Let \(X=\{\varphi :\varphi \in C^{n-1}[0,+\infty ), \sup_{x\in [0,+ \infty )}\frac{|\varphi ^{(j)}(x)|}{e^{x}}<\infty ,\varphi ^{(i)}(0)=0,i=1,2, \ldots ,n-3,j=0,1,2,\ldots ,n-1\}\) with the norm \(\|\varphi \|= \max \{\|\frac{\varphi ^{(i)}}{e^{x}}\|_{\infty }, i=0,1,2, \ldots ,n-1\}\), where \(\|\varphi \|_{\infty }=\sup_{t\in [0,+\infty )}|\varphi (x)|\), and \(Y=L[0,+\infty )\) with the norm \(\|u\|_{1}=\int _{0}^{+\infty }|u(x)|\,dx\).
It is easy to prove that \((X,\|\cdot \|)\) and \((Y,\|\cdot \|_{1})\) are Banach spaces.
We define an operator L as follows:
$$ L\varphi (x)=\varphi ^{(n)}(x) $$
with \(\operatorname{dom} L=\{\varphi \in X:\varphi ^{(n)}(x)\in Y,\varphi ^{(n-1)}(+ \infty )=0,\Gamma _{1}(\varphi (x))=0,\Gamma _{2}(\varphi (x))=0\}\).
An operator \(N:X\rightarrow Y\) is defined as follows:
$$ (N\varphi ) (x)=f\bigl(x,\varphi (x),\varphi '(x),\ldots ,\varphi ^{(n-1)}(x) \bigr),\quad \varphi \in X, x\in [0,+\infty ). $$
So, problem (1.1) becomes \(L\varphi =N\varphi \).
For convenience, we denote
- \((H_{2})\):
-
The linear functionals \(\Gamma _{1},\Gamma _{2}:X\rightarrow \mathbb{R}\) satisfy \(\Gamma _{2}(1)=b\), \(\Gamma _{2}(x^{n-2})=a\), \(\Gamma _{1}(1)=\alpha b\), \(\Gamma _{1}(x^{n-2})=\alpha a\), \((\Gamma _{1}-\alpha \Gamma _{2})(x^{n}e^{-x}) \neq 0\), where \(a^{2}+b^{2}\neq 0\), \(\alpha ,a,b\in \mathbb{R}\).
- \((H_{3})\):
-
The functionals \(\Gamma _{1},\Gamma _{2}:X\rightarrow \mathbb{R}\) are linear continuous with the respective norms \(\beta _{1}\), \(\beta _{2}\), that is, \(|\Gamma _{i}(\varphi )|\leq \beta _{i}\|\varphi \|\), \(i=1,2\).
Lemma 3.1
There must exist
\(g\in Y\)
such that
$$ (\Gamma _{1}-\alpha \Gamma _{2}) \biggl( \int _{0}^{+ \infty }k(x,y)g(y)\,dy \biggr)=1, $$
where
$$ k(x,y)= \textstyle\begin{cases} -\frac{x^{n-1}}{(n-1)!}, & 0\leq x\leq y< +\infty ; \\ -\frac{x^{n-1}-(x-y)^{n-1}}{(n-1)!}, & 0\leq y\leq x< +\infty . \end{cases} $$
(3.1)
Proof
Since we just need to find a specific g satisfying the equation, in particular, let \(\mathbf{x}=x^{n}e^{-x}\in X\), \(h(x)=(x^{n}e^{-x})^{(n)}\in Y\), \(x\in [0,+ \infty )\), we have
$$\begin{aligned} \int _{0}^{+\infty }k(x,y)h(y)\,dy&=\frac{1}{(n-1)!} \int _{0}^{x}(x-y)^{n-1}h(y)\,dy- \int _{0}^{+\infty }\frac{x^{n-1}}{(n-1)!}h(y)\,dy \\ &=\mathbf{x}(x)-\mathbf{x}(0)- \frac{\mathbf{x}^{(n-1)}(+\infty )}{(n-1)!}x^{n-1}- \frac{\mathbf{x}^{(n-2)}(0)}{(n-2)!}x^{n-2} \\ &=\mathbf{x}(x). \end{aligned}$$
Now, considering condition \((H_{2})\), take
$$g(y)= \frac{h(y)}{(\Gamma _{1}-\alpha \Gamma _{2}) (\int _{0}^{+\infty }k(x,y)\mathbf{x}^{(n)}(y)\,dy )},$$
thus,
$$\Gamma \biggl(\int _{0}^{+\infty }k(x,y)g(y)\,dy \biggr)=1.$$
□
Lemma 3.2
L is a Fredholm operator of index zero.
Proof
From \(\varphi ^{(n)}(x)=u(x)\) and \(\varphi ^{(i)}(0)=0\), \(1\leq i\leq n-3\), we have
$$\begin{aligned} \varphi (x)=\frac{1}{(n-1)!} \int _{0}^{x}(x-y)^{n-1}u(y)\,dy+\varphi (0)+ \frac{\varphi ^{(n-1)}(0)}{(n-1)!}x^{n-1} + \frac{\varphi ^{(n-2)}(0)}{(n-2)!}x^{n-2}. \end{aligned}$$
(3.2)
Now we will give KerL and ImL.
