In this section, we show our main results of the existence of at least three weak solutions for the problem (1).
To obtain our first result, we take the two positive constants θ and η in such a way that
$$\frac{(a_{2}+LTC^{2})T\eta^{2}}{ \sum_{k=1}^{m}F_{k}(\eta )}< \frac{(a_{3}LTC^{2})\theta^{2}}{ C^{2}\max_{t\leq \theta} [\sum_{k=1}^{m}F_{k}(t) ]} $$
and taking
$$\lambda\in\Lambda_{1}\, := \biggl]\frac{(a_{2}+LTC^{2})T\eta ^{2}}{ \sum_{k=1}^{m}F_{k}(\eta)}, \frac{(a_{3}LTC^{2})\theta^{2}}{ C^{2}\max_{t\leq\theta} [\sum_{k=1}^{m}F_{k}(t) ]} \biggr[, $$
set
$$ \delta_{\lambda}=\min \biggl\{ \frac{\theta ^{2}\frac{C^{2}\lambda}{(a_{3}LTC^{2})} \max_{t\leq \theta} [\sum_{k=1}^{m}F_{k}(t) ]}{\frac {C^{2}}{(a_{3}LTC^{2})}G^{\theta}}, \frac{\eta^{2} \frac{\lambda}{(a_{2}+LTC^{2})T} [\sum_{k=1}^{m}F_{k}(\eta) ]}{\frac {1}{(a_{2}+LTC^{2})T}G_{\eta}} \biggr\} $$
(3)
and
$$ \bar {\delta}_{\lambda}:=\min \biggl\{ \delta_{\lambda }, \frac{1}{ \max \{0,\frac{C^{2}}{(a_{3}LTC^{2})} \limsup_{t\to\infty}\frac{\sup\sum_{k=1}^{m}[G_{k}(t)]}{t^{2}} \}} \biggr\} , $$
(4)
where we say \({\rho}/{0}=+\infty\), so that, for example, \(\bar {\delta}_{\lambda}=+\infty\) when
$$\limsup_{t\to\infty}\frac{\sup\sum_{k=1}^{m}[G_{k}(t)]}{t^{2}}\leq0 $$
and \(G_{\eta}=G^{\theta}=0\).
Theorem 3.1
Suppose that
V
satisfies the assumptions (A1), (A2), and (A3). Assume that there exist two positive constants
θ
and
η
such that
\(\theta<\sqrt{T}C\eta\)
and

(A4)
\(\frac{ \max_{t\leq \theta} [\sum_{k=1}^{m}F_{k}(t) ]}{\theta^{2}}< \frac{a_{3}LTC^{2}}{C^{2}(a_{2}+LTC^{2})T}\frac{ \sum_{k=1}^{m}F_{k}(\eta)}{\eta^{2}}\), where
\(a_{3}=\min\{\frac {1}{2},a_{1}\}\);

(A5)
\(\limsup_{t\to+\infty }\frac{\sum_{k=1}^{m}[F_{k}(t)]}{t^{2}}\leq0\).
Then, for each
\(\lambda\in\Lambda_{1}\)
and for each arbitrary function
\(G_{k}\in {\mathrm{C}}^{1}({\mathbb{R}}^{N},\mathbb{R})\)
denoting
\(g_{k}(\xi)=\operatorname{grad}_{\xi}G_{k}(\xi)\)
for each
\(\xi\in\mathbb{R}^{N}\)
for
\(k=1,2,\ldots,m\), fulfilling the condition
$$\limsup_{t\to\infty}\frac{\sum_{k=1}^{m}[G_{k}(t)]}{t^{2}}< +\infty, $$
there exists
\(\bar {\delta}_{\lambda}>0\)
given by (4) such that, for each
\(\mu\in[0,\bar {\delta}_{\lambda}[\), the problem (1) admits at least three distinct weak solutions in
X.
