In this section, we shall prove our main results by using variational methods.

Throughout this section, *Ω* is a symmetric and positive semidefinite matrix, 0 is a simple eigenvalue with multiplicity one and with the eigenvector (1, 1,⋯,1)^{⊤}. We denote the eigenvalues of *Ω* by λ_{1}, λ_{2},⋯, λ_{T}.

Denote

$$\begin{array}{}{\displaystyle \underset{\_}{\lambda}=min\left\{{\lambda}_{i}|{\lambda}_{i}\ne 0,i=1,2,\cdots ,T\right\},}\end{array}$$(4.1)

and

$$\begin{array}{}{\displaystyle \overline{\lambda}=max\left\{{\lambda}_{i}|{\lambda}_{i}\ne 0,i=1,2,\cdots ,T\right\}.}\end{array}$$(4.2)

Set *X*_{2} = {(*d*, *d*,⋯, *d*)^{⊤} ∈ *X*∣*d* ∈ ℝ}. Obviously, *X*_{2} is an invariant subspace of *X*. We denote a subspace *X*_{1} of *X* by

$$\begin{array}{}{\displaystyle X={X}_{1}\oplus {X}_{2}.}\end{array}$$

#### Proof of Theorem 3.1

Let {*x*_{k}}_{k∈ℕ} ⊂ *X* be such that {*I*(*x*_{k})}_{k∈ℕ} is bounded and *I*′(*x*_{k}) → 0 as *k* → ∞. Accordingly, for any *k* ∈ ℕ, there is a number *c*_{7} > 0 such that

$$\begin{array}{}{\displaystyle -{c}_{7}\le I\left({x}_{k}\right)\le {c}_{7}.}\end{array}$$

Let $\begin{array}{}{x}_{k}={x}_{k}^{(1)}+{x}_{k}^{(2)}\in {X}_{1}\oplus {X}_{2}.\end{array}$ On one hand, for *k* large enough, since

$$\begin{array}{}{\displaystyle -\parallel x\parallel \le \left({I}^{\prime}\left({x}_{k}\right),x\right)=-\left(\mathit{\Omega}{x}_{k},x\right)+\sum _{t=1}^{T}f\left(t,{x}_{k}(t)\right)x(t),}\end{array}$$

combining with (*G*_{1}), we have

$$\begin{array}{}{\displaystyle \left(\mathit{\Omega}{x}_{k},{x}_{k}^{(1)}\right)\le \sum _{t=1}^{T}f\left(t,{x}_{k}(t)\right){x}_{k}^{(1)}(t)+\u2225{x}_{k}^{(1)}\u2225}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\le {c}_{1}\sum _{t=1}^{T}\left|{x}_{k}^{(1)}\right|+\u2225{x}_{k}^{(1)}\u2225}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\le \left({c}_{1}\sqrt{T}+1\right)\u2225{x}_{k}^{(1)}\u2225.}\end{array}$$

On the other hand, we have

$$\begin{array}{}{\displaystyle \left(\mathit{\Omega}{x}_{k},{x}_{k}^{(1)}\right)=\left(\mathit{\Omega}{x}_{k}^{(1)},{x}_{k}^{(1)}\right)\ge \underset{\_}{\lambda}{\u2225{x}_{k}^{(1)}\u2225}^{2}.}\end{array}$$

Consequently, we have

$$\begin{array}{}{\displaystyle \underset{\_}{\lambda}{\u2225{x}_{k}^{(1)}\u2225}^{2}\le \left({c}_{1}\sqrt{T}+1\right)\u2225{x}_{k}^{(1)}\u2225.}\end{array}$$(4.3)

(4.3) means that $\begin{array}{}{\left\{{x}_{k}^{(1)}\right\}}_{k=1}^{\mathrm{\infty}}\end{array}$ is bounded.

Then, we shall prove that $\begin{array}{}{\left\{{x}_{k}^{(2)}\right\}}_{k=1}^{\mathrm{\infty}}\end{array}$ is bounded.

