In this section, we will study the existence and multiplicities of problem (1.1). First, we give a Lemma.
Lemma 3.1
Assume that (H1) and (H2) hold. Then the function
\(\varphi:E^{\alpha,p}\rightarrow\mathbb{R}\)
defined by (2.5) is continuous differentiable and weakly sequentially lower semi-continuous. Moreover, it satisfies the Palais-Smale condition.
Proof
From the continuity of f and \(I_{j}\), we know that φ and \(\varphi^{\prime}\) are continuous and differentiable. Let \(u_{k}\rightharpoonup u\) in \(E^{\alpha,p}\). Then \(\|u\|_{E^{\alpha ,p}}\leq\underline{\lim}_{k\rightarrow\infty}\inf\|u_{k}\| _{E^{\alpha,p}}\), and \(\{u_{k}\}\) converges uniformly to u in \(C([0,T])\). So
$$ \begin{aligned}[b] \lim_{k\rightarrow\infty}\inf\varphi(u_{k})&=\lim _{k\rightarrow\infty} \Biggl\{ \frac{1}{p}\|u_{k} \|^{p}_{E^{\alpha,p}} +\sum_{j=1}^{m} \int_{0}^{u_{k}(t_{j})}I_{j}(s)\,ds- \int _{0}^{T}F \bigl(t,u_{k}(t) \bigr)\,dt \Biggr\} \\ &\geq\frac{1}{p}\|u\|^{p}_{E^{\alpha,p}}+\sum _{j=1}^{m} \int _{0}^{u(t_{j})}I_{j}(s)\,ds- \int_{0}^{T}F \bigl(t,u(t) \bigr)\,dt=\varphi(u), \end{aligned} $$
(3.1)
which implies that φ is weakly sequentially lower semi-continuous.
We will verify that φ satisfies the Palais-Smale condition. Assume that \(\{u_{k}\}_{k\in N}\subset E^{\alpha,p}\) is a sequence such that \(\{\varphi(u_{k})\}_{k\in N}\) is bounded and \(\lim_{k\rightarrow\infty}\varphi^{\prime}(u_{k})=0\). We firstly prove that \(\{u_{k}\}_{k\in N}\) is bounded in \(E^{\alpha,p}\). Obviously, there exists a constant \(c>0\) such that
$$ \big|\varphi(u_{k})\big|\leq c,\qquad \varphi^{\prime}(u_{k}) \rightarrow 0 \quad\text{as } k\rightarrow\infty. $$
(3.2)
Then we have
$$ \int_{0}^{T}f \bigl(t,u_{k}(t) \bigr)u_{k}(t)\,dt =\|u_{k}\|^{p}_{E^{\alpha,p}}+ \sum_{j=1}^{m}I_{j} \bigl(u_{k}(t_{j}) \bigr)-\varphi ^{\prime}(u_{k})u_{k}(t). $$
(3.3)
From condition (H1), we have
$$\vartheta \int_{0}^{u_{k}(t_{j})}I_{j}(s)\,ds-I_{j} \bigl(u_{k}(t_{j}) \bigr)u_{k}(t_{j}) \geq (\vartheta-\mu) \int_{0}^{u_{k}(t_{j})}I_{j}(s)\,ds\geq0\quad (\mu> \vartheta), $$
which together with (3.2), (3.3) and the condition (H2) makes
$$\begin{aligned} \vartheta\varphi(u_{k})-\varphi^{\prime}(u_{k})u_{k}(t) \geq{}& \biggl(\frac {\vartheta}{p}-1 \biggr)\|u_{k}\|^{p}_{E^{\alpha,p}} +\sum_{j=1}^{m} \biggl(\vartheta \int _{0}^{u_{k}(t_{j})}I_{j}(s) \,ds-I_{j} \bigl(u_{k}(t_{j}) \bigr)u_{k}(t_{j}) \biggr) \\ & + \int_{0}^{T} \bigl(f \bigl(t,u_{k}(t) \bigr)u_{k}(t)-\vartheta F \bigl(t,u_{k}(t) \bigr) \bigr) \,dt \\ \geq{}& \biggl(\frac{\vartheta}{p}-1 \biggr)\|u_{k}\|^{p}_{E^{\alpha ,p}} \quad( \text{since }\vartheta>p), \end{aligned} $$
which implies \(\{u_{k}\}\) is bounded in \(E^{\alpha,p}\).
