorname {Ker} L$ is an isomorphism.

Then the equation $Lx=Nx$ has at least one solution in $\operatorname{dom} L \cap\overline{\Omega}$.

We work in $X=C^{2}[0,1]$ with the norm $\Vert x \Vert = \max\{ \Vert x \Vert _{\infty }, \Vert x' \Vert _{\infty}, \Vert x'' \Vert _{\infty}\}$, where $\Vert x \Vert _{\infty}=\max_{t\in[0,1]} \vert x(t) \vert $. We define $Y=L^{1}[0,1]$ with the norm $\Vert y \Vert _{1}= {\int_{0}^{1} \vert y(t) \vert \,dt}$.

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

$(C)$ :

There exist constants $k_{i}>0$, $i=1,2,3$, such that $\vert \varphi_{i}(x) \vert \leq k_{i} \Vert x \Vert $, $x\in X$ and the function $f(t,u,v,w)$ satisfies the Carathéodory conditions, that is, $f(\cdot,u,v,w)$ is measurable for each fixed $(u,v,w)\in \mathbb{R}^{3}$, $f(t,\cdot,\cdot,\cdot)$ is continuous for a.e. $t\in[0,1]$.

Case I. $\varphi_{i}(1)=0$, $i=1,2,3$.

Clearly, $\Delta=0$. In this case, we assume that there exists $j\in\{ 1,2,3\}$ such that $\Delta_{j3}\neq0$. In what follows, we choose and fix such j.

Lemma 3.1

There exists a function $g_{3}\in Y$ such that $\Delta_{3}(g_{3})=1$.

Proof

Suppose the contrary. Then

Hence

It follows from $\Delta_{j3}\neq0$ and $\varphi_{i}(1)=0$, $i=1,2,3$, that there exist constants a and b such that

where $k,l\in\{1,2,3\}$, $k,l\neq j$, $k\neq l$. Hence $\varphi _{j}(x)=(a\varphi_{k}+b\varphi_{l})(x)$, $x\in X$. This is a contradiction because $\varphi_{1},\varphi_{2},\varphi_{3}$ are linearly independent on X. Hence, there exists a function $h\in Y$ with $\Delta_{3}(h)\neq 0$ and, as a result, $g_{3} = {\frac{1}{\Delta_{3}(h)}}h \in Y$ with $\Delta_{3}(g_{3})=1$. □

Define operators $L:\operatorname{dom} L\subset X\rightarrow Y$, $N:X\rightarrow Y$ as follows:

where $\operatorname{dom} L = \{x\in X: x''' \in Y, \varphi_{i}(x)=0, i =1,2,3\}$.

If $x\in\operatorname{dom} L$ with $Lx=0$, then $x=at^{2}+bt+c$, $a,b,c\in \mathbb{R}$ and $\varphi_{i}(x)=0$, $i=1,2,3$, that is,

Since $\Delta_{j3}\neq0$, we have $a=b=0$. So, $x\equiv c$, that is, $\operatorname{Ker} L = \{c:c\in\mathbb{R}\}$.

Lemma 3.2

$\operatorname{Im} L = \{y\in Y:\Delta_{3}(y)=0\}$.

Proof

If $x\in\operatorname{dom}L$, $Lx=y$, then there exist constants $a,b,c$ such that the following equalities hold:

So, y satisfies $\Delta_{3}(y)=0$.

Inversely, if $y\in Y$ with $\Delta_{3}(y)=0$, we let

Obviously, $x'''(t)=y(t)$. Considering $\Delta_{1}(y)_{j3}=-\Delta_{3}(y)_{j1}$, $\Delta_{2}(y)_{j3}=\Delta _{3}(y)_{j2}$, $\Delta_{j3}=\Delta_{3}(y)_{j3}$ and

we have

Clearly, $\varphi_{i}(x)=0$, $i\neq j$, $i\in\{1,2,3\}$, which implies that $x\in\operatorname{dom} L$ and, consequently, $y\in\operatorname{Im} L$. □

Define the operators $P_{3}:X\rightarrow X$, $Q_{3}:Y\rightarrow Y$ by

Clearly, $P_{3}$, $Q_{3}$ are projectors such that (2.3) hold.