Taking \(\varphi \in \operatorname{dom}L\) with \(L\varphi =0\), we obtain \(\varphi (x)=\varphi (0)+\frac{\varphi ^{(n-1)}(0)}{(n-1)!}x^{n-1} + \frac{\varphi ^{(n-2)}(0)}{(n-2)!}x^{n-2}\). This, together with \(\varphi ^{(n-1)}(+\infty )=0\), \(\Gamma _{1}(\varphi (x))=0\) and \(\Gamma _{2}(\varphi (x))=0\), implies that
$$\begin{aligned}& \Gamma _{1}\bigl(\varphi (x)\bigr)=\varphi (0)\Gamma _{1}(1)+ \frac{\varphi ^{(n-2)}(0)}{(n-2)!}\Gamma _{1}\bigl(x^{n-2} \bigr)=0, \\& \Gamma _{2}\bigl(\varphi (x)\bigr)=\varphi (0)\Gamma _{2}(1)+ \frac{\varphi ^{(n-2)}(0)}{(n-2)!}\Gamma _{2}\bigl(x^{n-2} \bigr)=0. \end{aligned}$$
Based on condition \((H_{2})\), we have
$$ \varphi (x)=c\bigl(a-bx^{n-2}\bigr),\quad c\in \mathbb{R}. $$
So,
$$ \operatorname{Ker} L=\bigl\{ c\bigl(a-bx^{n-2}\bigr)|a^{2}+b^{2} \neq 0,c\in \mathbb{R}\bigr\} , \qquad \operatorname{dim}\operatorname{Ker}L=1. $$
In order to prove
$$\begin{aligned} \operatorname{Im}L= \biggl\{ u\in Y:(\Gamma _{1}-\alpha \Gamma _{2}) \biggl( \int _{0}^{\infty }k(x,y)u(y)\,dy \biggr)=0 \biggr\} . \end{aligned}$$
(3.3)
To see this, let us suppose that \(u\in \operatorname{Im}L\), then there exists \(\varphi \in \operatorname{dom}L\) such that \(u=L\varphi \in Y\). From (3.1) and \(\varphi ^{(n-1)}(+\infty )=0\), we have
$$\begin{aligned} \varphi (x) =&\frac{1}{(n-1)!} \int _{0}^{x}(x-y)^{n-1}u(y)\,dy- \frac{1}{(n-1)!} \int _{0}^{+\infty }x^{n-1}u(y)\,dy \\ &{}+\varphi (0) + \frac{\varphi ^{(n-2)}(0)}{(n-2)!}x^{n-2}. \end{aligned}$$
(3.4)
It follows from \(\Gamma _{1}(\varphi (x))=0\), \(\Gamma _{2}(\varphi (x))=0\), and \((H_{2})\) that
$$ (\Gamma _{1}-\alpha \Gamma _{2}) \biggl( \int _{0}^{ \infty }k(x,y)u(y)\,dy \biggr)=0. $$
That is,
$$ \operatorname{Im}L\subseteq \biggl\{ u\in Y:(\Gamma _{1}-\alpha \Gamma _{2}) \biggl( \int _{0}^{\infty }k(x,y)u(y)\,dy \biggr)=0 \biggr\} . $$
If \(u\in \{ u\in Y:(\Gamma _{1}-\alpha \Gamma _{2}) (\int _{0}^{\infty }k(x,y)u(y)\,dy )=0 \} \), take
$$ \varphi (x)=-\frac{ax^{n-2}+b}{a^{2}+b^{2}}\Gamma _{2} \biggl( \int _{0}^{ \infty }k(x,y)u(y)\,dy \biggr)+ \int _{0}^{\infty }k(x,y)u(y)\,dy. $$
In view of \((H_{2})\), it is easy to deduce that \(L\varphi =\varphi ^{(n)}(x)=u(x)\), \(\varphi ^{(i)}(0)=0\), \(1\leq i\leq n-3\), \(\varphi ^{(n-1)}(+\infty )=0\), and \(\Gamma _{i}(\varphi (x))=0\), \(i=1,2\). That is, \(u\in \operatorname{Im}L\), i.e.,
$$ \biggl\{ u\in Y:(\Gamma _{1}-\alpha \Gamma _{2}) \biggl( \int _{0}^{ \infty }k(x,y)u(y)\,dy \biggr)=0 \biggr\} \subseteq \operatorname{Im}L. $$
Combining the above, we obtain (3.3).
Define \(Q:Y\rightarrow Y\) by
$$ Qu=(\Gamma _{1}-\alpha \Gamma _{2}) \biggl( \int _{0}^{\infty }k(x,y)u(y)\,dy \biggr)g(x), $$
where \(g(x)\) is introduced in Lemma 3.1.
Clearly, \(\operatorname{Ker}Q=\operatorname{Im}L\), \(\operatorname{Im}Q=\{u|u=cg(x),x\geq 0,c\in \mathbb{R}\}\), and \(Q:Y\rightarrow Y\) is a linear projector. In fact, for \(u\in Y\), we have
$$ \bigl(Q^{2}u\bigr) (x)=(\Gamma _{1}-\alpha \Gamma _{2}) \biggl( \int _{0}^{\infty }k(x,y)u(y)\,dy \biggr) (\Gamma _{1}-\alpha \Gamma _{2}) \biggl( \int _{0}^{\infty }k(x,y)g(y)\,dy \biggr)g(x)=Qu. $$
For \(u\in Y\), we have \(u=(u-Qu)+Qu\), \(Qu\in \operatorname{Im}Q\), \((I-Q)u\in \operatorname{Ker}Q= \operatorname{Im}L\). So, we obtain \(Y=\operatorname{Im}Q+\operatorname{Im}L\). Take \(u_{0}\in \operatorname{Im}Q\) means that \(u_{0}\) can be written \(u_{0}=cg(x)\), \(c\in \mathbb{R}\). At the same time, by \(u_{0}\in \operatorname{Im}L\) and Lemma 3.1, we get \(c=0\). This implies that \(u_{0}=0\). Thus, \(Y=\operatorname{Im}Q\oplus \operatorname{Im}L\) and \(\operatorname{dim}\operatorname{Ker}L=\operatorname{codim}\operatorname{Im}L<+\infty \). Observing that ImL is a closed subspace of Y; L is a Fredholm operator of index zero. □
Define \(P:X\rightarrow X\) as
$$ (P\varphi ) (x)=\frac{1}{a^{2}+b^{2}} \biggl(a\varphi (0)- \frac{b\varphi ^{(n-2)}(0)}{(n-2)!} \biggr) \bigl(a-bx^{n-2}\bigr), \quad \varphi \in X. $$
Clearly, \(P:X\rightarrow X\) is a linear continuous projector and \(\operatorname{Im}P=\{u|u(x)=c(a-bx^{n-2}),x\geq 0,c\in \mathbb{R}\}= \operatorname{Ker}L\). Thus, \(X=\operatorname{Im}P\oplus \operatorname{Ker}P=\operatorname{Ker}L\oplus \operatorname{Ker}P\).
Define \(K_{P}:\operatorname{Im}L\rightarrow \operatorname{dom}L\cap \operatorname{Ker}P\) by
$$ (K_{P}u) (x)=-\frac{ax^{n-2}+b}{a^{2}+b^{2}}\Gamma _{2} \biggl( \int _{0}^{ \infty }k(x,y)u(y)\,dy \biggr)+ \int _{0}^{\infty }k(x,y)u(y)\,dy,\quad u\in \operatorname{Im}L . $$
By simple calculations, for \(\varphi \in \operatorname{dom}L\cap \operatorname{Ker}P\),
$$\begin{aligned} (K_{P}L_{P})\varphi &=-\frac{ax^{n-2}+b}{a^{2}+b^{2}}\Gamma _{2} \biggl( \int _{0}^{\infty }k(x,y)\varphi ^{(n)}(y)\,dy \biggr)+ \int _{0}^{ \infty }k(x,y)\varphi ^{(n)}(y)\,dy \\ &=-\frac{ax^{n-2}+b}{a^{2}+b^{2}}\Gamma _{2} \biggl(\varphi (x)- \varphi (0)- \frac{\varphi ^{(n-2)}(0)x^{n-2}}{(n-2)!} \biggr) \\ &\quad{}+\varphi (x)-\varphi (0)-\frac{\varphi ^{(n-2)}(0)x^{n-2}}{(n-2)!} \\ &=\varphi (x)-\frac{ax^{n-2}+b}{a^{2}+b^{2}} \biggl(-b\varphi (0)- \frac{a\varphi ^{(n-2)}(0)}{(n-2)!} \biggr)-\varphi (0)- \frac{\varphi ^{(n-2)}(0)x^{n-2}}{(n-2)!} \\ &=\varphi (x)-\frac{1}{a^{2}+b^{2}} \biggl(a\varphi (0)- \frac{b\varphi ^{(n-2)}(0)}{(n-2)!} \biggr) \bigl(a-bx^{n-2}\bigr) \\ &=\varphi (x)-P\varphi (x)=\varphi (x) \end{aligned}$$
and \((L_{P}K_{P})u=u\), \(\forall u\in \operatorname{Im}L\). So, \(K_{P}=(L_{P})^{-1}\), where \(L_{P}=L|_{\operatorname{dom}L\cap \operatorname{Ker}P}:\operatorname{dom}L \cap \operatorname{Ker}P\rightarrow \operatorname{Im}L\).
