In this section, we begin to prove the existence of solutions to problem (1.1). Consider the functions \(\phi_{1}(z) = \sum_{i=1}^{\infty} a_{i}\xi_{i}^{z}\), \(\phi_{2}(z) = \sum_{i=1}^{\infty} b_{i}\eta _{i}^{z}\), \(z\in[0,\infty)\). According to \(\sum_{i=1}^{\infty} \vert a_{i} \vert <\infty\), \(\sum_{i=1}^{\infty} \vert b_{i} \vert <\infty\), one has the series are (uniformly) convergent and thus \(\phi _{1}\), \(\phi_{2}\) are continuous on \([0,\infty)\).
The following assumption will be used in our main results:
-
\((\mathrm{H}_{0})\)
:
-
There exist \(z_{0}\), \(\tilde{z}_{0}\) with \(z_{0}\geq\alpha_{1}\), \(\tilde{z}_{0}\geq \alpha_{2}\) such that \(\phi_{1}(z_{0})\cdot\phi_{2}(\tilde{z}_{0})\neq0\).
The following lemma is fundamental in the proofs of our main results.
Lemma 3.1
Problem (1.1) is equivalent to the following equation:
$$ \textstyle\begin{cases} D_{0+}^{\alpha_{1}} u(t)=\phi_{q} [I_{0+}^{\beta_{1}}f (t,v(t),D_{0 +}^{\alpha_{2}-1}v(t), \ldots,D_{0+}^{\alpha_{2}-(n-1)}v(t) ) ],\quad t\in(0,1), \\ D_{0+}^{\alpha_{2}} v(t)=\phi_{q} [I_{0+}^{\beta_{2}}g (t,u(t),D_{0 +}^{\alpha_{1}-1}u(t), \ldots,D_{0+}^{\alpha_{1}-(n-1)}u(t) ) ],\quad t\in(0,1), \\ u'(0)= \cdots=u^{(n-1)}(0)=0, \quad \quad u(0)=\sum_{i=1}^{\infty}a_{i}u(\xi _{i}),\\ v'(0)= \cdots=v^{(n-1)}(0)=0, \quad \quad v(0)=\sum_{i=1}^{\infty}b_{i}v(\eta_{i}). \end{cases} $$
(3.1)
Proof
By Lemma 2.1, \(D_{0+}^{\beta_{1}} \phi_{p}( D_{0+}^{\alpha_{1}} u(t))=f (t,v(t),D_{0+}^{\alpha_{2}-1}v(t), \ldots,D_{0+}^{\alpha_{2}-(n-1)}v(t) )\) has the following solution:
$$\phi_{p} \bigl( D_{0+}^{\alpha_{1}}u(t) \bigr)=I_{0+}^{\beta_{1}}f \bigl(t,v(t),D_{0+}^{\alpha_{2}-1}v(t), \ldots,D_{0+}^{\alpha_{2}-(n-1)}v(t) \bigr)+c, \quad c\in\mathbb{R}. $$
Substituting \(t=0\) into the above formula, by \(D_{0^{+}}^{\alpha _{1}}u(0)=0\), we obtain \(c=0\). Then we have
$$\begin{aligned} \phi_{p} \bigl( D_{0+}^{\alpha_{1}}u(t) \bigr)=I_{0+}^{\beta_{1}}f \bigl(t,v(t),D_{0+}^{\alpha_{2}-1}v(t), \ldots,D_{0+}^{\alpha_{2}-(n-1)}v(t) \bigr). \end{aligned}$$
(3.2)
Applying the operator \(\phi_{q}\) to the both sides of (3.2) respectively, we have
$$D_{0+}^{\alpha_{1}} u(t)=\phi_{q} \bigl[I_{0+}^{\beta_{1}}f \bigl(t,v(t),D_{0+}^{\alpha_{2}-1}v(t), \ldots,D_{0+}^{\alpha_{2}-(n-1)}v(t) \bigr) \bigr]. $$
By a similar argument, we have
$$D_{0+}^{\beta_{2}} \phi_{p} \bigl( D_{0+}^{\alpha_{2}} v(t) \bigr)=g \bigl(t,u(t),D_{0+}^{\alpha_{1}-1}u(t), \ldots,D_{0^{+}}^{\alpha_{1}-(n-1)}u(t) \bigr) $$
is equivalent to
$$D_{0+}^{\alpha_{2}} v(t)=\phi_{q} \bigl[I_{0+}^{\beta_{2}}g \bigl(t,u(t),D_{0^{+}}^{\alpha_{1}-1}u(t), \ldots,D_{0+}^{\alpha_{1}-(n-1)}u(t) \bigr) \bigr]. $$
Therefore, BVP (1.1) is rewritten by (3.1)
It is easy to verify that equation (1.1) has a solution \((u,v) \) if and only if \((u,v)\) solves equation (3.1). □
Let \(E=C[0,1]\) with the norm \(\Vert x \Vert _{\infty}=\max_{0 \le t \le1} \vert x(t) \vert \). Now, we set \(X_{1} =\{u(t): u(t), D_{0+}^{\alpha_{1}-i}u(t)\in E, i=1,2,\ldots,n-1\}\) with the norm
$${{ \Vert u \Vert }_{X_{1}}}=\max \bigl\{ \Vert u \Vert _{\infty}, \bigl\Vert D_{0+}^{\alpha _{1}-1}u \bigr\Vert _{\infty},\ldots, \bigl\Vert D_{0+}^{\alpha_{1}-(n-1)}u \bigr\Vert _{\infty}\bigr\} $$
and \(X_{2} =\{v(t): v(t), D_{0+}^{\alpha_{2}-i}v(t)\in E, i=1,2,\ldots,n-1\} \) with the norm
$${{ \Vert v \Vert }_{X_{2}}}=\max \bigl\{ \Vert v \Vert _{\infty}, \bigl\Vert D_{0+}^{\alpha_{2}-1}v \bigr\Vert _{\infty},\ldots, \bigl\Vert D_{0+}^{\alpha_{2}-(n-1)}v \bigr\Vert _{\infty}\bigr\} . $$
Let \(Y=X_{1}\times X_{2}\) with the norm \(\Vert (u,v) \Vert _{Y}=\max\{ \Vert u \Vert _{X_{1}}, \Vert v \Vert _{X_{2}} \} \) and \(Z=E\times E\) with the norm \(\Vert (x,y) \Vert _{Z}=\max \{ \Vert x \Vert _{\infty}, \Vert y \Vert _{\infty}\}\).
Clearly, X and Y are Banach spaces.
