1 Introduction
Let \(A=(a_{ij})_{n\times n}\) be a square matrix of order n. \(\alpha:[0,1]\rightarrow \mathbb{R}\) is a bounded variation function, and \(\int _{0}^{1}u(t)\,d\alpha (t)=(\int _{0}^{1}u_{1}(t)\,d\alpha (t),\int _{0}^{1}u_{2}(t)\,d\alpha (t),\ldots , \int _{0}^{1}u_{n}(t)\,d\alpha (t))^{T}\) where \(\int _{0}^{1}u_{i}(t)\,d \alpha (t)\) denotes the Riemann–Stieltjes integrals of \(u_{i}\) with respect to α and \(u^{T}\) denotes the vector transpose of the row vector u. Take \(h=\int _{0}^{1}t\,d\alpha (t)\) and \(B=hA\).
In this paper, we will study the existence of solutions for the following integral boundary value problem at resonance in \(\mathbb{R} ^{n}\):
under the following assumptions: If the condition (\(H_{1}\)) is considered, the associated linear problem \(-u''(t)=0\), \(u(0)=0\), \(u(1)=A\int _{0}^{1}u(t)\,d\alpha (t)\) has a nontrivial solution \(u(t)=\psi t\) with \(\psi \in \operatorname{Ker}(I-B)\). This means that this problem is a resonant integral boundary value problem (IBVP).
$$ \textstyle\begin{cases} -u''(t)=f(t,u(t),u'(t)),\quad t\in (0,1), \\ u(0)=0,\qquad u(1)=A\int _{0}^{1}u(t)\,d\alpha (t), \end{cases} $$
(1.1)
(H1)
B is a diagonalization matrix, and \(\det (I-B)=0\);
(H2)
\(\int _{0}^{1}t(1-t)\,d\alpha (t)\neq0\);
(H3)
\(f:[0,1]\times \mathbb{R}^{2n}\rightarrow \mathbb{R}^{n}\) satisfies the Carathéodory conditions.
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Integral boundary value problems of this form arise in different areas of applied mathematics and physics such as heat conduction, thermoelasticity, underground water flow, and plasma physics. Moreover, integral boundary value problems constitute a very important class of problems based on the fact that two-point, three-point, multi-point and nonlocal boundary value problems can be treated as special cases of Riemann–Stieltjes integral boundary value problems. As a result, the existence of solutions for such problems has received great attention (see [1, 5‐9, 11, 12, 14, 15, 23, 26‐28]). It is well known that, when \(n = 1\), the existence theory of integral boundary value problems for ordinary differential equations or fractional differential equations has been well studied; we refer the reader to [4, 10, 17, 20, 21, 24, 25, 29‐35, 37] for some recent results at non-resonance and to [2, 3, 16, 18, 22, 23, 27, 36] for results at resonance. When \(n\geq 2\) and A is not a diagonal matrix, IBVP (1.1) becomes a system of ordinary differential equations with coupled boundary conditions. Differential systems with coupled integral boundary conditions can be applied to reaction–diffusion phenomena, interaction problems and Lotka–Volterra models. Recently, there have been many papers addressing the existence of solutions for differential systems of coupled integral boundary value problems; see, for example, [1, 3, 5‐7, 9, 11‐14].
To the best of our knowledge, the solvability of problem (1.1) at resonance has not been considered before. The main purpose of this paper is to establish an existence result for problem (1.1) when \(n\geq 2\). Our main method is based on the coincidence degree theory of Mawhin and the theory of matrix diagonalization in linear algebra.
We end this section by recalling some notations and abstract results from coincidence degree theory.
Let X and Y be two real Banach spaces, \(L: \operatorname{dom}L \subset X \rightarrow Y\) be a linear Fredholm operator of index zero, and \(P: X\rightarrow X\) and \(Q: Y\rightarrow Y\) be two continuous projectors such that
It follows from the above equalities that the reduced operator
is invertible. We denote its inverse by \(K_{P}\) (the generalized inverse operator of L). If Ω is an open bounded subset of X such that \(\operatorname{dom}L\cap \varOmega \neq\emptyset \), the mapping \(N: X\rightarrow Y\) will be called L-compact on Ω̅ if \(QN(\overline{ \varOmega })\) is bounded and \(K_{P}(I-Q)N: \overline{\varOmega }\rightarrow X\) is compact.
$$ \operatorname{Im} P=\operatorname{ker} L,\qquad \operatorname{ker} Q=\operatorname{Im} L,\qquad X= \operatorname{ker} L\oplus \operatorname{ker} P,\qquad Y=\operatorname{Im} L\oplus \operatorname{Im} Q. $$
$$ L|_{\operatorname{dom}L\cap \operatorname{Ker}P}: \operatorname{dom}L\cap \operatorname{ker} P\rightarrow \operatorname{Im} L $$
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We make use of the following result from Mawhin [19].
Theorem 1.1
([19] (Mawhin continuation theorem))
Let
\(L: \operatorname{dom}L \subset X \rightarrow Y\)
be a Fredholm operator of index zero and
N
be L-compact on
Ω̅. The equation
\(L\varphi =N\varphi \)
has at least one solution in
\(\operatorname{dom}L \cap \overline{ \varOmega }\)
if the following conditions are satisfied:
(1)
\(L\varphi \neq \lambda N\varphi \)
for every
\((\varphi, \lambda ) \in [(\operatorname{dom}L\backslash \operatorname{ker}L)\cap \partial \varOmega ] \times (0,1)\);
(2)
\(N\varphi \notin \operatorname{Im}L\)
for every
\(\varphi \in \operatorname{ker}L \cap \partial \varOmega \);
(3)
\(\operatorname{deg} (JQN|_{\operatorname{ker}L},\varOmega \cap \operatorname{ker}L,0 ) \neq 0\), where
\(J:\operatorname{Im} Q\rightarrow \operatorname{Ker} L\)
is some isomorphism.
2 Preliminaries
We use the classical spaces \(X=C^{1}([0,1], \mathbb{R}^{n})\) and \(Y=L^{1}([0,1], \mathbb{R}^{n})\). For \(u\in X\), we use the norm \(\|u\|_{X}=\max \{\|u\|_{\infty }, \|u'\|_{\infty }\}\), where \(\|u\|_{\infty }=\max_{t\in [0,1]}\{|u_{1}(t)|, |u_{2}(t)|, \ldots,|u_{n}(t)|\}\), and denote the norm in \(L^{1}([0,1], \mathbb{R}^{n})\) by \(\|u\|_{1}=\max_{1\leq i\leq n}\int _{0} ^{1}|u_{i}(t)|\,dt\). We also use the Sobolev space defined by
We define L to be the linear operator from \(D(L)\subset X\) to Y with
and for \(u\in D(L)\), \(Lu=-u''\). Let \(N: X\rightarrow Y\) be the nonlinear operator defined by
Thus, problem (1.1) can be written as \(Lu=Nu\).
$$ X_{0}=\bigl\{ u\in X: u' \text{ is absolutely continuous on } [0,1] \text{ and } u''\in Y \bigr\} . $$
$$ \operatorname{dom}L= \biggl\{ u\in X_{0}: u(0)=0, u(1)=A \int _{0}^{1}u(t)\,d\alpha (t) \biggr\} $$
$$ (Nu) (t)=f\bigl(t,u(t),u'(t)\bigr),\quad t\in [0,1]. $$
Lemma 2.1
The following results hold:
(1)
\(\operatorname{Ker} L=\{u\in X: u(t)=\psi t, \psi \in \operatorname{Ker}(I-B)\subset \mathbb{R}^{n}\}\).
(2)
\(\operatorname{Im} L=\{v\in Y: \varphi (v)\in \operatorname{Im}(I-B)\}\), where
\(\varphi:Y\rightarrow \mathbb{R} ^{n}\)
is a linear operator defined by
where
$$ \varphi (v)=A \int _{0}^{1} \int _{0}^{1}G(t,s)v(s)\,ds\,d\alpha (t), $$
(2.1)
$$ G(t,s)=\textstyle\begin{cases} t(1-s),& 0\leq t\leq s\leq 1, \\ s(1-t),&0\leq s\leq t\leq 1. \end{cases} $$
(2.2)
Proof
(1) For \(u\in \operatorname{Ker} L\), we obtain \(-u''=0\). Then \(u(t)=\psi t+\psi _{1}\) with \(\psi,\psi _{1} \in \mathbb{R}^{n}\). With consideration of the boundary conditions \(u(0)=0\) and \(u(1)=A\int _{0} ^{1}u(t)\,d\alpha (t)\), we conclude that \(\psi _{1}=0\) and \(\psi =A\int _{0}^{1}\psi t\,d\alpha (t)=A(h\psi )=B\psi \). Thus, \(\psi \in \operatorname{Ker}(I-B)\). Again, if \(u(t)=\psi t\) with \(\psi \in \operatorname{Ker}(I-B)\), then \(u\in \operatorname{Ker} L\).
