Now, we are ready to prove our main results. Let \(X=\{u: u,u', u'', {}^{\mathrm{c}}D^{p} u\in C(J,\mathbb{R})\}\) endowed with the norm \(\|u\|=\sup_{t\in J} |u(t)|+\sup_{t\in J}|u'(t)| + \sup_{t\in J}|u''(t)| + \sup_{t\in J}|{}^{\mathrm{c}}D^{p} u(t)| \). Then \((X, {\Vert\cdot\Vert})\) is a Banach space [35].
Lemma 2.1
Let
\(y\in L^{1}(J,{\mathbb{R}})\). Then the integral solution of the linear problem
$$ \left \{ \textstyle\begin{array}{l} {}^{\mathrm{c}}D^{\alpha}u(t)=y(t ), \\ u(0)+u(1)=0, \qquad {}^{\mathrm{c}}D^{p}u( 1)= \gamma\int_{0}^{1} u(\tau )\, d\tau, \\ u'(0)+u''( \xi)= \eta\int_{0}^{1} u(\tau)\, d\tau, \end{array}\displaystyle \right . $$
(2.1)
is given by
$$\begin{aligned} u(t) =& I^{\alpha}y(t) + \biggl(\frac{1}{2}-t \biggr)I^{\alpha-2}y(\xi) + A(t) I^{\alpha}y(1) \\ &{}+ B(t)I^{\alpha-p}y(1) + C(t)I^{\alpha+1}y(1), \end{aligned}$$
(2.2)
where
$$\begin{aligned}& A(t) = -\frac{1}{2}\bigl(1+C(t)\bigr), \\& B(t) = \frac{1}{4} \Gamma(3-p) \bigl(-1+4t-2t^{2}\bigr)+ \frac{\Gamma (3-p)}{12}C(t), \\& C(t) = \frac{3}{12+\gamma\Gamma(3-p)} \bigl[\gamma\Gamma (3-p) \bigl(1-4t+2t^{2} \bigr)+2\eta(2t-1) \bigr]. \end{aligned}$$
(2.3)
Proof
It is well known that the solution of equation \({}^{\mathrm{c}}D^{\alpha}u(t)=y(t)\) can be written as
$$u( t)= I^{\alpha}y(t) + c_{0} + c_{1}t+c_{2}t^{2}, $$
where \(c_{0}, c_{1}, c_{2}\in\mathbb{R}\) are arbitrary constants. Then we get
$$u'( t)=I^{\alpha-1}y(t) +c_{1}+2c_{2}t, \qquad u''(t)=I^{\alpha-2}y(t) +2c_{2} $$
and
$${}^{\mathrm{c}}D^{p} u(t)=I^{\alpha-p}y(t)+c_{2} \frac{2t^{2-p}}{\Gamma(3-p)},\quad 1< p\leq2. $$
By using the boundary conditions \(u(0)+u(1)=0\), \({}^{\mathrm{c}}D^{p}u( 1)= \gamma \int_{0}^{1} u(\tau)\, d\tau\), \(u'(0)+u''( \xi)= \eta\int_{0}^{1} u(\tau)\, d\tau\), we obtain
$$\begin{aligned}& c_{0} = -\frac{1}{2}I^{\alpha}y(1) +I^{\alpha-2}y(\xi) - \frac{\Gamma(3-p)}{4} I^{\alpha-p}y(1) + \frac{\gamma\Gamma (3-p)-2\eta}{4 }\int _{0}^{1} u( \tau)\, d\tau, \\& c_{1} = - I^{\alpha-2}y(\xi) + \Gamma(3-p)I^{\alpha-p}y(1) + \bigl(\eta-\gamma\Gamma(3-p) \bigr)\int_{0}^{1} u( \tau)\, d\tau \end{aligned}$$
and
$$c_{2} = - \frac{\Gamma(3-p)}{2}I^{\alpha-p}y(1) + \frac{\gamma\Gamma (3-p)}{2} \int_{0}^{1} u( \tau)\, d\tau. $$
Then
$$\begin{aligned} u( t ) =& I^{\alpha}y(t) - \frac{1}{2} I^{\alpha}y(1) + \biggl(\frac {1}{2}-t \biggr)I^{\alpha-2}y(\xi) + \frac{\Gamma(3-p)( -1+4t-2t^{2})}{4}I^{\alpha-p}y(1) \\ &{}+\frac{1}{4} \bigl[\gamma\Gamma(3-p) \bigl(1-4t+2t^{2} \bigr)+2\eta(2t-1) \bigr]\int_{0}^{1} u( \tau)\, d \tau. \end{aligned}$$
(2.4)
Letting \(A= \int_{0}^{1} u( \tau)\, d\tau\), we have