Since \(E^{\alpha,p}\) is a reflexive Banach space, going if necessary to a subsequence, we can assume that \(u_{k}\rightharpoonup u\) in \(E^{\alpha,p}\), \(u_{k}\rightarrow u\) in \(L^{p}([0,T])\) and \(u_{k}\rightarrow u\) uniformly in \(C([0,T])\). Hence
$$ \textstyle\begin{cases} \int_{0}^{T}(f(t,u_{k}(t))-f(t,u(t)))(u_{k}(t)-u(t))\,dt\rightarrow0,\\ \sum_{j=1}^{m}(I_{j}(u_{k}(t_{j}))-I_{j}(t,u(t_{j})))(u_{k}(t_{j})-u(t_{j}))\rightarrow0, \end{cases} $$
(3.4)
as \(k\rightarrow\infty\). Moreover, by \(\varphi^{\prime }(u_{k})\rightarrow0\) as \(k\rightarrow\infty\), we have
$$ \bigl\langle \varphi^{\prime}(u_{k})-\varphi^{\prime}(u),u_{k}-u \bigr\rangle \leq \big\| \varphi^{\prime}(u_{k})\big\| _{(E^{\alpha,p})^{*}} \cdot \|u_{k}-u\|_{E^{\alpha,p}}- \bigl\langle \varphi^{\prime }(u),u_{k}-u \bigr\rangle \rightarrow0 $$
(3.5)
as \(k\rightarrow\infty\).
Let
$$\phi(p,\alpha,k):= \int_{0}^{T} \bigl(\big|{}^{c}_{0}D^{\alpha }_{t}u_{k}(t)\big|^{p-2} {^{c}_{0}D^{\alpha}_{t}u_{k}(t)}-\big|{}^{c}_{0}D^{\alpha }_{t}u(t)\big|^{p-2} {^{c}_{0}D^{\alpha}_{t}u(t)} \bigr) \bigl({{}_{0}^{c}D^{\alpha}_{t}}u_{k}(t)-{{}_{0}^{c}D^{\alpha}_{t}}u(t) \bigr)\,dt $$
and
$$\phi(p,k):= \int_{0}^{T} \bigl(\big|u_{k}(t)\big|^{p-2}u_{k}(t)-\big|u(t)\big|^{p-2}u(t) \bigr) \bigl(u_{k}(t)-u(t) \bigr)\,dt. $$
Notice that
$$\begin{aligned} \bigl\langle \varphi^{\prime}(u_{k})- \varphi^{\prime}(u),u_{k}-u \bigr\rangle ={}&\phi(p,\alpha,k)+ \phi(p,k)+\sum_{j=1}^{m} \bigl(I_{j} \bigl(u_{k}(t_{j}) \bigr)-I_{j} \bigl(t,u(t_{j}) \bigr) \bigr) \bigl(u_{k}(t_{j})-u(t_{j}) \bigr) \\ & - \int_{0}^{T} \bigl(f \bigl(t,u_{k}(t) \bigr)-f \bigl(t,u(t) \bigr) \bigr) \bigl(u_{k}(t)-u(t) \bigr)\,dt, \end{aligned} $$
which together with (3.4) and (3.5) yields
$$ \phi(p,\alpha,k)+\phi(p,k)\rightarrow0 \quad\text{as }k\rightarrow\infty. $$
(3.6)
From the well-known inequality (see [9], Lemma 4.2)