(2) For \(v\in \operatorname{Im} L\), there exists \(u\in \operatorname{dom}L\) such that \(-u''(t)=v(t)\). Thus,
Using the boundary conditions \(u(0)=0\) and \(u(1)=A\int _{0}^{1}u(t)\,d \alpha (t)\), it follows from (2.3) that
This implies that \(\varphi (v)=(I-B)u(1)\). Thus, \(v\in \{v\in Y: \varphi (v)\in \operatorname{Im}(I-B)\}\). Conversely, if \(v\in Y\) and \(\varphi (v)\in \operatorname{Im}(I-B)\), let
where \(\xi \in \mathbb{R}^{n}\) satisfies
Then \(-u''(t)=v(t)\), \(u(0)=0\), \(u(1)=\xi \) and
Thus, \(u(1)=A\int _{0}^{1}u(t)\,d\alpha (t)\) and \(v\in \operatorname{Im} L\). This completes the proof. □
$$ u(t)=u(0) (1-t)+u(1)t+ \int _{0}^{1}G(t,s)v(s)\,ds. $$
(2.3)
$$\begin{aligned} u(1)& =A \int _{0}^{1}u(t)\,d\alpha (t)=A \int _{0}^{1} \biggl(u(1)t+ \int _{0}^{1}G(t,s)v(s)\,ds \biggr)\,d \alpha (t) \\ & =A\bigl(h u(1)\bigr)+\varphi (v)=B u(1)+\varphi (v). \end{aligned}$$
$$ u(t)=\xi t+ \int _{0}^{1}G(t,s)v(s)\,ds, $$
$$ (I-B)\xi =\varphi (v). $$
(2.4)
$$ A \int _{0}^{1}u(t)\,d\alpha (t)=A\xi \int _{0}^{1} t\,d\alpha (t) +\varphi (v)=B\xi + \varphi (v)=\xi. $$
Recall that a matrix is diagonalizable over the field \(\mathbb{R}\) if and only if its minimal polynomial is a product of distinct linear factors over \(\mathbb{R}\). Thus, it follows from \((H_{1})\) that the minimal polynomial of the matrix B can be written as
where \(1-x\) and \(g(x)\) are two polynomials which are relatively prime. Hence, there exist two polynomials, \(a(t)\) and \(b(t)\), such that
From this, we conclude that \((I-B)a(B)+g(B)b(B)=I\). Thus,
Moreover, by (2.5), we have \((I-B)g(B)=0\), that is,
Consequently, we deduce that
where \(g(1)\neq0\) holds following from the fact that \(1-x\) and \(g(x)\) are two relatively prime polynomials.
$$ \mu _{B}(x)=(1-x)g(x), $$
(2.5)
$$ (1-x)a(x)+g(x)b(x)=1. $$
$$ \begin{aligned} &\operatorname{Im} (I-B) =\operatorname{Ker} g(B), \\ &\operatorname{Ker} (I-B) =\operatorname{Im} g(B), \\ &\mathbb{R}^{n} = \operatorname{Im} (I-B)\oplus \operatorname{Im} g(B). \end{aligned} $$
(2.6)
$$ g(B)B=g(B). $$
(2.7)
$$ g(B)g(B)=g(1)g(B), $$
(2.8)
Example 2.1
When \(B^{m}=B\) with \(2\leq m\leq n\), the minimal polynomial of the matrix, B, is
Thus, we have
When \(B^{m}=I\) with a \(2\leq m\leq n\), the minimal polynomial of the matrix, B, can be explicitly given by
Thus, we obtain
$$ \mu _{B}(x)=x-x^{m}=(1-x) \bigl(x+x^{2}+\cdots +x^{m-1}\bigr)=(1-x)g(x). $$
$$ g(1)=m-1. $$
$$ \mu _{B}(x)=1-x^{m}=(1-x) \bigl(1+x+x^{2}+\cdots +x^{m-1}\bigr)=(1-x)g(x). $$
$$ g(1)=m. $$
Lemma 2.2
L
is a Fredholm operator of index zero.
Proof
We define an operator \(Q: Y\rightarrow Y\) by
where φ is given in (2.1) and \(k=\frac{2h}{g(1)\int _{0}^{1}t(1-t)\,d\alpha (t)}\). Note that if \(w(t)=\psi \) with \(\psi \in \mathbb{R}^{n}\), we have
Hence,
Therefore, the map Q is a continuous linear projector. Moreover, by (2.6) and Lemma 2.1, we have
This means that \(\operatorname{Ker} Q=\operatorname{Im} L\). For \(v\in Y\), \(v-Qv \in \operatorname{Ker} Q=\operatorname{Im} L\). Therefore, \(Y=\operatorname{Im} L+ \operatorname{Im} Q\), and, again, \(\operatorname{Im} L\cap \operatorname{Im} Q=\{0\}\). Hence, \(Y=\operatorname{Im} L\oplus \operatorname{Im} Q\). Combining the previous results with the additional information that ImL is closed, we conclude that L is a Fredholm operator of index zero. □
$$ (Qv) (t)=kg(B)\varphi (v),\quad v\in Y, $$
$$\begin{aligned} (Qw) (t)&= kg(B)A\psi \int _{0}^{1} \int _{0}^{1}G(t,s)\,ds\,d\alpha (t) \\ &= kg(B)A\psi \int _{0}^{1}\frac{t(1-t)}{2}\,d\alpha (t)= \frac{h}{g(1)}g(B)A \psi \\ &= \frac{1}{g(1)}g(B)B\psi =\frac{1}{g(1)}g(B)\psi. \end{aligned}$$
$$\begin{aligned} \bigl(Q^{2}v\bigr) (t)&= \frac{1}{g(1)}g(B) \bigl(kg(B)\varphi (v)\bigr) \\ &= \frac{k}{g(1)}g(B)g(B)\varphi (v) \\ &= kg(B)\varphi (v)=(Qv) (t). \end{aligned}$$
$$ v\in \operatorname{Ker} Q\quad \Leftrightarrow\quad \varphi (v)\in \operatorname{Ker} g(B) \quad\Leftrightarrow\quad \varphi (v)\in \operatorname{Im}(I-B) \quad\Leftrightarrow\quad v\in \operatorname{Im} L. $$
In what follows, we make the following assumption on the matrix B:
(H4)
There exists \(l\in \mathbb{R}\) such that \(l(I-B)(I-B)=(I-B)\).
Example 2.2
When \(B^{2}=B\), we can take \(l=1\) such that
If \(B^{2}=I\), we can take \(l=\frac{1}{2}\) such that
$$ (I-B) (I-B)=I-2B+B^{2}=I-B. $$
$$ l(I-B) (I-B)=\frac{1}{2}\bigl(I-2B+B^{2}\bigr)= \frac{1}{2}(2I-2B)=I-B. $$
Lemma 2.3
Assuming that (H4) holds,
$$ l(I-B)\psi =\psi,\quad \forall \psi \in \operatorname{Im}(I-B). $$
Proof
Let \(\psi \in \operatorname{Im}(I-B)\) so that \(\psi =(I-B)\psi _{1}\) for \(\psi _{1}\in \mathbb{R}^{n}\). Using the condition (H4), we have
□
$$ l(I-B)\psi =l(I-B) (I-B)\psi _{1}=(I-B)\psi _{1}=\psi. $$
Lemma 2.4
If
Ω
is an open bounded subset such that
\(\operatorname{dom}L \cap \varOmega \neq\emptyset \), then
N
is
L-compact on
Ω̅.