Define the operator $K_{P_{3}}: Y\rightarrow X$ by

Lemma 3.3

$K_{P_{3}}=(L| _{\operatorname{dom}L\cap\operatorname{Ker}P_{3}})^{-1}$.

Proof

Let $x\in\operatorname{dom}L\cap\operatorname{Ker}P_{3}$. Then $\varphi_{i}(x)=0$, $i=1,2,3$, and $x(0)=0$. So, we get

It follows from (2.1), (2.2) that $\Delta _{1}(Lx)_{j3}=-x''(0)\Delta_{j3}$, $\Delta_{2}(Lx)_{j3}=-2x'(0)\Delta _{j3}$. So, $K_{P_{3}}Lx=x$.

Inversely, $y\in\operatorname{Im}L$ results in $\Delta_{3}(y)=0$. As the proof of Lemma 3.2, $\varphi_{i}(K_{P_{3}}y)=0$, $i=1,2,3$. Clearly, $(K_{P_{3}}y)'''=y$. Thus, $K_{P_{3}}y\in\operatorname{dom}L$ and $LK_{P_{3}}y=y$, $y\in\operatorname{Im}L$. □

We introduce the constants $l_{3} = k_{1} \vert \Delta_{13} \vert +k_{2} \vert \Delta _{23} \vert +k_{3} \vert \Delta_{33} \vert $ and

The latter is frequently used in the remainder of the paper.

The next assumption is fulfilled in the main results by virtue of appropriate assumptions on $f(t, \cdot,\cdot,\cdot)$:

$(H_{1})$ :

For any $r>0$, there exists a function $h_{r}\in Y $ such that $\vert f(t,x(t),x'(t),x''(t)) \vert \leq h_{r}(t)$, $x\in X$, $\Vert x \Vert \leq r$.

Lemma 3.4

If $(H_{1})$ holds and $\Omega\subset X$ is bounded, then N is L-compact on Ω̅.

Proof

Take $r\in\mathbb{R}$ large enough such that $\Vert x \Vert \leq r$, $x\in\overline{\Omega}$. Then

So, $\Vert Q_{3}Nx \Vert _{1}\leq l_{3} \Vert h_{r} \Vert _{1} \Vert g_{3} \Vert _{1}$, which shows that $Q_{3}N(\overline{\Omega})$ is bounded. For $y\in Y$, we have

where, for convenience, we define, using (3.1), the constant

Then

Thus, $K_{P_{3}}(I-Q_{3})N(\overline{\Omega})$ is bounded.

For $0\leq t_{1}< t_{2}\leq1$, $x\in\overline{\Omega}$, we have

that is, $(K_{P_{3}}(I-Q_{3})N)''(\overline{\Omega})$ is equicontinuous on $[0,1]$ as well as $(K_{P_{3}}(I-Q_{3})N)'(\overline{\Omega})$ and $(K_{P_{3}}(I-Q_{3})N)(\overline{\Omega})$ by the mean value theorem. Therefore, by the Arzela-Ascoli theorem, $K_{P_{3}}(I-Q_{3})N(\overline {\Omega})$ is compact. □

In order to obtain the main results, we impose the following conditions:

$(H_{2})$ :

There exist nonnegative functions $a,b,c,d \in Y$ such that $\vert f(t,u,v,w) \vert \leq a(t)+b(t) \vert u \vert +c(t) \vert v \vert +d(t) \vert w \vert $, $t\in[0,1]$, $u,v,w\in\mathbb{R}$;

$(H_{3})$ :

There exists a constant $M_{03}>0$ such that $\Delta _{3}(Nx) \neq0$ if $\vert x(t) \vert >M_{03}$, $t\in[0,1]$;

$(H_{4})$ :

There exists a constant $M_{13}>0$ such that if $\vert c \vert > M_{13}$, then one of the following two inequalities holds:

$$ c\Delta_{3}(Nc)>0, $$

(3.3)

or

$$ c\Delta_{3}(Nc)< 0. $$

(3.4)

(Here $Nc = f(t,c,0,0)$, $c \in\mathbb{R}$.)