Proof
Fix λ, \(G_{k}\) for \(k=1,2,\ldots,m\) and μ as in the conclusion. Our aim is applying Theorem 2.1 for the functionals \(\Phi, \Psi:X \to\mathbb{R}\), defined by
$$\Phi(u)= \int_{0}^{T} \biggl[\frac{1}{2} \bigl\vert \dot{u}(t) \bigr\vert ^{2}V \bigl(t,u(t) \bigr) \biggr]\,{ \mathrm{d}}t+ \int_{0}^{T}H \bigl(u(t) \bigr)\,{\mathrm{d}}t $$
and
$$\Psi(u)= \Biggl(\sum_{k=1}^{m}F_{k} \bigl(u(s_{k}) \bigr)+\frac{\mu }{\lambda} \sum _{k=1}^{m}G_{k} \bigl(u(s_{k}) \bigr) \Biggr). $$
It is easily observable that Ψ is a Gâteaux differentiable functional and sequentially weakly upper semicontinuous whose Gâteaux derivative at the point \(u\in X\) is the functional \(\Psi'(u)\in X^{*}\), given by
$$\Psi'(u)v=  \Biggl(\sum_{k=1}^{m} \bigl(f_{k} \bigl(u(s_{k}) \bigr),v(s_{k}) \bigr)+ \frac{\mu}{\lambda} \sum_{k=1}^{m} \bigl(g_{k} \bigl(u(s_{k}) \bigr),v(s_{k}) \bigr) \Biggr), $$
and \(\Psi':X \to X^{*}\) is a compact operator. Moreover, Φ is a Gâteaux differentiable functional whose Gâteaux derivative at the point \(u\in X\) is the functional \(\Phi'(u)\in X^{*}\), given by
$$\begin{aligned} \Phi'(u)v =& \int_{0}^{T} \bigl[ \bigl(\dot{u}(t),\dot{v}(t) \bigr) \bigl(V_{u} \bigl(t,u(t) \bigr),v(t) \bigr) \bigr]\,{ \mathrm{d}}t \\ &{}+ \int_{0}^{T} \bigl(h \bigl(u(t) \bigr),v(t) \bigr) \,{ \mathrm{d}}t \end{aligned}$$
for every \(v\in X\), while Proposition 2.3 shows that \(\Phi'\) admits a continuous inverse on \(X^{*}\). Furthermore, Φ is sequentially weakly lower semicontinuous. Indeed, let \(u_{n}\in X\) with \(u_{n}\to u\) weakly in X, taking weakly lower semicontinuity of the norm, we have \(\liminf_{n\to+\infty}\u_{n}\\geq\u\\) and \(u_{n}\to u\) uniformly on \([0,T]\). Hence, since V and H are continuous, we have
$$\begin{aligned}& \lim_{n\to+\infty}\frac{1}{2} \int_{0}^{T} \bigl[ \bigl\vert \dot{u_{n}}(t) \bigr\vert ^{2} V \bigl(t,u_{n}(t) \bigr) \bigr]\,{\mathrm{d}}t+ \int_{0}^{T}H \bigl(u_{n}(t) \bigr)\,{ \mathrm{d}}t \\& \quad \geq \frac{1}{2} \int_{0}^{T} \bigl[ \bigl\vert \dot{u}(t) \bigr\vert ^{2} V \bigl(t,u(t) \bigr) \bigr]\,{\mathrm{d}}t+ \int_{0}^{T}H \bigl(u(t) \bigr)\,{\mathrm{d}}t. \end{aligned}$$
Thus \(\liminf_{n\to+\infty}\Phi(u_{n})\geq\Phi(u)\), that is, Φ is sequentially weakly lower semicontinuous. Like the proof of Lemma 1 of [26], we observe that the weak solutions of the problem (1) are concisely the solutions of the equation \(\Phi'(u)\lambda\Psi'(u)=0\). Since \(L\xi\leqh(\xi) \leq L\xi\) for every \(\xi\in\mathbb{R}^{N}\), we have \(H(\xi)\leq L\xi^{2}\) for all \(\xi\in\mathbb{R}^{N}\). In parallel lines with the assumption (A1),
$$ \bigl(a_{3} LTC^{2} \bigr)\u \^{2} \leq\Phi(u)\leq \bigl(a_{4}+LTC^{2} \bigr) \u\^{2}, $$
(5)
