1 Introduction

A Hamiltonian system is in fact a mathematical formality introduced by Hamilton for studying the evolution equations in physical systems. The advantage of this explanation is that it uncover important insight about the dynamics. It plays important role in classical and celestial mechanics by studying periodic solutions of Hamiltonian systems.

Mawhin and Willem [1] studied the periodic solutions of the following convex Hamiltonian system

$$\begin{aligned} \left\{ \begin{array}{ll} \ddot{q}+\nabla H(t,q(t))=0,\ \ t\in (0,T) \\ q(0)-q(T)=\dot{q}(0)-\dot{q}(T)=0,\\ \end{array} \right. \end{aligned}$$
(1)

where \(H:[0, T]\times R^N\rightarrow R\) is measurable in t for every \(q\in R^n\) and continuously differentiable and convex in q for a.e. \(t\in [0, T]\). The readers are refereed to see [2,3,4,5,6,7,8,9,10,11] for more information about Hamiltonian systems and related problems. Ge and Zhao [12] have studied the existence of multiple solutions of the following four point boundary value problem:

$$\begin{aligned} \left\{ \begin{array}{ll} \big (P(t)x'(t)\big )'=\nabla H(t,x(t)),\ \ t\in (0,1) \\ x'(0)=\alpha x(\xi ), \ x'(1)=\beta x(\eta ),\\ \end{array} \right. \end{aligned}$$
(2)

where, the matrix P is symmetric. Also, Wen Lian et. al. [13] have studied the existence of multiple solutions to the following three-point differential Hamiltonian system

$$\begin{aligned} \left\{ \begin{array}{ll} \big (P(t)x'(t)\big )'+\nabla H(t,x(t))=0,\ \ a.e. t\in (0,1) \\ x(0)=0, \ x(1)=\beta x(\eta ),\\ \end{array} \right. \end{aligned}$$
(3)

with some conditions on P(t) (that is a matrix) and H(t, x(t)).

Research on integro-differential equations of the integer order and even fractional order in scientific sources shows that they are used in many fields. Readers can study [14, 15] and references therein. On the other hand, in general, there are different methods for checking the existence of the solutions to the boundary value problems, the most important of which are the fixed point method based on the Green function through the upper and lower solutions, and the variational methods [1, 16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32].

In this paper, we study the existence of weak solutions for the following system of integro-differential equations

$$\begin{aligned} \left\{ \begin{array}{ll} \big (A(t)u'(t)\big )'=\int _0^1K(t,s)\nabla H(s,u(t))ds, \\ u(0)=u(1)=0.\\ \end{array} \right. \end{aligned}$$
(4)

where \(t \in (0,1)\), \(A:[0,1]\rightarrow \mathbb {R}^{n\times n}\) is a continuously symmetric matrix, i.e., \(A^T(t)=A(t)\) being continuous in t, the kernel \(K(.,.)\in C^1(]0,1[)\), \(H:\mathbb {R}^n\rightarrow \mathbb {R}\) is differentiable scaler function such that its partial derivatives are continuously differentiable, and

$$\begin{aligned} \nabla H(t,x)=\begin{bmatrix} \partial _{x_1}H(t,x)\\ \partial _{x_2}H(t,x)\\ ...\\ \partial _{x_n}H(t,x)\\ \end{bmatrix}. \end{aligned}$$

First of all, to achieve our purpose, we recall one basic optimization principle for extreme eigenvalues of symmetric matrixes:

Lemma 1.1

([33]) If A is a real symmetric \(n\times n\) matrix, then

$$\begin{aligned} \lambda _1=\max \{x^TAx\ |\ \Vert x\Vert =1\}, \ \ \ \lambda _n=\min \{x^TAx\ |\ \Vert x\Vert =1\}. \end{aligned}$$

where \(\lambda _1\) and \(\lambda _n\) are respectively, its largest and smallest eigenvalues.