Define the linear isomorphism \(J: \operatorname{Ker}L\rightarrow \operatorname{Im}Q\) as \(J(c(a-bx^{n-2}))=cg(x)\), \(c\in \mathbb{R}\), where \(g(x)\) is also introduced in Lemma 3.1.
The next lemma provides norm estimates needed for the following result.
Lemma 3.3
For \(u\in Y\), \(\|K_{P}u\|\leq B\|u\|_{1}\), where
$$ B= \biggl(1+\frac{( \vert a \vert (n-2)!+ \vert b \vert )}{a^{2}+b^{2}}\beta _{2} \biggr). $$
Proof
Observe that due to \(|\Gamma _{2}(\varphi (x))|\leq \beta _{2}\|\varphi \|\),
$$\begin{aligned} \biggl\vert \Gamma _{2} \biggl( \int _{0}^{\infty }k(x,y)u(y)\,dy \biggr) \biggr\vert & \leq \beta _{2} \biggl\Vert \int _{0}^{\infty }k(x,y)u(y)\,dy \biggr\Vert \\ &\leq \beta _{2} \biggl\Vert \frac{1}{(n-1)!} \int _{0}^{x}(x-y)^{n-1}u(y)\,dy- \int _{0}^{\infty }\frac{x^{n-1}}{(n-1)!}u(y)\,dy \biggr\Vert \\ &\leq \beta _{2} \biggl\Vert \frac{1}{(n-1)!} \int _{0}^{\infty }(x-y)^{n-1}u(y)\,dy- \int _{0}^{\infty }\frac{x^{n-1}}{(n-1)!}u(y)\,dy \biggr\Vert \\ &\leq \beta _{2} \biggl\Vert \int _{0}^{\infty }\frac{x^{n-1}}{(n-1)!}u(y) \biggr\Vert \\ &\leq \beta _{2} \int _{0}^{\infty } \bigl\vert u(y) \bigr\vert \,dy \biggl\Vert \frac{x^{n-1}}{(n-1)!} \biggr\Vert \\ &\leq \beta _{2} \Vert u \Vert _{1}. \end{aligned}$$
Then
$$\begin{aligned} \Vert K_{P}u \Vert &= \biggl\Vert -\frac{ax^{n-2}+b}{a^{2}+b^{2}}\Gamma _{2} \biggl( \int _{0}^{\infty }k(x,y)u(y)\,dy \biggr)+ \int _{0}^{\infty }k(x,y)u(y)\,dy \biggr\Vert \\ &\leq \frac{ \vert \Gamma _{2} (\int _{0}^{\infty }k(x,y)u(y)\,dy ) \vert }{a^{2}+b^{2}} \bigl\Vert ax^{n-2}+b \bigr\Vert + \biggl\Vert \int _{0}^{\infty }k(x,y)u(y)\,dy \biggr\Vert \\ &\leq \frac{( \vert a \vert (n-2)!+ \vert b \vert )}{a^{2}+b^{2}}\beta _{2} \Vert u \Vert _{1}+ \Vert u \Vert _{1} \\ &= \biggl(1+\frac{( \vert a \vert (n-2)!+ \vert b \vert )}{a^{2}+b^{2}}\beta _{2} \biggr) \Vert u \Vert _{1}=B \Vert u \Vert _{1}. \end{aligned}$$
□
Lemma 3.4
N is L-compact on Ω̅ if \(\operatorname{dom}L\cap \overline{\Omega }\neq \emptyset \), where Ω is a bounded open subset of X.
Proof
For convenience, denote \(M_{r}:=\int _{0}^{+\infty }|h_{r}(y)|\,dy\).
We will prove that \(QN:X\rightarrow Y \) is continuous and bounded.
Since \(\Omega \subset X\) is bounded, for \(\varphi \in \overline{\Omega }\), there exists a constant \(r>0\) such that \(\|\varphi \|< r\).
$$\begin{aligned} \Vert QN\varphi \Vert _{1}&= \int _{0}^{+\infty } \biggl\vert (\Gamma _{1}- \alpha \Gamma _{2}) \biggl( \int _{0}^{+\infty }k(x,y)N\varphi (y)\,dy \biggr)g(x) \biggr\vert \,dx \\ &\leq \biggl\vert (\Gamma _{1}-\alpha \Gamma _{2}) \biggl( \int _{0}^{+ \infty }k(x,y)N\varphi (y)\,dy \biggr) \biggr\vert \int _{0}^{+\infty } \bigl\vert g(x) \bigr\vert \,dx \\ &\leq \bigl(\beta _{1}+ \vert \alpha \vert \beta _{2} \bigr) \biggl\Vert \int _{0}^{+ \infty }k(x,y)N\varphi (y)\,dy \biggr\Vert \int _{0}^{+\infty } \bigl\vert g(x) \bigr\vert \,dx. \end{aligned}$$
It is not difficult to verify that \(\sup_{x\in [0,+\infty )} \frac{|(\int _{0}^{+\infty }k(x,y)N\varphi (y)\,dy)^{(i)}|}{e^{x}}\leq \int _{0}^{+\infty }|N\varphi (y)|\,dy\), \(i=0,1,2,\ldots ,n-1\).
By condition \((H_{1})\), we have
$$\begin{aligned} \biggl\Vert \int _{0}^{+\infty }k(x,y)N\varphi (y)\,dy \biggr\Vert &=\sup_{x\in [0,+\infty )} \frac{ \vert (\int _{0}^{+\infty }k(x,y)N\varphi (y)\,dy)^{(i)} \vert }{e^{x}} \\ &\leq \int _{0}^{+\infty } \bigl\vert N\varphi (y) \bigr\vert \,dy\leq \int _{0}^{+\infty }h_{r}(y) \,dy=M_{r}. \end{aligned}$$
So, \(QN:X\rightarrow Y\) is bounded. By \((H_{1})\) 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.