where \(a_{4}=\min\{\frac{1}{2},a_{2}\}\). Put \(r:=\frac{\theta^{2}(a_{3}LTC^{2})}{C^{2}}\) and \(w(t):=\eta\) for every \(t\in[0,T]\). Because \(\min\{\frac{1}{2},a_{1}\}>{TLC}^{2}\), we have \(\min \{1,a_{1}\}>{TLC}^{2}\), which means \(a_{3}LTC^{2}>0\), and so \(r>0\). It is clear that \(w\in X\) and
$$\w\^{2}=T\eta^{2}. $$
Since \(\theta<\sqrt{T}C\eta\), using (5), we have \(0< r<\Phi(w)\). Taking (2) into account, from (5) we observe that
$$\begin{aligned} \Phi^{1} \bigl( ]{}\infty,r [ \bigr)&= \bigl\{ u\in X; \Phi(u)\leq r \bigr\} \\ &\subseteq \bigl\{ u\in X; \bigl(a_{3}LTC^{2} \bigr)\u \^{2}\leq r \bigr\} \\ &\subseteq \bigl\{ u\in X; \bigl\vert u(t) \bigr\vert \leq\theta\text{ for each } t\in [0,T] \bigr\} , \end{aligned}$$
and it follows that
$$\begin{aligned} \sup_{u\in\Phi^{1}(]\infty,r])}\Psi(u) &=\sup_{u\in\Phi^{1}(]\infty,r])} \Biggl[ \sum _{k=1}^{m}F_{k} \bigl(u(s_{k}) \bigr) \frac{\mu}{\lambda} \sum _{k=1}^{m}G_{k} \bigl(u(s_{k}) \bigr) \Biggr] \\ &\leq \max_{\xi\leq \theta} \Biggl[\sum_{k=1}^{m}F_{k}( \xi) \Biggr]+\frac{\mu}{\lambda }G^{\theta}. \end{aligned}$$
Moreover, we have
$$\begin{aligned} \Psi(w)&=\sum_{k=1}^{m}F_{k} \bigl(w(t) \bigr) \frac{\mu}{\lambda}\sum_{k=1}^{m}G_{k} \bigl(w(t) \bigr) \\ &\geq\sum_{k=1}^{m}F_{k}(\eta) +\frac{\mu}{\lambda}G_{\eta}. \end{aligned}$$
So, we obtain
$$\begin{aligned} \frac{ \sup_{u\in\Phi^{1}(]\infty,r])}\Psi(u)}{r} &=\frac{ \sup_{u\in\Phi^{1}(]\infty,r])} [\sum_{k=1}^{m} [F_{k}(u(s_{k}))+\frac{\mu}{\lambda}G_{k}(u(s_{k}))] ]}{r} \\ &\leq \frac{ \max_{\xi\leq \theta} [\sum_{k=1}^{m}F_{k}(\xi) ]+\frac{\mu}{\lambda }G^{\theta}}{ \frac{\theta^{2}(a_{3}LTC^{2})}{C^{2}}} \end{aligned}$$
(6)
and
$$\begin{aligned} \frac{\Psi(w)}{ \Phi(w)}&\geq\frac{ \sum_{k=1}^{m}F_{k}(w(t))\frac {\mu}{\lambda}\sum_{k=1}^{m}G_{k}(w(t))}{ \frac{(a_{2}+LTC^{2})\eta^{2}}{C^{2}}} \\ &\geq\frac{ \sum_{k=1}^{m}F_{k}(\eta)+\frac{\mu }{\lambda}G_{\eta}}{ \frac{(a_{2}+LTC^{2})\eta^{2}}{C^{2}}}. \end{aligned}$$
(7)
Since \(\mu<\delta_{\lambda}\), one has
$$\mu< \frac{\theta^{2}\frac{C^{2}}{a_{3}LTC^{2}} \lambda \max_{\xi\leq \theta} [\sum_{k=1}^{m}F_{k}(\xi) ]}{\frac {C^{2}}{(a_{3}LTC^{2})}G^{\theta}}, $$
this means
$$\frac{ \max_{\xi\leq \theta} [\sum_{k=1}^{m}F_{k}(\xi) ]+\frac{\mu}{\lambda} G^{\theta}}{ \frac{\theta^{2}(a_{3}LTC^{2})}{C^{2}}} < \frac{1}{\lambda}. $$
Furthermore,
$$\mu< \frac{\eta^{2}\frac{C^{2}}{a_{3}LTC^{2}} \lambda [\sum_{k=1}^{m}F_{k}(\eta) ]}{ \frac{C^{2}}{a_{3}LTC^{2}}G_{\eta}}, $$
this means
$$\frac{ \sum_{k=1}^{m}F_{k}(\eta)+\frac{\mu}{\lambda} G_{\eta}}{ \frac{\eta^{2}(a_{3}LTC^{2})}{C^{2}}}>\frac {1}{\lambda}. $$
Then
$$ \frac{ \max_{\xi\leq \theta} [\sum_{k=1}^{m}F_{k}(\xi) ]+\frac{\mu}{\lambda }G^{\theta}}{ \frac{\theta^{2}(a_{3}LTC^{2})}{C^{2}}} < \frac{1}{\lambda}< \frac{ \sum_{k=1}^{m}F_{k}(\eta )+\frac{\mu}{\lambda} G_{\eta}}{ \frac{\eta^{2}(a_{3}LTC^{2})}{C^{2}}}. $$