Lemma 3.5

Assume that $(H_{2})$, $(H_{3})$ hold and let

where $A_{P_{3}}$ satisfies (3.2). Then $\Omega_{13}=\{x\in \operatorname{dom}L\setminus\operatorname{Ker}L:Lx=\lambda Nx,\lambda\in(0,1)\} $ is bounded.

Proof

Since $x\in\Omega_{13}$, then $\Delta_{3}(Nx) = 0$. By $(H_{3})$, there exists $t_{0}\in[0,1]$ such that $\vert x(t_{0}) \vert \leq M_{03}$. Now,

and

Thus, $\Vert P_{3}x \Vert = \vert P_{3}x(t_{0}) \vert \leq M_{03}+A_{P_{3}} \Vert Lx \Vert _{1}$. It follows from $x = P_{3}x + (I-P_{3})x$ and $(H_{2})$ that

So,

Therefore, $\Omega_{13}$ is bounded by (3.5). □

Lemma 3.6

Assume that $(H_{4})$ holds. Then $\Omega_{23}=\{x\in\operatorname{Ker}L: Nx\in\operatorname{Im}L\}$ is bounded.

Proof

If $x\in\Omega_{23}$, then $x \equiv c$ and $Q_{3}(Nc)=0$, that is, $\Delta_{3}(Nc) = 0$. By $(H_{4})$, it follows that $\vert c \vert \leq M_{13}$. Thus, $\Omega_{23}$ is bounded. □

Lemma 3.7

Assume that $(H_{4})$ holds. Then

is bounded, where $\rho= \{ \scriptsize{ \begin{array}{l@{\quad}l} 1, &\textit{if } (3.3) \textit{ holds},\\ -1, &\textit{if } (3.4) \textit{ holds}. \end{array}} $

Proof

Let $x\in\Omega_{33}$. Then $x\equiv c\in\mathbb {R}$ and $\rho\lambda c +(1-\lambda)\Delta_{3}(Nc) =0$. If $\lambda =0$, then $\Delta_{3}(Nc) = 0$. By $(H_{4})$, $\vert c \vert \leq M_{13}$. If $\lambda=1$, then $c=0$. If $\lambda\in(0,1)$, then $c=- {\frac {1-\lambda}{\lambda\rho}} \Delta_{3}(Nc)$. Hence, $c^{2}= - {\frac {1-\lambda}{\lambda\rho}} c \Delta_{3}(Nc)$. If $\vert c \vert >M_{13}$, by $(H_{4})$, we obtain

which is a contradiction. Therefore, $\vert c \vert \leq M_{13}$ and $\Omega _{33}$ is bounded. □

Theorem 3.8

Assume that $(H_{2})$-$(H_{4})$ and (3.5) hold. Then problem (1.1) has at least one solution.

Proof

Let $\Omega\supset\overline{\Omega}_{13}\cup \overline{\Omega}_{23}\cup\overline{\Omega}_{33} $ be bounded. It follows from Lemmas 3.5 and 3.6 that $Lx \neq \lambda Nx$, $x\in(\operatorname{dom}L\setminus\operatorname{Ker}L)\cap \partial\Omega$, $\lambda\in(0,1)$ and $Nx\notin\operatorname{Im}L$, $x\in\operatorname{Ker}L\cap\partial\Omega$. Let

where $J_{3}: \operatorname{Im}Q_{3} \rightarrow \operatorname{Ker}L$ is an isomorphism defined by $J_{3}(c g_{3})=c$, $c\in\mathbb{R}$. By Lemma 3.7, we know $H(x,\lambda)\neq0$, $x\in\partial\Omega\cap \operatorname{Ker}L$, $\lambda\in[0,1]$. Since the degree is invariant under a homotopy,

By Theorem 2.1, $Lx=Nx$ has a solution in $\operatorname{dom}L\cap \overline{\Omega}$. □

Case II. $\varphi_{i}(t)=0$, $i=1,2,3$.

