Assume that the following conditions hold in this paper:
- (H1):
-
\(f:[0,1]\times\mathbb{R}^{2}\rightarrow\mathbb{R}\) is a continuous function.
- (H2):
-
There exist nonnegative functions \(u, v, w\in C[0,T]\) such that
$$\bigl\vert f (t,x_{1},x_{2} ) \bigr\vert \leq u(t) \vert x_{1} \vert +v(t) \vert x_{2} \vert +w(t),\quad t \in[0,T], x_{1},x_{2}\in \mathbb{R}. $$
- (H3):
-
There exists a constant \(M>0\) such that if \(|x(t)|+|x'(t)|>M\) for all \(t\in[0,T]\), then
$$v_{2}\bigl(\beta{}^{\rho}I^{p}J^{q}Nx( \xi)-J^{q}Nx(T)\bigr)+ \alpha v_{4} I^{\gamma,\delta}_{\eta}J^{q}Nx( \zeta)\neq0. $$
- (H3′):
-
There exists a constant \(M>0\) such that if \(|x'(t)|>M\) for all \(t\in[0,T]\), then
$$\beta{}^{\rho}I^{p}J^{q}Nx(\xi)-J^{q}Nx(T) \neq0. $$
- (H4):
-
There is a constant \(D>0\) such that either
$$ c v_{2}\bigl(\beta{}^{\rho}I^{p}J^{q}N \phi_{1}(\xi)-J^{q}N\phi_{1}(T)\bigr)+ c\alpha v_{4} I^{\gamma,\delta}_{\eta}J^{q}N \phi_{1}(\zeta)>0 $$
(5)
or
$$ c v_{2}\bigl(\beta{}^{\rho}I^{p}J^{q}N \phi_{1}(\xi)-J^{q}N\phi_{1}(T)\bigr)+ c\alpha v_{4} I^{\gamma,\delta}_{\eta}J^{q}N \phi_{1}(\zeta)< 0 $$
(6)
holds if \(|c|>D\), where \(\phi_{1}(t)=c\).
- (H4′):
-
There is a constant \(D>0\) such that either
$$c \beta{}^{\rho}I^{p}J^{q}N\phi_{2}( \xi)-cJ^{q}N\phi_{2}(T)>0 $$
or
$$c \beta{}^{\rho}I^{p}J^{q}N\phi_{2}( \xi)-cJ^{q}N\phi_{2}(T)< 0 $$
holds if \(|c|>D\), where \(\phi_{2}(t)=ct\).
- (H4″):
-
There is a constant \(D>0\) such that either
$$c v_{2}\bigl(\beta{}^{\rho}I^{p}J^{q}N \phi_{3}(\xi)-J^{q}N\phi_{3}(T)\bigr)+ c\alpha v_{4} I^{\gamma,\delta}_{\eta}J^{q}N \phi_{3}(\zeta)>0 $$
or
$$c v_{2}\bigl(\beta{}^{\rho}I^{p}J^{q}N \phi_{3}(\xi)-J^{q}N\phi_{3}(T)\bigr)+ c\alpha v_{4} I^{\gamma,\delta}_{\eta}J^{q}N \phi_{3}(\zeta)< 0 $$
holds if \(|c|>D\), where \(\phi_{3}(t)=c(1+kt)\), \(k=\frac{v_{1}}{v_{2}}\).
Then we can present the following theorem.
Theorem 3.1
Suppose that
\((A1)\)
and (H1)–(H4) are satisfied, then there must be at least one solution of problem (1) in
X
provided that
\(2T^{q}\|u\|_{\infty}+2T^{q-1}\|v\| _{\infty}<\Gamma(q)\).
To prove the theorem, we need the following lemmas.
Lemma 3.1
Assume that
\((A1)\)
holds, then
\(L:\operatorname{dom}L \subset X\rightarrow Y\)
is a Fredholm operator with index zero. And a linear continuous projector
\(P:X\rightarrow X \)
can be defined by
Furthermore, define the linear operator
\(K_{p}:\operatorname {Im}L\rightarrow\operatorname{dom}L\cap\operatorname{ker}P\)
as follows:
$$(K_{p}y) (t)= \frac{1}{\Gamma(q)} \int_{0}^{t}(t-s)^{q-1}y(s)\,ds - \frac{\alpha I^{\gamma,\delta}_{\eta}J^{q}y(\zeta)}{v_{2}}\cdot t $$
such that
\(K_{p}=(L|_{\operatorname{dom}L\cap\operatorname{ker}P})^{-1}\).