As a result, we can extend preceding lemma for \(\Vert x\Vert \ne 1\), by normalization in the form of below:

$$\begin{aligned} \lambda _1=\max \left\{\frac{x^TAx}{\Vert x\Vert ^2}\ | x\ne 0\right\},\ \ \ \lambda _n=\min \left\{\frac{x^TAx}{\Vert x\Vert ^2}\ | x\ne 0\right\}. \end{aligned}$$
(5)

2 Variational Structure on the Problem

Consider the sobolev Banach space

$$\begin{aligned} H^1([0,1],\mathbb {R}^n)=\big \{u\in L^2([0,1],\mathbb {R}^n)\mid \int _0^1|u(t)|^2dt+\int _0^1|u'(t)|^2dt<\infty \big \}, \end{aligned}$$

with respect to the norm

$$\begin{aligned} \Vert u\Vert _{H^1([0,1],\mathbb {R}^n)}=\left (\int _0^1|u(t)|^2dt+\int _0^1|u'(t)|^2dt\right )^\frac{1}{2}. \end{aligned}$$
(6)

In confronting with our boundary values, we consider its closed subspace

$$\begin{aligned} X:=H_0^1([0,1],\mathbb {R}^n)=\big \{u\in H^1([0,1],\mathbb {R}^n)\mid u(0)=u(1)=0\big \}. \end{aligned}$$

\(H^1([0,1],\mathbb {R}^n)\) is reflexive, so is X.

Lemma 2.1

([34]) If \(x\in C^1[0,1]\) and \(x(0)=0\), then we have

$$\begin{aligned} \int _0^1|x(t)|^2dt\le \frac{4}{\pi ^2}\int _0^1|x'(t)|^2. \end{aligned}$$

By preceding lemma, the norm (11) is equivalent to

$$\begin{aligned} \Vert u\Vert =\left (\int _0^1|u'(t)|^2dt\right)^\frac{1}{2}. \end{aligned}$$
(7)

After that, we apply this norm.

Lemma 2.2

([13]) X is compactly embedded into \(C([0,1],\mathbb {R}^n)\) there exists a positive embedding constant k such that \(\Vert u\Vert _{\infty }\le k \Vert u\Vert \ \big (\Vert u\Vert _{\infty }=\max _{t\in [0,1]}|u(t)|\big )\).

Definition 2.3

A function \(u:[0,1]\rightarrow \mathbb {R}^n\) is said to be a weak solution of problem (4), if \(u\in X\) and

$$\begin{aligned} \int _0^1\prec (A(t)u'(t))',\phi (t)\succ dt-\int _0^1\prec \int _0^1K(t,s)\nabla H(s,u(t))ds,\phi (t)\succ dt=0, \end{aligned}$$
(8)

for all \(C_0^{\infty }(]0,1[,\mathbb {R}^n)\).

Now define the functional \(\Psi (u):\mathbb {R}^n\rightarrow \mathbb {R}\) as

$$\begin{aligned} \Psi (u)=\int _0^1\big [\frac{1}{2}\prec A(t)u'(t),u'(t)\succ +\int _0^1K(t,s)H(s,u(t))ds\big ]dt. \end{aligned}$$
(9)

It is clear that \(\Psi \) is well defined and we can say more:

Lemma 2.4

The critical points of the functional \(\Psi (u)\) are exactly weak solutions of (4).

Proof

Since A as function is continuous at t, and K and H are continuously differentiable, so is \(\Psi (u)\), and whose Gâteaux derivative at point u in direction v, given by

$$\begin{aligned} \nonumber{} & {} \prec \Psi '(u),v\succ =\frac{d}{d\theta }[\Psi (u+\theta v)|_{\theta =0}=\frac{d}{d\theta }\big [\int _0^1\big (\frac{1}{2}\prec A(t)u'(t)+\theta v'(t)),(u'(t)+\theta v'(t))\succ \\ \nonumber{} & {} +\int _0^1K(t,s)H(s,u(t)+\theta v(t))ds\big )dt\big ]_{\theta =0}=\\ \nonumber{} & {} \prec \int _0^1\big [\frac{1}{2}\big (\prec A(t)v'(t),u'(t)+\theta v'(t)\succ +\prec A(t)(u'(t)+\theta v'(t)),v'(t)\succ \big )+\\ \nonumber{} & {} \int _0^1K(t,s)\nabla H(s,u(t)+\theta v(t)),v(t)\succ ds\big ]dt\mid _{\theta =0}=\\ \nonumber{} & {} \int _0^1\big [\frac{1}{2}\big (\prec A(t)v'(t),u'(t)\succ +\prec A(t)u'(t),v'(t)\succ \big )+\\ \nonumber{} & {} \int _0^1K(t,s)\prec \nabla H(s,u(t),v(t)\succ ds\big ]dt=\\{} & {} \int _0^1\big [\prec A(t)u'(t),v'(t)\succ +\int _0^1K(t,s)\prec \nabla H(s,u(t)),v(t)\succ ds\big ]dt. \end{aligned}$$
(10)