Proof
Define the linear operator \(P: X\rightarrow X\) by
Then we have
This shows that P is a continuous projection operator. In the following, we will assert that \(\operatorname{Im} P= \operatorname{Ker} L\). In fact, if \(v\in \operatorname{Im} P\), there is \(u\in X\), such that
Thus, it follows from (2.6) that \(v\in \operatorname{Ker} L\). Conversely, if \(v\in \operatorname{Ker} L\), we have
On account of the second identity in (2.6), there exists \(\psi _{1} \in \mathbb{R}^{n}\), such that \(\psi =g(B)\psi _{1}\). Taking \(u(t)=g(1)\psi _{1}\), we then have
which implies that \(v\in \operatorname{Im} P\). Thus, we conclude that \(\operatorname{Im} P=\operatorname{Ker} L\), and consequently,
Therefore, the generalized inverse \(K_{P}: \operatorname{Im} L\rightarrow \operatorname{dom}L\cap \operatorname{Ker} P\) can be given by
where the constant l is given in (H4). Note that, since \(G(1,s)=0\) for all \(s\in [0,1]\) and from (2.9), we have
Hence, \(K_{P}v\in \operatorname{Ker} P\). For \(v\in \operatorname{Im} L\), we know that
and
Therefore, \((K_{P}v)(1)=A\int _{0}^{1}(K_{P}v)(t)\,d\alpha (t)\), and consequently, \(K_{P}\) is well defined. Furthermore, if \(u\in \operatorname{dom}L \cap \operatorname{Ker} P\), then, using (2.7) and Lemma 2.3, we have
This shows that \(K_{P}=(L|_{\operatorname{dom}L\cap \operatorname{Ker} P})^{-1}\) and that \(LK_{P}v(t)=v(t)\), \(v\in \operatorname{Im} L\). For \(v\in \operatorname{dom}L\), by (2.8) we obtain
Notice that
and
where \(\bigvee_{0}^{t}(\alpha )\) denotes the total variation of α on \([0,t]\) defined by
where the supremum runs over the set of all partitions
Let \(\|\cdot \|_{*}\) be the max-norm of matrices defined by
and \(\|\cdot \|_{\mathbb{R}^{n}}\) be the maximum norm in \(\mathbb{R} ^{n}\). Then we have
Thus,
and
Consequently, we conclude that
where \(M=1+n|l| \|(I-\frac{1}{g(1)}g(B))A\|_{*}\int _{0}^{1}t(1-t)\,d (\bigvee_{0}^{t}(\alpha ) )\). It is easy to see that
and
By using the standard argument, we can show that \(QN((\overline{ \varOmega }))\) is bounded and \(K_{P}(I-Q)N(\overline{\varOmega })\) is compact. Thus, N is L-compact on Ω̅. □
$$ (Pu) (t)=\frac{1}{g(1)}g(B)u(1)t. $$
$$\begin{aligned} \bigl(P^{2}u\bigr) (t)&=\frac{1}{g(1)}g(B) \biggl( \frac{1}{g(1)}g(B)u(1) \biggr)t \\ &= \frac{1}{g^{2}(1)}g(B)g(B)u(1)t=\frac{1}{g(1)}g(B)u(1)t=(Pu) (t). \end{aligned}$$
$$ v(t)=Pu(t)=\frac{1}{g(1)}g(B)u(1)t. $$
$$ v(t)=\psi t,\quad \psi \in \operatorname{Ker}(I-B). $$
$$ Pu(t)=\frac{1}{g(1)}g(B)u(1)t=g(B)\psi _{1}t=\psi t, $$
$$ X=\operatorname{Ker} L\oplus \operatorname{Ker} P. $$
$$ (K_{P}v) (t)= \int _{0}^{1}G(t,s)v(s)\,ds+l\biggl(I- \frac{1}{g(1)}g(B)\biggr)\varphi (v)t, $$
(2.9)
$$ P(K_{P}v) (t)=\frac{l}{g(1)}g(B) \biggl(I-\frac{1}{g(1)}g(B) \biggr)\varphi (v)t=0. $$
$$ (K_{P}v) (1)=l\biggl(I-\frac{1}{g(1)}g(B)\biggr)\varphi (v) $$
$$\begin{aligned} & A \int _{0}^{1}(K_{P}v) (t)\,d\alpha (t) \\ &\quad= A \int _{0}^{1} \biggl( \int _{0}^{1}G(t,s)v(s)\,ds+l\biggl(I- \frac{1}{g(1)}g(B)\biggr) \varphi (v)t \biggr) \,d\alpha (t) \\ &\quad= \varphi (v)+lA\biggl(I-\frac{1}{g(1)}g(B)\biggr)\varphi (v) \int _{0}^{1} t\,d\alpha (t) \\ &\quad= \varphi (v)+lA\biggl(I-\frac{1}{g(1)}g(B)\biggr)\varphi (v)h =\varphi (v)+lB\biggl(I- \frac{1}{g(1)}g(B)\biggr)\varphi (v) \\ &\quad= \varphi (v)+l(B-I+I) \biggl(I-\frac{1}{g(1)}g(B)\biggr)\varphi (v) \\ &\quad= \varphi (v)-l(I-B) \biggl(I-\frac{1}{g(1)}g(B)\biggr)\varphi (v) +l \biggl(I- \frac{1}{g(1)}g(B)\biggr)\varphi (v) \\ &\quad= \varphi (v)-l(I-B)\varphi (v) +l\biggl(I-\frac{1}{g(1)}g(B)\biggr) \varphi (v) =l\biggl(I- \frac{1}{g(1)}g(B)\biggr)\varphi (v). \end{aligned}$$
$$\begin{aligned} &(K_{P}Lu) (t)\\ &\quad = - \int _{0}^{1}G(t,s)u''(s) \,ds+l\biggl(I-\frac{1}{g(1)}g(B)\biggr)\varphi (Lu)t \\ &\quad= u(t)-u(1)t+ l\biggl(I-\frac{1}{g(1)}g(B)\biggr)A \biggl( \int _{0}^{1} \bigl(u(t)-u(1)t\bigr)\,d \alpha (t) \biggr)t \\ &\quad= u(t)-u(1)t+ l\biggl(I-\frac{1}{g(1)}g(B)\biggr) \biggl(A \int _{0}^{1}u(t)\,d\alpha (t)-Ahu(1) \biggr)t \\ &\quad= u(t)-u(1)t+ l\biggl(I-\frac{1}{g(1)}g(B)\biggr) (I-B)u(1)t \\ &\quad= u(t)-u(1)t+ l(I-B)u(1)t \\ &\quad= u(t). \end{aligned}$$
$$ (K_{P}v)'(t)= \int _{t}^{1}(1-s)v(s)\,ds- \int _{0}^{t}sv(s)\,ds+l\biggl(I- \frac{1}{g(1)}g(B)\biggr)A \int _{0}^{1} \int _{0}^{1}G(t,s)v(s)\,ds\,d\alpha (t). $$
$$\begin{aligned} &\max_{t\in [0,1]} \biggl\vert \int _{0}^{1}G(t,s)v(s)\,ds \biggr\vert \leq \max _{t\in [0,1]} \int _{0}^{1}t(1-t) \bigl\vert v(s) \bigr\vert \,ds \leq \frac{ \Vert v \Vert _{1}}{4}, \\ &\max_{t\in [0,1]} \biggl\vert \int _{0}^{1}G'_{t}(t,s)v(s) \,ds \biggr\vert = \max_{t\in [0,1]} \biggl\vert - \int _{0}^{t}sv(s)\,ds + \int _{t}^{1}(1-s)v(s)\,ds \biggr\vert \\ &\phantom{\max_{t\in [0,1]} \biggl\vert \int _{0}^{1}G'_{t}(t,s)v(s) \,ds \biggr\vert }\leq \max_{t\in [0,1]} \biggl(t \int _{0}^{t} \bigl\vert v(s) \bigr\vert \,ds+(1-t) \int _{t} ^{1} \bigl\vert v(s) \bigr\vert \,ds \biggr) \\ &\phantom{\max_{t\in [0,1]} \biggl\vert \int _{0}^{1}G'_{t}(t,s)v(s) \,ds \biggr\vert }\leq \max_{t\in [0,1]} \biggl(t \int _{0}^{1} \bigl\vert v(s) \bigr\vert \,ds+(1-t) \int _{0} ^{1} \bigl\vert v(s) \bigr\vert \,ds \biggr)= \Vert v \Vert _{1}, \end{aligned}$$
$$\begin{aligned} \biggl\vert \int _{0}^{1} \int _{0}^{1}G(t,s)v(s)\,ds\,d\alpha (t) \biggr\vert &\leq \int _{0}^{1} \int _{0}^{1}t(1-t) \bigl\vert v(s) \bigr\vert \,ds\,d \Biggl(\bigvee_{0}^{t}(\alpha ) \Biggr)\\ & = \int _{0}^{1}t(1-t)\,d \Biggl(\bigvee _{0}^{t}(\alpha ) \Biggr) \cdot \Vert v \Vert _{1}, \end{aligned}$$