$$|x-y|^{p}\leq \textstyle\begin{cases} (|x|^{p-2}x-|y|^{p-2}y)(x-y), & \text{if } p\geq2,\\ ((|x|^{p-2}x-|y|^{p-2}y)(x-y))^{\frac{p}{2}}(|x|^{p}+|y|^{p})^{\frac {2-p}{2}}, &\text{if } 1< p< 2, \end{cases} $$
for all \(x,y\in\mathbb{R}\). Then there exist constants \(c_{i}>0\) (\(i=1,2,3,4\)) such that
$$ \phi(p,\alpha,k) \geq \textstyle\begin{cases} c_{1}\int_{0}^{T}|{{}_{0}^{c}D^{\alpha}_{t}}u_{k}(t)-{{}_{0}^{c}D^{\alpha}_{t}}u(t)|^{p}\,dt, & \text{if } p\geq2,\\ c_{2}\int_{0}^{T}\frac{|{{}_{0}^{c}D^{\alpha}_{t}}u_{k}(t)-{{}_{0}^{c}D^{\alpha}_{t}}u(t)|^{2}}{(|{{}_{0}^{c}D^{\alpha}_{t}}u_{k}(t)|+|{{}_{0}^{c}D^{\alpha}_{t}}u(t)|)^{2-p}}\,dt,& \text{if } 1< p< 2, \end{cases} $$
(3.7)
and
$$ \phi(p,k)\geq \textstyle\begin{cases} c_{3}\int_{0}^{T}|u_{k}(t)-u(t)|^{p}\,dt, &\text{if } p\geq2,\\ c_{4}\int_{0}^{T}\frac{|u_{k}(t)-u(t)|^{2}}{(|u_{k}(t)|+|u(t)|)^{2-p}}\,dt, & \text{if } 1< p< 2. \end{cases} $$
(3.8)
When \(1< p<2\), by the Hölder inequality, we have
$$ \begin{aligned}[b] \int_{0}^{T} \bigl(\big|{{}_{0}^{c}D^{\alpha}_{t}}u_{k}(t)-{{}_{0}^{c}D^{\alpha}_{t}}u(t)\big|^{p} \bigr)\,dt \leq{}& \biggl( \int_{0}^{T}\frac{|{{}_{0}^{c}D^{\alpha}_{t}}u_{k}(t)-{{}_{0}^{c}D^{\alpha}_{t}}u(t)|^{2}}{ (|{{}_{0}^{c}D^{\alpha}_{t}}u_{k}(t)|+|{{}_{0}^{c}D^{\alpha}_{t}}u(t)|)^{2-p}}\,dt \biggr)^{\frac{p}{2}} \\ & \cdot \biggl( \int_{0}^{T} \bigl(\big|{{}_{0}^{c}D^{\alpha}_{t}}u_{k}(t)\big|+\big|{{}_{0}^{c}D^{\alpha}_{t}}u(t)\big| \bigr)^{p}\,dt \biggr)^{\frac{2-p}{2}} \\ \leq{}& M \bigl(\big\| {{}_{0}^{c}D^{\alpha}_{t}}u_{k} \big\| ^{p}_{L^{p}}+\big\| {{}_{0}^{c}D^{\alpha}_{t}}u \big\| ^{p}_{L^{p}} \bigr)^{\frac{2-p}{2}} \\ & \cdot \biggl( \int_{0}^{T}\frac{|{{}_{0}^{c}D^{\alpha}_{t}}u_{k}(t)-{{}_{0}^{c}D^{\alpha}_{t}}u(t)|^{2}}{ (|{{}_{0}^{c}D^{\alpha}_{t}}u_{k}(t)|+|{{}_{0}^{c}D^{\alpha}_{t}}u(t)|)^{2-p}}\,dt \biggr)^{\frac{p}{2}}, \end{aligned} $$
(3.9)
where \(M=2^{(p-1)(2-p)/2}\) is a positive constant. Similarly, we have