As a matter of fact,

$$\begin{array}{}{\displaystyle {c}_{7}\ge I\left({x}_{k}\right)=-\frac{1}{2}{x}_{k}^{\mathrm{\top}}\mathit{\Omega}{x}_{k}+\sum _{t=1}^{T}G\left(t,{x}_{k}(t)\right)}\\ \\ \\ {\displaystyle \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=-\frac{1}{2}{\left({x}_{k}^{(1)}\right)}^{\mathrm{\top}}\mathit{\Omega}{x}_{k}^{(1)}+\sum _{t=1}^{T}\left[G\left(t,{x}_{k}(t)\right)-G\left(t,{x}_{k}^{(2)}(t)\right)\right]+\sum _{t=1}^{T}G\left(t,{x}_{k}^{(2)}(t)\right).}\end{array}$$

Thus,

$$\begin{array}{}{\displaystyle \sum _{t=1}^{T}G\left(t,{x}_{k}^{(2)}(t)\right)\le {c}_{7}+\frac{1}{2}{\left({x}_{k}^{(1)}\right)}^{\mathrm{\top}}\mathit{\Omega}{x}_{k}^{(1)}+\sum _{t=1}^{T}\left|G\left(t,{x}_{k}(t)\right)-G\left(t,{x}_{k}^{(2)}(t)\right)\right|}\\ \\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\le {c}_{7}+\frac{\overline{\lambda}}{2}{\u2225{x}_{k}^{(1)}\u2225}^{2}+\sum _{t=1}^{T}\left|f\left(t,{x}_{k}^{(1)}(t)+\xi {x}_{k}^{(2)}(t)\right)\right|\cdot \left|{x}_{k}^{(1)}(t)\right|}\\ \\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\le {c}_{7}+\frac{\overline{\lambda}}{2}{\u2225{x}_{k}^{(1)}\u2225}^{2}+{c}_{1}\sqrt{T}\u2225{x}_{k}^{(1)}(t)\u2225,}\end{array}$$

which means that $\begin{array}{}\left\{\sum _{t=1}^{T}G\left(t,{x}_{k}^{(2)}(t)\right)\right\}\end{array}$ is bounded. Here *ξ* ∈ (0, 1).

It comes from condition (*G*_{2}) that $\begin{array}{}{\left\{{x}_{k}^{(2)}\right\}}_{k=1}^{\mathrm{\infty}}\end{array}$ is bounded. If not, assume that $\begin{array}{}\u2225{x}_{k}^{(2)}\u2225\to +\mathrm{\infty}\end{array}$ as *k* → ∞. In that there are *c*_{k} ∈ ℝ, *k* ∈ ℕ, such that $\begin{array}{}{x}_{k}^{(2)}\end{array}$ = (*c*_{k}, *c*_{k},⋯, *c*_{k})^{⊤} ∈ *X*, then

$$\begin{array}{}{\displaystyle \u2225{x}_{k}^{(2)}\u2225={\left(\sum _{t=1}^{T}{\left|{x}_{k}^{(2)}(t)\right|}^{2}\right)}^{\frac{1}{2}}={\left(\sum _{t=1}^{T}{\left|{c}_{k}\right|}^{2}\right)}^{\frac{1}{2}}=\sqrt{T}\left|{c}_{k}\right|\to +\mathrm{\infty},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}k\to \mathrm{\infty}.}\end{array}$$

Since $\begin{array}{}G\left(t,{x}_{k}^{(2)}(t)\right)=G\left(t,{c}_{k}\right),\end{array}$ then $\begin{array}{}G\left(t,{x}_{k}^{(2)}(t)\right)\to +\mathrm{\infty}\end{array}$ as *k* → ∞. This contradicts the fact $\begin{array}{}\left\{\sum _{t=1}^{T}G\left(t,{x}_{k}^{(2)}(t)\right)\right\}\end{array}$ is bounded. Therefore, the functional *I*(*x*) satisfies the Palais-Smale condition. Therefore, it suffices to prove that *I*(*x*) satisfies the conditions (*I*_{1}) and (*I*_{2}) of Saddle Point Theorem.