$$ \bigvee_{0}^{t}(\alpha )=\sup _{\mathcal{P}}\sum_{i=1}^{n} \bigl\vert \alpha (t_{i})-\alpha (t_{i-1}) \bigr\vert , $$
$$ \mathcal{P}=\bigl\{ P=\{t_{0}, \ldots, t_{n}\} | P \text{ is a partition of } [0,t] \bigr\} . $$
$$ \Vert A \Vert _{*}=\max_{1\leq i\leq n, 1\leq j\leq m} \vert a_{ij} \vert ,\quad \text{for } A=(a_{ij})_{n\times m}, $$
$$ \Vert Av \Vert _{\mathbb{R}^{n}}\leq m \Vert A \Vert _{*} \Vert v \Vert _{\mathbb{R}^{m}},\quad \text{for } A=(a_{ij})_{n\times m} \text{ and } v\in \mathbb{R} ^{m}. $$
$$ \Vert K_{P}v \Vert _{\infty }\leq \Biggl( \frac{1}{4}+n \vert l \vert \cdot \biggl\Vert \biggl(I- \frac{1}{g(1)}g(B)\biggr)A \biggr\Vert _{*} \int _{0} ^{1}t(1-t)\,d \Biggl(\bigvee _{0}^{t}(\alpha ) \Biggr) \Biggr) \Vert v \Vert _{1} $$
$$ \bigl\Vert K'_{P}v \bigr\Vert _{\infty }\leq \Biggl(1+n \vert l \vert \cdot \biggl\Vert \biggl(I-\frac{1}{g(1)}g(B) \biggr)A \biggr\Vert _{*} \int _{0}^{1}t(1-t) \,d \Biggl(\bigvee _{0}^{t}(\alpha ) \Biggr) \Biggr) \Vert v \Vert _{1}. $$
$$ \Vert K_{P}v \Vert _{X}\leq M \Vert v \Vert _{1}, $$
$$ (QNu) (t)=kg(B)A \int _{0}^{1} \int _{0}^{1}G(t,s)f\bigl(s,u(s), u'(s) \bigr)\,ds\,d \alpha (t)\in \mathbb{R}^{n} $$
$$\begin{aligned} & K_{P}(I-Q)Nu(t)\\ &\quad=K_{P}Nu(t)-K_{P}QNu(t) \\ &\quad= \int _{0}^{1}G(t,s)Nu(s)\,ds+l\biggl(I- \frac{1}{g(1)}g(B)\biggr)\varphi (Nu)t \\ &\qquad{} + \int _{0}^{1}G(t,s)\,ds\cdot QNu(t)+l\biggl(I- \frac{1}{g(1)}g(B)\biggr)\varphi (QNu)t \\ &\quad= \int _{0}^{1}G(t,s)Nu(s)\,ds+\frac{t(1-t)}{2} QNu(t) \\ &\qquad{} +l\biggl(I-\frac{1}{g(1)}g(B)\biggr) \bigl(\varphi (Nu)+\varphi (QNu) \bigr)t. \end{aligned}$$
The above results (Lemmas 2.1, 2.2, and 2.4) may be concrete for a specific matrix. In the following, we suppose that a diagonalizable matrix B satisfy \(B^{2}=I\) and \(\dim \operatorname{Ker}(I-B)=k\). So, there exist a set of linearly independent vectors \(\{\eta _{1},\eta _{2},\ldots,\eta _{n}\} \) such that
where \(\eta _{i}\)
\((i=1,2,\ldots, k)\) is an eigenvector of B with eigenvalue 1 and \(\eta _{i}\)
\((i=k+1,k+2,\ldots, n)\) is an eigenvector of B with eigenvalue −1. Moreover, we shall suppose that
Take \(C^{-1}=(c_{ij})_{n\times n}\),
$$ BC= C \begin{pmatrix} I_{k} &0 \\ 0&-I_{n-k} \end{pmatrix},\quad C=(\eta _{1},\eta _{2},\ldots, \eta _{n}), \eta _{i}= \begin{pmatrix} \eta _{1i} \\ \eta _{2i} \\ \vdots \\ \eta _{ni} \end{pmatrix}, $$
$$ \begin{pmatrix} \eta _{11} & \eta _{12} &\cdots & \eta _{1k} \\ \eta _{21} & \eta _{22} &\cdots & \eta _{2k} \\ \vdots & \vdots &\cdots & \vdots \\ \eta _{k1} & \eta _{k2} &\cdots & \eta _{kk} \end{pmatrix}=I_{k}, \qquad \begin{pmatrix} \eta _{k+1, k+1} & \eta _{k+1,k+2} &\cdots & \eta _{k+1,n} \\ \eta _{k+2,k+1} & \eta _{k+2,k+2} &\cdots & \eta _{k+2,n} \\ \vdots & \vdots &\cdots & \vdots \\ \eta _{n,k+1} & \eta _{n,k+2} &\cdots & \eta _{n, n} \end{pmatrix}=I_{n-k}. $$
(2.10)
$$ D_{1}= \begin{pmatrix} c_{k+1,1} & c_{k+1,2} &\cdots & c_{k+1,k} \\ c_{k+2,1} & c_{k+2,2} &\cdots & c_{k+2,k} \\ \vdots & \vdots & \cdots & \vdots \\ c_{n,1} & c_{n,2} &\cdots & c_{n,k} \end{pmatrix},\qquad D_{2}= \begin{pmatrix} c_{k+1,k+1} & c_{k+1,k+2} &\cdots & c_{k+1,n} \\ c_{k+2,k+1} & c_{k+2,k+2} &\cdots & c_{k+2,n} \\ \vdots & \vdots & \cdots & \vdots \\ c_{n,k+1} & c_{n,k+2} &\cdots & c_{n,n} \end{pmatrix}. $$
It follows from \(C^{-1}C=I\) and (2.10) that, for \(l,m\in \{1, 2, \ldots,k\}\),
Based on the notation above, (2.4) can be rewritten as
Thus,
$$ \sum_{i=1}^{n}c_{li}\eta _{im}=\delta _{lm}=\textstyle\begin{cases} 1 & \text{if } l=m, \\ 0 & \text{if } l\neq m. \end{cases} $$
(2.11)
$$ C \begin{pmatrix} 0 &0 \\ 0&2I_{n-k} \end{pmatrix}C^{-1}\xi =\frac{1}{h}C \begin{pmatrix} I_{k} &0 \\ 0&-I_{n-k} \end{pmatrix}C^{-1} \int _{0}^{1} \int _{0}^{1}G(t,s)v(s)\,ds\,d\alpha (t). $$
$$ \begin{pmatrix} 0 &0 \\ 0&2I_{n-k} \end{pmatrix}C^{-1}\xi =\frac{1}{h} \begin{pmatrix} I_{k} &0 \\ 0&-I_{n-k} \end{pmatrix}C^{-1} \int _{0}^{1} \int _{0}^{1}G(t,s)v(s)\,ds\,d\alpha (t). $$
Then the above matrix equation reduces to
or
Consequently,
By (2.10), we know \(\operatorname{det}(D_{2})\neq0\). From the second part of (2.12), we infer that
$$ \textstyle\begin{cases} \int _{0}^{1}\int _{0}^{1}G(t,s)\sum_{j=1}^{n}c_{i j}v_{j}(s)\,ds\,d \alpha (t)=0,\quad i=1,2,\ldots, k, \\ -\int _{0}^{1}\int _{0}^{1}G(t,s)\sum_{j=1}^{n}c_{i j}v_{j}(s)\,ds\,d \alpha (t)=2h\sum_{j=1}^{n}c_{i j}\xi _{j},\quad i=k+1,k+2,\ldots, n, \end{cases} $$
$$ \textstyle\begin{cases} \int _{0}^{1}\int _{0}^{1}G(t,s)\sum_{j=1}^{n}c_{i j}v_{j}(s)\,ds\,d \alpha (t)=0,\quad i=1,2,\ldots, k, \\ -\int _{0}^{1}\int _{0}^{1}G(t,s)(D_{1}, D_{2})v(s)\,ds\,d\alpha (t)=2h(D _{1}, D_{2})\xi. \end{cases} $$
(2.12)
$$ \operatorname{Im} L= \Biggl\{ v\in Y: \int _{0}^{1} \int _{0}^{1}G(t,s) \sum _{j=1}^{n}c_{i j}v_{j}(s)\,ds\,d \alpha (t)=0, i=1,2,\ldots , k \Biggr\} . $$
$$ \begin{pmatrix} \xi _{k+1} \\ \xi _{k+2} \\ \vdots \\ \xi _{n} \end{pmatrix} =-\frac{D_{2}^{-1}D_{1}}{2h} \begin{pmatrix} \xi _{1} \\ \xi _{2} \\ \vdots \\ \xi _{k} \end{pmatrix} - \frac{1}{2h} \int _{0}^{1} \int _{0}^{1}G(t,s) \bigl(D_{2}^{-1}D_{1}, I_{n-k}\bigr)v(s)\,ds\,d\alpha (t). $$
(2.13)
Define the linear operator \(P: X\rightarrow X\) by
It follows from the left formula in (2.10) that P is a continuous projection operator with \(\operatorname{Im} P= \operatorname{Ker} L\).