(8)
Hereupon, from (6)(8) we infer that the condition (a_{1}) of Theorem 2.1 is achieved. Eventually, since \(\mu<\bar {\delta}_{\lambda}\), we can fix \(l>0\) in such a manner that
$$\limsup_{\xi\to\infty}\frac{\sum_{k=1}^{m} [G_{k}(\xi) ]}{\xi^{2}}< l $$
and \(\mu l<\frac{a_{3}LTC^{2}}{ C^{2}}\). Therefore, there exists a constant q such that
$$ \sum_{k=1}^{m} \bigl[G_{k}(u) \bigr]\leq l u^{2}+q\quad \text{for all } u\in\mathbb{R}^{N} $$
(9)
for \(k=1,2,\ldots,m\). Now, fix \(0<\varepsilon< \frac{a_{3}LTC^{2}}{C^{2} \lambda}\frac {\mu l}{\lambda}\). Owing to the assumption (A4) there is a constant \(q_{\varepsilon}\) such that
$$ \sum_{k=1}^{m} \bigl[F_{k}(u) \bigr]\leq\varepsilon u^{2}+q_{\varepsilon} \quad \text{for all } u\in\mathbb{R}^{N} $$
(10)
for \(k=1,2,\ldots,m\). Due to (5), (9), and (10) we have
$$\begin{aligned} \Phi(u)\lambda \Psi(u) =& \int_{0}^{T} \biggl[\frac{1}{2} \bigl\vert \dot {u}(t) \bigr\vert ^{2}V \bigl(t,u(t) \bigr)+H \bigl(u(t) \bigr) \biggr]\,{\mathrm{d}}t \\ &{} \lambda \Biggl[\sum_{k=1}^{m} \biggl[F_{k} \bigl(u(s_{k}) \bigr)+\frac{\mu}{\lambda }G \bigl(u(s_{k}) \bigr) \biggr] \Biggr] \\ \geq& \bigl(a_{3}LTC^{2} \bigr)\u\^{2}\lambda \varepsilon u^{2}\lambda q_{\varepsilon}\mu l u^{2} \mu q \\ \geq& \bigl(a_{3}LTC^{2}\lambda C^{2} \varepsilon\mu C^{2} l \bigr)\u\^{2} \lambda q_{\varepsilon}\mu q. \end{aligned}$$
This means that the functional \(\Phi\lambda\Psi\) is coercive, and the assumption (a_{2}) of Theorem 2.1 is verified. From (6) and (8),
$$\lambda\in \, \biggl]\frac{\Phi(w)}{\Psi(w)}, \frac{r}{\sup_{\Phi(u)\leq r}\Psi(u)} \biggr[ $$
and Theorem 2.1 (with \(\bar {v}=w\)) ensures that the problem (1) possesses at least three weak solutions in X. □
We now offer another version of Theorem 3.1 within which no asymptotic condition on the nonlinear term is necessary; contrarily, each constituent of \(f_{k}\) and \(g_{k}\) for \(k=1,2,\ldots,m\) is considered to be negative.
Fix positive constants \(\theta_{1}\), \(\theta_{2}\), and η in such a way that
$$\begin{aligned}& \frac{3}{2}\frac{(a_{2}+LTC^{2})T\eta^{2}}{ [\sum_{k=1}^{m}F(\eta) ]} \\& \quad < \frac{a_{3}LTC^{2}}{C^{2}}\min \biggl\{ \frac{\theta_{1}^{2}}{ \max_{\xi\leq \theta_{1}} [\sum_{k=1}^{m}F_{k}(\xi) ]}, \frac{\theta_{2}^{2}}{2 \max_{\xi\leq \theta_{2}} [\sum_{k=1}^{m}F_{k}(\xi) ]} \biggr\} \end{aligned}$$
and put
$$\begin{aligned} \Lambda_{2} :=& \, \biggl] \frac{3}{2}\frac{(a_{2}+LTC^{2}) T\eta^{2}}{ [\sum_{k=1}^{m}F(\eta) ]}, \\ &\frac{a_{3}LTC^{2}}{C^{2}}\min \biggl\{ \frac{\theta_{1}^{2}}{ \max_{\xi\leq \theta_{1}} [\sum_{k=1}^{m}F_{k}(\xi) ]}, \frac{\theta_{2}^{2}}{2 \max_{\xi\leq \theta_{2}} [\sum_{k=1}^{m}F_{k}(\xi) ]} \biggr\} \biggr[. \end{aligned}$$
By the above symbolization, we obtain the following multiplicity result.