Define the linear operator \(L_{1}:\operatorname{dom}L_{1}\rightarrow E\) by setting
$$\operatorname{dom}L_{1}= \Biggl\{ u\in X_{1}\bigg|u'(0)= \cdots=u^{(n-1)}(0)=0, u(0)=\sum_{i=1}^{\infty}a_{i}u(\xi_{i}) \Biggr\} $$
and
$$L_{1}u=D_{0 +}^{\alpha_{1}}u, \quad u\in\operatorname{dom}L_{1}. $$
Define the linear operator \(L_{2}\) from \(\operatorname{dom}L_{2} \rightarrow E\) by setting
$$\operatorname{dom}L_{2}= \Biggl\{ v\in X_{2}\bigg|v'(0)= \cdots=v^{(n-1)}(0))=0, v(0)=\sum_{i=1}^{\infty}b_{i}v(\eta_{i}) \Biggr\} $$
and
$$L_{2}v=D_{0 +}^{\alpha_{2}}v, \quad v\in\operatorname{dom}L_{2}. $$
Define the operator \(L: \operatorname{dom}L \rightarrow Z\) with
$$\operatorname{dom}L= \bigl\{ (u,v)\in Y|u\in\operatorname{dom}L_{1}, v \in\operatorname{dom}L_{2} \bigr\} $$
and
$$L(u,v)=(L_{1}u,L_{2}v). $$
Let \(N:Y\to Z\) be the Nemytskii operator
$$N(u,v)=(N_{1}v,N_{2}u), $$
where \(N_{1}:X\to E\) is defined by
$$N_{1}v(t)=\phi_{q} \bigl[I_{0+}^{\beta_{1}}f \bigl(t,v(t),D_{0+}^{\alpha_{2}-1}v(t), \ldots,D_{0+}^{\alpha_{2}-(n-1)}v(t) \bigr) \bigr] $$
and \(N_{2}:X\to E\) is defined by
$$N_{2}u(t)=\phi_{q} \bigl[I_{0+}^{\beta_{2}}g \bigl(t,u(t),D_{0 +}^{\alpha_{1}-1}u(t), \ldots,D_{0+}^{\alpha_{1}-(n-1)}u(t) \bigr) \bigr]. $$
Then BVP (3.1) can be written as \(L(u,v)=N(u,v)\).
Lemma 3.2
L
is defined as above, then
$$\begin{aligned} &\operatorname{Ker}L= \bigl\{ (u,v)\in X: (u,v)=(c_{0},d_{0}), c_{0},d_{0} \in \mathbb{R} \bigr\} , \end{aligned}$$
(3.3)
$$\begin{aligned} &\operatorname{Im} L = \Biggl\{ (x,y)\in Z: \sum_{i=1}^{\infty}a_{i}I_{0+}^{\alpha _{1}}x(\xi_{i})=0; \sum _{i=1}^{\infty}b_{i}I_{0+}^{\alpha_{2}}y( \eta_{i})=0 \Biggr\} . \end{aligned}$$
(3.4)
Proof
For \((u,v)\in\operatorname{Ker}L\), then \(L_{1}u=L_{2}v=0\). By Lemma 2.1, the equation \(D_{0 +}^{\alpha_{1}}u(t)=0\) has solution
$$u(t)=c_{0}+ c_{1}t+\cdots+c_{n-1}t^{n-1} . $$
In view of \(u^{(i)}(0)=0\), \(i=1,2,\ldots,n-1\), we get \(c_{i}=0\), \(i=1,2,\ldots,n-1\). Then \(u(t)=c_{0}\). Similarly, for \(v\in\operatorname{Ker}L_{2}\), we have \(v(t)=d_{0}\in\mathbb{R}\). Thus, we obtain (3.3).
Next we prove that (3.4) holds. Let \((x,y)\in\operatorname{Im}L\), so there exists \((u,v)\in\operatorname{dom}L\) such that \(x(t)=D_{0 +}^{\alpha _{1}}u(t)\), \(y(t)=D_{0 +}^{\alpha_{2}}v(t)\). By Lemma 2.1, we have
$$u(t) =I_{0 +}^{\alpha_{1} }x(t)+\sum_{i=0}^{n-1}{{c}_{i}}t^{i}, \quad\quad v(t) =I_{0 +}^{\alpha_{2}}y(t)+\sum _{i=0}^{n-1}{{d}_{i}}t^{i},\quad c_{i},d_{i} \in \mathbb{R}. $$
In view of \(u^{(i)}(0)=v^{(i)}(0)=0\), \(i=1,2,\ldots,n-1\), we get \(c_{i}=d_{i}=0\), \(i=1,2,\ldots, n-1\). Hence,
$$u(t)=I_{0 +}^{\alpha_{1}}x(t)+c_{0},\quad\quad v(t)=I_{0 +}^{\alpha _{2}}y(t)+d_{0}. $$
According to \(u(0)=\sum_{i=1}^{\infty}a_{i}u(\xi_{i})\) and \(v(0)=\sum_{i=1}^{\infty}b_{i}v(\eta_{i})\), we have
$$ \begin{gathered} u(0)= I_{0 +}^{\alpha_{1}}x(0)+c_{0}= \sum_{i=1}^{\infty}a_{i}u( \xi_{i})=\sum_{i=1}^{\infty}a_{i} \bigl(I_{0 +}^{\alpha_{1} }x(\xi_{i})+{{c}_{0}} \bigr)=\sum_{i=1}^{\infty}a_{i}I_{0 +}^{\alpha_{1}}x( \xi_{i})+{c}_{0}, \\ v(0)=I_{0 +}^{\alpha_{2} }y(0)+d_{0}=\sum _{i=1}^{\infty}b_{i}v(\xi_{i})=\sum _{i=1}^{\infty}b_{i} \bigl(I_{0 +}^{\alpha_{2}}y(\eta_{i})+{{c}_{0}} \bigr)=\sum_{i=1}^{\infty}b_{i}I_{0 +}^{\alpha_{2}}y( \eta_{i})+{d}_{0}, \end{gathered} $$
that is,
$$\sum_{i=1}^{\infty}a_{i}I_{0 +}^{\alpha_{1}}x( \xi_{i})=0,\qquad\sum_{i=1}^{\infty}b_{i}I_{0 +}^{\alpha_{2}}y(\eta_{i})=0. $$
On the other hand, suppose that \((x,y)\) satisfies the above equations. Let \(u(t)=I_{0 +}^{\alpha_{1}}x(t)\) and \(v(t)=I_{0 +}^{\alpha_{2}}y(t)\), we can prove \((u ,v )\in\operatorname{dom} L\) and \(L (u ,v )=(x,y)\). Then (3.4) holds. □
Lemma 3.3
The mapping
\(L:\operatorname{dom} L \subset Y\rightarrow Z\)
is a Fredholm operator of index zero.
Proof
The linear continuous projector operator \(P(u,v)=(P_{1}u,P_{2}v)\) can be defined as
$${{P}_{1}}u=u(0), \qquad {{P}_{2}}v=v(0). $$
Obviously, \(P_{1}^{2}=P_{1}\) and \(P_{2}^{2}=P_{2}\).