$$ (Pu) (t)=u_{1}(1)\eta _{1}t+u_{2}(1)\eta _{2}t+\cdots +u_{k}(1)\eta _{k}t,\quad u\in X, $$
Based on the assumption (H2), we define linear operators \(Q: Y\rightarrow Y\) by
where \(\gamma = \frac{2}{\int _{0}^{1}t(1-t)\,d\alpha (t)}\). For given \(v\in Y\), we take
Then (2.13) reduces to
$$ Qv=\sum_{i=1}^{k}\gamma \int _{0}^{1} \int _{0}^{1}G(t,s)\sum _{j=1}^{n}c_{i j}v_{j}(s)\,ds\,d \alpha (t)\cdot \eta _{i}, \quad v\in Y, $$
(2.14)
$$ \alpha _{i}= \int _{0}^{1} \int _{0}^{1}G(t,s)\sum _{j=1}^{n}c_{i j}v _{j}(s)\,ds\,d \alpha (t),\quad i=1,2,\ldots,k. $$
$$ (Qv) (t)=\gamma \Biggl(\alpha _{1},\alpha _{2},\ldots,\alpha _{k}, \sum_{m=1}^{k}\alpha _{m}\eta _{k+1,m}, \sum_{m=1}^{k} \alpha _{m}\eta _{k+2,m},\ldots,\sum _{m=1}^{k}\alpha _{m}\eta _{n,m} \Biggr). $$
With the help of (2.11), we have
Consequently,
This implies that Q is a continuous projection operator. Obviously, \(\operatorname{Ker} Q=\operatorname{Im} L\) holds from the linear independence of the vectors \(\{\eta _{1},\eta _{2},\ldots,\eta _{k}\}\).
$$\begin{aligned} & \frac{1}{\gamma }\sum_{j=1}^{n}c_{i j}(Qv)_{j}(s) \\ &\quad= c_{i 1}\alpha _{1}+c_{i 2}\alpha _{2}+ \cdots +c_{i k}\alpha _{k}+c_{i k+1} \sum _{m=1}^{k}\alpha _{m}\eta _{k+1,m} \\ &\qquad{} +c_{i k+2} \sum_{m=1}^{k}\alpha _{m}\eta _{k+2,m}+\cdots +c_{i n} \sum _{m=1}^{k}\alpha _{m}\eta _{n,m} \\ &\quad= \alpha _{1}(c_{i 1}+c_{i k+1}\eta _{k+1,1}+c_{i k+2}\eta _{k+2,1} + \cdots +c_{i n} \eta _{n,1}) \\ &\qquad{} +\alpha _{2}(c_{i 2}+c_{i k+1}\eta _{k+1,2}+c_{i k+2}\eta _{k+2,2} + \cdots +c_{i n} \eta _{n,2}) \\ &\qquad{} +\cdots + \alpha _{k}(c_{i k}+c_{i k+1}\eta _{k+1,k}+c_{i k+2}\eta _{k+2,k} + \cdots +c_{i n} \eta _{n,k}) \\ &\quad= \alpha _{1}\delta _{i1}+\alpha _{2}\delta _{i2}+\cdots + \alpha _{k}\delta _{ik},\quad i=,1,2, \ldots,k. \end{aligned}$$
$$\begin{aligned} \bigl(Q^{2}v\bigr) (t)&= \sum_{i=1}^{k} \gamma \int _{0}^{1} \int _{0}^{1}G(t,s)\sum _{j=1}^{n}c_{i j}(Qv)_{j}(s)\,ds \,d\alpha (t)\cdot \eta _{i} \\ &= \gamma \int _{0}^{1} \int _{0}^{1}G(t,s)\,ds\,d\alpha (t) \Biggl(\sum _{i=1}^{k}\gamma \int _{0}^{1} \int _{0}^{1}G(t,s)\sum _{j=1}^{n}c_{i j}v_{j}(s)\,ds\,d \alpha (t)\cdot \eta _{i} \Biggr) \\ &= \gamma \int _{0}^{1}\frac{t(1-t)}{2}\,d\alpha (t) \Biggl( \sum_{i=1}^{k}\gamma \int _{0}^{1} \int _{0}^{1}G(t,s)\sum _{j=1}^{n}c_{i j}v_{j}(s)\,ds\,d \alpha (t)\cdot \eta _{i} \Biggr) \\ &= (Qv) (t). \end{aligned}$$
From (2.12), it follows that the generalized inverse \(K_{P}: \operatorname{Im} L\rightarrow \operatorname{dom}L\cap \operatorname{Ker} P\) can be defined by
where \(\delta \in \mathbb{R}^{n}\) is given by
Similar to the proof of Lemma 2.4, we obtain
$$ (K_{P}v) (t)= \int _{0}^{1}G(t,s)v(s)\,ds+ \delta \cdot t, $$
$$ \delta = \biggl(\underbrace{0,\ldots,0} _{k}, \frac{1}{2h} \int _{0}^{1} \int _{0}^{1}G(t,s) \bigl(D_{2}^{-1}D_{1}, I_{n-k}\bigr)v(s)\,ds\,d\alpha (t) \biggr) ^{T}. $$
$$ \Vert K_{P}v \Vert _{X}\leq \Biggl(1+ \frac{n}{2h} \bigl\Vert \bigl(D_{2}^{-1}D_{1}, I_{n-k}\bigr) \bigr\Vert _{*} \int _{0}^{1}t(1-t)\,d \Biggl(\bigvee _{0}^{t} ( \alpha ) \Biggr) \Biggr) \Vert v \Vert _{1}. $$
(2.15)
3 Main result
In this section, we use Theorem 1.1 to prove the existence of solutions to (1.1). For this purpose, we use the following assumptions:
(H5)
There exist nonnegative functions \(a, b, c\in L^{1}[0,1]\) such that, for all \(u,v\in \mathbb{R}^{n}\) and \(t\in [0,1]\),
$$ \bigl\vert f(t,u,v) \bigr\vert \leq a(t) \Vert u \Vert _{\mathbb{R}^{n}}+b(t) \Vert v \Vert _{\mathbb{R}^{n}}+c(t). $$
(H6)
There exists a constant \(\varLambda >0\) such that, for each \(u\in \operatorname{dom}L\), if \(\|u'(t)\|_{\mathbb{R}^{n}}>\varLambda \) for all \(t\in [0,1]\), then
$$ g(B)A \int _{0}^{1} \int _{0}^{1}G(t,s)f\bigl(s,u(s),u'(s) \bigr)\,ds\,d\alpha (t)\neq0. $$
(H7)
There exists a constant \(\varLambda _{1}>0\) such that either for any \(\psi \in \mathbb{R}^{n}\) with \(\psi =B\psi \) and \(\|\psi \|_{ \mathbb{R}^{n}}>\varLambda _{1}\),
or for any \(\psi \in \mathbb{R}^{n}\) with \(\psi =B\psi \) and \(\|\psi \|_{\mathbb{R}^{n}}>\varLambda _{1}\),
where \(u(t)=\psi t\), and \((\cdot,\cdot )\) denotes the scalar product in \(\mathbb{R}^{n}\).
$$ (\psi,QNu)\leq 0, $$
(3.1)
$$ (\psi,QNu)\geq 0, $$
(3.2)
Theorem 3.1
Let the assumptions (H1)–(H7) hold. Then (1.1) has at least one solution in
X
provided that
\((n\|g(B)\|_{*}+M|g(1)|)( \|a\|_{1}+\|b\|_{1})<|g(1)|\).