Proof
Let \(\varphi(t)=1\), \(\psi(t)=t\). From \((A1)\) and Lemma 2.4, we can easily get
$$\operatorname{ker}L=\{c, c\in\mathbb{R}\}. $$
Moreover, we can obtain that
$$\operatorname{Im}L=\bigl\{ y\in Y:v_{2}\bigl(\beta{}^{\rho}I^{p}J^{q}y( \xi)-J^{q}y(T)\bigr)+ \alpha v_{4} I^{\gamma,\delta}_{\eta}J^{q}y( \zeta)=0\bigr\} . $$
On the one hand, suppose \(y\in\operatorname{Im}L\), then there exists \(x \in\operatorname{dom}L\) such that
Then we have
$$ x(t)=J^{q}y(t)+c_{0}+c_{1}t= \frac{1}{\Gamma(q)} \int_{0}^{t}(t-s)^{q-1}y(s) \,ds+c_{0}+c_{1}t, $$
(7)
where \(c_{0}, c_{1}\in\mathbb{R}\). Furthermore, for \(x \in\operatorname{dom}L\),
$$\begin{aligned} x(0)&=\alpha I^{\gamma,\delta}_{\eta}x(\zeta)= \alpha I^{\gamma,\delta }_{\eta}J^{q}y(\zeta)+ c_{0} \alpha I^{\gamma,\delta}_{\eta}\varphi(\zeta) +c_{1}\alpha I^{\gamma,\delta}_{\eta}\psi(\zeta) \\ &= \alpha I^{\gamma,\delta}_{\eta}J^{q}y(\zeta)+c_{0} +c_{1}v_{2}, \end{aligned} $$
and
$$x(0)={J^{q}y(t)+c_{0}+c_{1}t} \big|_{t=0}=c_{0}. $$
The above two equalities imply that
$$ \alpha I^{\gamma,\delta}_{\eta}J^{q}y( \zeta)+c_{1}v_{2}=0. $$
(8)
Using (3) and (7), we get the system
$$\begin{gathered} \begin{aligned} x(T)&=\beta{}^{\rho}I^{p}x(\xi)= \beta{}^{\rho}I^{p}J^{q}y(\xi )+c_{0} \beta{}^{\rho}I^{p}\varphi(\xi)+c_{1} \beta{}^{\rho}I^{p}\psi(\xi) \\ &=\beta{}^{\rho}I^{p}J^{q}y(\xi)+c_{0}+c_{1}(T-v_{4}), \end{aligned} \\ x(T)={J^{q}y(t)+c_{0}+c_{1}t} \big|_{t=T}=J^{q}y(T)+c_{0}+c_{1}T.\end{gathered} $$
From this together with the second boundary value condition of (1), we can get
$$ \beta{}^{\rho}I^{p}J^{q}y( \xi)-J^{q}y(T)=c_{1}v_{4}. $$
(9)
By using the eliminated element method, equalities (8) and (9) are changed into the equality
$$v_{2}\bigl(\beta{}^{\rho}I^{p}J^{q}y( \xi)-J^{q}y(T)\bigr)+ \alpha v_{4} I^{\gamma,\delta}_{\eta}J^{q}y( \zeta)=0. $$
So we obtain that
$$\operatorname{Im}L\subset\bigl\{ y\in Y: v_{2}\bigl( \beta{}^{\rho}I^{p}J^{q}y(\xi )-J^{q}y(T) \bigr)+ \alpha v_{4} I^{\gamma,\delta}_{\eta}J^{q}y( \zeta)=0\bigr\} . $$
On the other hand, if \(y\in Y\) satisfies \(v_{2}(\beta{}^{\rho }I^{p}J^{q}y(\xi)-J^{q}y(T))+ \alpha v_{4} I^{\gamma,\delta}_{\eta}J^{q}y(\zeta)=0\), we let
$$x(t)=J^{q}y(t)-\frac{\alpha I^{\gamma,\delta}_{\eta}J^{q}y(\zeta )}{v_{2}}\cdot t. $$
Then we conclude that
$$Lx(t)={}^{c}D^{q}x(t)=y(t), $$
and
$$\begin{gathered} x(0)=0, \\ \begin{aligned} \alpha I^{\gamma,\delta}_{\eta}x(\zeta) & = \alpha I^{\gamma,\delta}_{\eta}J^{q}y(\zeta)-\frac{\alpha I^{\gamma ,\delta}_{\eta}J^{q}y(\zeta)}{v_{2}} \cdot\alpha I^{\gamma,\delta}_{\eta }\psi(\zeta) \\ &= \alpha I^{\gamma,\delta}_{\eta}J^{q}y(\zeta)- \frac{\alpha I^{\gamma ,\delta}_{\eta}J^{q}y(\zeta)}{v_{2}}\cdot v_{2}=0. \end{aligned} \end{gathered}$$