$$\begin{aligned}[b] \int_{0}^{T} \bigl(\big|u_{k}(t)-u(t)\big|^{p} \bigr)\,dt\leq{}& M \bigl(\|u_{k}\|^{p}_{L^{p}}+\|u\| ^{p}_{L^{p}} \bigr)^{\frac{2-p}{2}}\\&\cdot \biggl( \int_{0}^{T}\frac{|u_{k}(t)-u(t)|^{2}}{ (|u_{k}(t)|+|u(t)|)^{2-p}}\,dt \biggr)^{\frac{p}{2}}. \end{aligned} $$
(3.10)
From (3.7) and (3.9), we have
$$ \begin{aligned}[b] \phi(p,\alpha,k) &\geq c_{2}M^{-\frac{2}{p}} \bigl( \|u_{k}\|^{p}_{L^{p}}+\|u\|^{p}_{L^{p}} \bigr)^{\frac{p-2}{p}} \cdot \biggl( \int_{0}^{T}\big|{{}_{0}^{c}D^{\alpha}_{t}}u_{k}(t)-{{}_{0}^{c}D^{\alpha}_{t}}u(t)\big|^{p}\,dt \biggr)^{\frac{2}{p}} \\ &=c_{2}M^{-\frac{2}{p}} \bigl(\|u_{k}\|^{p}_{L^{p}}+ \|u\|^{p}_{L^{p}} \bigr)^{\frac {p-2}{p}}\cdot \big\| {{}_{0}^{c}D^{\alpha}_{t}}(u_{k}-u) \big\| ^{2}_{L^{p}}. \end{aligned} $$
(3.11)
It follows from (3.8) and (3.10) that
$$ \phi(p,k)\geq c_{4}M^{-\frac{2}{p}} \bigl(\|u_{k} \|^{p}_{L^{p}}+\|u\| ^{p}_{L^{p}} \bigr)^{\frac{p-2}{p}}\cdot\|u_{k}-u\|^{2}_{L^{p}}. $$
(3.12)
When \(1< p<2\), by (3.11) and (3.12), we get
$$ \phi(p,\alpha,k)+\phi(p,k)\geq M_{1} \bigl(\big\| {{}_{0}^{c}D^{\alpha}_{t}}(u_{k}-u) \big\| ^{2}_{L^{p}}+\|u_{k}-u\|^{2}_{L^{p}} \bigr)=M_{1}\|u_{k}-u\|^{2}_{E^{\alpha,p}}, $$
(3.13)
where
$$M_{1}:=M^{-\frac{2}{p}}\min \bigl\{ c_{2} \bigl( \|u_{k}\|^{p}_{L^{p}}+\|u\|^{p}_{L^{p}} \bigr)^{\frac {p-2}{p}}, c_{4} \bigl(\|u_{k} \|^{p}_{L^{p}}+\|u\|^{p}_{L^{p}} \bigr)^{\frac{p-2}{p}} \bigr\} . $$
When \(p\geq2\), in view of (3.7) and (3.8), we have
$$ \phi(p,\alpha,k)+\phi(p,k)\geq M_{2} \bigl(\big\| {{}_{0}^{c}D^{\alpha}_{t}}(u_{k}-u) \big\| ^{p}_{L^{p}}+\|u_{k}-u\|^{p}_{L^{p}} \bigr)=M_{2}\|u_{k}-u\|^{p}_{E^{\alpha,p}}, $$
(3.14)
where \(M_{2}=\min\{c_{1},c_{3}\}\). Therefore, it follows from (3.6), (3.13) and (3.14) that \(\|u_{k}-u\|_{E^{\alpha,p}}\rightarrow0\) as \(k\rightarrow+\infty\). That is, \(u_{k}\rightarrow u\) in \(E^{\alpha ,p}\). Hence, φ satisfies the Palais-Smale condition. □
Now we prove Theorem 1.1 and Theorem 1.2.