Proof
Set
Suppose that \(u\in \varOmega _{1}\), and \(Lu=\lambda Nu\). Then \(\lambda \neq0\) and \(Nu\in \operatorname{Im}L=\operatorname{Ker}Q\) so that
Thus, from (H6), there is \(t_{0}\in [0,1]\) such that \(\|u'(t_{0})\|_{\mathbb{R}^{n}}\leq \varLambda \). By the absolute continuity of u, for \(t\in [0,1]\), we have
and
This yields
Again, if \(u\in \operatorname{dom}L\), then \((I-P)u\in \operatorname{dom}L\cap \operatorname{Ker} P\) and \(LPu=0\). Then, by (2.7) and Lemma 2.3,
Using (3.3), (3.4) and (3.5), we conclude that
The last inequality allows us to deduce that
Thus, \(\varOmega _{1}\) is bounded. Let
For \(u\in \varOmega _{2}\) and from the definition of ImL, \(u(t)=\psi t\), where \(\psi \in \mathbb{R}^{n}\). Since \(QNu=0\), we have
Hence, from (H6), we can show that
Therefore, \(\varOmega _{2}\) is a bounded set in X.
$$ \varOmega _{1}=\bigl\{ u\in \operatorname{dom}L\backslash \operatorname{Ker} L: Lu=\lambda Nu \text{ for some } \lambda \in [0,1]\bigr\} . $$
$$ g(B)A \int _{0}^{1} \int _{0}^{1}G(t,s)f\bigl(s,u(t),u'(t) \bigr)=0, \quad\text{for all } t\in [0,1]. $$
$$ \bigl\vert u(t) \bigr\vert = \biggl\vert u(0)- \int _{0}^{t}u'(s)\,ds \biggr\vert \leq \bigl\Vert u' \bigr\Vert _{\infty } $$
$$ \bigl\vert u'(t) \bigr\vert = \biggl\vert u'(t_{0})- \int _{t_{0}}^{t}u''(s)\,ds \biggr\vert \leq \varLambda + \bigl\Vert u'' \bigr\Vert _{1}. $$
(3.3)
$$ \Vert Pu \Vert _{X}=\max \bigl\{ \Vert Pu \Vert _{\infty }, \bigl\Vert (Pu)' \bigr\Vert _{\infty } \bigr\} \leq \frac{n \Vert g(B) \Vert _{*}}{ \vert g(1) \vert } \bigl\Vert u(1) \bigr\Vert _{\mathbb{R}^{n}}\leq \frac{n \Vert g(B) \Vert _{*}}{ \vert g(1) \vert } \bigl\Vert u' \bigr\Vert _{\infty }. $$
(3.4)
$$\begin{aligned} \bigl\Vert (I-P)u \bigr\Vert _{X}&= \bigl\Vert K_{P}L(I-P)u \bigr\Vert _{X}\leq M \bigl\Vert L(I-P)u \bigr\Vert _{1} \\ &= M \Vert Lu \Vert _{1}=M \bigl\Vert u'' \bigr\Vert _{1}\leq M \Vert Nu \Vert _{1}. \end{aligned}$$
(3.5)
$$\begin{aligned} \Vert u \Vert _{X}&= \bigl\Vert Pu+(I-P)u \bigr\Vert _{X}\leq \Vert Pu \Vert _{X}+ \bigl\Vert (I-P)u \bigr\Vert _{X} \\ &\leq \frac{n \Vert g(B) \Vert _{*}}{ \vert g(1) \vert } \bigl(\varLambda + \bigl\Vert u'' \bigr\Vert _{1}\bigr)+M \bigl\Vert u'' \bigr\Vert _{1} \\ &= \frac{n \Vert g(B) \Vert _{*}}{ \vert g(1) \vert } \varLambda + \biggl(\frac{n \Vert g(B) \Vert _{*}}{ \vert g(1) \vert } +M \biggr) \bigl\Vert u'' \bigr\Vert _{1} \\ &\leq \frac{n \Vert g(B) \Vert _{*}}{ \vert g(1) \vert } \varLambda + \biggl(\frac{n \Vert g(B) \Vert _{*}}{ \vert g(1) \vert } +M \biggr) \int _{0}^{1} \bigl\vert f\bigl(s, u(s),u'(s)\bigr) \bigr\vert \,ds \\ &\leq \frac{n \Vert g(B) \Vert _{*}}{ \vert g(1) \vert } \varLambda + \biggl(\frac{n \Vert g(B) \Vert _{*}}{ \vert g(1) \vert } +M \biggr) \bigl( \Vert a \Vert _{1} \Vert u \Vert _{\infty }+ \Vert b \Vert _{1} \bigl\Vert u' \bigr\Vert _{\infty }+ \Vert c \Vert _{1}\bigr) \\ &\leq \frac{n \Vert g(B) \Vert _{*}}{ \vert g(1) \vert } \varLambda + \biggl(\frac{n \Vert g(B) \Vert _{*}}{ \vert g(1) \vert } +M \biggr) \Vert c \Vert _{1}+ \biggl(\frac{n \Vert g(B) \Vert _{*}}{ \vert g(1) \vert } +M \biggr) \bigl( \Vert a \Vert _{1}+ \Vert b \Vert _{1}\bigr) \Vert u \Vert _{X}. \end{aligned}$$
$$ \Vert u \Vert _{X}\leq \frac{ n \Vert g(B) \Vert _{*}\varLambda + (n \Vert g(B) \Vert _{*}+M \vert g(1) \vert ) \Vert c \Vert _{1}}{ \vert g(1) \vert -(n \Vert g(B) \Vert _{*}+M \vert g(1) \vert )( \Vert a \Vert _{1}+ \Vert b \Vert _{1})}. $$
$$ \varOmega _{2}=\{u\in \operatorname{Ker} L: Nu\in \operatorname{Im}L\}. $$
$$ g(B)A \int _{0}^{1} \int _{0}^{1}G(t,s)f\bigl(s,u(s),u'(s) \bigr)\,ds\,d\alpha (t)=0. $$
$$ \Vert u \Vert _{X}= \Vert u \Vert _{\infty }= \Vert \psi \Vert _{\mathbb{R}^{n}}\leq \varLambda. $$
Let \(J(\psi )=\psi t\) be the isomorphism. Then we want to show that the set of u in KerL such that
with \(\lambda \in [0,1]\) is bounded if (3.1) holds. This means that (with \(\psi = B\psi \) and \(u=\psi t\))
or
If \(\lambda =0\), we have \(QN(\psi t)=0\), that is,
Thus, we deduce that \(\|\psi \|_{\mathbb{R}^{n}}\leq \varLambda \) follows from (H5). Otherwise, if \(\|\psi \|_{\mathbb{R}^{n}}>\varLambda _{1}\), in view of (H7), we have
Thus, \(\|u\|_{X}=\|\psi \|_{\mathbb{R}^{n}}\leq \varLambda _{1}\). Using the same argument as above, we can conclude that the set of u in KerL such that
with \(\lambda \in [0,1]\) is bounded if (3.2) holds. Therefore, the set
is bounded if conditions (H6) and (H7) are satisfied, where
$$ -\lambda u+(1-\lambda )JQNu=0 $$
$$ -\lambda \psi t+(1-\lambda )QN(\psi t)t=0, $$
$$ -\lambda \psi +(1-\lambda )QN(\psi t)=0. $$
$$ g(B)A \int _{0}^{1} \int _{0}^{1}G(t,s)f(s,\psi s,\psi )\,ds\,d\alpha (t)=0. $$
$$ 0< \lambda \Vert \psi \Vert ^{2}_{\mathbb{R}^{n}}=(1-\lambda ) \bigl( \psi,QN(\psi t)\bigr) \leq 0. $$
$$ \lambda u+(1-\lambda )JQNu=0 $$
$$ \varOmega _{3}=\bigl\{ u \in \operatorname{Ker} L: \mu \lambda u+(1- \lambda )JQNu=0, \lambda \in [0,1] \bigr\} $$
$$ \mu =\textstyle\begin{cases} -1 & \text{if (3.1) holds}; \\ 1 & \text{if (3.2) holds}. \end{cases} $$
Finally, the proof of this theorem is now an easy consequence of Lemmas 2.1, 2.2, and 2.3 and Theorem 1.1. Let Ω be a bounded and open subset of X
\(\bigcup_{i=1}^{3}\overline{\varOmega _{i}}\subset \varOmega \). Then, by the above argument, we have Then, by Theorem 1.1, \(Lu=Nu\) has at least one solution in \(\operatorname{dom}L\cap \overline{\varOmega }\) so that the IBVP (1.1) has at least one solution. □
(i)
\(Lx\neq \lambda Nx\), for every \((x,\lambda )\in [( \operatorname{dom}L\backslash \operatorname{Ker} L)\cap \partial \varOmega ]\times (0,1)\),
(ii)
\(Nx\notin \operatorname{Im}L\) for \(x\in \operatorname{Ker} L\cap \partial \varOmega \),
(iii)
\(H(x,\lambda )=\mu \lambda x+(1-\lambda )JQNx\). By the homotopy property of degree,
$$ \deg (JQN|_{\operatorname{Ker} L}, \operatorname{Ker} L\cap \varOmega,0)=\deg ( \mu I, \operatorname{Ker} L\cap \varOmega,0)\neq 0. $$
For the special case that a diagonalizable matrix B satisfy \(B^{2}=I\) and \(\dim \operatorname{Ker}(I-B)=k\), we make the following assumptions:
(H8)
There exists a constant \(\varLambda >0\) such that, for each \(u\in \operatorname{dom}L\), if \(|u_{1}'(t)|>\varLambda \) for all \(t\in [0,1]\) or \(|u_{2}'(t)|>\varLambda \) for all \(t\in [0,1]\), or … , or \(|u_{k}'(t)|>\varLambda \) for all \(t\in [0,1]\), then
$$ (QNu) (t)=\sum_{i=1}^{k}\gamma \int _{0}^{1} \int _{0}^{1}G(t,s)\sum _{j=1}^{n}c_{i j}f_{j} \bigl(s,u(s),u'(s)\bigr)\,ds\,d\alpha (t)\cdot \eta _{i} \neq0. $$
Theorem 3.2
Let the assumptions (H2), (H3), (H5), (H7), and (H8) hold. Then (1.1) has at least one solution in
X
provided that
\((\|\sum_{i=1}^{k}\eta _{i}\|_{\mathbb{R}^{n}}+M _{1})(\|a\|_{1}+\|b\|_{1})<1\), where
\(M_{1}= (1+\frac{n}{2h} \|(D _{2}^{-1}D_{1}, I_{n-k})\|_{*}\int _{0}^{1}t(1-t)\,d (\bigvee_{0}^{t}(\alpha ) ) )\).