In this case, assume there exists $j\in\{1,2,3\}$ such that $\Delta _{j2}\neq0$. With an adjustment of the method of Lemma 3.1, we can assert the existence of a function $g_{2}\in Y$ such that $\Delta _{2}(g_{2})=1$.

Clearly, $\Delta=0$ and $\operatorname{Ker}L=\{ct:c\in\mathbb{R}\}$. Similar to the proof of Lemma 3.2, we can show that $\operatorname {Im}L=\{y\in Y:\Delta_{2}(y)=0\}$.

Define the operators $P_{2}:X\rightarrow X$, $Q_{2}:Y\rightarrow Y$ by

Obviously, $P_{2}$ and $Q_{2}$ are continuous linear projectors satisfying (2.3).

Define the operator $K_{P_{2}}: Y\rightarrow X$ as

As above, we can obtain that $K_{P_{2}}=(L| _{\operatorname{dom}L\cap\operatorname {\operatorname{Ker}}P_{2}})^{-1}$ and $\Vert K_{P_{2}}y \Vert \leq A_{P_{2}} \Vert y \Vert _{1}$, where

Suppose that the following conditions hold:

$(H_{5})$ :

There exists $M_{02}>0$ such that $\Delta_{2}(Nx)\neq 0$, if $\vert x'(t) \vert >M_{02}$, $t\in[0,1]$;

$(H_{6})$ :

There exists $M_{12}>0$ such that if $\vert c \vert >M_{12}$, then either

$$ c \Delta_{2} \bigl(N(ct) \bigr)>0, $$

(3.7)

or

$$ c\Delta_{2} \bigl(N(ct) \bigr)< 0. $$

(3.8)

Lemma 3.9

Assume that conditions $(H_{2})$, $(H_{5})$ hold and let

where $A_{P_{2}}$ satisfies (3.6). Then the set

is bounded.

Proof

If $x\in\Omega_{12}$, then $\Delta_{2}(Nx)=0$. By $(H_{5})$, there exists a constant $t_{1}\in[0,1]$ such that $\vert x'(t_{1}) \vert \leq M_{02}$. Since $x(t)=P_{2}x(t)+(I-P_{2})x(t)$, $x'(t_{1})=x'(0)+((I-P_{2})x)'(t_{1})$ and

we have

So,

Thus,

which proves that $\Omega_{12}$ is bounded. □

Lemma 3.10

Assume that $(H_{6})$ holds. Then the set

is bounded.

Proof

Since $x\in\Omega_{22}$, $x=ct$, $c\in\mathbb{R}$ and $\Delta_{2}(N(ct))=0$. By $(H_{6})$, we have $\vert c \vert \leq M_{12}$. So, $\Vert x \Vert = \vert c \vert \leq M_{12}$, that is, $\Omega_{22}$ is bounded. □

Lemma 3.11

Assume that $(H_{6})$ holds. Then the set

is bounded, where $J_{2}: \operatorname{Im}Q_{2}\rightarrow\operatorname{Ker}L$, $J_{2}(cg_{2})(t) = ct$, $c\in\mathbb{R}$, and $\rho= \{ \scriptsize{ \begin{array}{l@{\quad}l} 1, &\textit{if } (3.7) \textit{ holds},\\ -1,& \textit{if } (3.8) \textit{ holds}. \end{array}} $

Proof

If $x\in\Omega_{32}$, then $x=ct$, $c\in\mathbb {R}$ and $\lambda\rho c+(1-\lambda)J_{2}Q_{2} (N(ct))=0$. So,

If $\lambda=0$, then $\Delta_{2}(N(ct))=0$. By $(H_{6})$, $\vert c \vert \leq M_{12}$. If $\lambda=1$, then $c=0$. If $\lambda\in(0,1)$, $c=- \frac{1-\lambda}{\lambda\rho} \Delta_{2}(N(ct))$. So,

If $\vert c \vert >M_{12}$, by $(H_{6})$, we obtain $c^{2}<0$, a contradiction. So, $\vert c \vert \leq M_{12}$, that is, $\Omega_{32}$ is bounded. □

Under assumption $(H_{1})$, N is L-compact on a bounded set Ω̅ as in the proof of Lemma 3.4.