Proof of Theorem 1.1
Step I: Obviously, \(\varphi(0)=0\), and Lemma 3.1 has shown that φ satisfies the Palais-Smale condition. For any \(r>0\), take \(\Omega_{r}=\{u\in E^{\alpha,p}:\|u\| _{E^{\alpha,p}}< r\}\). It is easy to show that \(\overline{\Omega}_{r}\) is bounded and weakly sequentially closed. Indeed, if we let \(\{u_{n}\} \subseteq\overline{\Omega}_{r}\) and \(u_{n}\rightharpoonup u\) as \(n\rightarrow\infty\), by the Mazur Theorem [10], there is a sequence of convex combinations \(v_{n}=\sum_{i=1}^{n}\beta_{n_{i}}u_{i}\) with \(\sum_{i=1}^{n}\beta_{n_{i}}=1\), \(\beta_{n_{i}}\geq0\), \(i\in\textbf{N}\) such that \(v_{n}\rightarrow u\) in \(E^{\alpha,p}\). Since \(\overline{\Omega}_{r}\) is a closed convex set, we have \(\{v_{n}\}\subseteq\overline{\Omega}_{r}\) and \(u\in\overline{\Omega}_{r}\).
From Lemma 3.1 we know that φ is weakly sequentially lower semi-continuous on \(\overline{\Omega}_{r}\). Besides, \(E^{\alpha ,p}\) is a reflexive Banach space, so by Lemma 2.2 we see that φ has a local minimum \(u_{0}\in\overline{\Omega}_{r}\). Without loss of generality, we assume that \(\varphi(u_{0})=\min\{\varphi(u):u\in\overline{\Omega}_{r}\}\). Now we will prove that \(\varphi(u_{0})\leq\inf\{\varphi(u):u\in\overline {\Omega}_{r}\}\) for some \(r=r_{0}\). Indeed, from (H3) we may choose \(r_{0},\varepsilon>0\) satisfying
$$ F(t,u)\leq\delta|u|^{\vartheta}, \quad\text{for }\|u\|_{E^{\alpha,p}}\leq r_{0},\qquad 0< \varepsilon< \frac{r_{0}^{p}}{p}-r_{0}^{\mu}\sum _{j=1}^{m}\delta_{j} A_{\infty}^{\mu}-r_{0}^{\vartheta}\delta A_{p}^{\vartheta}. $$
(3.15)
For every \(u\in \partial\Omega_{r_{0}}\) with \(\|u\|_{E^{\alpha ,p}}=r_{0}\), from (2.2), (2.3), (2.5), (H4) and (3.15) we have
$$\begin{aligned} \varphi(u)&\geq\frac{1}{p}\|u\|_{E^{\alpha,p}}^{p}- \sum_{j=1}^{m}\delta_{j}\big|u(t_{j})\big|^{\mu}- \delta \int_{0}^{T}\big|u(t)\big|^{\vartheta}\,dt \\ &\geq\frac{r_{0}^{p}}{p}-r_{0}^{\mu}\sum _{j=1}^{m}\delta_{j} A_{\infty}^{\mu}-r_{0}^{\vartheta}\delta A_{p}^{\vartheta}> \varepsilon, \end{aligned} $$
which implies that \(\varphi(u)>\varepsilon\) for every \(u\in \partial \Omega_{r_{0}}\). Moreover, \(\varphi(u_{0})\leq\varphi(0)=0\). Then \(\varphi(u_{0})\leq\varphi(0)<\varepsilon<\varphi(u)\) for every \(u\in \partial\Omega_{r_{0}}\). Hence \(\varphi(u_{0})<\inf\{\varphi (u):u\in \partial\Omega_{r_{0}}\}\). So φ has a local minimum \(u_{0}\in \partial\Omega_{r_{0}}\).
Step II: We will verify that there exists \(u_{1}\) with \(\| u_{1}\|_{E^{\alpha,p}}>r_{0}\) such that \(\varphi(u_{1})<\inf\{\varphi (u):u\in \partial\Omega_{r_{0}}\}\), where \(r_{0}\) is given above.