Proof
For \(u\in \varOmega _{1}\), \(QNu=0\). Then, from (H8), there is \(t_{i}\in [0,1]\) (\(i=1,2,\ldots,k\)) such that \(|u_{i}'(t_{i})|\leq \varLambda \). By the absolute continuity of \(u_{i}\), for \(t\in [0,1]\), we have
and
This yields
Again, if \(u\in \operatorname{dom}L\), then \((I-P)u\in \operatorname{dom}L\cap \operatorname{Ker} P\) and \(LPu=0\). Then, by (2.15),
Using the above three inequalities, we conclude that
The last inequality allows us to deduce that
Thus, \(\varOmega _{1}\) is bounded. The rest of the proof repeats that of Theorem 3.1. □
$$ \bigl\vert u_{i}(t) \bigr\vert = \biggl\vert u_{i}(0)- \int _{0}^{t}u_{i}'(s)\,ds \biggr\vert \leq \bigl\Vert u' \bigr\Vert _{\infty },\quad i=1,2,\ldots,k, $$
$$ \bigl\vert u_{i}'(t) \bigr\vert = \biggl\vert u_{i}'(t_{i})- \int _{t_{i}}^{t}u_{i}''(s) \,ds \biggr\vert \leq \varLambda + \bigl\Vert u'' \bigr\Vert _{1},\quad i=1,2,\ldots,k. $$
$$\begin{aligned} \Vert Pu \Vert _{X}&= \max \bigl\{ \Vert Pu \Vert _{\infty }, \bigl\Vert (Pu)' \bigr\Vert _{\infty }\bigr\} = \Vert Pu \Vert _{\infty } \\ &\leq \max_{1\leq i\leq k}\bigl\{ \bigl\vert u_{i}(1) \bigr\vert \bigr\} \cdot \Biggl\Vert \sum_{i=1} ^{k}\eta _{i} \Biggr\Vert _{\mathbb{R}^{n}} \leq \Biggl\Vert \sum_{i=1}^{k}\eta _{i} \Biggr\Vert _{\mathbb{R}^{n}}\cdot \bigl\Vert u' \bigr\Vert _{\infty }. \end{aligned}$$
$$\begin{aligned} \bigl\Vert (I-P)u \bigr\Vert _{X}&= \bigl\Vert K_{P}L(I-P)u \bigr\Vert _{X}\leq M_{1} \bigl\Vert L(I-P)u \bigr\Vert _{1} \\ &= M_{1} \Vert Lu \Vert _{1}=M_{1} \bigl\Vert u'' \bigr\Vert _{1}\leq M_{1} \Vert Nu \Vert _{1}. \end{aligned}$$
$$\begin{aligned} \Vert u \Vert _{X}&= \bigl\Vert Pu+(I-P)u \bigr\Vert _{X}\leq \Vert Pu \Vert _{X}+ \bigl\Vert (I-P)u \bigr\Vert _{X} \\ &\leq \Biggl\Vert \sum_{i=1}^{k}\eta _{i} \Biggr\Vert _{\mathbb{R}^{n}}\bigl(\varLambda + \bigl\Vert u'' \bigr\Vert _{1}\bigr)+M_{1} \bigl\Vert u'' \bigr\Vert _{1} = \Biggl\Vert \sum_{i=1}^{k}\eta _{i} \Biggr\Vert _{\mathbb{R}^{n}} \varLambda \\ &{}+ \Biggl( \Biggl\Vert \sum _{i=1}^{k}\eta _{i} \Biggr\Vert _{\mathbb{R}^{n}} +M_{1} \Biggr) \bigl\Vert u'' \bigr\Vert _{1} \\ &\leq \Biggl\Vert \sum_{i=1}^{k}\eta _{i} \Biggr\Vert _{\mathbb{R}^{n}} \varLambda + \Biggl( \Biggl\Vert \sum_{i=1}^{k}\eta _{i} \Biggr\Vert _{\mathbb{R}^{n}} +M_{1} \Biggr) \int _{0}^{1} \bigl\vert f\bigl(s, u(s),u'(s)\bigr) \bigr\vert \,ds \\ &\leq \Biggl\Vert \sum_{i=1}^{k}\eta _{i} \Biggr\Vert _{\mathbb{R}^{n}} \varLambda + \Biggl( \Biggl\Vert \sum_{i=1}^{k}\eta _{i} \Biggr\Vert _{\mathbb{R}^{n}} +M_{1} \Biggr) \Vert c \Vert _{1}\\ &{}+ \Biggl( \Biggl\Vert \sum_{i=1}^{k}\eta _{i} \Biggr\Vert _{\mathbb{R}^{n}} +M _{1} \Biggr) \bigl( \Vert a \Vert _{1}+ \Vert b \Vert _{1}\bigr) \Vert u \Vert _{X}. \end{aligned}$$
$$ \Vert u \Vert _{X}\leq \frac{ \Vert \sum_{i=1}^{k}\eta _{i} \Vert _{\mathbb{R} ^{n}} \varLambda + ( \Vert \sum_{i=1}^{k}\eta _{i} \Vert _{\mathbb{R} ^{n}} +M_{1} ) \Vert c \Vert _{1}}{1- ( \Vert \sum_{i=1}^{k}\eta _{i} \Vert _{\mathbb{R}^{n}} +M_{1} )( \Vert a \Vert _{1}+ \Vert b \Vert _{1})}. $$
Example 3.1
Consider the differential system
Here \(f_{i}:[0,1]\times \mathbb{R}^{6}\rightarrow \mathbb{R}\), \(i=1,2,3\) are defined, respectively, by
where \(x=(x_{1},x_{2},x_{3})^{T}, y=(y_{1},y_{2},y_{3})^{T}\in \mathbb{R}^{3}\).