$$\begin{aligned} & \biggl\vert \frac{K_{P}(I-Q)N\varphi (x)}{e^{x}} \biggr\vert \\ &\quad = \biggl\vert -\frac{ax^{n-2}+b}{e^{x}} \frac{1}{a^{2}+b^{2}}\Gamma _{2} \biggl( \int _{0}^{+\infty }k(x,y) (I-Q)N \varphi (y)\,dy \biggr)\\ &\qquad {}+ \frac{\int _{0}^{+\infty }k(x,y)(I-Q)N\varphi (y)\,dy}{e^{x}} \biggr\vert \\ &\quad \leq \frac{\beta _{2}}{a^{2}+b^{2}} \biggl\vert \frac{ax^{n-2}+b}{e^{x}} \biggr\vert \bigl\Vert (I-Q)N\varphi \bigr\Vert _{1} + \biggl\vert \frac{\int _{0}^{+\infty }k(x,y)(I-Q)N\varphi (y)\,dy}{e^{x}} \biggr\vert \\ &\quad \leq M_{r} \biggl(\frac{ \vert a \vert + \vert b \vert }{a^{2}+b^{2}}\beta _{2}+1 \biggr) \biggl(1+\bigl(\beta _{1}+ \vert \alpha \vert \beta _{2}\bigr) \int _{0}^{+\infty } \bigl\vert g(x) \bigr\vert \,dx \biggr)< +\infty , \\ & \biggl\vert \frac{(K_{P}(I-Q)N\varphi (x))'}{e^{x}} \biggr\vert \\ &\quad = \biggl\vert -\frac{a(n-2)x^{n-3}}{e^{x}} \frac{1}{a^{2}+b^{2}}\Gamma _{2} \biggl( \int _{0}^{+\infty }k(x,y) (I-Q)N \varphi (y)\,dy \biggr)\\ &\qquad {}+ \frac{(\int _{0}^{+\infty }k(x,y)(I-Q)N\varphi (y)\,dy)'}{e^{x}} \biggr\vert \\ &\quad \leq \frac{\beta _{2}}{a^{2}+b^{2}} \biggl\vert \frac{a(n-2)x^{n-3}}{e^{x}} \biggr\vert \biggl\Vert \int _{0}^{+\infty }k(x,y) (I-Q)N \varphi (y)\,dy \biggr\Vert \\ &\qquad {}+ \biggl\Vert \int _{0}^{+\infty }k(x,y) (I-Q)N \varphi (y)\,dy \biggr\Vert \\ &\quad \leq M_{r} \biggl(\frac{ \vert a \vert (n-2)}{a^{2}+b^{2}}\beta _{2}+1 \biggr) \biggl(1+\bigl(\beta _{1}+ \vert \alpha \vert \beta _{2}\bigr) \int _{0}^{+\infty } \bigl\vert g(x) \bigr\vert \,dx \biggr)< +\infty . \end{aligned}$$
Similarly, \(\{ \frac{ \vert (K_{P}(I-Q)N\varphi )^{(i)} \vert }{e^{x}}<+\infty : \varphi \in \overline{\Omega },i=2,3,\ldots ,n-1 \}\), i.e., \(K_{P}(I-Q)N: \overline{\Omega }\rightarrow X\) is bounded.
Secondly, for \(\varphi \in \overline{\Omega }\), \(0< x_{1}< x_{2}< T<+\infty \),
$$\begin{aligned}& \biggl\vert \frac{K_{P}(I-Q)N\varphi (x_{2})}{e^{x_{2}}}- \frac{K_{P}(I-Q)N\varphi (x_{1})}{e^{x_{1}}} \biggr\vert \\& \quad = \biggl\vert -\frac{{ax_{2}}^{n-2}+b}{e^{x_{2}}} \frac{1}{a^{2}+b^{2}}\Gamma _{2} \biggl( \int _{0}^{+\infty }k(x,y) (I-Q)N \varphi (y)\,dy \biggr)\\& \qquad {}+ \frac{\int _{0}^{+\infty }k(x_{2},y)(I-Q)N\varphi (y)\,dy}{e^{x_{2}}} \\& \qquad {}+\frac{{ax_{1}}^{n-2}+b}{e^{x_{1}}}\frac{1}{a^{2}+b^{2}}\Gamma _{2} \biggl( \int _{0}^{+\infty }k(x,y) (I-Q)N\varphi (y)\,dy \biggr)\\& \qquad {}- \frac{\int _{0}^{+\infty }k(x_{1},y)(I-Q)N\varphi (y)\,dy}{e^{x_{1}}} \biggr\vert \\& \quad \leq \biggl\vert \frac{A_{r}}{a^{2}+b^{2}} \biggl( \frac{a{x_{1}}^{n-2}+b}{e^{x_{1}}}- \frac{a{x_{2}}^{n-2}+b}{e^{x_{2}}} \biggr) \biggr\vert \\& \qquad {} + \int _{0}^{+\infty } \biggl\vert \frac{k(x_{2},y)}{e^{x_{2}}} - \frac{k(x_{1},y)}{e^{x_{1}}} \biggr\vert \bigl\vert (I-Q)N\varphi (y)\,dy \bigr\vert \,dy, \\& \biggl\vert \frac{(K_{P}(I-Q)N\varphi )'(x_{2})}{e^{x_{2}}}- \frac{(K_{P}(I-Q)N\varphi )'(x_{1})}{e^{x_{1}}} \biggr\vert \\& \quad = \biggl\vert -\frac{{a(n-2)x_{2}}^{n-3}}{e^{x_{2}}} \frac{\Gamma _{2} (\int _{0}^{+\infty }k(x,y)(I-Q)N\varphi (y)\,dy )}{a^{2}+b^{2}}\\& \qquad {}+ \frac{(\int _{0}^{+\infty } k(x,y)(I-Q)N\varphi (y)\,dy)'|_{x=x_{2}}}{e^{x_{2}}} \\& \qquad {}+\frac{{a(n-2)x_{1}}^{n-3}}{e^{x_{1}}} \frac{\Gamma _{2} (\int _{0}^{+\infty }k(x,y)(I-Q)N\varphi (y)\,dy )}{a^{2}+b^{2}}\\& \qquad {}- \frac{(\int _{0}^{+\infty }k(x,y)(I-Q)N\varphi (y)\,dy)'|_{x=x_{1}}}{e^{x_{1}}} \biggr\vert \\& \quad \leq \biggl\vert \frac{a(n-2)A_{r}}{a^{2}+b^{2}} \biggl( \frac{{x_{1}}^{n-3}}{e^{x_{1}}}- \frac{{x_{2}}^{n-3}}{e^{x_{2}}} \biggr) \biggr\vert + \biggl\vert \frac{(\int _{0}^{+\infty }k(x,y)(I-Q)N\varphi (y)\,dy)'|_{x=x_{2}}}{e^{x_{2}}} \\& \qquad {}- \frac{(\int _{0}^{+\infty }k(x,y)(I-Q)N\varphi (y)\,dy)'|_{x=x_{1}}}{e^{x_{1}}} \biggr\vert , \end{aligned}$$
where
$$\biggl\Vert \Gamma _{2} \biggl(\int _{0}^{+\infty }k(x,y)(I-Q)N \varphi (y)\,dy \biggr) \biggr\Vert \leq M_{r}\beta _{2} \biggl(1+(\beta _{1}+ \alpha \beta _{2})\int _{0}^{+\infty }|g(x)|\,dx \biggr):=A_{r}. $$
Define \(F_{1}(x)=\int _{0}^{+\infty }k(x,y)(I-Q)N\varphi (y)\,dy\), \(x\in (0,T)\), \(T< \infty \). Obviously,