Theorem 3.2
Order the Banach space
X
by the positive cone
\(X^{+}\) (see Section 5.4 of [51]), and suppose that
V
satisfies in the assumptions (A1), (A2), and (A3), \(F_{k}\in C^{1}({\mathbb{R}}^{N},\mathbb{R})\), each component of
\(f_{k}(\xi)= \operatorname{grad}_{\xi}F_{k}(\xi)\)
for
\(k=1,2,\ldots,m\)
is negative and there exist three positive constants
\(\theta_{1}\), \(\theta_{2}\), and
η
such that
\(\theta_{1}< C\sqrt{\frac{T}{2}}\eta<\frac{\theta_{2}}{2}\sqrt{\frac {a_{3}LTC^{2}}{a_{2}+LTC^{2}}}\)
where
\(a_{3}=\min\{\frac{1}{2},a_{1}\}\)
and
$$\begin{aligned}& (\mathrm{B}1)\quad \max \biggl\{ \frac{ \max_{\xi\leq \theta_{1}} [\sum_{k=1}^{m}F_{k}(\xi) ]}{\theta _{1}^{2}}, \frac{ 2\max_{\xi\leq \theta_{2}} [\sum_{k=1}^{m}F_{k}(\xi) ]}{\theta _{2}^{2}} \biggr\} \\& \hphantom{(\mathrm{B}1)\quad}\quad < \frac{2}{3}\frac{a_{3}LTC^{2}}{C^{2}(a_{2}+LTC^{2})T}\frac{ [\sum_{k=1}^{m}F(\eta) ]}{\eta^{2}}. \end{aligned}$$
Then, for each
\(\lambda\in\Lambda_{2}\)
and for every arbitrary function
\(G_{k}\in{\mathrm{C}}^{1}({\mathbb{R}}^{N},\mathbb{R})\)
such that each component of
\(g_{k}(\xi)=\operatorname{grad}_{\xi}G_{k}(\xi)\)
for every
\(\xi\in\mathbb{R}^{N}\)
is negative for
\(k=1,2,\ldots,m\), there exists
\(\delta^{*}_{\lambda}>0\)
defined by
$$\begin{aligned}& \min \biggl\{ \frac{(a_{3}LTC^{2})\theta_{1}^{2}C^{2}\lambda \max_{\xi\leq \theta_{1}} [\sum_{k=1}^{m}F(\xi) ]}{C^{2}G^{\theta _{1}}}, \\& \quad \frac{ (a_{3}LTC^{2})\theta_{2}^{2}2C^{2}\lambda \max_{\xi\leq \theta_{2}} [\sum_{k=1}^{m}F_{k}(\xi) ]}{2C^{2}G^{\theta _{2}}} \biggr\} \end{aligned}$$
such that, for each
\(\mu\in[0,\delta^{*}_{\lambda}[\), the problem (1) possesses at least three weak solutions
\(u_{1}\), \(u_{2}\), and
\(u_{3}\)
such that
\(u_{i}(t)\in X^{+}\) (or
\(u_{i}(t)\geq0\)) for all
\(t\in[0,T]\)
and
\(i=1,2,3\).
Proof
Fix λ, \(G_{k}\) for \(k=1,2,\ldots,m\) and μ as in the conclusion and take X, Φ, and Ψ as in the proof of Theorem 3.1. Obviously, the regularity assumptions of Theorem 2.2 on Φ and Ψ are satisfied. Our goal is to check (b_{1}) and (b_{2}). For this purpose, put \(w(t)=\eta\) for every \(t\in[0,T]\),
$$r_{1}:= \frac{(a_{3}LTC^{2})\theta_{1}^{2}}{C^{2}} $$
and
$$r_{2}:= \frac{(a_{3}LTC^{2})\theta_{2}^{2}}{C^{2}}. $$
According to condition \(\theta_{1}< C\sqrt{\frac{T}{2}}\eta<\frac{\theta_{2}}{2} \sqrt{\frac{a_{3}LTC^{2}}{a_{2}+LTC^{2}}}\), and from (5), we get
$$2r_{1}< \Phi(w)< \frac{r_{2}}{2}. $$
Since \(\mu<\delta^{*}_{\lambda}\) and \(G_{\eta}=0\), one has
