Nonlinear second-order impulsive differential problems with dependence on the derivative via variational structure

Abstract

In this paper we use variational methods and fixed point theorems to study the existence of solutions to a boundary value problem for impulsive differential equations with nonlinear dependence on the derivative.

Appendix A: Proof of Lemma 4.1

We prove first that S is continuous.Let $\{w_n\}_{n\in {\mathbb {N}}}$ a sequence on $H_0^1(0,T)$, $w_n\longrightarrow w$. Denote $u_n=S(w_n)$ for all $n\in {\mathbb {N}}$. Take an arbitrary subsequence of $\{u_n\}_{n\in {\mathbb {N}}}$ (for simplicity denoted again by $\{u_n\}_{n\in {\mathbb {N}}}$).We know that $\Vert u_n\Vert \le M$, hence there exists a subsequence $\{u_{n_k}\}_{k\in {\mathbb {N}}}$ such that $u_{n_k}$ converges weakly to some $u\in H_0^1(0,T)$. Then the sequence $\{u_{n_k}\}_{k\in {\mathbb {N}}}$ converges to u in C([0, T]).Using that $w_{n_k}\longrightarrow w$ in $H_0^1(0,T)$, we have that $w_{n_k}'\longrightarrow w'$ in $L^2(0,T)$. So there exists a subsequence $\{w_{n_{k_l}}\}_{l\in {\mathbb {N}}}$ such that

$$\begin{aligned} w_{n_{k_l}}'(t)\longrightarrow w'(t)\text { a.e. }t\in [0,T]. \end{aligned}$$

The function f is continuous, so

$$\begin{aligned} f(t,u_{n_{k_l}}(t),w_{n_{k_l}}'(t))\longrightarrow f(t,u(t),w'(t))\text { a.e. }t\in [0,T]. \end{aligned}$$

Furthermore, the functions $I_j$ are continuous, so

$$\begin{aligned} I_j(u_{n_{k_l}}(t_j))\longrightarrow I_j(u(t_j)). \end{aligned}$$

We will check that $u_{n_{k_l}}\longrightarrow u$. For any $v\in H_0^1(0,T)$ and using Lebesgue’s dominated convergence theorem we have

$$\begin{aligned} \begin{array}{ccccccc} (u_{n_{k_l}},v) &{} + &{} \displaystyle \sum _{j=1}^p I_j(u_{n_{k_l}}(t_j))v(t_j) &{} - &{} \displaystyle \int _0^T f(t,u_{n_{k_l}}(t),w_{n_{k_l}}'(t))v(t)\,{\text {d}}t &{} = &{} 0\\ \Big \downarrow &{}&{} \Big \downarrow &{}&{} \Big \downarrow &{}&{} \Big \downarrow \\ (u,v) &{} + &{} \displaystyle \sum _{j=1}^p I_j(u(t_j))v(t_j) &{} - &{} \displaystyle \int _0^T f(t,u(t),w'(t))v(t)\,{\text {d}}t &{} = &{} 0 \end{array} \end{aligned}$$

Then $Sw=u$ (by uniqueness of the critical point), and we know that the sequence $u_{n_{k_l}}=Sw_{n_{k_l}}$ converges weakly to Sw.Furthermore, taking $v=u_{n_{k_l}}$ we get $\Vert u_{n_{k_l}}\Vert \longrightarrow \Vert u\Vert$. Then we have that $u_{n_{k_l}}\longrightarrow u$ in $H_0^1(0,T)$, i.e., $S_{w_{n_{k_l}}}\longrightarrow Sw$. So any arbitrary subsequence of $\{u_n\}_{n\in {\mathbb {N}}}$ has a subsequence which converges to Sw. This implies that $u_n\longrightarrow Sw$.

We prove that S is compact. Let $\{w_n\}_{n\in {\mathbb {N}}}$ be a bounded sequence in $H_0^1(0,T)$. We have to prove that $\{Sw_n\}_{n\in {\mathbb {N}}}$ has a convergent subsequence. Fix $j\in \{0,\ldots ,p\}$ and take $v\in H_0^1(0,T)$ with $v(t)=0\quad \forall \,t\notin (t_j,t_{j+1})$. The functions $Sw_n$ are a weak solution of (1.2) with $w=w_n$, so

$$\begin{aligned} \int _{t_j}^{t_{j+1}} (Sw_n)'(t)v'(t)+ a(t)(Sw_n)(t)v(t)\,{\text {d}}t - \int _{t_j}^{t_{j+1}} f(t,Sw_n(t),w_n'(t))v(t)\,{\text {d}}t. \end{aligned}$$

is equal to 0. Then we can say that

$$\begin{aligned} \int _{t_j}^{t_{j+1}} (Sw_n)'(t)v'(t) = \int _{t_j}^{t_{j+1}} [f(t,Sw_n(t),w_n'(t))-a(t)(Sw_n)(t)]v(t)\,{\text {d}}t = 0. \end{aligned}$$