$$\begin{aligned}& \biggl\vert \frac{F_{1}(x_{2})}{e^{x_{2}}}- \frac{F_{1}(x_{1})}{e^{x_{1}}} \biggr\vert \\& \quad = \biggl\vert \int _{0}^{+\infty } \biggl( \frac{k(x_{2},y)}{e^{x_{2}}}- \frac{k(x_{1},y)}{e^{x_{1}}} \biggr) (I-Q)N \varphi (y)\,dy \biggr\vert \\& \quad \leq \biggl\vert \int _{0}^{x_{1}} \biggl( \frac{x_{1}^{n-1}-(x_{1}-y)^{n-1}}{e^{x_{1}}(n-1)!} - \frac{x_{2}^{n-1}-(x_{2}-y)^{n-1}}{e^{x_{2}}(n-1)!} \biggr) (I-Q)N \varphi (y)\,dy \biggr\vert \\& \qquad {}+ \biggl\vert \int _{x_{1}}^{x_{2}} \biggl( \frac{x_{2}^{n-1}-(x_{2}-y)^{n-1}}{e^{x_{2}}(n-1)!} - \frac{x_{1}^{n-1}}{e^{x_{1}}(n-1)!} \biggr) (I-Q)N\varphi (y)\,dy \biggr\vert \\& \qquad {}+ \biggl\vert \int _{T}^{+\infty } \biggl( \frac{x_{1}^{n-1}}{e^{x_{1}}(n-1)!} - \frac{x_{2}^{n-1}}{e^{x_{2}}(n-1)!} \biggr) (I-Q)N\varphi (y)\,dy \biggr\vert , \\& \biggl\vert \frac{F_{1}'(x_{2})}{e^{x_{2}}}- \frac{F_{1}'(x_{1})}{e^{x_{1}}} \biggr\vert \\& \quad \leq \biggl\vert \int _{0}^{x_{1}} \biggl( \frac{x_{1}^{n-2}-(x_{1}-y)^{n-2}}{e^{x_{1}}(n-2)!} - \frac{x_{2}^{n-2}-(x_{2}-y)^{n-2}}{e^{x_{2}}(n-2)!} \biggr) (I-Q)N \varphi (y)\,dy \biggr\vert \\& \qquad {}+ \biggl\vert \int _{x_{1}}^{x_{2}} \biggl( \frac{x_{2}^{n-2}-(x_{2}-y)^{n-2}}{e^{x_{2}}(n-2)!}- \frac{x_{1}^{n-2}}{e^{x_{1}}(n-2)!} \biggr) (I-Q)N\varphi (y)\,dy \biggr\vert \\& \qquad {}+ \biggl\vert \int _{x_{2}}^{T} \biggl( \frac{x_{1}^{n-2}}{e^{x_{1}}(n-2)!}- \frac{x_{2}^{n-2}}{e^{x_{2}}(n-2)!} \biggr) (I-Q)N\varphi (y)\,dy \biggr\vert \\& \qquad {}+ \biggl\vert \int _{T}^{+\infty } \biggl( \frac{x_{1}^{n-2}}{e^{x_{1}}(n-2)!}- \frac{x_{2}^{n-2}}{e^{x_{2}}(n-2)!} \biggr) (I-Q)N\varphi (y)\,dy \biggr\vert . \end{aligned}$$
By the uniform continuity of \(\{ \frac{{x}^{n-j}}{e^{x}},j=1,2,3,\ldots ,n \} \) in \([0,T]\) and the absolute continuity of the integral, we see that \(\{\frac{(K_{P}(I-Q)N\varphi )^{(i)}}{e^{x}}:\varphi \in \overline{\Omega },i=0,1,\ldots ,n-1 \} \) is equicontinuous on \([0,T]\), \(\forall T<\infty \). Thirdly, for \(\varepsilon >0\), there exists a constant \(T_{1}>0\) such that if \(x>T_{1}\), then
$$ \biggl\vert \frac{x^{n-2}}{e^{x}} \biggr\vert < \frac{a^{2}+b^{2}}{6 \vert a \vert A_{r}}\varepsilon , \qquad \frac{1}{e^{x}}< \frac{a^{2}+b^{2}}{6 \vert b \vert A_{r}}\varepsilon . $$
For \(\varepsilon >0\), there exists a constant \(T_{2}>0\) such that if \(x>T_{2}\), then
$$ \biggl\vert \frac{k(x,y)}{e^{x}} \biggr\vert < \frac{x^{n-1}}{(n-1)!e^{x}}< \frac{\beta _{2}}{6A_{r}}\varepsilon . $$
Take \(\widetilde{T}=\max \{T_{1},T_{2}\}\), for \(\varphi \in \overline{\Omega }\), \(\widetilde{T}\leq x_{1}< x_{2}\), we have
$$\begin{aligned}& \biggl\vert \frac{(K_{P}(I-Q)N\varphi (x_{2})}{e^{x_{2}}}- \frac{(K_{P}(I-Q)N\varphi (x_{1})}{e^{x_{1}}} \biggr\vert \\& \quad = \biggl\vert -\frac{{ax_{2}}^{n-2}+b}{e^{x_{2}}} \frac{1}{a^{2}+b^{2}}\Gamma _{2} \biggl( \int _{0}^{+\infty }k(x_{2},y) (I-Q)N \varphi (y)\,dy \biggr)\\& \qquad {}+ \frac{\int _{0}^{+\infty }k(x_{2},y)(I-Q)N\varphi (y)\,dy}{e^{x_{2}}} \\& \qquad {}+\frac{{ax_{1}}^{n-2}+b}{e^{x_{1}}}\frac{1}{a^{2}+b^{2}}\Gamma _{2} \biggl( \int _{0}^{+\infty }k(x_{1},y) (I-Q)N\varphi (y) \,dy \biggr) \\& \qquad {}- \frac{\int _{0}^{+\infty }k(x_{1},y)(I-Q)N\varphi (y)\,dy}{e^{x_{1}}} \biggr\vert \\& \quad \leq \frac{A_{r}}{a^{2}+b^{2}} \biggl\vert \biggl( \frac{a{x_{1}}^{n-2}}{e^{x_{1}}}- \frac{a{x_{2}}^{n-2}}{e^{x_{2}}} \biggr)+ \biggl(\frac{b}{e^{x_{1}}}-\frac{b}{e^{x_{2}}} \biggr) \biggr\vert \\& \qquad {} + \int _{0}^{+\infty } \biggl\vert \frac{k(x_{2},y)}{e^{x_{2}}}- \frac{k(x_{1},y)}{e^{x_{1}}} \biggr\vert \bigl\vert (I-Q)N\varphi (y) \bigr\vert \,dy \\& \quad \leq \frac{A_{r}}{a^{2}+b^{2}} \biggl(2 \vert a \vert \frac{a^{2}+b^{2}}{6 \vert a \vert A_{r}} \varepsilon +2 \vert b \vert \frac{a^{2}+b^{2}}{6 \vert b \vert A_{r}}\varepsilon \biggr) +2 \frac{A_{r}}{\beta _{2}}\frac{\beta _{2}}{6A_{r}}\varepsilon < \frac{\varepsilon }{3}+ \frac{\varepsilon }{3}+\frac{\varepsilon }{3}= \varepsilon . \end{aligned}$$
For \(\varepsilon >0\), there exists a constant \(T_{3}>0\) such that if \(x>T_{3}\), then