By applying divergence theorem, we conclude:

$$\begin{aligned} \nonumber{} & {} \prec \Psi '(u),y\succ =\\ \nonumber{} & {} -\int _0^1\prec (A(t)u'(t))',v'(t)\succ dt+\int _0^1\int _0^1K(t,s)\prec \nabla H(s,u(t),v(t)\succ dsdt=\\ \nonumber{} & {} -\left ( \int _0^1\prec (A(t)u'(t))',v'(t)\succ dt-\int _0^1\prec \int _0^1K(t,s)\nabla H(s,u(t),v(t)ds\succ dt\right ). \end{aligned}$$

On the other hand the assumption, u is a critical point of \(\Psi \), implies

$$\begin{aligned} \prec \Psi '(u),y\succ =0 \end{aligned}$$

This means u is exactly a weak solution of (4).\(\Box \)

From now on, we will focus on finding critical points for functional \(\Psi \). Our main tool is a famous min–max Theorem named Mountain Pass as follows:

Theorem 2.5

([35]) Assume X be a real banach space and \(I\in C^1(X,R)\) satisfies the Palais–Smale condition. Suppose also

  1. 1.

    \(I(0)=0,\)

  2. 2.

    there exist constants \(r,\alpha >0\) such that \(I(u)>\alpha \) if \(\Vert u\Vert =r,\)

  3. 3.

    there exists an element \(v\in X\) with \(\Vert v\Vert >r\), \(I(v)\le 0.\)

define

$$\begin{aligned} \Gamma :=\{g\in (C[0,1],X)|\ g(0)=0,g(1)=v\}. \end{aligned}$$

Then

$$\begin{aligned} c=\inf \limits _{g\in \Gamma }\max \limits _{0\le t\le 1}I(g(t)) \end{aligned}$$

is a critical value of I.

3 Main Result

Now we can turn to the main theorem:

Theorem 3.1

Assume that the following conditions hold:

\(\mathbf {(H_1)}\) \(\{\lambda _i(t)\}\) is the eigenvalue of A(t) and \(0<a\le \min \limits _{0\le t\le 1}\min \limits _{1\le j\le n}\lambda _{j}(t)\le \max \limits _{0\le t\le 1}\max \limits _{1\le j\le n}\lambda _{j}(t)\le b;\)

\(\mathbf {(H_2)}\) for partial derivatives \(H_{x_i}\), we have \(\lim \limits _{|x|\rightarrow 0}\min \limits _{0\le t\le 1}\frac{\int _0^1K(t,s)H_{x_i}(s,x)ds}{|x|}=0;\)

\(\mathbf {(H_3)}\) there is constant \(\mu >0\) such that \(\frac{a}{2}-\frac{b}{\mu }>0\) and \(\alpha \ge 0\) such that for \(|x|\ge \alpha \), we have

$$\begin{aligned} \int _0^1\mu K(t,s)H(s,x(t))ds-\int _0^1K(t,s)(x,\nabla H(s,x(t)))ds\ge 0. \end{aligned}$$

Then problem (4) has a unique nontrivial weak solution.

Proof

It is sufficient to show that functional \(\Psi \), defined in (9), satisfies all conditions of Theorem 2.5. The first step is to show (PS)-condition holds. In this order, assume \(\{u_k\}\subset X\) be a sequence such that \(\{\Psi (u_k)\}\) is bounded, i.e. there exists \(L>0\) such that \(|\Psi (u_k)|\le L\), and \(\Psi '(u_k)\rightarrow 0\) as \(k\rightarrow \infty \). We should prove that \(\{u_k\}\) admits a convergence subsequence in X. By definition of symmetric matrix, \(\prec A(t)u'(t),u'(t)\succ =\prec u'(t),A(t)u'(t)\succ \). So by definition of \(\Psi \), and taking into account \((H_1)\) and (5) one has

$$\begin{aligned} \nonumber \Psi (u)= & {} \int _0^1\big [\frac{1}{2}\prec A(t)u'(t),u'(t)\succ +\int _0^1K(t,s)H(s,u(t))ds\big ]dt\\\ge & {} \frac{a}{2}\Vert u\Vert ^2-\int _0^1\int _0^1K(t,s)H(s,u(t))dsdt; \end{aligned}$$
(11)