It is clear that
$$\operatorname{Ker}P= \bigl\{ (u,v): u(0)=0, v(0)=0 \bigr\} . $$
It follows from \((u,v)=(u,v)-P(u,v)+P(u,v)\) that \(Y=\operatorname{Ker}P+ \operatorname{Ker}L\). For \((u,u)\in\operatorname{Ker} L\cap\operatorname{Ker}P\), then \(u=c_{0}\), \(v=d_{0}\), \(c_{0},d_{0}\in\mathbb{R}\). Furthermore, by the definition of KerP, we have \(c_{0}= d_{0}=0\). Thus, we get
$$Y=\operatorname{Ker}L\oplus\operatorname{Ker}P. $$
By \((\mathrm{H}_{0})\), the linear operator \(Q(x,y)=(Q_{1}x,Q_{2}y)\) can be defined as
$$\begin{aligned}& Q_{1}x(t) =t^{\theta_{1}}\cdot\frac{\sum_{i=1}^{\infty}a_{i}I_{0 +}^{\alpha_{1}}x(\xi_{i})}{\sum_{i=1}^{\infty}a_{i}(I_{0 +}^{\alpha_{1}}t^{\theta _{1}})(\xi_{i})}=t^{\theta_{1}} \cdot\frac{\Gamma(1+\alpha_{1}+\theta_{1} )}{\phi _{1}(z_{0})\Gamma(1+\theta_{1} )}\sum_{i=1}^{\infty}a_{i}I_{0 +}^{\alpha_{1}}x(\xi _{i}), \\& Q_{2}y(t) =t^{\theta_{2}}\cdot\frac{\sum_{i=1}^{\infty}b_{i}I_{0 +}^{\alpha _{2}}y(\eta_{i})}{\sum_{i=1}^{\infty}b_{i}(I_{0 +}^{\alpha_{2}}t^{\theta_{2}})(\eta _{i})}=t^{\theta_{2}} \cdot\frac{\Gamma(1+\alpha_{2}+\theta_{2} )}{\phi_{2}(\tilde {z}_{0})\Gamma(1+\theta_{2} )}\sum_{i=1}^{\infty}b_{i}I_{0 +}^{\alpha_{2}}y(\eta_{i}), \end{aligned}$$
where \(\theta_{1}=z_{0}-\alpha_{1}\), \(\theta_{2}=\tilde{z}_{0}-\alpha_{2}\).
Obviously, \(Q(x,y)= (Q_{1}x(t),Q_{2}y(t) )\cong\mathbb{R}^{2}\).
For \(x(t)\in E\), we have
$$\begin{aligned} {{Q}_{1}} \bigl({{Q}_{1}}x(t) \bigr) =&\frac{\Gamma(1+\alpha_{1}+\theta_{1} )}{\phi_{1}(z_{0})\Gamma(1+\theta_{1} )} \sum_{i=1}^{\infty}a_{i}I_{0 +}^{\alpha _{1}}x( \xi_{i})\cdot Q_{1} \bigl(t^{\theta_{1}} \bigr) \\ =&\frac{\Gamma(1+\alpha_{1}+\theta_{1} )}{\phi_{1}(z_{0})\Gamma(1+\theta_{1} )}\sum_{i=1}^{\infty}a_{i}I_{0 +}^{\alpha_{1}}x(\xi_{i})\cdot t^{\theta_{1}}\cdot \frac{\Gamma(1+\alpha_{1}+\theta_{1} )}{\phi_{1}(z_{0})\Gamma(1+\theta_{1} )} \cdot\sum_{i=1}^{\infty}a_{i} \bigl(I_{0 +}^{\alpha_{1}}t^{\theta_{1}} \bigr) ( \xi_{i}) \\ =&\frac{\Gamma(1+\alpha_{1}+\theta_{1} )}{\phi_{1}(z_{0})\Gamma(1+\theta_{1} )}\sum_{i=1}^{\infty}a_{i}I_{0 +}^{\alpha_{1}}x(\xi_{i})\cdot t^{\theta_{1}} \\ &{}\cdot \frac{\Gamma(1+\alpha_{1}+\theta_{1} )}{\phi_{1}(z_{0})\Gamma(1+\theta_{1} )}\cdot\sum_{i=1}^{\infty}a_{i}\frac{\Gamma(1+\theta_{1} ){\xi_{i}}^{\alpha _{1}+\theta_{1}}}{{\Gamma(1+\alpha_{1}+\theta_{1} )}} \\ =&\frac{\Gamma(1+\alpha_{1}+\theta_{1} )}{\phi_{1}(z_{0})\Gamma(1+\theta_{1} )}\sum_{i=1}^{\infty}a_{i}I_{0 +}^{\alpha_{1}}x(\xi_{i})\cdot t^{\theta_{1}}\cdot \frac{\Gamma(1+\alpha_{1}+\theta_{1} )}{\phi_{1}(z_{0})\Gamma(1+\theta_{1} )}\cdot\sum_{i=1}^{\infty}a_{i}\frac{\Gamma(1+\theta_{1} ){\xi _{i}}^{z_{0}}}{{\Gamma(1+\alpha_{1}+\theta_{1} )}} \\ =&\frac{\Gamma(1+\alpha_{1}+\theta_{1} )}{\phi_{1}(z_{0})\Gamma(1+\theta_{1} )}\sum_{i=1}^{\infty}a_{i}I_{0 +}^{\alpha_{1}}x(\xi_{i})\cdot t^{\theta_{1}}\cdot \frac{\Gamma(1+\alpha_{1}+\theta_{1} )}{\phi_{1}(z_{0})\Gamma(1+\theta_{1} )} \cdot\frac{\phi_{1}(z_{0})\Gamma(1+\theta_{1} )}{\Gamma(1+\alpha_{1}+\theta _{1} )} \\ =&t^{\theta_{1}}\cdot\frac{\Gamma(1+\alpha_{1}+\theta_{1} )}{\phi _{1}(z_{0})\Gamma(1+\theta_{1} )}\sum_{i=1}^{\infty}a_{i}I_{0 +}^{\alpha_{1}}x(\xi _{i}) \\ =&{{Q}_{1}}x(t). \end{aligned}$$
Similarly, \(Q_{2}^{2}=Q_{2}\), that is to say, the operator Q is idempotent. It follows from \((x,y)=(x,y)-Q(x,y)+Q(x,y)\) that \(Z=\operatorname{Im}L+ \operatorname{Im}Q\). Moreover, by \(\operatorname{Ker}Q=\operatorname{Im}L \) and \(Q_{2}^{2}=Q_{2}\), we get \(\operatorname{Im}L\cap\operatorname{Im}Q=\{(0,0)\}\). Hence,
$$Z=\operatorname{Im}L\oplus\operatorname{Im}Q. $$
Now, \(\operatorname{Ind}L = \operatorname{\operatorname{dim} \operatorname{Ker}} L - \operatorname{\operatorname{codim} \operatorname{Im}}L = 0\), so L is a Fredholm mapping of index zero. □
For every \((u,v)\in Y\),
$$\begin{aligned} \bigl\Vert P(u,v) \bigr\Vert _{Y} = \max \bigl\{ \Vert P_{1}u \Vert _{X_{1}}; \Vert P_{2}v \Vert _{X_{2}} \bigr\} =\max \bigl\{ \bigl\vert {u} (0) \bigr\vert ; \bigl\vert {v} (0) \bigr\vert \bigr\} . \end{aligned}$$
(3.5)
Furthermore, the operator \(K_{P}:\operatorname{Im}L\to \operatorname {dom}L \cap\operatorname{Ker} P\) can be defined
$$K_{P}(x,y)= \bigl(I_{0 +}^{\alpha_{1}} x,I_{0 +}^{\alpha_{2}}y \bigr). $$
For \((x,y)\in\operatorname{Im} L \), we have
$$\begin{aligned} LK_{P}(x,y)=L \bigl(I_{0 +}^{\alpha_{1}}x,I_{0 +}^{\alpha_{2}}y \bigr)= \bigl(D_{0 +}^{\alpha_{1}}I_{0 +}^{\alpha_{1}}x, D_{0 +}^{\alpha_{2}}I_{0 +}^{\alpha_{2}}y \bigr) =(x,y). \end{aligned}$$
(3.6)
On the other hand, for \((u,v)\in\operatorname{dom} L \cap\operatorname{Ker} P \), according to Lemma 2.1, we have
$$\begin{aligned}& I_{0 +}^{\alpha_{1}}L_{1}u(t)=I_{0 +}^{\alpha_{1}} D_{0 +}^{\alpha_{1}}u(t)=u(t)+{{c}_{0}}+{{c}_{1}}t+ \cdots +{{c}_{n-1}} {{t}^{n-1}}, \\& I_{0 +}^{\alpha_{2}}L_{2}v(t)=I_{0 +}^{\alpha_{2}}D_{0 +}^{\alpha_{2}} v(t)=v(t)+{{d}_{0}}+{{d}_{1}}t+\cdots+{{d}_{n-1}} {{t}^{n-1}}. \end{aligned}$$
By the definitions of domL and KerP, one has \(u^{(i)}(0)=v^{(i)}(0)\), \(i=0,1,\ldots,n-1\), which implies that \(c_{i}=d_{i}\), \(i=0,1,\ldots,n-1\). Thus, we obtain
$$\begin{aligned} K_{p}L(x,y)= \bigl(I_{0 +}^{\alpha_{1}}D_{0 +}^{\alpha_{1}} x,I_{0 +}^{\alpha_{2}}D_{0 +}^{\alpha_{2}}y \bigr)=(x,y). \end{aligned}$$
(3.7)
Combining (3.6) and (3.7), we get \(K_{P}=(L_{\operatorname{dom} L \cap\operatorname{Ker} P })^{-1}\).