First, we shall prove the condition (*I*_{2}). For any *x*^{(2)} ∈ *X*_{2}, *x*^{(2)} = (*x*^{(2)}(1), *x*^{(2)}(2),⋯, *x*^{(2)}(*T*))^{⊤}, there is *c* ∈ ℝ such that

$$\begin{array}{}{\displaystyle {x}^{(2)}(i)=c,\mathrm{\forall}i\in [1,T{]}_{\mathbb{Z}}.}\end{array}$$

It comes from (*G*_{2}) that there is a constant *c*_{8} > 0 such that *G*(*t*, *c*) > 0 for *t* ∈ ℤ and ∣*c*∣ > *c*_{8}.

Set

$$\begin{array}{}{\displaystyle {c}_{9}=\underset{n\in [1,T{]}_{\mathbb{Z}},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}|c|\le {c}_{8}}{min}G(t,c),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{c}_{10}=min\{0,{c}_{9}\}.}\end{array}$$

Thus,

$$\begin{array}{}{\displaystyle G(t,c)\ge {c}_{10},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall}(t,c)\in [1,T{]}_{\mathbb{Z}}\times \mathbb{R}.}\end{array}$$

Then

$$\begin{array}{}{\displaystyle I\left({x}^{(2)}\right)=\sum _{t=1}^{T}G(t,{x}^{(2)}(t))=\sum _{t=1}^{T}G(t,c)\ge T{c}_{10},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall}{x}^{(2)}\in {X}_{2}.}\end{array}$$

Next, we shall prove the condition (*I*_{1}). For any *x*^{(1)} ∈ *X*_{1}, by $\begin{array}{}({G}_{1}^{\prime}),\end{array}$ we have

$$\begin{array}{}{\displaystyle I\left({x}^{(1)}\right)=-\frac{1}{2}{\left({x}^{(1)}\right)}^{\mathrm{\top}}\mathit{\Omega}{x}^{(1)}+\sum _{t=1}^{T}G\left(t,{x}^{(1)}(t)\right)}\\ \\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\le -\frac{\underset{\_}{\lambda}}{2}{\u2225{x}^{(1)}\u2225}^{2}+{c}_{1}\sum _{t=1}^{T}\left|{x}^{(1)}(t)\right|+T{c}_{2}}\\ \\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\le -\frac{\underset{\_}{\lambda}}{2}{\u2225{x}^{(1)}\u2225}^{2}+{c}_{1}\sqrt{T}\u2225{x}^{(1)}\u2225+T{c}_{2}.\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}}\end{array}$$

Take

$$\begin{array}{}{\displaystyle \omega =T{c}_{10}.}\end{array}$$

Then, there exists a constant *ρ* > 0 large enough such that

$$\begin{array}{}{\displaystyle I\left({x}^{(1)}\right)\le \omega -1=\sigma <\omega ,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall}{x}^{(1)}\in {X}_{1},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\u2225{x}^{(1)}\u2225=\rho .}\end{array}$$

The conditions of (*I*_{1}) and (*I*_{2}) of Saddle Point Theorem are satisfied. In the light of Saddle Point Theorem, Theorem 3.1 holds. □

#### Proof of Theorem 3.3

Let {*x*_{k}}_{k∈ℕ} ⊂ *X* be such that {*I*(*x*_{k})}_{k∈ℕ} is bounded and *I*′(*x*_{k}) → 0 as *k* → ∞. Accordingly, for any *k* ∈ ℕ, there is a number *c*_{11} > 0 such that

$$\begin{array}{}{\displaystyle -{c}_{11}\le I\left({x}_{k}\right)\le {c}_{11}.}\end{array}$$

For *k* large enough, it comes from $\begin{array}{}\underset{k\to \mathrm{\infty}}{lim}{I}^{\prime}\left({x}_{k}\right)=0\end{array}$ that

$$\begin{array}{}{\displaystyle \left|\left({I}^{\prime}\left({x}_{k}\right),{x}_{k}\right)\right|\le \u2225{x}_{k}\u2225.}\end{array}$$

Since

$$\begin{array}{}{\displaystyle \left({I}^{\prime}\left({x}_{k}\right),{x}_{k}\right)=-{x}_{k}^{\mathrm{\top}}\mathit{\Omega}{x}_{k}+\sum _{t=1}^{T}f\left(t,{x}_{k}(t)\right){x}_{k}(t).}\end{array}$$