Besides,
$$\begin{gathered} x(T)=J^{q}y(T)-\frac{\alpha I^{\gamma,\delta}_{\eta}J^{q}y(\zeta )}{v_{2}}\cdot T, \\ \begin{aligned} \beta{}^{\rho}I^{p}x(\xi) & = \beta{}^{\rho}I^{p} J^{q}y(\xi)-\frac{\alpha I^{\gamma,\delta }_{\eta}J^{q}y(\zeta)}{v_{2}}\cdot \beta{}^{\rho}I^{p}\psi(\xi) \\ &= \beta{}^{\rho}I^{p} J^{q}y(\xi)-\frac{\alpha I^{\gamma,\delta }_{\eta}J^{q}y(\zeta)}{v_{2}} \cdot(T-v_{4}) \\ &= -\frac{\alpha v_{4} I^{\gamma,\delta}_{\eta}J^{q}y(\zeta )}{v_{2}}+J^{q}y(T)-\frac{\alpha I^{\gamma,\delta}_{\eta}J^{q}y(\zeta )}{v_{2}} \cdot(T-v_{4}) \\ &= J^{q}y(T)-\frac{\alpha I^{\gamma,\delta}_{\eta}J^{q}y(\zeta )}{v_{2}}\cdot T, \end{aligned} \end{gathered}$$
therefore
$$x(0)=\alpha I^{\gamma,\delta}_{\eta}x(\zeta),\qquad x(T)= \beta{}^{\rho }I^{p}x( \xi). $$
That is, \(x \in\operatorname{dom}L\), then \(y\in\operatorname{Im}L\). In conclusion,
$$\operatorname{Im}L=\bigl\{ y\in Y: v_{2}\bigl(\beta{}^{\rho}I^{p}J^{q}y( \xi)-J^{q}y(T)\bigr)+ \alpha v_{4} I^{\gamma,\delta}_{\eta}J^{q}y( \zeta)=0\bigr\} . $$
We define the linear operator \(P:X\rightarrow X\) as
It is obvious that \(P^{2}x=Px\) and \(\operatorname{Im}P=\operatorname {ker}L\). For any \(x\in X\), together with \(x=(x-Px)+Px\), we have \(X=\operatorname {ker}P+\operatorname{ker}L\). It is easy to obtain that \(\operatorname{ker}L\cap\operatorname{ker}P=\emptyset\), which implies
$$X=\operatorname{ker}P\oplus\operatorname{ker}L. $$
Next the operator \(Q:Y\rightarrow Y\) is defined as follows:
$$\begin{aligned} (Qy) (t)&= \bigl(v_{2}\bigl( \beta{}^{\rho}I^{p}J^{q}y(\xi)-J^{q}y(T) \bigr)+ \alpha v_{4} I^{\gamma,\delta}_{\eta}J^{q}y( \zeta) \bigr) z(t) \\ &= B\bigl(J^{q}y\bigr) z(t), \end{aligned} $$
where B is given by (4) and \(z\in Y\) satisfying \(B(J^{q}z)=1\).
Obviously, Q is a projection operator such that \(\operatorname {ker}Q=\operatorname{Im}L\) and \(\operatorname{Im} L=\{cz(t): c\in \mathbb{R}\}\). For any \(y\in Y\), because \(y=(y-Qy)+Qy\), we have \(Y=\operatorname{Im}L+ \operatorname{Im}Q\). Moreover, by a simple calculation, we can get \(\operatorname{Im}Q\cap\operatorname {Im}L=\emptyset\). Above all, \(Y=\operatorname{Im}L\oplus\operatorname{Im}Q\).
To sum up, we can get that ImL is a closed subspace of Y; \(\operatorname{dim}\operatorname{ker}L=\operatorname{co} \operatorname{dim}\operatorname{Im}L<+\infty\); that is, L is a Fredholm operator of index zero.