$$ \biggl\vert \frac{x^{n-3}}{e^{x}} \biggr\vert < \frac{a^{2}+b^{2}}{4 \vert a \vert (n-2)A_{r}}\varepsilon . $$
For \(\varepsilon >0\), there exists a constant \(T_{4}>0\) such that if \(x>T_{4}\), then
$$ \biggl\vert \frac{\frac{\partial k(x,y)}{\partial x}}{e^{x}} \biggr\vert < \frac{x^{n-2}}{(n-2)!e^{x}}< \frac{\beta _{2}}{4A_{r}}\varepsilon . $$
Take \(T^{*}=\max \{T_{3},T_{4}\}\), for \(\varphi \in \overline{\Omega }\), \(T^{*}\leq x_{1}< x_{2}\), we have
$$\begin{aligned}& \biggl\vert \frac{(K_{P}(I-Q)N\varphi )'(x_{2})}{e^{x_{2}}}- \frac{(K_{P}(I-Q)N\varphi )'(x_{1})}{e^{x_{1}}} \biggr\vert \\& \quad = \biggl\vert -\frac{{a(n-2)x_{2}}^{n-3}}{e^{x_{2}}} \frac{\Gamma _{2} (\int _{0}^{+\infty }k(x,y)(I-Q)N\varphi (y)\,dy )}{a^{2}+b^{2}}\\& \qquad {}+ \frac{\int _{0}^{+\infty }\frac{\partial k(x,y)}{\partial x}|_{x=x_{2}}(I-Q)N\varphi (y)\,dy}{e^{x_{2}}} \\& \qquad {}+\frac{{a(n-2)x_{1}}^{n-3}}{e^{x_{1}}} \frac{\Gamma _{2} (\int _{0}^{+\infty }k(x,y)(I-Q)N\varphi (y)\,dy )}{a^{2}+b^{2}}\\& \qquad {}- \frac{\int _{0}^{+\infty }\frac{\partial k(x,y)}{\partial x}|_{x=x_{1}}(I-Q)N\varphi (y)\,dy}{e^{x_{1}}} \biggr\vert \\& \quad \leq \biggl\vert \frac{a(n-2)A_{r}}{a^{2}+b^{2}} \biggl( \frac{{x_{1}}^{n-3}}{e^{x_{1}}}- \frac{{x_{2}}^{n-3}}{e^{x_{2}}} \biggr) \biggr\vert \\& \qquad {} + \int _{0}^{+\infty } \biggl\vert \frac{\frac{\partial k(x,y)}{\partial x}|_{x=x_{2}}}{e^{x_{2}}} - \frac{\frac{\partial k(x,y)}{\partial x}|_{x=x_{1}}}{e^{x_{1}}} \biggr\vert \bigl\vert (I-Q)N\varphi (y) \bigr\vert \,dy \\& \quad < \frac{2 \vert a \vert (n-2)A_{r}}{a^{2}+b^{2}} \frac{a^{2}+b^{2}}{4 \vert a \vert (n-2)A_{r}}\varepsilon +2 \frac{A_{r}}{\beta _{2}} \frac{\beta _{2}}{4A_{r}}\varepsilon = \frac{\varepsilon }{2}+\frac{\varepsilon }{2}=\varepsilon . \end{aligned}$$
For the above two arguments, take \(T=\max \{\widetilde{T},T^{*}\}\), we hold that \(\{(K_{P}(I-Q)N\varphi )^{(i)}:\varphi \in \overline{\Omega },i=0,1 \}\) is equiconvergent for any \(x_{1},x_{2}\geq T\). Similarly, \(\{(K_{P}(I-Q)N\varphi )^{(i)}:\varphi \in \overline{\Omega },i=2,3,\ldots ,n-1 \}\) is also equiconvergent at infinity. Therefore, by the Ascoli–Arzela theorem, \(K_{P}(I-Q)N:\overline{\Omega }\rightarrow X\) is compact and \(QN(\overline{\Omega })\) is bounded, i.e., N is L-compact. □
Theorem 3.5
Assume that \((H_{1})\)–\((H_{3})\) with \(b\neq 0\) and the following conditions hold:
- \((H_{4})\):
-
There exist functions \(r, q_{i}\in L[0,+\infty )\) with \((\|K_{P}\|+\frac{|a|+|b|}{|b|(n-2)!} ) \sum_{i=1}^{n} \|q_{i}\|_{1}<1\) such that
$$\bigl\vert f(x,y_{1},y_{2},\ldots ,y_{n}) \bigr\vert \leq \sum_{i=1}^{n}q_{i}(x)e^{-x} \vert y_{i} \vert +r(x), $$
where \(x\in [0,+\infty )\), \(y_{i}\in \mathbb{R}\).
- \((H_{5})\):
-
There exists a constant \(M>0\) such that if \(|\varphi ^{(n-2)}(x)|>M\) for all \(x\in [0,+\infty )\), then
$$(\Gamma _{1}-\alpha \Gamma _{2}) \biggl( \int _{0}^{+\infty }k(x,y) f\bigl(y,\varphi (y), \varphi '(y),\varphi ''(y),\ldots ,\varphi ^{(n-1)}(y)\bigr)\,dy \biggr) \neq 0. $$
- \((H_{6})\):
-
There is a constant \(a_{0}>0\) such that if \(|c|>a_{0}\) either
$$\begin{aligned} c\Gamma \biggl( \int _{0}^{+\infty }k(x,y)N\bigl(c\bigl(a-by^{n-2} \bigr)\bigr)\,dy \biggr)< 0 \end{aligned}$$
(3.5)
or
$$\begin{aligned} c\Gamma \biggl( \int _{0}^{+\infty }k(x,y)N\bigl(c\bigl(a-by^{n-2} \bigr)\bigr)\,dy \biggr)>0. \end{aligned}$$
(3.6)
Then boundary value problem (1.1) has at least one solution.
The following results will play a very important role in our subsequent analysis.
Lemma 3.6
The set
$$ \Omega _{1}=\bigl\{ \varphi \in \operatorname{dom}L\setminus \operatorname{Ker}L:L \varphi =\lambda N\varphi ,\lambda \in [0,1]\bigr\} $$
is bounded if \((H_{1})\)–\((H_{5})\) are satisfied.
Proof
Take \(\varphi \in \Omega _{1}\), then \(N\varphi \in \operatorname{Im}L\), thus we have
$$\begin{aligned} (\Gamma _{1}-\alpha \Gamma _{2}) \biggl( \int _{0}^{+ \infty }k(x,y)f\bigl(y,\varphi (y),\varphi '(y),\varphi ''(y),\ldots , \varphi ^{(n-1)}(y)\bigr)\,dy \biggr)=0. \end{aligned}$$
(3.7)
This, together with \((H_{5})\), means that there exists \(x_{0}\in [0,\infty )\) such that \(|\varphi ^{(n-2)}(x_{0})|\leq M\).