In view of (H3), we choose a sufficiently large \(r_{1}\) such that for all \(\|u\|_{E^{\alpha,p}}\geq r_{1}>r_{0}\)
$$ F(t,u)\geq\gamma|u|^{\vartheta}. $$
(3.16)
From (H1), we have the following:
$$\begin{aligned}& \frac{\mu}{u}\leq\frac{I_{j}(u)}{\int_{0}^{u}I_{j}(s)\,ds},\quad \text{for }u>0, \end{aligned}$$
(3.17)
$$\begin{aligned}& \frac{\mu}{u}\geq\frac{I_{j}(u)}{\int_{0}^{u}I_{j}(s)\,ds}, \quad\text{for }u< 0. \end{aligned}$$
(3.18)
Integrating (3.17) and (3.18) from T to u and u to −T, respectively, we get
$$\begin{gathered} \int_{0}^{u}I_{j}(s)\,ds\leq \frac{u^{\mu}}{T^{\mu}} \int_{0}^{T}I_{j}(s)\,ds, \quad\text{for }u>T, \\\int_{0}^{u}I_{j}(s)\,ds\leq \frac{(-u)^{\mu}}{T^{\mu}} \int_{0}^{-T}I_{j}(s)\,ds,\quad \text{for }u< -T. \end{gathered}$$
Note that \(\int_{0}^{T}I_{j}(s)\,ds<0\) and \(\int_{0}^{-T}I_{j}(s)\,ds<0\). We take
$$\begin{aligned} \gamma_{j}&=&T^{-\mu}\cdot\min \biggl\{ \bigg| \int_{0}^{T}I_{j}(s)\,ds\bigg|, \bigg| \int _{0}^{-T}I_{j}(s)\,ds\bigg| \biggr\} >0, \end{aligned} $$
and we get
$$ \begin{aligned} \int_{0}^{u}I_{j}(s)\,ds\leq- \gamma_{j}|u|^{\mu},\quad \forall u\in (-\infty ,-T]\cup[T,+ \infty ). \end{aligned} $$
(3.19)
Clearly, \(\int_{0}^{u}I_{j}(s)\,ds\) is continuous in \([-T,T]\), so there is a constant \(K>0\) such that
$$ \begin{aligned} \int_{0}^{u}I_{j}(s)\,ds\leq K,\quad \forall u \in[-T,T]. \end{aligned} $$
(3.20)
Combining (3.19) and (3.20), we get
$$ \begin{aligned} \int_{0}^{u}I_{j}(s)\,ds\leq- \gamma_{j}|u|^{\mu}+K, \quad\forall u\in R. \end{aligned} $$
(3.21)
For any \(u\in E^{\alpha,p}\) with \(u\neq0\), \(\tau>0\), it follows from (3.16) and (3.21) that
$$\varphi(\tau u)\leq\frac{\tau^{p}}{p}\|u\|_{E^{\alpha,p}}^{p}-\tau ^{\mu}\sum_{j=1}^{m} \gamma_{j}\big|u(t_{j})\big|^{\mu}-\tau^{\vartheta}\gamma\|u\|^{\vartheta}_{L^{\vartheta}}+K_{*}\rightarrow -\infty \quad(\mu> \vartheta>p), $$
as \(\tau\rightarrow\infty\), where \(K_{*}\) is a positive constant. So there exists a sufficiently large \(\tau_{0}\) such that \(\varphi(\tau_{0} u)\leq0\). That is to say, we can choose \(u_{1}\) with \(\|u_{1}\|_{E^{\alpha ,p}}\geq r_{1}\) sufficiently large such that \(\varphi(u_{1})<0\). Therefore, from Step I and Step II we get \(\max\{ \varphi(u_{0}),\varphi(u_{1})\}<\inf\{\varphi(u): u\in \partial\Omega _{r_{0}}\}\). Then Lemma 2.1 admits the critical point \(u_{*}\). So, \(u_{0}\) and \(u_{*}\) are two different critical points of φ, and they are weak solutions of (1.1). □