In similar way, by (10) and \((H_1)\) we get

$$\begin{aligned} \nonumber \frac{1}{\mu }\prec \Psi '(u),u\succ= & {} \frac{1}{\mu }\int _0^1\big [\prec A(t)u'(t),u'(t)\succ +\int _0^1K(t,s)\prec \nabla H(s,u(t),u(t)\succ ds\big ]dt\\ \nonumber\le & {} \frac{1}{\mu }\big [\int _0^1 b|u'(t)|^2dt-\int _0^1\int _0^1K(t,s)\prec \nabla H(s,u(t),u(t)\succ dsdt\big ]\\= & {} \frac{1}{\mu }\big [b\Vert u\Vert ^2-\int _0^1\int _0^1K(t,s)\prec \nabla H(s,u(t),u(t)\succ dsdt\big ]. \end{aligned}$$
(12)

For k large enough, put \(u:=u_k\), and \(E:=\{t\in [0,1]|\ |u(t)|\ge \alpha \}\), combining (11) and (12) implies:

$$\begin{aligned} \nonumber L\ge & {} \Psi (u)\\ \nonumber\ge & {} \frac{1}{\mu }\prec \Psi '(u),u\succ +( \frac{a}{2}-\frac{b}{\mu })\Vert u\Vert ^2+\\ \nonumber{} & {} \frac{1}{\mu }\int _0^1\int _0^1\big [K(t,s)\big (H(s,u(t))-\prec \nabla H(s,u(t)),u(t)\succ \big )\big ]dsdt\\ \nonumber= & {} \frac{1}{\mu }\prec \Psi '(u),u\succ +( \frac{a}{2}-\frac{b}{\mu })\Vert u\Vert ^2+\\ \nonumber{} & {} \int _{E}\frac{1}{\mu }\int _0^1\big [K(t,s)\big (H(s,u(t))-\prec \nabla H(s,u(t)),u(t)\succ \big )\big ]dsdt\\{} & {} +\int _{[0,1]\setminus E}\int _0^1\big [K(t,s)\big (H(s,u(t))-\prec \nabla H(s,u(t)),u(t)\succ \big )\big ]dsdt. \end{aligned}$$
(13)

The first term on the right side of (13) tend to zero, by \(\mathbf {(H_3)}\), \(\frac{a}{2}-\frac{b}{\mu }>0\) and the third term is positive. For fourth term, since \(|u(t)|<\alpha \), it is clear that is bounded by a constant independently of k. So \(\{u_k\}\) is bounded in X. Whereas X is reflexive, \(\{u_k\}\) has a weakly convergence subsequence. Going to a subsequence, Assume \(u_k\overset{w}{\rightarrow }u\) in X. According to compactly embedding in lemma 2.2, \(u_k\rightarrow u\) in \(C_0^{\infty }(]0,1[,\mathbb {R}^n)\). By properties of inner product, we conclude

$$\begin{aligned} \prec \Psi '(u_k)-\Psi '(u),u_k-u\succ =\prec \Psi '(u_k),u_k-u\succ -\prec \Psi '(u),u_k-u\succ \rightarrow 0, \ (k\rightarrow \infty ) \end{aligned}$$
(14)

Due to \(\mathbf {(H_1)}\),

$$\begin{aligned} \nonumber \prec \Psi '(u_k)-\Psi '(u),u_k-u\succ =\prec A(t)u_k'(t)-u'(t),u'(t)\succ dt\\ \nonumber +\int _0^1\int _0^1K(t,s)\prec \nabla H(s,u_k(t))-\nabla H(s,u(t)),u_k(t)-u(t)\succ dsdt\\ \nonumber \ge a\int _0^1|u_k'(t)-u'(t)|^2dt\\ +\int _0^1\int _0^1K(t,s)\prec \nabla H(s,u_k(t))-\nabla H(s,u(t)),u_k(t)-u(t)\succ dsdt. \end{aligned}$$
(15)