$$\begin{aligned} \frac{ \sup_{u\in\Phi^{1}(]\infty,r_{1}])}\Psi(u)}{r_{1}} &=\frac{ \sup_{u\in\Phi^{1}(]\infty,r_{1}])} [\sum_{k=1}^{m} [F_{k}(u(s_{k}))+\frac{\mu}{\lambda}G_{k}(u(s_{k}))] ]}{r_{1}} \\ &\leq\frac{ \max_{\xi\leq \theta_{1}} [\sum_{k=1}^{m}F_{k}(\xi) ]+\frac{\mu }{\lambda}G^{\theta_{1}}}{ \frac{(a_{3}LTC^{2})\theta_{1}^{2}}{C^{2}}} \\ &< \frac{1}{\lambda} < \frac{2}{3}\frac{ [\sum_{k=1}^{m}F_{k}(\eta ) ]+\frac{\mu}{\lambda} G\mathcal{\eta}}{ \frac{(a_{2}+LTC^{2})T\eta^{2}}{C^{2}}} \\ &\leq\frac{2}{3}\frac{\Psi(w)}{\Phi(w)} \end{aligned}$$
and
$$\begin{aligned} \frac{2 \sup_{u\in\Phi^{1}(]\infty,r_{2}])}\Psi (u)}{r_{2}}&=\frac{2 \sup_{u\in\Phi^{1}(]\infty ,r_{2}])} [\sum_{k=1}^{m} [F_{k}(u(s_{k}))+\frac{\mu}{\lambda}G_{k}(u(s_{k}))] ]}{r_{2}} \\ &\leq\frac{2 \sup_{\xi\leq\theta_{2}} [\sum_{k=1}^{m}F_{k}(\xi) ]+2\frac{\mu}{\lambda}G^{\theta_{2}}}{ \frac{(a_{3}LTC^{2})\theta_{2}^{2}}{C^{2}}} \\ &< \frac{1}{\lambda} < \frac{2}{3}\frac{ [\sum_{k=1}^{m}F_{k}(\eta ) ]+\frac{\mu}{\lambda}G_{\eta}}{ \frac{(a_{2}+LTC^{2})T\eta^{2}}{C^{2}}} \\ &\leq\frac{2}{3}\frac{\Psi(w)}{\Phi(w)}. \end{aligned}$$
Therefore, (b_{1}) and (b_{2}) of Theorem 2.2 are fulfilled. In the following, we show that \(\Phi\lambda\Psi\) satisfies the assumption 2 of Theorem 2.2. Let \(u_{1}\) and \(u_{2}\) be two local minima for \(\Phi\lambda\Psi\). Then \(u_{1}\) and \(u_{2}\) are critical points for \(\Phi\lambda\Psi\), and, thus, they are weak solutions for the problem (1). We want to show that they are nonnegative. Let \(u_{0}\) be a nontrivial weak solution of problem (1). Arguing by a contradiction, assume that the set \(\mathcal{A}= \{t \in[0,T] : u_{0}(t)<0 \}= \{t \in[0,T] : 0u_{0}(t)\in X^{+}, u_{0}(t)\neq0 \}\) is nonempty and its measure is positive. Put
$$\bar{v}(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} 0,& 0\leq u_{0}(t), \\ u_{0}(t), & u_{0}(t)< 0 \end{array}\displaystyle \right . $$
for all \(t \in[0,T]\). Clearly, \(\bar{v} \in X\). Since \(u_{0}\) is a weak solution of (1) we have
$$\begin{aligned}& \int_{0}^{T} \bigl[ \bigl(\dot{u}_{0}(t), \dot{\bar{v}}(t) \bigr) \bigl(V_{u_{0}} \bigl(t,u_{0}(t) \bigr), \bar{v}(t) \bigr)+\bigl(h \bigl(u_{0}(t),\bar{v}(t) \bigr)\bigr) \bigr]\,{ \mathrm{d}}t \\& \quad =\lambda\sum_{k=1}^{m} \bigl(f_{k} \bigl(u_{0}(s_{k}) \bigr), \bar{v}(s_{k}) \bigr)\mu \sum_{k=1}^{m} \bigl(g_{k} \bigl(u_{0}(s_{k}) \bigr), \bar{v}(s_{k}) \bigr). \end{aligned}$$
Thus, from our sign assumptions on the data, since \(L\xi^{2}\leq(h(\xi),\xi) \leq L\xi^{2}\) for every \(\xi\in\mathbb{R}^{N}\), we have
$$\begin{aligned} \begin{aligned} 0&\leq \bigl(\min \{1, a_{1}\}{TLC}^{2} \bigr) \u_{0}\_{X(\mathcal{A})}^{2}\leq \int_{\mathcal{A}} \bigl( \bigl\vert \dot{u}_{0}(t) \bigr\vert ^{2}+a_{1} \bigl\vert u_{0}(t) \bigr\vert ^{2}L \bigl\vert u_{0}(t) \bigr\vert ^{2} \bigr)\,{\mathrm{d}}t \\ &\leq \int_{0}^{T} \bigl[ \bigl(\dot{u}_{0}(t), \dot{u}_{0}(t) \bigr) \bigl(V_{u_{0}} \bigl(t,u_{0}(t) \bigr),u_{0}(t) \bigr)+ \bigl(h \bigl(u_{0}(t) \bigr),u_{0}(t) \bigr) \bigr]\,{\mathrm{d}}t\leq0. \end{aligned} \end{aligned}$$