$$ \textstyle\begin{cases} -x''_{1}(t)=\sin x_{2}(t)x'_{2}(t)+\frac{1}{5}\arctan x_{2}(t)+ \frac{1}{5}x'_{1}(t)+t, \\ -x''_{2}(t)=\sin x_{1}(t)x_{3}(t)-\cos x'_{3}(t) +\frac{1}{20}x'_{2}(t)- \frac{1}{20}x_{2}(t)+e^{t}, \\ -x''_{3}(t)=\arctan x_{3}^{2}(t)+\cos x_{1}(t)+\cos x_{2}(t)+ \frac{1}{20}x_{2}(t)+\frac{1}{10}x'_{2}(t), \\ x_{1}(0)=x_{2}(0)=x_{3}(0)=0,\qquad x_{1}(1)=2\int _{0}^{1}x_{1}(t)\,dt, \\ x_{2}(1)=2\int _{0}^{1}x_{3}(t)\,dt,\qquad x_{3}(1)=2\int _{0}^{1}x_{2}(t)\,dt. \end{cases} $$
(3.6)
$$\begin{aligned} \begin{aligned} &f_{1}(t,x,y)= \sin x_{2}y_{2}+\frac{1}{5}\arctan x_{2}+ \frac{1}{5}y_{1}+t, \\ &f_{2}(t,x,y)= \sin x_{1}x_{3}-\cos y_{3} -\frac{1}{20}y_{2}-\frac{1}{20}x_{2}+e^{t}, \\ &f_{3}(t,x,y)= \arctan x_{3}^{2}+\cos x_{1}+\cos x_{2}+ \frac{1}{20}x_{2}+ \frac{1}{10}y_{2}, \end{aligned} \end{aligned}$$
(3.7)
Take \(\alpha (t)=t\),
and
Then \(h=\int _{0}^{1}t\,d\alpha (t)=\frac{1}{2}\), \(B=Ah=\frac{1}{2}A\), \(B^{2}=I\),
It follows from (3.7) that
$$ A= \begin{pmatrix} 2&0&0 \\ 0&0 &2 \\ 0&2&0 \end{pmatrix} $$
$$ f(t,x,y)= \begin{pmatrix} f_{1}(t,x,y) \\ f_{2}(t,x,y) \\ f_{3}(t,x,y) \end{pmatrix}. $$
$$\begin{aligned} &C= \begin{pmatrix} 1&0&0 \\ 0&1 &-1 \\ 0&1&1 \end{pmatrix}=(\eta _{1},\eta _{2},\eta _{3}), \qquad C^{-1}= \begin{pmatrix} 1&0&0 \\ 0&\frac{1}{2} &\frac{1}{2} \\ 0&-\frac{1}{2}&\frac{1}{2} \end{pmatrix}=(c_{ij})_{3\times 3}, \\ &D_{2}=\frac{1}{2},\qquad D_{1}=\biggl(0, - \frac{1}{2}\biggr),\qquad n=3, \qquad k=2, \\ &(Qv) (t)= 12 \int _{0}^{1} \int _{0}^{1}G(t,s)v_{1}(s)\,ds\,d\alpha (t)\cdot \eta _{1} \\ &\phantom{(Qv) (t)=}{}+ 6 \int _{0}^{1} \int _{0}^{1}G(t,s) \bigl(v_{2}(s)+v_{3}(s) \bigr)\,ds\,d\alpha (t)\cdot \eta _{2} \\ &\phantom{(Qv) (t)}= \begin{pmatrix} 12\int _{0}^{1}\int _{0}^{1}G(t,s)v_{1}(s)\,ds\,d\alpha (t) \\ 6\int _{0}^{1}\int _{0}^{1}G(t,s)(v_{2}(s)+v_{3}(s))\,ds\,d\alpha (t) \\ 6\int _{0}^{1}\int _{0}^{1}G(t,s)(v_{2}(s)+v_{3}(s))\,ds\,d\alpha (t) \end{pmatrix}. \end{aligned}$$
$$ \bigl\vert f(t,x,y) \bigr\vert \leq \frac{1}{20} \vert x \vert + \frac{1}{5} \vert y \vert +4. $$
Note that \(M_{1}= (1+\frac{n}{2h} \|(D_{2}^{-1}D_{1}, I_{n-k})\| _{*}\int _{0}^{1}t(1-t)\,d (\bigvee_{0}^{t}(\alpha ) ) )\), we have \(M_{1}=\frac{3}{2}\). Then we obtain \((\|\sum_{i=1}^{2}\eta _{i} \|_{\mathbb{R}^{3}}+M_{1})(\|a\|_{1}+\|b\|_{1})=\frac{5}{8}<1\). Therefore, condition (H5) is satisfied.
Take \(\varLambda =200\). Then, for \(|y_{1}(t)|\geq \varLambda \) for all \(t\in [0,1]\), we have
and for \(|y_{2}(t)|\geq \varLambda \) for all \(t\in [0,1]\), we have
Thus, for each \(u\in X\), if \(|v_{1}(t)|\geq \varLambda \) for all \(t\in [0,1]\) or \(|v_{2}(t)|\geq \varLambda \) for all \(t\in [0,1]\), we conclude that
Hence, (H8) holds.
$$\begin{aligned} & \bigl\vert f_{1}(t,x,y) \bigr\vert \\ &\quad= \biggl\vert \sin x_{2}y_{2}+\frac{1}{5}\arctan x_{2}+\frac{1}{5}y_{1}+t \biggr\vert \geq \frac{1}{5} \vert y_{1} \vert -4\geq 36 \end{aligned}$$
$$\begin{aligned} & \bigl\vert (f_{2}+f_{3}) (t,x,y) \bigr\vert \\ &\quad=\biggl\vert \sin x_{1}x_{3}-\cos y_{3}+e^{t}+ \arctan x_{3}^{2}+\cos x_{1}+ \cos x_{2}+\frac{1}{10}y_{2} \biggr\vert \geq \frac{1}{20} \vert y_{2} \vert -6\geq 4. \end{aligned}$$
$$ (QNv) (t) = \begin{pmatrix} 12\int _{0}^{1}\int _{0}^{1}G(t,s)v_{1}(s)\,ds\,d\alpha (t) \\ 6\int _{0}^{1}\int _{0}^{1}G(t,s)(v_{2}(s)+v_{3}(s))\,ds\,d\alpha (t) \\ 6\int _{0}^{1}\int _{0}^{1}G(t,s)(v_{2}(s)+v_{3}(s))\,ds\,d\alpha (t) \end{pmatrix}\neq0. $$
Let \(\psi =(c_{1},c_{2},c_{2})^{T}\in \operatorname{Ker} (I-B)=\{ c_{1}(1,0,0)+ c _{2}(0,1,1): c_{1},c_{2}\in \mathbb{R}\}\). Then we have
and
Note that \(\|\psi \|_{\mathbb{R}^{3}}=\max \{|c_{1}|, |c_{2}|\}\) and
we see that condition (H7) is satisfied. It follows from Theorem 3.2 that the problem (3.7) has at least one solution.
$$\begin{aligned} &f_{1}(t,\psi t,\psi )= \sin c_{2}^{2}t+ \frac{1}{5}\arctan c_{2}t+\frac{1}{5}c_{1}+t, \\ &f_{2}(t,\psi t,\psi )= \sin c_{1}c_{2}t^{2}- \cos c_{2} -\frac{1}{20}c_{2}-\frac{t}{20}c_{2}+e ^{t}, \\ &f_{3}(t,\psi t,\psi )= \arctan c_{2}^{2}t^{2}+ \cos c_{1}t+\cos c_{2}t+ \frac{t}{20}c_{2}+ \frac{1}{10}c_{2}, \end{aligned}$$
$$\begin{aligned} & c_{1}f_{1}(t,\psi t, \psi )+c_{2} \bigl(f_{2}(t,\psi t, \psi )+f_{3}(t,\psi t, \psi )\bigr) \\ &\quad= c_{1} \biggl( \sin c_{2}^{2}t+ \frac{1}{5}\arctan c_{2}t+\frac{1}{5}c_{1}+t \biggr) \\ &\qquad{} +c_{2} \biggl(\frac{1}{10}c_{2}+\sin c_{1}c_{2}t^{2}-\cos c_{2} +e^{t} +\arctan c_{2}^{2}t^{2}+\cos c_{1}t+\cos c_{2}t \biggr) \\ &\quad\geq \frac{1}{5}c_{1}^{2}-3 \vert c_{1} \vert +\frac{1}{10}c_{2}^{2}-10 \vert c_{2} \vert \geq \frac{1}{10}\max \bigl\{ c_{1}^{2}, c_{2}^{2}\bigr\} -13\max \bigl\{ \vert c_{1} \vert , \vert c_{2} \vert \bigr\} . \end{aligned}$$
$$ \bigl(\psi,QN(\psi t)\bigr)=12 \int _{0}^{1} \int _{0}^{1}G(t,s)\bigl[c_{1}f_{1}+c_{2}(f_{2}+f_{3}) \bigr]\,ds\,dt\geq 156, \quad\text{if } \Vert \psi \Vert _{\mathbb{R}^{3}}\geq 131, $$
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