Accordingly, for *k* large enough, we have

$$\begin{array}{}{\displaystyle {c}_{11}+\frac{1}{2}\u2225{x}_{k}\u2225\ge I\left({x}_{k}\right)-\frac{1}{2}\left({I}^{\prime}\left({x}_{k}\right),{x}_{k}\right)}\\ \\ \\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\sum _{t=1}^{T}\left[G\left(t,{x}_{k}(t)\right)-\frac{1}{2}f\left(t,{x}_{k}(t)\right){x}_{k}(t)\right].}\end{array}$$

Denote

$$\begin{array}{}{\displaystyle {\mathit{\Gamma}}_{1}=\left\{t\in [1,T{]}_{\mathbb{Z}}:\left|{x}_{k}(t)\right|\ge \delta \right\};}\\ \\ {\displaystyle {\mathit{\Gamma}}_{2}=\left\{t\in [1,T{]}_{\mathbb{Z}}:\left|{x}_{k}(t)\right|<\delta \right\}.}\end{array}$$

Combining with (*G*_{3}), we have

$$\begin{array}{}{\displaystyle {c}_{11}+\frac{1}{2}\u2225{x}_{k}\u2225\ge \sum _{t=1}^{T}G\left(t,{x}_{k}(t)\right)-\frac{1}{2}\sum _{t\in {\mathit{\Gamma}}_{1}}^{T}f\left(t,{x}_{k}(t)\right){x}_{k}(t)-\frac{1}{2}\sum _{t\in {\mathit{\Gamma}}_{2}}^{T}f\left(t,{x}_{k}(t)\right){x}_{k}(t)}\\ \\ \\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\ge \sum _{t=1}^{T}G\left(t,{x}_{k}(t)\right)-\frac{\mu}{2}\sum _{t\in {\mathit{\Gamma}}_{1}}^{T}G\left(t,{x}_{k}(t)\right)-\frac{1}{2}\sum _{t\in {\mathit{\Gamma}}_{2}}^{T}f\left(t,{x}_{k}(t)\right){x}_{k}(t)}\\ \\ \\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\left(1-\frac{\mu}{2}\right)\sum _{t=1}^{T}G\left(t,{x}_{k}(t)\right)+\frac{1}{2}\sum _{t\in {\mathit{\Gamma}}_{2}}^{T}\left[\mu G\left(t,{x}_{k}(t)\right)-f\left(t,{x}_{k}(t)\right){x}_{k}(t)\right].}\end{array}$$

Set

$$\begin{array}{}{\displaystyle L(t,x)=\mu G\left(t,x\right)-f\left(t,x\right)x.}\end{array}$$

By the continuity of *L*(*t*, *x*) with respect to the first and second variables, we have that there is a constant *c*_{12} > 0 such that

$$\begin{array}{}{\displaystyle L(t,x)\ge -{c}_{12},}\end{array}$$

for all *t* ∈ [1, *T*]_{ℤ} and ∣*x*∣ ≥ *δ*. Hence,

$$\begin{array}{}{\displaystyle {c}_{11}+\frac{1}{2}\u2225{x}_{k}\u2225\ge \left(1-\frac{\mu}{2}\right)\sum _{t=1}^{T}G\left(t,{x}_{k}(t)\right)-\frac{1}{2}T{c}_{12},\left|x\right|\ge \delta .}\end{array}$$

It follows from (*G*_{4}) and (2.2) that

$$\begin{array}{}{\displaystyle {c}_{11}+\frac{1}{2}\u2225{x}_{k}\u2225\ge \left(1-\frac{\mu}{2}\right){c}_{3}\sum _{t=1}^{T}{\left|{x}_{k}(t)\right|}^{\nu}-\left(1-\frac{\mu}{2}\right){c}_{4}T-\frac{1}{2}T{c}_{12}}\\ \\ \\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\ge \left(1-\frac{\mu}{2}\right){c}_{3}{\tau}_{1}^{\nu}{\u2225{x}_{k}\u2225}^{\nu}-\left(1-\frac{\mu}{2}\right){c}_{4}T-\frac{1}{2}T{c}_{12}.}\end{array}$$