Theorem 3.12

Assume that $(H_{2})$, $(H_{5})$, $(H_{6})$ and (3.9) hold. Then FBVP (1.1) has at least one solution.

The proof is similar to that of Theorem 3.8.

Case III. $\varphi_{i}(t^{2})=0$, $i=1,2,3$.

In this case, assume that there exists $j\in\{1,2,3\}$ such that $\Delta_{j1}\neq0$.

Similarly, there exists a function $g_{1}\in Y$ such that $\Delta_{1}(g_{1})=1$.

Obviously, $\Delta=0$ and $\operatorname{Ker}L=\{ct^{2}:c\in\mathbb{R}\}$. Similar to the proof of Lemma 3.2, we can obtain $\operatorname {Im}L=\{y\in Y:\Delta_{1}(y)=0\}$.

Define the operators $P_{1}:X\rightarrow X$, $Q_{1}:Y\rightarrow Y$ as

Clearly, $P_{1}$ and $Q_{1}$ are continuous linear projectors. Introduce the operator $K_{P_{1}}:Y\rightarrow X$ by

As above, it is easy to show that $K_{P_{1}}=(L| _{\operatorname{dom}L\cap \operatorname{Ker}P_{1}})^{-1}$ and $\Vert K_{P_{1}}y \Vert \leq A_{P_{1}} \Vert y \Vert _{1}$, where

By the same method we used in Lemma 3.4, we can show that N is L-compact on Ω̅.

To prove the main result, we need the following hypotheses:

$(H_{7})$ :

There exists $M_{01}>0$ such that $\Delta_{1}(Nx)\neq0$ if $\vert x''(t) \vert >M_{01}$, $t\in[0,1]$;

$(H_{8})$ :

There exists $M_{11}$ such that if $\vert c \vert >M_{11}$, then either $c \Delta_{1}(N(ct^{2}))>0$ or $c \Delta_{1}(N(ct^{2}))<0$.

Lemma 3.13

Assume that $(H_{2})$, $(H_{7})$ hold. In addition, assume that

where $A_{P_{1}}$ is given by (3.10). Then the set

is bounded.

Proof

For $x\in\Omega_{11}$, we have $\Delta_{1}(Nx)=0$. By $(H_{7})$, there exists $t_{2}\in[0,1]$ such that $\vert x''(t_{2}) \vert \leq M_{01}$. Since $x=P_{1}x+(I-P_{1})x$, $\Vert (I-P_{1})x \Vert \leq A_{P_{1}} \Vert Lx \Vert _{1} < A_{P_{1}} \Vert Nx \Vert _{1}$,

and $(P_{1}x)''(t_{2})=x''(0)$, we get

Combining the inequalities above, we get

Thus,

In view of (3.11), $\Omega_{11}$ is bounded. □

Similarly, if $(H_{7})$ and $(H_{8})$ hold, we can prove that $\Omega _{21} = \{x\in\operatorname{Ker}L: Nx\in\operatorname{Im}L\}$ and $\Omega_{31} = \{x\in\operatorname{Ker}L: \rho\lambda x+(1-\lambda)J_{1} Q_{1}Nx=0, \lambda\in[0,1]\}$, with an isomorphism $J_{1}: \operatorname{Im}Q \to \operatorname{Ker}L$, $J_{1}(cg_{1})(t) =ct^{2}$, $c\in\mathbb{R}$, are bounded.

Theorem 3.14

Assume that $(H_{2})$, $(H_{7})$, $(H_{8})$ and (3.11) hold. Then FBVP (1.1) has at least one solution.