Since \(\varphi ^{(n-1)}(x)=-\int _{x}^{+\infty }\varphi ^{(n)}(y)\,dy+\varphi ^{(n-1)}(+ \infty )\) and \(\varphi ^{(n-2)}(x)=\int _{x_{0}}^{x}\varphi ^{(n-1)}(y)\,dy+\varphi ^{(n-2)}(x_{0})\), we get \(|\varphi ^{(n-1)}(x)|\leq \|N\varphi \|_{1}\) and
$$\begin{aligned} \frac{ \vert \varphi ^{(n-2)}(x) \vert }{e^{x}}\leq \Vert N\varphi \Vert _{1}+M. \end{aligned}$$
(3.8)
Write \(\varphi (x)=\varphi _{1}(x)+\varphi _{2}(x)\), where \(\varphi _{1}(x)=(I-P)\varphi (x)\in \operatorname{dom}L\cap \operatorname{Ker}P\) and \(\varphi _{2}(x)=P\varphi (x)\in \operatorname{Im} P\). Then since \(\varphi _{1}(x)=(I-P)\varphi (x)\in \operatorname{dom}L\cap \operatorname{Ker}P\), we have
$$ \varphi _{1}(x)=K_{P}L_{P}\varphi _{1}(x)=K_{P}L_{P}(I-P)\varphi (x)=K_{P}L_{P} \varphi (x)=\lambda K_{P}N\varphi (x). $$
As is the proof of Lemma 3.3,
$$\begin{aligned} \Vert \varphi _{1} \Vert \leq \Vert K_{P} \Vert \Vert N\varphi \Vert _{1}. \end{aligned}$$
(3.9)
Now, \(\varphi _{2}(x)=P\varphi (x)=c(\varphi )(a-bx^{n-2})\), where
$$ c(\varphi )=\frac{1}{a^{2}+b^{2}} \biggl(a\varphi (0)- \frac{b\varphi ^{(n-2)}(0)}{(n-2)!} \biggr) $$
is introduced for the sake of brevity. Hence
$$ \frac{ \vert \varphi ^{(n-2)}(x) \vert }{e^{x}}= \frac{ \vert -b(n-2)!c(\varphi ) \vert }{e^{x}}\leq \Vert N\varphi \Vert _{1}+M $$
by (3.8). That is,
$$ \bigl\vert c(\varphi ) \bigr\vert \leq \frac{ \Vert N\varphi \Vert _{1}+M}{ \vert b \vert (n-2)!}e^{x}. $$
Thus,
$$\begin{aligned} \Vert \varphi _{2} \Vert \leq \bigl\Vert c(\varphi ) \bigr\Vert \cdot \bigl\Vert a-bx^{n-2} \bigr\Vert \leq \frac{ \Vert N\varphi \Vert _{1}+M_{0}}{ \vert b \vert (n-2)!}\bigl( \vert a \vert + \vert b \vert \bigr). \end{aligned}$$
(3.10)
By (3.9) and (3.10), \(\|\varphi \|\leq \|\varphi _{1}\|+\|\varphi _{2}\|\leq C_{1}\|N \varphi \|_{1}+C_{2}\), where
$$ C_{1}= \Vert K_{P} \Vert +\frac{ \vert a \vert + \vert b \vert }{ \vert b \vert (n-2)!},\qquad C_{2}= \frac{M_{0}}{ \vert b \vert (n-2)!}\bigl( \vert a \vert + \vert b \vert \bigr). $$
Finally, it follows from \((H_{4})\) that
$$ \Vert \varphi \Vert \leq \frac{C_{1} \Vert r \Vert _{1}+C_{2}}{1-C_{1}\sum_{i=1}^{n} \Vert q_{i} \Vert _{1}}. $$
Therefore, \(\Omega _{1}\) is bounded. □
Lemma 3.7
The set
$$ \Omega _{2}=\{\varphi :\varphi \in \operatorname{Ker}L, N\varphi \in \operatorname{Im}L\} $$
is bounded if \((H_{1})\)–\((H_{3})\), \((H_{6})\) hold.
Proof
Let \(\varphi \in \Omega _{2}\), then \(\varphi (x)\equiv c(a-bx^{n-2})\), and \(N\varphi \in \operatorname{Im}L\). So, we can get
$$\begin{aligned} (\Gamma _{1}-\alpha \Gamma _{2})\Gamma \biggl( \int _{0}^{+\infty }k(x,y)f\bigl(y,c\bigl(a-by^{n-2} \bigr), \ldots ,-bc(n-2)!,0\bigr)\,dy \biggr)=0. \end{aligned}$$
According to \((H_{6})\), we have \(|c|\leq a_{0}\), i.e., \(\|\varphi \| \leq a_{0}(|a|+|b|(n-2)!)\). The proof of Lemma 3.7 is completed. □
Lemma 3.8
The set
$$ \Omega _{3}=\bigl\{ \varphi \in \operatorname{Ker} L: \lambda J\varphi +\alpha (1-\lambda )QN \varphi =0, \lambda \in [0,1]\bigr\} $$
is bounded if conditions \((H_{1})\)–\((H_{3})\), \((H_{6})\) are satisfied, where \(J: \operatorname{Ker}L\rightarrow \operatorname{Im}Q\) is a linear isomorphism given by \(J(c(a-bx^{n-2}))=cg(x)\), \(c\in \mathbb{R}\) and
$$ \alpha = \textstyle\begin{cases} -1,&\textit{if }(3.5)\textit{ holds}; \\ 1,&\textit{if }(3.6)\textit{ holds}. \end{cases} $$
(3.11)
Proof
Suppose that \(\varphi \in \Omega _{3}\), we have \(\varphi (x)=c(a-bx^{n-2})\) and
$$ \lambda c=-\alpha (1-\lambda ) (\Gamma _{1}-\alpha \Gamma _{2}) \biggl( \int _{0}^{+\infty }k(x,y)f\bigl(y,c\bigl(a-by^{n-2} \bigr),\ldots ,-bc(n-2)!,0\bigr)\,dy \biggr). $$
If \(\lambda =0\), by \((H_{6})\), we have \(|c|\leq a_{0}\). If \(\lambda =1\), then \(c=0\). If \(\lambda \in (0,1)\), we suppose \(|c|> a_{0}\), then
$$\begin{aligned} \lambda c^{2}=-\alpha (1-\lambda )c(\Gamma _{1}-\alpha \Gamma _{2}) \biggl( \int _{0}^{+\infty }k(x,y)f\bigl(y,c\bigl(a-by^{n-2} \bigr),\ldots ,-bc(n-2)!,0\bigr)\,dy \biggr)< 0, \end{aligned}$$
which contradicts \(\lambda c^{2}>0\). So, Lemma 3.8 holds. □
Theorem 3.5 can be proved next.
Proof of Theorem 3.5
Let Ω be a bounded open subset of X such that \(\{0\}\cup \bigcup_{j=1}^{3} \overline{\Omega }_{j} \subset \Omega \). From Lemma 3.4, we know that N is L-compact on Ω̅. In view of Lemmas 3.6 and 3.7, we can get:
-
(i)
\(L\varphi \neq \lambda N\varphi \) for every \((\varphi ,\lambda )\in [(\operatorname{dom} L\setminus \operatorname{Ker} L) \cap \partial \Omega ]\times (0,1)\);
-
(ii)
\(N\varphi \notin \operatorname{Im} L\) for every \(\varphi \in \operatorname{Ker} L \cap \partial \Omega \).