We now define the operator \(K_{p}y:Y\rightarrow X\) as follows:
$$\begin{aligned} (K_{p}y) (t)& =J^{q}y(t)- \frac{\alpha I^{\gamma,\delta}_{\eta }J^{q}y(\zeta)}{v_{2}}\cdot t \\ & = \frac{1}{\Gamma(q)} \int_{0}^{t}(t-s)^{q-1}y(s)\,ds - \frac{\alpha I^{\gamma,\delta}_{\eta}J^{q}y(\zeta)}{v_{2}}\cdot t. \end{aligned} $$
For any \(y\in\operatorname{Im}L\), we have
$$(K_{p}y) (0)=\alpha I^{\gamma,\delta}_{\eta}(K_{p}y) ( \zeta),\qquad (K_{p}y) (T)=\beta{}^{\rho}I^{p}(K_{p}y) (\xi), $$
then \((K_{p}y)(t)\in\operatorname{dom}L\). In addition, \((K_{p}y)(0)=0\), which means \(K_{p}y\in \operatorname{ker}P\). Therefore
$$K_{p}y\in\operatorname{dom}L\cap\operatorname{ker}P,\quad y\in \operatorname{Im} L. $$
Next we will prove that \(K_{p}\) is the inverse of \(L|_{\operatorname {dom}L\cap\operatorname{ker}P}\). It is clear that
$$(LK_{p}y) (t)=y(t),\quad y\in\operatorname{Im}L. $$
By Lemma 2.1, for each \(x\in\operatorname{dom}L\cap\operatorname {ker}P\), we have \(x(0)=0\) and
$$\begin{aligned} (K_{p}Lx) (t)&= J^{q}D^{q}x(t)- \frac{\alpha I^{\gamma,\delta}_{\eta }J^{q}D^{q}x(\zeta)}{v_{2}}\cdot t \\ &=x(t)-x(0)-x'(0)t-\frac{ \alpha I^{\gamma,\delta}_{\eta}x(\zeta)- x(0) \alpha I^{\gamma,\delta}_{\eta}\varphi(\zeta) -x'(0)\alpha I^{\gamma,\delta}_{\eta}\psi(\zeta)}{ v_{2}}\cdot t \\ &=x(t)-x(0)-x'(0)t-\frac{ x(0)- x(0) \alpha I^{\gamma,\delta}_{\eta }\varphi(\zeta) -x'(0)\alpha I^{\gamma,\delta}_{\eta}\psi(\zeta)}{ v_{2}}\cdot t \\ &=x(t)-x'(0)t+x'(0)t \\ &=x(t). \end{aligned}$$
This implies that \(K_{p}Lx=x\). So \(K_{p}=(L|_{\operatorname{dom}L\cap \operatorname{ker}P})^{-1}\). Thus the lemma holds. □
Lemma 3.2
N
is
L-compact on Ω̅ if
\(\operatorname{dom}L\cap \overline{\Omega}\neq\emptyset\), where Ω is a bounded open subset of X.
Proof
It follows from the continuity of f in condition (H1) and \(z\in Y\) that \((I-Q)N(\overline{\Omega})\) is bounded. In addition,
$$\biggl\{ \frac{1}{\Gamma(q)} \int_{0}^{t}(t-s)^{q-1}y(s)\,ds - \frac{\alpha I^{\gamma,\delta}_{\eta}J^{q}y(\zeta)}{v_{2}}\cdot t : y\in(I-Q)N(\overline{\Omega}) \biggr\} $$
and
$$\biggl\{ \frac{1}{\Gamma(q-1)} \int_{0}^{t}(t-s)^{q-2}y(s)\,ds - \frac{\alpha I^{\gamma,\delta}_{\eta}J^{q}y(\zeta)}{v_{2}} : y\in(I-Q)N(\overline{\Omega}) \biggr\} $$
are equi-continuous and uniformly bounded. By Ascoli–Arzela theorem, we get \(K_{p}(I-Q)N:\overline{\Omega}\rightarrow X\) is compact. Thus, N is L-compact. The proof is completed. □
Lemma 3.3
The set
\(\Omega_{1}=\{x \in\operatorname{dom}L \setminus \operatorname{ker} L:Lx=\lambda Nx, \lambda\in[0,1]\}\)
is bounded if (H1)–(H3) are satisfied.