Example 3.1
Let \(\alpha=0.6\), \(T>0\), \(p=\frac{5}{2}\), \(t_{1}\in (0,T)\), \(a_{0}>0\), \(a(t)\in C([0,T])\) with \(a(t)>0\). Consider the following fractional boundary value problem:
$$ \textstyle\begin{cases} {{}_{t}D}_{T}^{0.6}\Phi_{\frac{5}{2}}({}_{0}^{c}D_{t}^{0.6} u(t))+|u(t)|^{\frac {1}{2}}u(t)=(2+\sin t)a(t)u^{4}(t), \quad 0< t< T,t\neq t_{1},\\ \Delta({}_{t}D^{-0.4}_{T}\Phi_{\frac{5}{2}}({}^{c}_{0}D^{0.6}_{t} u))(t_{1})=-a_{0}u^{5}(t_{1}), \\ u(0)=u(T)=0. \end{cases} $$
(3.22)
Obviously, \(f(t,u)=(2+\sin t)a(t)u^{4}(t)\), \(I_{1}(u)=-a_{0}u^{5}(t_{1})\), \(\frac {1}{p}=0.4<\alpha=0.6\). Let \(\mu=6\), \(\vartheta=5\), \(\delta_{1}=\frac {a_{0}}{6}\), and \(\delta=\frac{1}{5}\max\{(2+\sin t)a(t):t\in[0,T]\} \), \(\gamma=\frac{1}{5}\min\{(2+\sin t)a(t):t\in[0,T]\}\). By simple computation, the conditions (H1)-(H4) are satisfied. From Theorem 1.1, problem (3.22) has at least two weak solutions.
Proof of Theorem 1.2
We will apply Lemma 2.3 to finish the proof. Obviously, \(\varphi\in C^{1}(E^{\alpha,p},\mathbb{R})\) is even and \(\varphi(0)=0\). Moreover, Lemma 3.1 shows that φ satisfies the Palais-Smale condition.
As \(E^{\alpha,p}\) is a reflexive and separable Banach space, then there are \(e_{i}\in E^{\alpha,p}\) such that \(E^{\alpha,p}=\overline {\operatorname{span}\{e_{i}:i=1,2,\ldots\}}\). For \(k=1,2,\ldots\) , denote
$$X_{i}:=\operatorname{span}\{e_{i}\},\qquad Y_{k}:=\bigoplus _{i=1}^{k}X_{i},\qquad Z_{k}:= \overline{\bigoplus_{i=k}^{\infty}X_{i}}. $$
Then \(E^{\alpha,p}=Y_{k}\oplus Z_{k}\).
For any \(u\in Z_{k}\) with \(\|u\|_{E^{\alpha,p}}\leq r_{0}\), combining (2.5) and the conditions (H1)-(H4), we have
$$\begin{aligned} \varphi(u)\geq\frac{1}{p}\|u\|_{E^{\alpha,p}}^{p}- \sum_{j=1}^{m}\delta_{j} A_{\infty}^{\mu}\|u\|^{\mu}_{E^{\alpha,p}}-\delta A_{p}^{\vartheta}\|u\|^{\vartheta}_{E^{\alpha,p}}, \quad\|u \|_{E^{\alpha,p}}\leq r_{0}, \end{aligned} $$
which implies that there exists \(\rho>0\) small enough such that \(\varphi(u)\geq\eta>0\) with \(\|u\|_{E^{\alpha,p}}=\rho\).