For \((x,y)\in\operatorname{Im} L\), we have
$$\begin{aligned} \bigl\Vert K_{P}(x,y) \bigr\Vert _{Y} &= \bigl\Vert \bigl(I_{0 +}^{\alpha_{1}}x,I_{0 +}^{\alpha_{2}}y \bigr) \bigr\Vert _{Y} =\max \bigl\{ \bigl\Vert I_{0 +}^{\alpha_{1}} x \bigr\Vert _{X_{1}}; \bigl\Vert I_{0 +}^{\alpha_{2}} y \bigr\Vert _{X_{2}} \bigr\} \\ &\leq\max \bigl\{ \max \bigl\{ \bigl\Vert I_{0 +}^{\alpha_{1}}x \bigr\Vert _{\infty}, \bigl\Vert D_{0+}^{\alpha_{1}-1}I_{0 +}^{\alpha_{1}}x \bigr\Vert _{\infty},\ldots, \bigl\Vert D_{0+}^{\alpha_{1}-(n-1)}I_{0^{+}}^{\alpha _{1}}x \bigr\Vert _{\infty}\bigr\} ; \\ &\quad{} \max \bigl\{ \bigl\Vert I_{0 +}^{\alpha_{2}} y \bigr\Vert _{\infty}, \bigl\Vert D_{0+}^{\alpha_{2}-1}I_{0 +}^{\alpha_{2}} y \bigr\Vert _{\infty},\ldots, \bigl\Vert D_{0+}^{\alpha_{2}-(n-1)}I_{0 +}^{\alpha_{2}} y \bigr\Vert _{\infty}\bigr\} \bigr\} \\ &=\max \bigl\{ \Vert x \Vert _{\infty}; \Vert y \Vert _{\infty}\bigr\} . \end{aligned}$$
(3.8)
Again, for \((u,v)\in\Omega_{1}\), \((u,v)\in\operatorname{dom}(L)\setminus \operatorname{Ker}(L)\), then \((I-P)(u,v)\in\operatorname{dom}L\cap\operatorname{Ker}P\) and \(LP(u,v)=(0,0)\), thus from (3.8) we have
$$\begin{aligned} \bigl\Vert (I-P) (u,v) \bigr\Vert _{Y} &= \bigl\Vert K_{P}L(I-P) (u,v) \bigr\Vert _{Y}= \bigl\Vert K_{P}(L_{1}u,L_{2}v) \bigr\Vert _{Y} \\ &\leq\max \bigl\{ \Vert N_{1}v \Vert _{\infty}; \Vert N_{2}u \Vert _{\infty}\bigr\} . \end{aligned}$$
(3.9)
By similar arguments as in [11] or [12], we have the following lemma. We omit the proof of it.
Lemma 3.4
\(K_{P}(I-Q)N:Y\rightarrow Y \)
is completely continuous.
For simplicity of notation, we set
$$a=\frac{1}{\Gamma(\alpha_{1}+1)}; \quad\quad b= \biggl[\frac{1}{\Gamma(\beta _{1}+1)} \biggr]^{q-1};\quad\quad \tilde{a}=\frac{1}{\Gamma(\alpha_{2}+1)}; \quad\quad \tilde{b}= \biggl[\frac{1}{\Gamma(\beta_{2}+1)} \biggr]^{q-1}. $$
Theorem 3.1
Assume that
\((\mathrm{H}_{0})\)
and the following conditions hold.