Proof
Take \(x \in\Omega_{1}\), then \(x\notin\operatorname{ker}L\), so \(\lambda\neq0\) and \(Nx\in\operatorname{Im}L\). Thus we have
$$\bigl(v_{2}\bigl(\beta{}^{\rho}I^{p}J^{q}Nx( \xi)-J^{q}Nx(T)\bigr)+ \alpha v_{4} I^{\gamma,\delta}_{\eta}J^{q}Nx( \zeta)\bigr)z(t)=0, $$
where \(z\in Y\) satisfying \(B(J^{q}z)=1\). So we get
$$ v_{2}\bigl(\beta{}^{\rho}I^{p}J^{q}Nx( \xi)-J^{q}Nx(T)\bigr)+ \alpha v_{4} I^{\gamma,\delta}_{\eta}J^{q}Nx( \zeta)=0. $$
(10)
According to (H3), there exists at least a point \(t_{0}\in[0,T]\) such that
$$\bigl\vert x(t_{0}) \bigr\vert + \bigl\vert x'(t_{0}) \bigr\vert \leq M. $$
Using the Newton–Leibnitz formula, we have
$$ \Vert x \Vert _{\infty}=\max_{t\in[0,T]} \bigl\vert x(t) \bigr\vert =\max_{t\in[0,T]} \biggl\vert x(t_{0})+ \int _{t_{0}}^{t}x'(s)\,ds \biggr\vert \leq M+T \bigl\Vert x' \bigr\Vert _{\infty}. $$
(11)
In addition, for \(Lx=\lambda Nx\) and \(x\in\operatorname{dom}L\), we have
$$x(t)= \frac{\lambda}{\Gamma(q)} \int_{0}^{t}(t-s)^{q-1}f \bigl(s,x(s),x'(s)\bigr)\, ds+x(0)+x'(0)t $$
and
$$ x'(t)= \frac{\lambda}{\Gamma(q-1)} \int _{0}^{t}(t-s)^{q-2}f \bigl(s,x(s),x'(s)\bigr)\,ds+x'(0). $$
(12)
Take \(t=t_{0}\) in (12), we get
$$x'(t_{0})= \frac{\lambda}{\Gamma(q-1)} \int _{0}^{t_{0}}(t_{0}-s)^{q-2}f \bigl(s,x(s),x'(s)\bigr)\,ds+x'(0). $$
This together with \(|x'(t_{0})|\leq M\) and (11) implies that
$$\begin{aligned} \bigl\vert x'(0) \bigr\vert &\leq \bigl\vert x'(t_{0}) \bigr\vert +\frac{\lambda}{\Gamma(q-1)} \int _{0}^{t_{0}}(t_{0}-s)^{q-2} \bigl\vert f\bigl(s,x(s),x'(s)\bigr) \bigr\vert \,ds \\ &\leq M+\frac{\lambda}{\Gamma(q-1)} \int _{0}^{t_{0}}(t_{0}-s)^{q-2} \bigl[u(s) \bigl\vert x(s) \bigr\vert +v(s) \bigl\vert x'(s) \bigr\vert +w(s)\bigr]\,ds \\ &\leq M+\frac{T^{q-1}}{\Gamma(q)}\bigl( \Vert u \Vert _{\infty} \Vert x \Vert _{\infty}+ \Vert v \Vert _{\infty} \bigl\Vert x' \bigr\Vert _{\infty}+ \Vert w \Vert _{\infty} \bigr) \\ &\leq M+\frac{T^{q-1}}{\Gamma(q)}\bigl( \Vert u \Vert _{\infty}\bigl(M+T \bigl\Vert x' \bigr\Vert _{\infty }\bigr)+ \Vert v \Vert _{\infty} \bigl\Vert x' \bigr\Vert _{\infty}+ \Vert w \Vert _{\infty}\bigr) \\ &= M+\frac{T^{q-1}}{\Gamma(q)}\bigl(M \Vert u \Vert _{\infty}+ \Vert w \Vert _{\infty}\bigr)+ \frac {T^{q} \Vert u \Vert _{\infty}+T^{q-1} \Vert v \Vert _{\infty}}{\Gamma(q)} \bigl\Vert x' \bigr\Vert _{\infty}. \end{aligned} $$
Then we conclude that