We define, for convenience,

By the same method as we used in the proof of Lemma 3.1 (see also [6]), there exists $g \in Y$ such that

It is easy to see that

Lemma 4.1

Proof

In fact, if $x\in\operatorname{dom}L$, $Lx=y$, then

and $\varphi_{i}(x)=0, i=1,2,3$. So, we have

In view of (4.1),

On the other hand, if $y\in Y$ satisfies the identity on the right-hand side of (4.2), we choose

Obviously, $Lx=y$. If $\Delta_{11}\neq0$, then

Considering $\Delta_{11}=\Delta_{1}(y)_{11}$, $\Delta_{2}(y)_{11}=\Delta _{1}(y)_{12}$, $\Delta_{3}(y)_{11}=-\Delta_{1}(y)_{13}$ and $\Delta _{1}(y)=0$, we get

Similarly, $\varphi_{1}(x)= -\frac{1}{2\Delta_{12}}\Delta_{2}(y)=0$, if $\Delta_{12}\neq0$ and $\varphi_{1}(x)= \frac{1}{2\Delta_{13}}\Delta _{3}(y)=0$, if $\Delta_{13}\neq0$. It is easy to check $\varphi _{2}(x)=\varphi_{3}(x)=0$.

Thus, $x\in\operatorname{dom}L$, that is, $y\in\operatorname{Im}L$. So, (4.2) holds. □

Define operators $P:X\rightarrow X$, $Q:Y\rightarrow Y$ by

where g is introduced at the beginning of the section. Moreover, $J: \operatorname{Im}Q \to\operatorname{Ker}L$ is defined by

We define $K_{P}:Y\rightarrow X$ as follows:

Lemma 4.2

$K_{P}=(L| _{\operatorname{dom}L\cap\operatorname{Ker}P})^{-1}$ and

Proof

If $\Delta_{11}\neq0$, for $x\in\operatorname{dom}L\cap \operatorname{Ker}P$, considering $\Delta_{2}(Lx)_{11}=-x''(0)\Delta_{12}-2x'(0)\Delta_{11}$, $\Delta _{3}(Lx)_{11}=x''(0)\Delta_{13}-2x(0)\Delta_{11}$, $\Delta _{11}x''(0)-\Delta_{12}x'(0)+\Delta_{13}x(0)=0$, we have

Inversely, if $y\in\operatorname{Im}L$, then $\Delta_{i}(y)=0$, $i=1,2,3$. This, together with $\Delta_{2}(y)_{11}=\Delta_{1}(y)_{12}$, $\Delta _{3}(y)_{11}=-\Delta_{1}(y)_{13}$, $\Delta_{11}=\Delta_{1}(y)_{11}$, $\Delta=0$, implies that

Obviously, $\varphi_{2}(K_{P}y)=\varphi_{3}(K_{P}y)=0$. So, $K_{P}y\in\operatorname {dom}L$. By a simple calculation, we can obtain $K_{P}y\in\operatorname {Ker}P$ and $LK_{P}y=y$. Therefore, $K_{P}=(L| _{\operatorname{dom}L\cap\operatorname {Ker}P})^{-1}$. If $\Delta_{11}=0$, $\Delta_{12}\neq0$ or $\Delta _{11}=\Delta_{12}=0$, $\Delta_{13}\neq0$, we can similarly get the result.

If $\Delta_{11}\neq0$, then

If $\Delta_{11}=0$, $\Delta_{12}\neq0$, then

Similarly, if $\Delta_{11}=\Delta_{12}=0$, $\Delta_{13}\neq0$, then

 □

Lemma 4.3

Assume that $(H_{1})$ holds and $\Omega\subset X$ is bounded. Then N is L-compact on Ω.

Proof

For $x\in\overline{\Omega}$, there exists a constant $r>0$ such that $\Vert x \Vert \leq r$. By $(H_{1})$, we get

that is, $QN(\overline{\Omega})$ is bounded. Hence,

So, $K_{P}(I-Q)N(\overline{\Omega})$ is bounded.