At last, we will prove that (iii) of Theorem 2.2 is satisfied.
Let \(H (\varphi ,\lambda )=\lambda J\varphi +\alpha (1-\lambda )QN \varphi \). Noting \(\Omega _{3}\subset \Omega \), we know \(H (\varphi ,\lambda )\neq 0\) for every \(\varphi \in \partial \Omega \cap \operatorname{Ker} L\). Thus, by the homotopic property of degree, we have
$$ \deg (QN|_{\operatorname{Ker}L},\Omega \cap \operatorname{Ker}L,0)=\deg ( \alpha J, \Omega \cap \operatorname{Ker}L,0 )\neq 0. $$
By Theorem 2.2, the functional boundary value problem (1.1) has at least one solution in X. The proof of Theorem 3.5 is completed. □
Example
Let us consider the following boundary value problem at resonance:
$$ \textstyle\begin{cases} \varphi ^{(3)}(x)=f(x,\varphi (x),\varphi '(x),\varphi ''(x)), \\ \varphi ''(+\infty )=0, \\ \Gamma _{1}(\varphi (x))=\varphi (0)+2\varphi '(1)=0, \\ \Gamma _{2}(\varphi (x))=2\varphi '(0)+2\varphi (1)=0, \end{cases} $$
where
$$ \begin{aligned} &f\bigl(x,\varphi (x),\varphi '(x),\varphi ''(x) \bigr)\\ &\quad = \textstyle\begin{cases} e^{-17x}\sin \varphi (x)+e^{-17x}\varphi '(x)+e^{-17x}\sin \varphi ''(x)+e^{-x};\quad x \in [0,A], \\ \operatorname{sgn}\{\varphi '(x)\}[e^{-17x}\operatorname{sgn}\{\varphi (x)\}\sin \varphi (x)+e^{-17x}\operatorname{sgn} \{\varphi ''(x)\} \\ \quad{}\times \sin \varphi ''(x)+e^{-x}], \quad x \in [A,+\infty ). \end{cases}\displaystyle \end{aligned} $$
Corresponding to problem (1.1), we have that \(n=3\), \(\Gamma _{1}(1)=1\), \(\Gamma _{1}(x)=2\), \(\Gamma _{2}(1)=2\), \(\Gamma _{2}(x)=4\). By simple calculation, we obtain \(\alpha =\frac{1}{2}\), \(a=4\), \(b=2\), and \(\operatorname{Ker} L=\{c(2-x),c\in \mathbb{R}\}\). It is not difficult to verify that \((\Gamma _{1}-\alpha \Gamma _{2}) (x^{3}e^{-x} )=\frac{3}{e}\neq 0\), \((\Gamma _{1}-\alpha \Gamma _{2}) (\int _{0}^{+\infty }k(x,y)g(y)\,dy )=1\), where \(g(y)=\frac{2e^{1-y}}{6-7e}\). Moreover,
$$ \bigl\vert \Gamma _{2}\bigl(\varphi (x)\bigr) \bigr\vert = \bigl\vert 2\varphi '(0)+2\varphi (1) \bigr\vert \leq 4 \Vert \varphi \Vert , $$
that is, \(\beta _{2}=4\), and \(\|K_{p}\|=(1+\frac{|a|+|b|}{a^{2}+b^{2}}\beta _{2})= \frac{11}{5}\).
Take \(h_{r}(x)=2e^{-17x}+e^{-x}+re^{-16x}\), \(q_{i}(x)=e^{-16x}\), \(i=1, 2,3\), \(r(x)=e^{-x}\); moreover, \((\|K_{P}\|+\frac{|a|+|b|}{|b|})\sum_{i=1}^{n}\|q_{i}\|_{1}= \frac{39}{40}<1\). Obviously, \((H_{1})\) and \((H_{4})\) are satisfied.
When \(A>0\) is fixed, we take \(M=3e^{17A}>0\). If \(\varphi '(x)>M\), then \(f(x,\varphi (x),\varphi '(x),\varphi ''(x))>Me^{-17A}-2>0\), \(x\in [0,A]\), \(f(x,\varphi (x),\varphi '(x),\varphi ''(x))>0\), \(x\in [A,+\infty )\), i.e., \(f(x,\varphi (x),\varphi '(x),\varphi ''(x))>0\), \(x\in [0,+\infty )\).
If \(\varphi '(x)<-M\), then \(f(x,\varphi (x),\varphi '(x),\varphi ''(x))<3-Me^{-A}<0\), \(x\in [0,A]\), \(f(x,\varphi (x),\varphi '(x), \varphi ''(x))<0\), \(x\in [A,+\infty )\), i.e., \(f(x,\varphi (x),\varphi '(x),\varphi ''(x))<0\), \(x\in [0,+\infty )\).
For convenience, we denote \(F(x)=N\varphi (x)\). Hence
$$\begin{aligned} &\biggl(\Gamma _{1}-\frac{1}{2}\Gamma _{2}\biggr) \biggl( \int _{0}^{\infty }k(x,y)F(y)\,dy\biggr)= \int _{0}^{1}\frac{-(y^{2}+2y)}{2}F(y)+ \int _{1}^{\infty }-\frac{1}{2}F(y)\,dy \neq 0 \end{aligned}$$
provided \(\varphi (x)\in \operatorname{dom} L\setminus \operatorname{Ker} L\) satisfies \(|\varphi '(x)|>M\). This means that condition \((H_{5})\) is satisfied.
Finally, take \(\varphi \in \operatorname{Ker} L\) and \(\varphi (x)=c(2-x)\); similarly, one can choose \(a_{0}=2e^{17A}>0\) such that
$$ \begin{gathered} c\biggl(\Gamma _{1}-\frac{1}{2}\Gamma _{2}\biggr) \biggl( \int _{0}^{\infty }k(x,y)N\bigl(c(2-y)\bigr)\,dy\biggr)\\ \quad =c \biggl[ \int _{0}^{1}\frac{-(y^{2}+2y)}{2}N\bigl(c(2-y)\bigr) \,dy+ \int _{1}^{\infty }- \frac{1}{2}N\bigl(c(2-y) \bigr)\,dy\biggr]>0 \end{gathered} $$
provided \(|c|>a_{0}\).
Since if \(A\geq 1\), \(cN(c(2-y))=ce^{-17y}\sin (c(2-y))+e^{-17y}(-c^{2})+ce^{-y}\leq 2|c|-c^{2}e^{-17}=|c|(2-|c|e^{-17})<0\), \(y\in [0,1]\), \(cN(c(2-y))\leq |c|e^{-17}-c^{2}e^{-17}+|c|e^{-1}<0\), \(y\in [1,A]\), and \(cN(c(2-y))<0\), \(y\in [A,+\infty )\).
Similarly, if \(A<1\), \(cN(c(2-y))<|c|(2-|c|e^{-17})<0\), \(y\in [0,A]\), and \(cN(c(2-y))<0\), \(y\in [A,+\infty )\).
So, condition \((H_{6})\) holds. It follows from Theorem 3.5 that there must be at least one solution in X.