On the other hand \(u_k\rightarrow u\) in \(C_0^{\infty }(]0,1[,\mathbb {R}^n)\). Hence \(|u_k(t)-u(t)|\rightarrow 0\) for \(t=0,1\) and

$$\begin{aligned} \int _0^1\int _0^1K(t,s)\prec \nabla H(s,u_k(t))-\nabla H(s,u(t)),u_k(t)-u(t)\succ dsdt\rightarrow 0 \end{aligned}$$

as \(k\rightarrow \infty \). Consequently, (15) implies

$$\begin{aligned} \int _0^1|u_k'(t)-u'(t)|^2dt\rightarrow 0 \ \ (k\rightarrow \infty ) \end{aligned}$$

But \(u_k\in X\), so \(u_k(0)=u_k(1)=0\), hence \(u(0)=u(1)=0\), i.e. \(u\in X\). That is \(\Vert u_k-u\Vert \rightarrow 0\) in X. Then (PS)-condition holds. It is obvious that \(\Psi (0)=0\). For the second step, according to \((H_2)\), \(\lim \limits _{u\rightarrow 0}\frac{\int _0^1K(t,s)H_{u_i}(s,u)ds}{|u|}=0\) that means for all \(\epsilon _i>0 (1\le i\le n)\) there is a \(0<\delta \le 1\) such that \(|u|\le \delta \) concludes \(|\int _0^1K(t,s)H_{u_i}(s,u)ds|\le {\epsilon }_i|u|\). Equivalently, for all \(\epsilon =\max \limits _{1\le i\le n}\epsilon _i>0\) we have

$$\begin{aligned} |\int _0^1 K(t,s)\nabla H(s,u(t))ds|\le \frac{\epsilon }{2}|u|^2 \end{aligned}$$
(16)

By integrating both side of above inequality and applying lemma 2.1 we have

$$\begin{aligned} \int _0^1\int _0^1 K(t,s)\nabla H(s,u(t))dsdt\le \frac{\epsilon }{2}\int _0^1|u(t)|^2dt\le \frac{2\epsilon }{{\pi }^2}\int _0^1|u'(t)|^2dt, \end{aligned}$$
(17)

Therefore

$$\begin{aligned} \Psi (u)\ge \frac{a}{2}\Vert u\Vert ^2-\frac{2\epsilon }{{\pi }^2}\int _0^1|u'(t)|^2dt=\big (\frac{a}{2}-\frac{2\epsilon }{{\pi }^2}\big )\Vert u\Vert ^2\ge \big (\frac{a}{2}-\frac{2\epsilon }{{\pi }^2}\big )\Vert u\Vert _{\infty }^2. \end{aligned}$$
(18)

By setting \(\epsilon =\frac{a{\pi }^2}{8}\), \(r=\delta \), \(\alpha =\frac{a{\delta }^2}{4}\), we conclude that \(\Psi (u)\ge \alpha \). So condition (2) in theorem 2.5 holds. To complete the proof, it is sufficient to show that condition (3) holds. Notice that by \((H_3)\), there exist constants \(c,M>0\) such that

$$\begin{aligned} \int _0^1 K(t,s)H(s,x(t))ds\le M-c|x|^\mu \ \ \forall t\in [0,1] , x\in \mathbb {R}^n. \end{aligned}$$
(19)

Thus

$$\begin{aligned}{} & {} \int _0^1\int _0^1 K(t,s)H(s,u(t))dsdt\le M-c\int _0^1|u(t)|^\mu dt \end{aligned}$$
(20)
$$\begin{aligned}{} & {} \Psi (uz)=\int _0^1\big [\frac{1}{2}\prec A(t)zu'(t),zu'(t)\succ +\int _0^1K(t,s)H(s,zu(t)),\phi (t)ds\big ]dt \nonumber \\{} & {} \quad \le \frac{bz^2}{2}\int _0^1|u'(t)|^2dt+\int _0^1\int _0^1K(t,s)H(s,zu(t)),\phi (t)dsdt\nonumber \\{} & {} \quad \le \frac{bz^2}{2}\Vert u\Vert ^2+M-cz^\mu \int _0^1|u(t)|^\mu dt \end{aligned}$$
(21)

which yields \(\Psi (uz)\rightarrow -\infty \) as \(z\rightarrow \infty \). So the desired result is achieved.\(\Box \)

Now consider the following condition:

\(\mathbf {(H_4)}\) \(\nabla H\) is locally Lipschitz continuous function in \((0,1)\times \mathbb {R}\).