$$\begin{aligned} \bigl\vert x'(t) \bigr\vert &\leq \frac{\lambda}{\Gamma(q-1)} \int _{0}^{t}(t-s)^{q-2} \bigl\vert f \bigl(s,x(s),x'(s)\bigr) \bigr\vert \,ds+ \bigl\vert x'(0) \bigr\vert \\ &\leq \frac{T^{q-1}}{\Gamma(q)}\bigl( \Vert u \Vert _{\infty} \Vert x \Vert _{\infty}+ \Vert v \Vert _{\infty} \bigl\Vert x' \bigr\Vert _{\infty}+ \Vert w \Vert _{\infty} \bigr) \\ &\quad{}+ M+\frac{T^{q-1}}{\Gamma(q)}\bigl(M \Vert u \Vert _{\infty}+ \Vert w \Vert _{\infty}\bigr)+ \frac {T^{q} \Vert u \Vert _{\infty}+T^{q-1} \Vert v \Vert _{\infty}}{\Gamma(q)} \bigl\Vert x' \bigr\Vert _{\infty} \\ &\leq \frac{T^{q-1}}{\Gamma(q)}\bigl( \Vert u \Vert _{\infty}\bigl(M+T \bigl\Vert x' \bigr\Vert _{\infty}\bigr)+ \Vert v \Vert _{\infty} \bigl\Vert x' \bigr\Vert _{\infty}+ \Vert w \Vert _{\infty}\bigr) \\ &\quad{} + M+\frac{T^{q-1}}{\Gamma(q)}\bigl(M \Vert u \Vert _{\infty}+ \Vert w \Vert _{\infty}\bigr)+ \frac {T^{q} \Vert u \Vert _{\infty}+T^{q-1} \Vert v \Vert _{\infty}}{\Gamma(q)} \bigl\Vert x' \bigr\Vert _{\infty} \\ &= M+2\frac{T^{q-1}}{\Gamma(q)}\bigl(M \Vert u \Vert _{\infty}+ \Vert w \Vert _{\infty}\bigr)+ 2\frac {T^{q} \Vert u \Vert _{\infty}+T^{q-1} \Vert v \Vert _{\infty}}{\Gamma(q)} \bigl\Vert x' \bigr\Vert _{\infty}. \end{aligned} $$
Therefore, we can obtain that
$$\bigl\Vert x' \bigr\Vert _{\infty} \leq \frac{ M\Gamma(q)+2MT^{q-1} \Vert u \Vert _{\infty}+2T^{q-1} \Vert w \Vert _{\infty} }{\Gamma(q)-2T^{q} \Vert u \Vert _{\infty}-2T^{q-1} \Vert v \Vert _{\infty}}=M_{1}. $$
Combining this with (11), we have
$$\Vert x \Vert _{\infty}\leq M+T \bigl\Vert x' \bigr\Vert _{\infty}\leq M+TM_{1}. $$
Then \(\Omega_{1}\) is bounded. The proof of the lemma is completed. □
Lemma 3.4
The set
\(\Omega_{2}=\{x:x\in\operatorname{ker}L, Nx\in\operatorname{Im}L\}\)
is bounded if (H1), (H4) hold.
Proof
Let \(x\in\Omega_{2}\), then \(x(t)\equiv c\) and \(Nx\in\operatorname{Im}L\), so we can get
$$v_{2}\bigl(\beta{}^{\rho}I^{p}J^{q}Nx( \xi)-J^{q}Nx(T)\bigr)+ \alpha v_{4} I^{\gamma,\delta}_{\eta}J^{q}Nx( \zeta)=0. $$
According to (H4), we have \(|c|\leq D\), that is to say, \(\Omega_{2}\) is bounded. We complete the proof. □
Lemma 3.5
The set
\(\Omega_{3}=\{x\in\operatorname {ker}L:\lambda x+\alpha(1-\lambda)JQNx=0, \lambda\in[0,1]\}\)
is bounded if conditions (H1), (H4) are satisfied, where
\(J :\operatorname{Im}Q\rightarrow\operatorname{ker}L\)
is a linear isomorphism defined by
$$J(cz_{1})=c,\quad c\in\mathbb{R}, $$
and
$$\alpha= \left \{ \textstyle\begin{array}{l@{\quad}l} -1,&\textit{if } (5) \textit{ holds};\\ 1,&\textit{if } (6) \textit{ holds}, \end{array}\displaystyle \right . $$
where
\(z_{1}\)
is introduced in Lemma 2.4.