By the same method as used in the proof of Lemma 3.3, we can demonstrate that $K_{P}(I-Q)N(\overline{\Omega})$ is compact. Thus N is L-compact. □

When $\Delta_{11}\neq0$, we assume that the following conditions hold:

$(H_{9})$ :

There exists a constant $M_{0}>0$ such that if $\vert x''(t) \vert > M_{0}$, then

$$(\varphi_{1}-k\varphi_{2}) \biggl( \int_{0}^{t}(t-s)^{2}Nx(s) \,ds \biggr) \neq0; $$

$(H_{10})$ :

There exists a constant $M_{1}>0$ such that if $\vert c \vert > M_{1}$, either

$$ c(\varphi_{1}-k\varphi_{2}) \biggl( \int_{0}^{t}(t-s)^{2}N \bigl(c \bigl( \Delta _{11}s^{2}-\Delta_{12}s+\Delta_{13} \bigr) \bigr) \,ds \biggr) > 0, $$

(4.5)

or

$$ c(\varphi_{1}-k\varphi_{2}) \biggl( \int_{0}^{t}(t-s)^{2}N \bigl(c \bigl( \Delta _{11}s^{2}-\Delta_{12}s+\Delta_{13} \bigr) \bigr) \,ds \biggr) < 0. $$

(4.6)

Lemma 4.4

Assume that $\Delta_{11}\neq0$, $(H_{2}), (H_{9})$ and

hold. Then the set

is bounded.

Proof

Since $x\in\Omega_{1}$, then $QNx=0$. By $(H_{9})$, there exists $t_{0}\in[0,1]$ such that $\vert x''(t_{0}) \vert \leq M_{0}$. Since $x= Px +(I-P)x$,

and $x''(t)=(Px)''(t)+((I-P)x)''(t)$, it follows that

Considering

we have

This, together with (4.8) and $(H_{2})$, means

So,

Therefore, $\overline{\Omega}_{1}$ is bounded due to (4.7). □

Lemma 4.5

Assume that $\Delta_{11}\neq0$ and $(H_{10})$ holds. Then the set $\Omega_{2}=\{ x\in\operatorname{Ker}L:Nx\in\operatorname{Im}L\}$ is bounded.

Proof

For $x\in\Omega_{2}$, we have $x=c(\Delta _{11}t^{2}-\Delta_{12}t+\Delta_{13})$ and $QNx=0$. By $(H_{10})$, we get that $\vert c \vert \leq M_{1}$. So, $\Omega_{2}$ is bounded. □

Let $\rho= \{ \scriptsize{ \begin{array}{l@{\quad}l} 1, &\mbox{if } (4.5) \mbox{ holds},\\ -1, &\mbox{if } (4.6) \mbox{ holds}. \end{array}} $

Lemma 4.6

Assume that $\Delta_{11}\neq0$ and $(H_{10})$ holds. The set

is bounded, where $J: \operatorname{Im}Q \rightarrow \operatorname{Ker}L$ is defined by (4.3).

Proof

If $x\in\Omega_{3}$, then $\lambda\rho x +(1-\lambda )JQNx=0$, $x=c(\Delta_{11}t^{2}-\Delta_{12}t+\Delta_{13})$.

If $\lambda=0$, then $QNx=0$. It follows from $(H_{10})$ that $\Vert x \Vert \leq M_{1}(2 \vert \Delta_{11} \vert + \vert \Delta_{12} \vert + \vert \Delta_{13} \vert )$. If $\lambda=1$, then $x \equiv0$. For $\lambda\in(0,1)$, we have

That is,

By $(H_{10})$, we know that $\vert c \vert \leq M_{1}$. So, $\Omega_{3}$ is bounded. □

Theorem 4.7

Assume that $\Delta_{11}\neq0$ and $(H_{2})$, $(H_{9})$, $(H_{10})$ hold and

where $A_{p}$ is given by (4.4). Then FBVP (1.1) has at least one solution.

The proof is similar to that of Theorem 3.8.

The example below illustrates Theorem 4.7.

Consider

where $A =1/113{,}414$, along with the functional conditions

In this case $\phi_{1}(t^{2}) = \phi_{2}(t^{2}) =8$, $\phi_{3}(t^{2})=1$, $\phi _{1}(t) = \phi_{2}(t) =10$, $\phi_{3}(t)=1$, and $\phi_{1}(1) = \phi_{2}(1) =19$, $\phi_{3}(1)=2$, so that $k=1$. Also, $\triangle= 0$ and $\triangle_{11}=1$, $\triangle_{12} = -3$, $\triangle_{13} = -2$, and

Subsequently,

and

We can easily check $(\phi_{1}-k\phi_{2})(-t^{5})=1$.