For any \(u\in Y_{k}\), let
$$ \|u\|_{*}:= \biggl( \int_{0}^{T}\big|u(t)\big|^{\vartheta}\,dt \biggr)^{\frac{1}{\vartheta}}, $$
(3.23)
and it is easy to show that \(\|\cdot\|_{*}\) defined by (3.23) is a norm of \(Y_{k}\). Since all the norms of a finite dimensional normed space are equivalent, there is a positive constant \(A_{1}\) such that
$$ A_{1}\|u\|_{E^{\alpha,p}}\leq\|u\|_{*}, \quad\text{for }u\in Y_{k}. $$
(3.24)
From (H1), we have
$$\sum_{j=1}^{m} \int_{0}^{u(t_{j})}I_{j}(s)\,ds\leq0. $$
The above inequality, (3.17) and (3.24) imply that, for any finite dimensional space \(W\subset E^{\alpha,p}\),
$$\begin{aligned} \varphi(u)&=\frac{1}{p}\|u\|_{E^{\alpha,p}}^{p}+ \sum_{j=1}^{m} \int_{0}^{u(t_{j})}I_{j}(s)\,ds- \int_{0}^{T}F \bigl(t,u(t) \bigr)\,dt \\ &\leq\frac{1}{p}\|u\|_{E^{\alpha,p}}^{p}- \int_{\Omega _{1}}F \bigl(t,u(t) \bigr)\,dt- \int_{\Omega_{2}}F \bigl(t,u(t) \bigr)\,dt \\ &\leq\frac{1}{p}\|u\|_{E^{\alpha,p}}^{p}-\gamma \int_{\Omega _{1}}\big|u(t)\big|^{\vartheta}\,dt- \int_{\Omega_{2}}F \bigl(t,u(t) \bigr)\,dt \\ &=\frac{1}{p}\|u\|_{E^{\alpha,p}}^{p}-\gamma \int_{0}^{T}\big|u(t)\big|^{\vartheta}\,dt+\gamma \int_{\Omega_{1}}\big|u(t)\big|^{\vartheta}\,dt - \int_{\Omega_{2}}F \bigl(t,u(t) \bigr)\,dt \\ &\leq\frac{1}{p}\|u\|_{E^{\alpha,p}}^{p}-A_{1}^{\vartheta}\gamma\|u\| _{E^{\alpha,p}}^{\vartheta}+M^{*}, \quad\forall u\in W, \end{aligned} $$
where \(\Omega_{1}:=\{t\in[0,T]:|u(t)|\geq A_{\infty}r_{1}\}\) (\(r_{1}\) is given in (3.17)), \(\Omega_{2}:=[0,T]\setminus{\Omega_{1}}\) and \(M^{*}\) is a positive constant. Since \(\vartheta>p\), the above inequality implies that \(\varphi(u)\rightarrow-\infty\) as \(\|u\|_{E^{\alpha ,p}}\rightarrow+\infty\). That is, there exists \(r>0\) such that \(\varphi(u)\leq0\) for \(u\in W\setminus{B_{r(W)}}\). By Lemma 2.3, the functional \(\varphi(u)\) possesses infinitely many critical points, i.e., the fractional impulsive problem (1.1) admits infinitely many weak solutions. The proof is complete. □
Example 3.2
Let \(\alpha=0.75\), \(T>0\), \(p=\frac{3}{2}\), \(t_{1}\in (0,T)\), \(a_{0}>0\), \(a(t)\in C([0,T])\) with \(a(t)>0\). Consider the following fractional boundary value problem:
$$ \textstyle\begin{cases} {{}_{t}D}_{T}^{0.75}\Phi_{\frac{3}{2}}({}_{0}^{c}D_{t}^{0.75} u(t))+|u(t)|^{-\frac {1}{2}}u(t)=(1+t^{2})a(t)u^{\frac{5}{3}}(t), \quad 0< t< T,t\neq t_{1},\\ \Delta({}_{t}D^{-0.25}_{T}\Phi_{\frac{3}{2}}({}^{c}_{0}D^{0.75}_{t} u))(t_{1})=-a_{0}u^{9}(t_{1}), \\ u(0)=u(T)=0. \end{cases} $$
(3.25)
Obviously, \(\frac{1}{p}=\frac{2}{3}<\alpha=0.75 \) and \(f(t,u)=(1+t^{2})a(t)u^{\frac{5}{3}}(t)\), \(I_{1}(u)=-a_{0}u^{9}(t_{1})\) are odd about u. Let \(\mu=10\), \(\vartheta=\frac{8}{3}\), \(\delta=\frac {3}{8}\max\{(1+t^{2})a(t):t\in[0,T]\}\), \(\gamma=\frac{3}{8}\min\{ (1+t^{2})a(t):t\in[0,T]\}\), and \(\delta_{1}=\frac{a_{0}}{10}\). Then by simple computation, the conditions in Theorem 1.2 are satisfied. Hence, problem (3.25) has infinitely many weak solutions.