So there exists $(Sw_n)''\in L^2(t_j,t_{j+1})$, which implies $Sw_n\in H^2(t_j,t_{j+1})$.We will check that $Sw_n$ is a bounded sequence in $H^2(t_j,t_{j+1})$.

$$\begin{aligned}&{\Vert Sw_n\Vert _{H^2}}^2={\Vert Sw_n\Vert _{H^1}}^2+{\Vert (Sw_n)''\Vert _{L^2}}^2 \\&\quad \le M^2 + \int _{t_j}^{t_{j+1}} (a(t)Sw_n(t)-f(t,Sw_n(t),w_n'(t)))^2\,{\text {d}}t \\&\quad \le M^2 + \int _{t_j}^{t_{j+1}} (a(t)Sw_n(t))^2\,{\text {d}}t + \int _{t_j}^{t_{j+1}} (f(t,Sw_n(t),w_n'(t)))^2\,{\text {d}}t \\&\qquad + 2 \int _{t_j}^{t_{j+1}} |a(t) f(t,Sw_n(t),w_n'(t)) Sw_n(t)|\,{\text {d}}t \\&\quad \le M^2 + {M_1}^2 + \int _{t_j}^{t_{j+1}} (\mu (|Sw_n(t)|)\beta (t))^2\,{\text {d}}t \\&\qquad + 2\int _{t_j}^{t_{j+1}} |a(t)|\mu (|Sw_n(t)|)\beta (t)|Sw_n(t)|\,{\text {d}}t. \end{aligned}$$

We know that $\Vert Sw_n\Vert _\infty \le c\Vert Sw_n\Vert \le cM$. Take

$$\begin{aligned} M_2=\max \{\mu (s):s\in [0,cM]\}. \end{aligned}$$

Then we have that

$$\begin{aligned} {\Vert Sw_n\Vert _{H^2}}^2&\le \cdots \le M^2+{M_1}^2 + \int _{t_j}^{t_{j+1}} (M_2 \beta (t))^2\,{\text {d}}t \\&\quad +2\int _{t_j}^{t_{j+1}} |a(t)|M_2cM\beta (t)\,{\text {d}}t<K \end{aligned}$$

and $K<\infty$. Then $Sw_n$ is a bounded sequence in $H^2(t_j,t_{j+1})$.Next we use a diagonal argument.For $j=0$, $Sw_n$ is a bounded sequence in $H^2(0,t_1)$, so there exists a subsequence $Sw_{n_{k_0}}$ weakly convergent. Let $u^0$ be its limit. Furthermore, the inclusion $H^2(0,t_1)\subset H^1(0,t_1)$ is compact, so $Sw_{n_{k_0}}\longrightarrow u^0$ in $H^1(0,t_1)$.For $j=1$, $Sw_{n_{k_0}}$ is a bounded sequence in $H^2(t_1,t_2)$, so there exists a subsequence $Sw_{n_{k_1}}$ weakly convergent. Let $u^1$ be its limit. The inclusion $H^2(t_1,t_2)\subset H^1(t_1,t_2)$ is compact, so $Sw_{n_{k_1}}\longrightarrow u^1$ in $H^1(t_1,t_2)$.Using this argument for $j=2,\ldots ,p$ we construct a subsequence $Sw_{n_k}$ such that

$$\begin{aligned} {Sw_{n_k}}|_{(t_j,t_{j+1})}\longrightarrow u^j \text { in } H^1(t_j,t_{j+1})\subset {\mathcal {C}}([t_j,t_{j+1}]). \end{aligned}$$

We define the function u as follows:

$$\begin{aligned} u(t)={\left\{ \begin{array}{ll} u^0(t),&{} t\in [0,t_1]\\ u^1(t),&{} t\in (t_1,t_2]\\ \cdots \\ u^p(t),&{} t\in (t_p,T]. \end{array}\right. } \end{aligned}$$

Furthermore,

$$\begin{aligned} (u^j)'\in H^2(t_j,t_{j+1})\subset {\mathcal {C}}^1([t_j,t_{j+1}]), \end{aligned}$$

so there exist $(u^j)'(t_j^+)$ and $Sw_{n_k}|_{(t_j,t_{j+1})}$ converges weakly to $u^j$ in $H^2(t_j,t_{j+1})$. This implies that $Sw_{n_k}|_{(t_j,t_{j+1})}\longrightarrow u^j$ in ${\mathcal {C}}^1([t_j,t_{j+1}])$, so

$$\begin{aligned} (u^j)'(t_j^+)=\lim \limits _{k\rightarrow \infty } (Sw_{n_k})'(t_j^+),\qquad (u^j)'(t_j^-)=\lim \limits _{k\rightarrow \infty } (Sw_{n_k})'(t_j^-). \end{aligned}$$

As a consequence, $\Delta u'(t_j)=I_j(u(t_j))$.Finally, $Sw_{n_k}\longrightarrow u$ in $H^1(0,T)$, and finally $u(0)=u(T)$, so $Sw_{n_k}\longrightarrow u$ in $H_0^1(0,T)$.

This implies that $\{Sw_n\}_{n\in {\mathbb {N}}}$ has a convergent subsequence, so S is compact.