Let

$$\begin{array}{}{\displaystyle {c}_{13}=-\left(1-\frac{\mu}{2}\right){c}_{4}T-\frac{1}{2}T{c}_{12}.}\end{array}$$

Then,

$$\begin{array}{}{\displaystyle {c}_{11}+\frac{1}{2}\u2225{x}_{k}\u2225\ge \left(1-\frac{\mu}{2}\right){c}_{3}{\tau}_{1}^{\nu}{\u2225{x}_{k}\u2225}^{\nu}+{c}_{13}.}\end{array}$$

Thus,

$$\begin{array}{}{\displaystyle \left(1-\frac{\mu}{2}\right){c}_{3}{\tau}_{1}^{\nu}{\u2225{x}_{k}\u2225}^{\nu}-\frac{1}{2}\u2225{x}_{k}\u2225\le {c}_{11}-{c}_{13},}\end{array}$$

which means that $\begin{array}{}{\left\{\u2225{x}_{k}\u2225\right\}}_{k=1}^{\mathrm{\infty}}\end{array}$ is bounded. For the reason that *X* is a finite dimensional space, $\begin{array}{}{\left\{{x}_{k}\right\}}_{k=1}^{\mathrm{\infty}}\end{array}$ possesses a convergent subsequence. Accordingly, the Palais-Smale condition is proved.

To exploit the Saddle Point Theorem, we shall prove that the functional *I* satisfies the conditions (*I*_{1}) and (*I*_{2}).

First, we shall prove that the functional *I* satisfies the condition (*I*_{2}). For any *x*^{(2)} ∈ *X*_{2}, in that *Ω* *x*^{(2)} = 0, we have

$$\begin{array}{}{\displaystyle I\left({x}^{(2)}\right)=\sum _{t=1}^{T}G\left(t,{x}^{(2)}(t)\right).}\end{array}$$

On account of (*G*_{4}),

$$\begin{array}{}{\displaystyle I\left({x}^{(2)}\right)\ge {c}_{3}\sum _{t=1}^{T}{\left|{x}^{(2)}(t)\right|}^{\nu}-{c}_{4}T\ge -{c}_{4}T.}\end{array}$$

Then, we shall prove the condition (*I*_{1}). For any *x*^{(1)} ∈ *X*_{1}, combining with (*G*_{4}), we have

$$\begin{array}{}{\displaystyle I\left({x}^{(1)}\right)=-\frac{1}{2}{\left({x}^{(1)}\right)}^{\mathrm{\top}}\mathit{\Omega}{x}^{(1)}+\sum _{t=1}^{T}G\left(t,{x}^{(1)}(t)\right)}\\ \\ \\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\le -\frac{\underset{\_}{\lambda}}{2}{\u2225{x}^{(1)}\u2225}^{2}+{c}_{3}\sum _{t=1}^{T}{\left|{x}^{(1)}(t)\right|}^{\nu}+T{c}_{4}}\\ \\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\le -\frac{\underset{\_}{\lambda}}{2}{\u2225{x}^{(1)}\u2225}^{2}+{c}_{3}{\tau}_{2}^{\nu}\sqrt{T}{\u2225{x}^{(1)}\u2225}^{\nu}+T{c}_{4}.}\end{array}$$

Take

$$\begin{array}{}{\displaystyle \omega =-{c}_{4}T.}\end{array}$$

Owing to 1 < *ν* < 2, there exists a constant *ρ* > 0 large enough such that

$$\begin{array}{}{\displaystyle I\left({x}^{(1)}\right)\le \omega -1=\sigma <\omega ,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall}{x}^{(1)}\in {X}_{1},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\u2225{x}^{(1)}\u2225=\rho .}\end{array}$$

Hence, the condition (*I*_{1}) is satisfied.

As a result of Saddle Point Theorem, the BVP (1.1),(1.2) possesses at least one solution. The proof is complete. □

## Comments (0)

General note:By using the comment function on degruyter.com you agree to our Privacy Statement. A respectful treatment of one another is important to us. Therefore we would like to draw your attention to our House Rules.