-
(H1)
There exist nonnegative functions
\(\psi(t),\tilde{\psi}(t),\varphi _{i}(t),\tilde{\varphi}_{i}(t) \in E\), \(i=1,2,\ldots,n-1\), such that for
\(t\in[0,1]\), \((u_{1},u_{2},\ldots,u_{n}),(v_{1},v_{2},\ldots,v_{n}) \in\mathbb {R}^{n}\), one has
$$\begin{aligned}& \bigl\vert f(t,u_{1},u_{2},\ldots,u_{n}) \bigr\vert \leq \psi(t)+ \varphi_{1}(t) \vert u_{1} \vert ^{p-1} +\cdots+\varphi _{n-1}(t) \vert u_{n} \vert ^{p-1}, \\& \bigl\vert g(t,v_{1},v_{2},\ldots,v_{n}) \bigr\vert \leq\tilde{\psi }(t)+ \tilde{\varphi}_{1}(t) \vert v_{1} \vert ^{p-1} +\cdots +\tilde{\varphi}_{n-1}(t) \vert v_{n} \vert ^{p-1}. \end{aligned}$$
-
(H2)
There exists
\(A>0\)
such that if
\(\vert u \vert >A \)
or
\(\vert v \vert >A\), \(\forall t\in[0,1] \), one has
$$\begin{aligned} &u\cdot \Biggl[ \sum_{i=1}^{\infty}a_{i}\phi_{q} \bigl[I_{0+}^{\beta_{1}}f \bigl(t,v(t),D_{0+}^{\alpha_{2}-1}v(t), \ldots,D_{0^{+}}^{\alpha_{2}-(n-1)}v(t) \bigr) \bigr]\big| _{t=\xi_{i}} \Biggr]>0, \\ &v\cdot \Biggl[ \sum_{i=1}^{\infty}b_{i}\phi_{q} \bigl[I_{0+}^{\beta_{2}}g \bigl(t,u(t),D_{0+}^{\alpha_{1}-1}u(t), \ldots,D_{0^{+}}^{\alpha_{1}-(n-1)}u(t) \bigr) \bigr]\big|_{t=\eta_{i}} \Biggr]>0, \end{aligned}$$
or
$$\begin{aligned} &u\cdot \Biggl[ \sum_{i=1}^{\infty}a_{i}\phi_{q} \bigl[I_{0+}^{\beta_{1}}f \bigl(t,v(t),D_{0+}^{\alpha_{2}-1}v(t), \ldots,D_{0^{+}}^{\alpha_{2}-(n-1)}v(t) \bigr) \bigr]\big|_{t=\xi_{i}} \Biggr]< 0, \\ &v\cdot \Biggl[ \sum_{i=1}^{\infty}b_{i}\phi_{q} \bigl[I_{0+}^{\beta_{2}}g \bigl(t,u(t),D_{0+}^{\alpha_{1}-1}u(t), \ldots,D_{0^{+}}^{\alpha_{1}-(n-1)}u(t) \bigr) \bigr]\big|_{t=\eta_{i}} \Biggr]< 0. \end{aligned}$$
Then BVP (3.1) has at least a solution in
X
provided that
$$\begin{aligned}& \max \bigl\{ 2^{q-1} \tilde{a}\tilde{b}\tilde{c}+2^{q-1}bc , 2^{q-1}abc +2^{q-1} \tilde{b}\tilde{c}, \\& \quad 2^{q-1}abc +2^{q-1}bc, 2^{q-1} \tilde{a}\tilde{b}\tilde{c}+2^{q-1} \tilde{b}\tilde{c} \bigr\} < 1 \quad \textit{for } p< 2, \end{aligned}$$
(3.10)
$$\begin{aligned}& \max \{ \tilde{a}\tilde{b}\tilde{c}+bc , abc+\tilde{b}\tilde{c}, abc+bc, \tilde{a}\tilde{b}\tilde{c}+\tilde{b}\tilde{c} \} < 1 \quad \textit{for }p\geq2, \end{aligned}$$
(3.11)
where
\(c= (\sum_{i=1}^{n-1} \Vert \varphi_{i}(t) \Vert _{\infty})^{q-1}\)
and
\(\tilde{c}= (\sum_{i=1}^{n-1} \Vert \tilde{\varphi}_{i}(t) \Vert _{\infty})^{q-1}\).
Proof
According to the definitions of \(N_{1}\) and \(N_{2}\), we have the following inequalities.
For \(1< p\leq2\), one has
$$\begin{aligned} \Vert N_{1}v \Vert _{\infty}&= \bigl\Vert \phi_{q} \bigl[I_{0+}^{\beta_{1}}f \bigl(t,v(t),D_{0+}^{\alpha_{2}-1}v(t), \ldots,D_{0+}^{\alpha_{2}-(n-1)}v(t) \bigr) \bigr] \bigr\Vert _{\infty} \\ &=\max \bigl\vert I_{0+}^{\beta_{1}}f \bigl(s,v(s),D_{0+}^{\alpha_{2}-1}v(s), \ldots,D_{0 +}^{\alpha_{2}-(n-1)}v(s) \bigr) \bigr\vert ^{q-1} \\ &\leq \Biggl\vert \frac{1}{\Gamma(\beta_{1}+1)} \Biggl[ \Vert \psi \Vert \infty+ \Vert v \Vert ^{p-1}_{X_{2}}\cdot\sum _{i=1}^{n-1} \bigl\Vert \varphi_{i}(t) \bigr\Vert _{\infty}\Biggr] \Biggr\vert ^{q-1} \\ & \leq2^{q-1}b \Vert \psi \Vert ^{q-1} _{\infty}+ 2^{q-1}b \Biggl(\sum_{i=1}^{n-1} \bigl\Vert \varphi_{i}(t) \bigr\Vert _{\infty}\Biggr)^{q-1}\cdot \Vert v \Vert _{X_{2}} \\ &=2^{q-1}b \Vert \psi \Vert ^{q-1} _{\infty}+ 2^{q-1}bc\cdot \Vert v \Vert _{X_{2}} \end{aligned}$$
(3.12)
and
$$\begin{aligned} \Vert N_{2}u \Vert _{\infty}&= \bigl\Vert \phi_{q} \bigl[I_{0+}^{\beta_{2}}g \bigl(t,u(t),D_{0+}^{\alpha_{1}-1}u(t), \ldots,D_{0+}^{\alpha_{1}-(n-1)}u(t) \bigr) \bigr] \bigr\Vert _{\infty} \\ &=\max \bigl\vert I_{0+}^{\beta_{2}}g \bigl(t,u(t),D_{0+}^{\alpha_{1}-1}u(t), \ldots,D_{0+}^{\alpha_{1}-(n-1)}u(t) \bigr) \bigr\vert ^{q-1} \\ &\leq \Biggl\vert \frac{1}{\Gamma(\beta_{2}+1)} \Biggl[ \Vert \tilde {\psi} \Vert _{\infty}+ \Vert u \Vert ^{p-1}_{X_{1}}\cdot\sum _{i=1}^{n-1} \bigl\Vert \varphi_{i}(t) \bigr\Vert _{\infty}\Biggr] \Biggr\vert ^{q-1} \\ & \leq2^{q-1}\tilde{b} \Vert \tilde{\psi} \Vert ^{q-1} _{\infty}+ 2^{q-1} \tilde{b} \Biggl(\sum _{i=1}^{n-1} \bigl\Vert \tilde{\varphi}_{i}(t) \bigr\Vert _{\infty}\Biggr)^{q-1}\cdot \Vert u \Vert _{X_{1}} \\ &= 2^{q-1}\tilde{b} \Vert \tilde{\psi} \Vert ^{q-1} _{\infty}+ 2^{q-1} \tilde{b}\tilde{c}\cdot \Vert u \Vert _{X_{1}}. \end{aligned}$$
(3.13)
By the similar proof of (3.12) and (3.13), one has
$$\begin{aligned}& \Vert N_{1}v \Vert _{\infty}\leq b \Vert \psi \Vert ^{q-1} _{\infty}+ bc\cdot \Vert v \Vert _{X_{2}} \quad \text{for }p \geq2 , \end{aligned}$$
(3.14)
$$\begin{aligned}& \Vert N_{2}u \Vert _{\infty}\leq \tilde{b} \Vert \tilde {\psi} \Vert ^{q-1} _{\infty}+ \tilde{b}\tilde{c}\cdot \Vert u \Vert _{X_{1}} \quad \text{for }p\geq 2. \end{aligned}$$
(3.15)
Let
$$\Omega_{1}= \bigl\{ (u,v)\in\operatorname{dom}L\setminus \operatorname{Ker}L: L(u,v)= \lambda N(u,v),\lambda\in(0,1) \bigr\} . $$
First, we give a proof that for \(1< p\leq2\), \(\Omega_{1}\) is bounded.