Theorem 3.2

Under the hypothesis of theorem 3.1, if H also satis \((H_4)\), then system (4) possess a positive and a negative solution.

Proof

Define

$$\begin{aligned} \widetilde{H}(t,x)=\left\{ \begin{array}{ll} 0, &{} x\le 0; \\ H(t,x),\ {} &{} x>0. \end{array}\right. \end{aligned}$$
(22)

The arguments of theorem 3.1 proof show that

$$\begin{aligned} \widetilde{\Psi (u)}=\frac{1}{2}\Vert u\Vert ^2-\int _0^1\int _0^1 K(t,s)H(s,u(t))dsdt \end{aligned}$$

satisfies the hypotheses of Mountain Pass theorem. Indeed \(\nabla H\) satisfies \((H_2)\). Also \((H_3)\) holds for \(x>0\) while \(\widetilde{H}=\nabla \widetilde{H}=0\) for \(x\ge 0\). Assumption \((H_3)\) was required to verify (PS) and (3) conditions. But \(\widetilde{H}\ge \frac{1}{\mu }\nabla \widetilde{H}\) for large enough |x| and this is sufficient to get (PS)-condition. Notice that (19) holds for \(x\ge 0\) so choose \(u\in X\backslash \{0\}\) in (21) to be nonnegative function, condition (3) in theorem 2.5 holds. Consequently by the Mountain Pass theorem the system

$$\begin{aligned} \left\{ \begin{array}{ll} \big (A(t)u'(t)\big )'=\int _0^1K(t,s)\nabla \widetilde{H}(s,u(t))ds, \\ u(0)=u(1)=0.\\ \end{array} \right. \end{aligned}$$
(23)

has a weak solution, \(u\not \equiv 0\). By \((H_4)\) and [remark 2.22,[27]], u is a classical solution of (4). Let \(\mathcal {A}=\{t\in [0,1]|u(t)<0\}\). Then by definition of \(\widetilde{H}\) we have

$$\begin{aligned} \big (A(t)u'(t)\big )'=0 \ \ \forall t\in [0,1] \end{aligned}$$

Therefore the maximum principle shows \(u\equiv 0\) in \(\mathcal {A}\). This means \( \mathcal {A}=\emptyset \). Hence \(u\ge 0\) in [0, 1]. The negative solution of (4) is produced in similar way. \(\Box \)

Theorem 3.3

Assume that conditions \(H_1\) and \(H_3\) holds. Furthermore the following hypotheses holds:

\(\mathbf {(H_5)}\) \(\int _0^1K(t,s)H(s,-x)ds=\int _0^1K(t,s)H(s,x)ds;\)

\(\mathbf {(H_6)}\) \(\lim \limits _{|x|\rightarrow 0}\min \limits _{0\le t\le 1}\frac{\int _0^1K(t,s)H_{x_i}(s,x)ds}{|x|^2}=-\infty .\)

Then BVP (4) has infinity many pairs of nontrivial weak solutions.

Proof

Condition \((H_6)\) means for any \(m\in \mathbb {N}\) there is \(\delta >0\) such that

$$\begin{aligned} \prec \int _0^1K(t,s)\nabla H(s,x)ds,x\succ \le -4b(m+1)^2\pi ^2|x(t)|^2<-4bm^2\pi ^2|x(t)|^2, \ \ 0<|x|\le \delta . \end{aligned}$$

So the conclusion comes from [theorem 1.1,[12]].

4 Concluding Remarks

In this paper, we have studied and researched on a class of important model namely Hamiltonian integro-differential system of Fredholm type with Dirichlet boundary conditions at the integral bounds (also bounds of the problem domain). It have been given and proved some good results on the existence and multiplicity of weak solutions to a Hamiltonian integro-differential system by applying the theory of min–max based on Mountain-Pass theorem. Furthermore, we showed that it is possible to extract also classical solutions through weak results.