Proof
Suppose that \(x\in\Omega_{3}\), we have \(x(t)=c\) and
$$\lambda x+\alpha(1-\lambda) JQNx=0, $$
thus we have
$$\lambda c=-\alpha(1-\lambda) \bigl(v_{2}\bigl(\beta{}^{\rho}I^{p}J^{q}Nx( \xi)-J^{q}Nx(T)\bigr)+ \alpha v_{4} I^{\gamma,\delta}_{\eta}J^{q}Nx( \zeta)\bigr). $$
If \(\lambda=0\), by condition (H4) we have \(|c|\leq D\). If \(\lambda=1\), then \(c=0\). If \(\lambda\in(0,1)\), we suppose \(|c|> D\), then
$$\lambda c^{2}=-\alpha(1-\lambda)c\bigl(v_{2}\bigl( \beta{}^{\rho}I^{p}J^{q}Nx(\xi )-J^{q}Nx(T) \bigr)+ \alpha v_{4} I^{\gamma,\delta}_{\eta}J^{q}Nx( \zeta)\bigr)< 0 $$
which contradicts with \(\lambda c^{2}>0\), so \(|c|\leq D\). Then the lemma holds. □
Theorem 3.1 can be proved now.
Proof of Theorem 3.1
Suppose that \(\Omega\supset \bigcup_{i=1}^{3}\overline{\Omega_{i}}\cup\{0\}\) is a bounded open subset of X, from Lemma 3.2 we know that N is L-compact on Ω̅. In view of Lemmas 3.3 and 3.4, we can get:
-
(i)
\(Lx\neq\lambda Nx\) for every \((x, \lambda) \in [(\operatorname{dom}L \setminus\operatorname{ker}L) \cap\partial \Omega] \times(0,1)\);
-
(ii)
\(Nx \notin\operatorname{Im}L\) for every \(x\in \operatorname{ker}L \cap \partial\Omega\).
Set \(H(x,\lambda)=\lambda Jx+\alpha(1-\lambda)QNx\). It follows from Lemma 3.5 that we have \(H(x,\lambda)\neq0\) for any \(x\in \partial\Omega\cap\operatorname{ker}L\). So, by the homotopic property of degree, we have
$$\operatorname{deg} (JQN|_{\operatorname{ker}L},\Omega\cap\operatorname {ker}L,0 )=\operatorname{deg} (\alpha I,\Omega \cap\operatorname{ker}L,0 )\neq0. $$
All the conditions of Theorem 2.1 are satisfied. So there must be at least one solution of problem (1) in X. The proof of Theorem 3.1 is completed. □
Theorem 3.2
Suppose that
\((A2)\)
and (H1), (H2), (H3′), (H4′) are satisfied, then there must be at least one solution of problem (1) in
X
provided that
\(2T^{q}\|u\|_{\infty}+2T^{q-1}\|v\| _{\infty}<\Gamma(q)\).
To prove the theorem, we need the following lemmas.
Lemma 3.6
Assume that
\((A2)\)
holds, then
\(L:\operatorname{dom}L \subset X\rightarrow Y\)
is a Fredholm operator with index zero. And the linear continuous projector
\(P:X\rightarrow X \)
can be defined by
$$(Px) (t)=\frac{x(T)}{T}t. $$
Furthermore, define the linear operator
\(K_{p}:\operatorname {Im}L\rightarrow\operatorname{dom}L\cap\operatorname{ker}P\)
as follows:
$$(K_{p}y) (t)=\frac{1}{\Gamma(q)} \int^{t}_{0}(t-s)^{q-1}y(s)\,ds $$
such that
\(K_{p}=(L|_{\operatorname{dom}L\cap\operatorname{ker}P})^{-1}\).
Proof
Let \(\varphi(t)=1\), \(\psi(t)=t\). In view of \((A2)\) we know
$$ \begin{gathered} \alpha=0, \\ \varphi(0)-\alpha I^{\gamma,\delta}_{\eta}\varphi(\zeta)=1,\qquad \psi(0)-\alpha I^{\gamma,\delta}_{\eta} \psi(\zeta)=0, \\ \varphi(T)-\beta{}^{\rho}I^{p}\varphi(\xi)=v_{3},\qquad \psi(T)-\beta{}^{\rho}I^{p} \psi(\xi)=0, \end{gathered} $$
(13)
and we get
$$\operatorname{ker}L=\{ct, c\in\mathbb{R}\} $$
and
$$\operatorname{Im}L=\bigl\{ y\in Y: J^{q}y(T)=\beta{}^{\rho}I^{p}J^{q}y( \xi)\bigr\} . $$
Besides, operators \(P:X\rightarrow X\), \(Q:Y\rightarrow Y\) can be defined as follows:
$$(Px) (t)=\frac{x(T)}{T}t $$
and
$$(Qy) (t)=\bigl(J^{q}y(T)-\beta{}^{\rho}I^{p}J^{q}y( \xi)\bigr)z_{1}(t), $$
where \(z_{1}\in Y\) satisfying \(J^{q}z_{1}(T)-\beta{}^{\rho }I^{p}J^{q}z_{1}(\xi)=1\). In addition, for each \(x\in\operatorname{dom}L\cap\operatorname {ker}P\), we have \(x(0)=0\), \(x(T)=0\), then we get the generalized inverse operator of L as follows:
$$(K_{p}y) (t)=J^{q}y(t)=\frac{1}{\Gamma(q)} \int^{t}_{0}(t-s)^{q-1}y(s)\,ds. $$
The detailed proof of Lemma 3.6 is similar to that of Lemma 3.1, so we omit it. □
Proof of Theorem 3.2
The proof of Theorem 3.2 is similar to that of Theorem 3.1, we omit it. □
Theorem 3.3
Suppose that
\((A3)\)
and (H1), (H2), (H3), and (H4″) are satisfied, then there must be at least one solution of problem (1) in
X
provided that
\(2T^{q}\|u\|_{\infty}+2T^{q-1}\|v\| _{\infty}<\Gamma(q)\).