Let \(L(u,v)=\lambda N(u,v)\in\operatorname{Im}L=\operatorname{Ker}Q\), that is, \(L_{1}u=\lambda N_{1}v\in\operatorname{Ker}Q_{1}\) and \(L_{2}v=\lambda N_{2}u\in \operatorname{Ker}Q_{2}\). By the definition of \(\operatorname{Ker}Q_{1}\) and \(\operatorname{Ker}Q_{2}\), we have
$$\begin{aligned} &\sum_{i=1}^{\infty}a_{i}\cdot \lambda \phi_{q} \bigl[I_{0+}^{\beta_{1}}f \bigl(t,v(t),D_{0+}^{\alpha_{2}-1}v(t), \ldots,D_{0^{+}}^{\alpha_{2}-(n-1)}v(t) \bigr) \bigr]_{t=\xi_{i}} =0, \\ & \sum_{i=1}^{\infty}b_{i}\cdot \lambda\phi_{q} \bigl[I_{0+}^{\beta_{2}}g \bigl(t,u(t),D_{0+}^{\alpha_{1}-1}u(t), \ldots,D_{0^{+}}^{\alpha_{1}-(n-1)}u(t) \bigr) \bigr]_{t=\eta_{i}} =0. \end{aligned}$$
According to (H2), there exist \(t_{0},t_{1}\in(0,1)\) such that \(\vert u(t_{0}) \vert \leq A\) and \(\vert v(t_{1}) \vert \leq A\). Again, \(L_{1}u=\lambda N_{1}v\), \(u\in\operatorname{dom}L_{1}\setminus\operatorname{Ker}L_{1}\), that is, \(D_{0^{+}}^{\alpha_{1}}u=\lambda N_{1}v\), we have
$$u(t)=\frac{\lambda}{\Gamma ( \alpha_{1} )} \int_{0}^{t}{{{ ( t-s )}^{\alpha_{1} -1}} \phi_{q} \bigl[I_{0+}^{\beta_{1}} f \bigl(s,v(s),D_{0+}^{\alpha_{2}-1}v(s), \ldots,D_{0^{+}}^{\alpha_{2}-(n-1)}v(s) \bigr) \bigr]}\,ds+c_{0}. $$
Substituting \(t=t_{0} \) into the above equation, we get
$$u(t_{0})=\frac{\lambda}{\Gamma ( \alpha_{1} )} \int _{0}^{t_{0}}{{{ ( t_{0}-s )}^{\alpha_{1} -1}}\phi_{q} \bigl[I_{0+}^{\beta_{1}} f \bigl(s,v(s),D_{0+}^{\alpha_{2}-1}v(s), \ldots,D_{0^{+}}^{\alpha_{2}-(n-1)}v(s) \bigr) \bigr]}\,ds+c_{0}. $$
So, we obtain
$$\begin{aligned} u(t)-u(t_{0})&=\frac{\lambda}{\Gamma ( \alpha_{1} )} \int _{0}^{t}{{{ ( t-s )}^{\alpha_{1} -1}} \phi_{q} \bigl[I_{0+}^{\beta_{1}} f \bigl(s,v(s),D_{0+}^{\alpha_{2}-1}v(s), \ldots,D_{0^{+}}^{\alpha_{2}-(n-1)}v(s) \bigr) \bigr]}\,ds \\ &\quad{} -\frac{\lambda}{\Gamma ( \alpha_{1} )} \int_{0}^{t_{0}}{{{ ( t_{0}-s )}^{\alpha_{1} -1}}\phi_{q} \bigl[I_{0+}^{\beta_{1}} f \bigl(s,v(s),D_{0+}^{\alpha_{2}-1}v(s), \ldots,D_{0^{+}}^{\alpha_{2}-(n-1)}v(s) \bigr) \bigr]}\,ds . \end{aligned}$$
Together with \(\vert u (t_{0}) \vert \leq A\) and (3.12), we have
$$\begin{aligned} \bigl\vert {{u} }(0) \bigr\vert &\le \bigl\vert u(t_{0}) \bigr\vert + \biggl\vert \frac{\lambda}{\Gamma ( \alpha_{1} )} \int_{0}^{t_{0}}{{{ ( t_{0}-s )}^{\alpha_{1} -1}} \phi_{q} \bigl[I_{0+}^{\beta_{1}}f \bigl(s,v(s),D_{0+}^{\alpha_{2}-1}v(s), \ldots,D_{0^{+}}^{\alpha_{2}-(n-1)}v(s) \bigr) \bigr] }\,ds \biggr\vert \\ &\le A+\frac{1}{\Gamma ( \alpha_{1} )} \int _{0}^{{{t}_{0}}}{{ ( {{t}_{0}}-s )}^{\alpha_{1} -1}} \bigl\vert \phi_{q} \bigl[I_{0+}^{\beta _{1}}f \bigl(s,v(s),D_{0+}^{\alpha_{2}-1}v(s), \ldots,D_{0^{+}}^{\alpha_{2}-(n-1)}v(s) \bigr) \bigr] \bigr\vert \,ds \\ &= A+\frac{1}{\Gamma ( \alpha_{1} )}\cdot \bigl(2^{q-1}b \Vert \psi \Vert ^{q-1} _{\infty}+ 2^{q-1}bc\cdot \Vert v \Vert _{X_{2}} \bigr) \cdot \int_{0}^{{{t}_{0}}} (t_{0}-s )^{\alpha_{1}-1} \,ds \\ &\leq A+ 2^{q-1}ab \Vert \psi \Vert ^{q-1} _{\infty}+2^{q-1}abc\cdot \Vert v \Vert _{X_{2}}. \end{aligned}$$
(3.16)
Similarly, by (3.13), we obtain
$$\begin{aligned} \bigl\vert {{v} }(0) \bigr\vert \le& \bigl\vert v(t_{0}) \bigr\vert \\ &{}+ \biggl\vert \frac{\lambda}{\Gamma ( \alpha_{2} )} \int_{0}^{t_{0}}{{{ ( t_{0}-s )}^{\alpha_{2} -1}} \phi_{q} \bigl[I_{0+}^{\beta_{2}}g \bigl(s,u(s),D_{0+}^{\alpha_{1}-1}u(s), \ldots,D_{0^{+}}^{\alpha_{1}-(n-1)}u(s) \bigr) \bigr] }\,ds \biggr\vert \\ \le &A+\frac{1}{\Gamma ( \alpha_{2} )} \int _{0}^{{{t}_{0}}}{{ ( {{t}_{0}}-s )}^{\alpha_{2} -1}} \bigl\vert \phi_{q} \bigl[I_{0+}^{\beta _{2}}g \bigl(s,u(s),D_{0+}^{\alpha_{1}-1}u(s), \ldots,D_{0^{+}}^{\alpha_{1}-(n-1)}u(s) \bigr) \bigr] \bigr\vert \,ds \\ =& A+\frac{1}{\Gamma ( \alpha_{2} )}\cdot \bigl( 2^{q-1}\tilde{b} \Vert \tilde{ \psi} \Vert ^{q-1} _{\infty}+ 2^{q-1} \tilde{b}\tilde{c} \cdot \Vert u \Vert _{X_{1}} \bigr) \cdot \int_{0}^{{{t}_{0}}} (t_{0}-s )^{\alpha_{2}-1} \,ds \\ \leq &A+2^{q-1}\tilde{a}\tilde{b} \Vert \tilde{\psi} \Vert ^{q-1} _{\infty}+ 2^{q-1} \tilde{a}\tilde{b}\tilde{c} \cdot \Vert u \Vert _{X_{1}}. \end{aligned}$$
(3.17)
For \((u,v)\in\Omega_{1}\), by (3.5) and (3.9), we have
$$\begin{aligned} \bigl\Vert (u,v) \bigr\Vert _{Y} &= \bigl\Vert P(u,v)+(I-P) (u,v) \bigr\Vert _{Y} \leq \bigl\Vert P(u,v) \bigr\Vert _{Y}+ \bigl\Vert (I-P) (u,v) \bigr\Vert _{Y} \\ &\leq\max \bigl\{ \bigl\vert {{u}}(0) \bigr\vert + \Vert N_{1}v \Vert _{\infty}; \bigl\vert {{u}}(0) \bigr\vert + \Vert N_{2}u \Vert _{\infty}; \\ & \quad{} \bigl\vert {{v}}(0) \bigr\vert + \Vert N_{1}v \Vert _{\infty}; \bigl\vert {{v}}(0) \bigr\vert + \Vert N_{2}u \Vert _{\infty}\bigr\} . \end{aligned}$$
The following proof is divided into four cases.