Hence, \(u_{0}=0\) in \(\mathcal{A}\) and this is absurd. Then we conclude \(u_{1}(t)\geq0\) and \(u_{2}(t)\geq0\) for every \(t\in[0,T]\). Thus, it follows that \(su_{1}+(1s)u_{2}\geq0\) for all \(s\in[0,1]\), and that
$$(\lambda f_{k}+\mu g_{k}) \bigl(su_{1}+(1s)u_{2} \bigr)\geq0 \quad \text{for } k=1,2,\ldots,m, $$
and consequently, \(\Psi(su_{1}+(1s)u_{2})\geq0\), for every \(s\in [0,1]\). Hence, since all the hypotheses of Theorem 2.2 are satisfied, it follows that, for every
$$\lambda\in \, \biggl]\frac{3}{2}\frac{\Phi(w)}{\Psi(w)}, \min \biggl\{ \frac{r_{1}}{ \sup_{u\in\Phi^{1}(]\infty,r_{1}[)}\Psi (u)}, \frac{{r_{2}}/{2}}{ \sup_{u\in\Phi^{1} (]\infty,r_{2}[)}\Psi(u)} \biggr\} \biggr[, $$
the functional \(\Phi\lambda\Psi\) has at least three distinct critical points \(u_{i}\) for \(i=1,2,3\), such that \(0\leq u_{i}(t)<\theta_{2}\) for all \(t\in[0,T]\) and \(i=1,2,3\), which are the weak solutions of the problem (1), and the favorable result is achieved. □
In the following, we present a special case of Theorem 3.1.
Corollary 3.3
Suppose that
V
satisfies the assumptions (A1), (A2), and (A3), and
$$\liminf_{\xi\to0}\frac{ \max_{t\leq \xi} [\sum_{k=1}^{m}F_{k}(t) ]}{\xi^{2}}= \limsup_{\xi \to +\infty} \frac{\sum_{k=1}^{m}[F_{k}(\xi)]}{\xi^{2}}=0. $$
Then there is
\(\lambda^{*}>0\)
such that for each
\(\lambda>\lambda^{*}\)
and every arbitrary function
\(G_{k}\in{\mathrm{C}}^{1}({\mathbb{R}}^{N},\mathbb{R})\), denoting
\(g_{k}(\xi)=\operatorname{grad}_{\xi}G_{k}(\xi)\)
for every
\(\xi\in\mathbb{R}^{N}\)
for
\(k=1,2,\ldots,m\), satisfying the asymptotical condition
$$\limsup_{t\to\infty}\frac{\sum_{k=1}^{m}[G_{k}(t)]}{ t^{2}}< +\infty, $$
there exists
\(\delta^{*}_{\lambda}>0\)
such that, for each
\(\mu\in[0,\delta^{*}_{\lambda}[\), the problem (1) admits at least three distinct weak solutions in
X.
Proof
Fix \(\lambda>\lambda^{*}:=\frac{(a_{2}+LTC^{2})T\eta^{2}}{ \sum_{k=1}^{m}F_{k}(\eta)}\) for some \(\eta>0\). Recalling
$$\liminf_{\xi\to0}\frac{ \max_{t\leq \xi} [\sum_{k=1}^{m}F_{k}(t) ]}{\xi^{2}}=0, $$
there exists a sequence \(\{\theta_{n}\}\subset\, ]0,+\infty[\) with this feature that \(\lim_{n\to\infty} \theta_{n}=0\) and
$$\lim_{n\to\infty}\frac{ \max_{t\leq \theta_{n}} [\sum_{k=1}^{m}F_{k}(t) ]}{\theta_{n}^{2}}=0. $$
Hence, there exists \(\bar {\theta}>0\) such that
$$\frac{ \max_{t\leq \bar {\theta}} [\sum_{k=1}^{m}F_{k}(t) ]}{\bar {\theta}^{2}}< \min \biggl\{ \frac{a_{3}LTC^{2}}{C^{2}(a_{2}+LTC^{2})T}\frac{ \sum_{k=1}^{m}F_{k}(\eta)}{\eta^{2}}; \frac{a_{3}LTC^{2}}{ \lambda C^{2}} \biggr\} $$
and \(\bar {\theta}<\sqrt{T}C\eta\). The conclusion follows from Theorem 3.1. □
Now, as an example, we present the following consequence of Theorem 3.2 with \(m=T=N=1\).