To prove the theorem, we need the following lemmas.
Lemma 3.7
Assume that
\((A3)\)
holds, then
\(L:\operatorname{dom}L \subset X\rightarrow Y\)
is a Fredholm operator with index zero. And a linear continuous projector
\(P:X\rightarrow X \)
can be defined by
$$(Px) (t)=x(0) (1+kt), $$
where
\(k=\frac{v_{1}}{v_{2}}=-\frac{v_{3}}{v_{4}}\). Furthermore, define the linear operator
\(K_{p}y:\operatorname{Im}L\rightarrow\operatorname {dom}L\cap\operatorname{ker}P\)
as follows:
$$(K_{p}y) (t)=\frac{1}{\Gamma(q)} \int^{t}_{0}(t-s)^{q-1}y(s)\,ds $$
such that
\(K_{p}=(L|_{\operatorname{dom}L\cap\operatorname{ker}P})^{-1}\).
Proof
Let \(\varphi_{1}(t)=1+kt\). In view of \((A3)\) we know
$$\varphi_{1}(0)-\alpha I^{\gamma,\delta}_{\eta} \varphi_{1}(\zeta)=0,\quad\quad \varphi_{1}(T)-\beta{}^{\rho}I^{p} \varphi_{1}(\xi)=0, $$
we can easily get
$$\operatorname{ker}L=\bigl\{ c(1+kt): c\in\mathbb{R}\bigr\} . $$
Moreover, we can obtain that
$$\operatorname{Im}L=\bigl\{ y\in Y: v_{1}\bigl(\beta{}^{\rho}I^{p}J^{q}y( \xi)-J^{q}y(T)\bigr)= \alpha v_{3} I^{\gamma,\delta}_{\eta}J^{q}y( \zeta)\bigr\} . $$
We define the linear operator \(P:X\rightarrow X\) as
$$(Px) (t)=x(0) (1+kt), $$
and the operator \(Q:Y\rightarrow Y\) as
$$(Qy) (t)= \bigl( v_{1}\bigl(\beta{}^{\rho}I^{p}J^{q}y( \xi)-J^{q}y(T)\bigr)- \alpha v_{3} I^{\gamma,\delta}_{\eta}J^{q}y( \zeta)\bigr)z_{2}(t), $$
where \(z_{2}\in Y\) satisfying \(v_{2}(\beta{}^{\rho}I^{p}J^{q}z_{2}(\xi )-J^{q}z_{2}(T))+\alpha v_{4}I^{\gamma,\delta}_{\eta}J^{q}z_{2}(\zeta)=1\).
In addition, for each \(x\in\operatorname{dom}L\cap\operatorname {ker}P\), we have \(x(0)=0\), then we get the generalized inverse operator of L as follows:
$$\begin{aligned} (K_{p}y) (t)& =J^{q}y(t)- \frac{\alpha I^{\gamma,\delta}_{\eta }J^{q}y(\zeta)}{v_{2}}\cdot t \\ & = \frac{1}{\Gamma(q)} \int_{0}^{t}(t-s)^{q-1}y(s)\,ds - \frac{\alpha I^{\gamma,\delta}_{\eta}J^{q}y(\zeta)}{v_{2}}\cdot t. \end{aligned} $$
The detailed proof of Lemma 3.7 is similar to that of Lemma 3.1, so we omit it. □
Proof of Theorem 3.3
The proof of Theorem 3.3 is similar to that of Theorem 3.1, we omit it. □