Case 1. \(\Vert (u,v) \Vert _{Y} \leq \vert {{u} }(0) \vert + \Vert N_{1}v \Vert _{\infty}\).
By (3.12) and (3.16), we have
$$\begin{aligned} \Vert v \Vert _{X_{2}}&\leq \bigl\Vert (u,v) \bigr\Vert _{Y} \leq \bigl\vert {{u} }(0) \bigr\vert + \Vert N_{1}v \Vert _{\infty} \\ & \leq A+ 2^{q-1}ab \Vert \psi \Vert ^{q-1} _{\infty}+2^{q-1}abc\cdot \Vert v \Vert _{X_{2}} +2^{q-1}b \Vert \psi \Vert ^{q-1} _{\infty}+ 2^{q-1}bc\cdot \Vert v \Vert _{X_{2}} \\ & = A+ 2^{q-1}ab \Vert \psi \Vert ^{q-1} _{\infty}+2^{q-1}b \Vert \psi \Vert ^{q-1} _{\infty}+ \bigl(2^{q-1}abc + 2^{q-1}bc \bigr)\cdot \Vert v \Vert _{X_{2}}. \end{aligned}$$
According to (3.10), we can derive
$$\Vert v \Vert _{X_{2}} \leq\frac{A+ 2^{q-1}ab \Vert \psi \Vert ^{q-1} _{\infty}+2^{q-1}b \Vert \psi \Vert ^{q-1} _{\infty}}{1-(2^{q-1}abc + 2^{q-1}bc)}:=M_{1}. $$
Thus, \(\Omega_{1}\) is bounded.
Case 2. \(\Vert (u,v) \Vert _{Y} \leq \vert {{u} }(0) \vert + \Vert N_{2}u \Vert _{\infty}\).
By (3.13) and (3.16), we have
$$\begin{aligned} \bigl\Vert (u,v) \bigr\Vert _{Y} &\leq \bigl\vert {{u} }(0) \bigr\vert + \Vert N_{2}u \Vert _{\infty} \\ &\leq A+ 2^{q-1}ab \Vert \psi \Vert ^{q-1} _{\infty}+2^{q-1}abc\cdot \Vert v \Vert _{X_{2}} +2^{q-1} \tilde{b} \Vert \tilde{\psi} \Vert ^{q-1} _{\infty}+ 2^{q-1} \tilde{b}\tilde{c}\cdot \Vert u \Vert _{X_{1}} \\ &= A+ 2^{q-1}ab \Vert \psi \Vert ^{q-1} _{\infty}+2^{q-1}\tilde{b} \Vert \tilde{\psi} \Vert ^{q-1} _{\infty}+2^{q-1}abc\cdot \Vert v \Vert _{X_{2}} + 2^{q-1} \tilde{b}\tilde{c}\cdot \Vert u \Vert _{X_{1}} \\ &\leq A+ 2^{q-1}ab \Vert \psi \Vert ^{q-1} _{\infty}+2^{q-1}\tilde{b} \Vert \tilde{\psi} \Vert ^{q-1} _{\infty}+ \bigl(2^{q-1}abc + 2^{q-1} \tilde{b}\tilde{c} \bigr)\cdot \bigl\Vert (u,v) \bigr\Vert _{Y}. \end{aligned}$$
By (3.10), we can derive
$$\bigl\Vert (u,v) \bigr\Vert _{Y}\leq\frac{A+ 2^{q-1}ab \Vert \psi \Vert ^{q-1} _{\infty}+2^{q-1}\tilde{b} \Vert \tilde{\psi} \Vert ^{q-1} _{\infty}}{1-2^{q-1}abc- 2^{q-1} \tilde{b}\tilde{c} }:=M_{2}. $$
Then \(\Omega_{1}\) is bounded.
Case 3. \(\Vert (u,v) \Vert _{Y} \leq \vert {{v}}(0) \vert + \Vert N_{1}v \Vert _{\infty}\).
According to (3.12) and (3.17), we have
$$\begin{aligned} \bigl\Vert (u,v) \bigr\Vert _{Y} &\leq \bigl\vert {{v}}(0) \bigr\vert + \Vert N_{1}v \Vert _{\infty} \\ &\leq A+2^{q-1}\tilde{a}\tilde{b} \Vert \tilde{\psi} \Vert ^{q-1} _{\infty}+ 2^{q-1} \tilde{a}\tilde{b}\tilde{c} \cdot \Vert u \Vert _{X_{1}} +2^{q-1}b \Vert \psi \Vert ^{q-1} _{\infty}+ 2^{q-1}bc\cdot \Vert v \Vert _{X_{2}} \\ &= A+2^{q-1}\tilde{a}\tilde{b} \Vert \tilde{\psi} \Vert ^{q-1} _{\infty}+ 2^{q-1}b \Vert \psi \Vert ^{q-1} _{\infty}+2^{q-1} \tilde {a}\tilde{b}\tilde{c} \cdot \Vert u \Vert _{X_{1}} + 2^{q-1}bc\cdot \Vert v \Vert _{X_{2}} \\ &\leq A+2^{q-1}\tilde{a}\tilde{b} \Vert \tilde{\psi} \Vert ^{q-1} _{\infty}+ 2^{q-1}b \Vert \psi \Vert ^{q-1} _{\infty}+ \bigl( 2^{q-1} \tilde{a}\tilde{b} \tilde{c} + 2^{q-1}bc \bigr) \cdot \bigl\Vert (u,v) \bigr\Vert _{Y}. \end{aligned}$$
By (3.10), we have
$$\bigl\Vert (u,v) \bigr\Vert _{Y} \leq \frac{A+2^{q-1}\tilde{a}\tilde {b} \Vert \tilde{\psi} \Vert ^{q-1} _{\infty}+ 2^{q-1}b \Vert \psi \Vert ^{q-1} _{\infty}}{1- ( 2^{q-1} \tilde{a}\tilde{b}\tilde{c} + 2^{q-1}bc ) }:=M_{3}. $$
Then \(\Omega_{1}\) is bounded.
Case 4. \(\Vert (u,v) \Vert _{Y} \leq \vert {{v}}(0) \vert + \Vert N_{2}u \Vert _{\infty}\).