Corollary 3.4
Suppose that
V
satisfies the assumptions (A1), (A2), and (A3), \(f_{1}:{\mathbb{R}}\to\mathbb{R}\)
is a negative continuous function and
\(h:{\mathbb{R}}\to\mathbb{R}\)
is a Lipschitz continuous function with the Lipschitz constant
\(L > 0\)
such that
\(h(0)=0\), \(\min\{1,a_{1}\}>2L\), and
\(a_{2}+2L<16(a_{3}2L)\)
where
\(a_{3}=\min\{\frac{1}{2},a_{1}\}\). Furthermore, assume that
$$\lim_{\xi\to0^{+}}\frac{f_{1}(\xi)}{\xi} =0 $$
and
$$\int_{0}^{1/2}f_{1}(x)\, \mathrm{d}x< \frac{3}{32}\frac{a_{2}+2L}{a_{3}2L} \int _{0}^{4}f_{1}(x)\, \mathrm{d}x. $$
Then, for every
\(\lambda\in\, ] \frac{3}{8}\frac{a_{2}+2L}{ \int _{0}^{1/2}f_{1}(x)\,\mathrm{d}x}, \frac{4(a_{3}2L) }{\int_{0}^{4}f_{1}(x)\,\mathrm{d}x} [\), and for every arbitrary negative continuous function
\(g_{1}:{\mathbb{R}}\to\mathbb{R}\), there exists
\(\delta^{*}_{\lambda}>0\)
such that, for each
\(\mu\in[0,\delta^{*}_{\lambda}[\), the problem
$$ \left \{ \textstyle\begin{array}{l} u''(t)+V_{u}(t,u(t))=h(u(t)),\quad t\neq s_{1}, \\ \Delta{u'(s_{1})}=\lambda f_{1}(u(s_{1}))+ \mu g_{1}(u(s_{1})), \\ u(0) u(1)= u'(0) u'(1)=0, \end{array}\displaystyle \right . $$
(11)
possesses at least three weak solutions
\(u_{1}\), \(u_{2}\), and
\(u_{3}\)
such that
\(0\leq u_{i}(t)<4\)
for all
\(t\in[0,T]\)
and
\(i=1,2,3\).
Proof
Our goal is to use Theorem 3.2 by choosing \(m=T=N=1\), \(\theta_{2}=4\) and \(\eta=\frac{1}{2}\). Since \(c=\sqrt{2}\), we observe that
$$\frac{3}{2}\frac{(a_{2}+LTC^{2})T\eta^{2}}{ [\sum_{k=1}^{m}F_{k}(\eta) ]}= \frac {3}{8}\frac{a_{2}+2L}{ \int_{0}^{1/2}f_{1}(x)\,\mathrm{d}x} $$
and
$$\frac{a_{3}LTC^{2}}{C^{2}} \frac{\theta_{2}^{2}}{2 \max_{\xi\leq \theta_{2}} [\sum_{k=1}^{m}F_{k}(\xi) ]}=\frac{4(a_{3}2L) }{\int_{0}^{4}f_{1}(x)\,\mathrm{d}x}. $$
Moreover, since \(\lim_{\xi\to 0^{+}}\frac{f_{1}(\xi)}{\xi}=0\), one has
$$\lim_{\xi\to0^{+}} \frac{ \int_{0}^{\xi}f_{1}(x)\,\mathrm{d}x}{\xi^{2}}=0. $$
Then there exists a positive constant \(\theta_{1}<\frac{1}{2}\) such that
$$\frac{ \int_{0}^{\theta_{1}}f_{1}(x)\,\mathrm{d}x}{\theta_{1}^{2}}> \frac{4}{3}\frac{a_{3}2L}{a_{2}+2L} \int_{0}^{\frac{1}{2}}f_{1}(x)\,\mathrm{d}x $$
and
$$\frac{\theta_{1}^{2}}{ \int_{0}^{\theta _{1}}f_{1}(x)\,\mathrm{d}x}< \frac{8}{ \int_{0}^{4}f_{1}(x)\,\mathrm{d}x}. $$
Finally, an easy calculation shows that all hypotheses of Theorem 3.2 are fulfilled, and the conclusion follows. □
Remark 3.1
From Assumptions (A1), (A2), and (A3), we can show, by the same reasoning as given in Theorem 4 of [26], that the problem (1) when \(h\equiv0\) does not possess any nonzero weak solution in the cases where impulsive terms are zero. Consequently, the ensured weak solutions for the problem (1) when \(h\equiv0\) in Theorems 3.1 and 3.2 and in Corollary 3.3 are generated by impulses when impulsive terms \(f_{k},g_{k}\neq0\) for some \(1\leq k\leq m\), as well as for the problem (11) when \(h\equiv0\) in Corollary 3.4 are generated by impulses when impulsive terms \(f_{1},g_{1}\neq0\).
Remark 3.2
The methods used here can be applied studying discrete boundary value problems as in [52].