The following estimates hold for $x''(t) < -M_{0}$:

and hence

provided $M_{0} > 2 + \frac{128}{35A}$.

If $x''(t) > M_{0}$, then

and hence

provided $M_{0} > 2$.

Therefore, if we choose $M_{0} > 2 + \frac{64}{35A}$, then ($H_{9}$) holds.

Let now $c\in\mathbb{R}$ and $x_{c}(t)= c(t^{2}+3t-2)$. Then

and

Then, repeating the computation leading to the choice of $M_{0}$, we obtain that $\vert c \vert > M_{1} = M_{0}/2$ results in

that is, ($H_{10}$) holds.

Finally, note that $k_{1} = 25$, $k_{2}=21$, $k_{3} = 25/2$, so that $l= 525$. Hence $A_{P} = 1 +16l/ \vert \triangle_{11} \vert =8{,}401$. Then

All the conditions of Theorem 4.7 are verified.

The following corollaries are stated without proof.

Corollary 4.8

Let $\Delta_{11}=0$, $\Delta_{12}\neq0$ and assume that ($H_{2}$) and the following conditions hold:

$(H_{11})$ :

There exists a constant $M'_{0}>0$ such that if $\vert x'(t) \vert \geq M'_{0}$, then

$$(\varphi_{1}-k\varphi_{2}) \biggl( \int_{0}^{t}(t-s)^{2}Nx(s) \,ds \biggr) \neq0; $$

$(H_{12})$ :

There exists a constant $M'_{1} > 0$ such that if $\vert c \vert > M'_{1}$, then either

$$c(\varphi_{1}-k\varphi_{2}) \biggl( \int_{0}^{t}(t-s)^{2}N \bigl(c(- \triangle _{12}s+\triangle_{13}) \bigr) \,ds \biggr) > 0, $$

or

$$c(\varphi_{1}-k\varphi_{2}) \biggl( \int_{0}^{t}(t-s)^{2}N \bigl(c(- \triangle _{12}s+\triangle_{13}) \bigr) \,ds \biggr) < 0. $$

Corollary 4.9

Let $\Delta_{11}=\Delta_{12}=0$, $\Delta _{13}\neq0$ and assume that ($H_{2}$) and the following conditions hold:

$(H_{13})$ :

There exists a constant $M''_{0}>0$ such that if $\vert x(t) \vert \geq M''_{0}$, then

$$(\varphi_{1}-k\varphi_{2}) \biggl( \int_{0}^{t}(t-s)^{2}Nx(s) \,ds \biggr) \neq0; $$

$(H_{14})$ :

There exists a constant $M''_{1} > 0$ such that if $\vert c \vert > M''_{1}$, then either

$$c(\varphi_{1}-k\varphi_{2}) \biggl( \int_{0}^{t}(t-s)^{2}N(c \triangle _{13}) \,ds \biggr) > 0, $$

or

$$c(\varphi_{1}-k\varphi_{2}) \biggl( \int_{0}^{t}(t-s)^{2}N(c \triangle _{13}) \,ds \biggr) < 0. $$

This paper is a study of third-order functional boundary value problems at resonance; it improves and generalizes some of the existent results. We present several generalizations to the existing results and improvements to the method based on Mawhin’s coincidence degree theory.

The authors are grateful to anonymous reviewers for carefully reading this paper and for their comments and suggestions which have improved the paper. The work WJ is supported by 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

2. Department of Mathematics and Statistics, University of Arkansas at Little Rock, Little Rock, AR, 72204-1099, USA

   • Nickolai Kosmatov

Corresponding author

Correspondence to Weihua Jiang.

Conflict of interest

The authors declare that there is no conflict of interest regarding the publication of this paper.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

Both authors contributed equally in writing this paper. They both read and approved the final manuscript.

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