According to (3.13) and (3.17), we have
$$\begin{aligned} \Vert u \Vert _{X_{1}} &\leq \bigl\Vert (u,v) \bigr\Vert _{Y}\leq \bigl\vert {{v}}(0) \bigr\vert + \Vert N_{2}u \Vert _{\infty}\\ &\leq A+2^{q-1}\tilde{a}\tilde{b} \Vert \tilde{\psi} \Vert ^{q-1} _{\infty}+ 2^{q-1} \tilde{a}\tilde{b}\tilde{c} \cdot \Vert u \Vert _{X_{1}} +2^{q-1}\tilde{b} \Vert \tilde{ \psi} \Vert ^{q-1} _{\infty}+ 2^{q-1} \tilde{b} \tilde{c} \cdot \Vert u \Vert _{X_{1}} \\ &= A+2^{q-1}\tilde{a}\tilde{b} \Vert \tilde{\psi} \Vert ^{q-1} _{\infty}+2^{q-1}\tilde{b} \Vert \tilde{\psi} \Vert ^{q-1} _{\infty}+ \bigl(2^{q-1} \tilde{a} \tilde{b}\tilde{c} + 2^{q-1} \tilde{b}\tilde{c} \bigr)\cdot \Vert u \Vert _{X_{1}}. \end{aligned}$$
By (3.10), we get
$$\Vert u \Vert _{X_{1}} \leq\frac{A+2^{q-1}\tilde{a}\tilde {b} \Vert \tilde{\psi} \Vert ^{q-1} _{\infty}+2^{q-1}\tilde {b} \Vert \tilde{\psi} \Vert ^{q-1} _{\infty}}{ 1- (2^{q-1} \tilde{a}\tilde{b}\tilde{c} + 2^{q-1} \tilde{b}\tilde{c} )}:=M_{4}. $$
Then \(\Omega_{1}\) is bounded.
Therefore, we have proved that \(\Omega_{1}\) is bounded for \(1< p\leq2\). By similar arguments as the above proof, according to (3.11), (3.14) and (3.15), we can prove that \(\Omega_{1}\) is also bounded for \(p>2\). We omit the proof of it.
Let
$$\Omega_{2}= \bigl\{ (u,v)\in\operatorname{Ker}L:N(u,v)\in \operatorname{Im}L \bigr\} . $$
Let \((u,v)\in\operatorname{Ker}L\), so we have \(u=c_{0} \), \(v=d_{0}\). In view of \(N(u,v)=(N_{1}v,N_{2}u)\in\operatorname{Im}L=\operatorname{Ker}Q\), we have \(Q_{1}(N_{1}v)=0\), \(Q_{2}(N_{2}u)=0\), that is,
$$\begin{aligned} & \sum_{i=1}^{\infty}a_{i} \phi_{q} \bigl[I_{0+}^{\beta_{1}}f \bigl(t,v(t),D_{0+}^{\alpha_{2}-1}v(t), \ldots,D_{0^{+}}^{\alpha_{2}-(n-1)}v(t) \bigr) \bigr]\big|_{t=\xi_{i}} =0, \\ & \sum_{i=1}^{\infty}b_{i} \phi_{q} \bigl[I_{0+}^{\beta_{2}}g \bigl(t,u(t),D_{0+}^{\alpha_{1}-1}u(t), \ldots,D_{0^{+}}^{\alpha_{1}-(n-1)}u(t) \bigr) \bigr]\big|_{t=\eta_{i}} =0. \end{aligned}$$
By (H2), there exist constants \(t_{0},t_{1}\in[0,1]\) such that
$$\bigl\vert u(t_{0}) \bigr\vert = \vert c_{0} \vert \leq A,\qquad \bigl\vert v( t_{1}) \bigr\vert = \vert d_{0} \vert \leq A. $$
Therefore, \(\Omega_{2}\) is bounded.
Let
$$\Omega_{3}= \bigl\{ (u,v)\in\operatorname{Ker}L:\lambda(u,v)+(1- \lambda) QN(u,v)=(0,0),\lambda\in[0,1] \bigr\} . $$
For \((u,v)\in\operatorname{Ker}L\), so we have \(u=c_{0}\) and \(v=d_{0}\). By the definition of the set \(\Omega_{3}\), we have
$$\begin{aligned} \lambda c_{0} +(1-\lambda)Q_{1}N_{1}(d_{0})=0, \qquad \lambda d_{0} +(1-\lambda)Q_{2}N_{2}(c_{0})=0. \end{aligned}$$
(3.18)
If \(\lambda=0\), similar to the proof of the boundedness of \(\Omega_{2}\), we have \(\vert c_{0} \vert \leq A\) and \(\vert d_{0} \vert \leq A\). If \(\lambda=1\), we have \(c_{0}=d_{0}=0\). If \(\lambda\in(0,1)\), we also have \(\vert c_{0} \vert \leq A\) and \(\vert d_{0} \vert \leq A\). Otherwise, if \(\vert c_{0} \vert >A\) or \(\vert d_{0} \vert >A\), in view of the first part of (H2), we obtain
$$\begin{aligned} \lambda c^{2}_{0} +(1-\lambda)c _{0}\cdot Q_{1}N_{1}(d_{0})>0,\quad \quad \lambda d^{2}_{0} +(1-\lambda)d_{0}\cdot Q_{2}N_{2}(c_{0})>0, \end{aligned}$$
which contradict (3.18). Thus, \(\Omega_{3}\) is bounded.
If the second part of (H2) holds, then we can prove that the set
$$\Omega_{3}'= \bigl\{ (u,v)\in\operatorname{Ker}L:- \lambda(u,v)+(1-\lambda) QN(u,v)=(0,0),\lambda\in[0,1] \bigr\} $$
is bounded.
Finally, let Ω to be a bounded open set of Y such that \(\bigcup_{i=1}^{3}{\overline{\Omega}_{i}}\subset\Omega\). By Lemma 3.4, N is L-compact on Ω. Then, by the above arguments, we get
-
(1)
\(L(u,v)\neq\lambda N(u,v)\), for every \((u,v)\in[(\operatorname{dom}L\setminus{Ker}L)\cap\partial\Omega]\times(0,1)\);
-
(2)
\(N(u,v)\notin\operatorname{Im}L\) for every \((u,v)\in\operatorname{Ker}L\cap \partial\Omega\);
-
(3)
Let \(H((u,v),\lambda)=\pm\lambda I(u,v)+(1-\lambda)JQN(u,v)\), where I is the identical operator. Via the homotopy property of degree, we obtain that
$$\begin{aligned} \deg (JQ N|_{\operatorname{Ker} L},\Omega\cap\operatorname{Ker} L,0 ) &= \deg \bigl(H( \cdot,0), \Omega\cap\operatorname{Ker} L,0 \bigr) \\ &= \deg \bigl(H(\cdot,1),\Omega\cap\operatorname{Ker} L,0 \bigr) \\ &= \deg (I,\Omega\cap\operatorname{Ker} L,0 ) \\ &=1\neq0. \end{aligned}$$
Applying Theorem 2.1, we conclude that \(Lu=Nu\) has at least one solution in \(\operatorname{dom}L\cap\overline{\Omega}\). □