$$\begin{aligned} A =& \int_{0}^{1} u( \tau)\, d\tau=\int _{0}^{1}\int_{0}^{t} \frac{(t-s)^{\alpha -1}}{\Gamma(\alpha)}y(s)\, ds\, dt+c_{0}+\frac{1}{2}c_{1}+ \frac{1}{3}c_{2} \\ =&I^{\alpha+1}y(1)+c_{0}+\frac{1}{2}c_{1}+ \frac{1}{3}c_{2}, \end{aligned}$$
or after substituting \(c_{0}\), \(c_{1}\), and \(c_{2}\),
$$A = \frac{12}{12+\gamma\Gamma(3-p)} \biggl[I^{\alpha+1}y(1)-\frac {1}{2}I^{\alpha}y(1) +\frac{\Gamma(3-p)}{12}I^{\alpha-p}y(1) \biggr]. $$
Substituting the value A in (2.4), we get (2.2). The proof is completed. □
Remark 2.2
Throughout this paper, the following relations hold:
$$\begin{aligned}& \bigl\vert A(t)\bigr\vert \le \frac{1}{2}(1+C_{1}),\qquad \bigl\vert B(t)\bigr\vert \le\Gamma(3-p) \biggl( \frac{7}{4}+ \frac{1}{12}C_{1} \biggr), \\& \bigl\vert C(t)\bigr\vert \le \frac{3}{12+\gamma\Gamma(3-p)}\bigl[7\gamma\Gamma(3-p)+2 \eta \bigr]:=C_{1}, \\& \bigl\vert A'(t)\bigr\vert \le \frac{1}{2}C'_{1}, \qquad \bigl\vert B'(t)\bigr\vert \le\Gamma(3-p) \biggl( 1 + \frac{1}{12}C_{1}' \biggr), \\& \bigl\vert C'(t)\bigr\vert \le\frac{12[\gamma\Gamma(3-p)+\eta]}{12+\gamma\Gamma (3-p)}:=C'_{1}, \\& \bigl\vert A''(t)\bigr\vert \le \frac{1}{2}C''_{1},\qquad \bigl\vert B''(t)\bigr\vert \le \Gamma(3-p) \biggl( 1 + \frac{1}{12}C''_{1} \biggr), \\& \bigl\vert C''(t)\bigr\vert \le\frac{12\gamma\Gamma(3-p)}{12+\gamma\Gamma(3-p)}:=C''_{1}, \\& \bigl\vert {}^{\mathrm{c}}D^{p}A(t)\bigr\vert \le D_{1}, \qquad \bigl\vert {}^{\mathrm{c}}D^{p}B(t)\bigr\vert \le1+ \frac{\Gamma(3-p)}{12} D_{1}, \\& \bigl\vert {}^{\mathrm{c}}D^{p}C(t)\bigr\vert \le\frac{12\gamma}{12+\gamma\Gamma(3-p)}:=D_{1}. \end{aligned}$$
Definition 2.3
A function \(u\in C^{2}(J, {\mathbb{R}})\) is called a solution for the problem (1.1)-(1.2) if there exists a function \(v\in L^{1}(J, {\mathbb{R}})\) with \(v(t)\in F(t, u(t), u'(t), u''(t), {}^{\mathrm{c}}D^{p}u(t))\) for almost all \(t\in J\), \(u(0)+u(1)=0\), \({}^{\mathrm{c}}D^{p}u( 1)= \gamma \int_{0}^{1} u(\tau)\, d\tau\), \(u'(0)+u''( \xi)= \eta\int_{0}^{1} u(\tau)\, d\tau\), and
$$ u(t) = I^{\alpha}v(t)+ \biggl(\frac{1}{2}-t \biggr)I^{\alpha-1}v(\xi) + A(t) I^{\alpha}v(1)+ B(t)I^{\alpha-p}v(1) + C(t)I^{\alpha+1}v(1) $$
(2.5)
for all \(t\in J\).
For the sake of brevity, we set
$$\begin{aligned}& \Lambda_{1} =\frac{1}{\Gamma(\alpha+1)}+\frac{1}{2} \frac{\xi ^{\alpha-2}}{\Gamma(\alpha-1)}+\frac{1}{2}(1+C_{1})\frac{1}{\Gamma(\alpha +1)} \\ & \hphantom{\Lambda_{1} =}{}+\Gamma(3-p) \biggl(\frac{7}{4}+\frac{1}{12}C_{1} \biggr)\frac{1}{\Gamma(\alpha-p+1)} +\frac{1}{2} C_{1}\frac{1}{\Gamma(\alpha+2)}, \end{aligned}$$
(2.6)
$$\begin{aligned}& \Lambda_{2} = \frac{1}{\Gamma(\alpha)}+ \frac{\xi^{\alpha -2}}{\Gamma(\alpha-1)}+ \frac{1}{2}C'_{1}\frac{1}{\Gamma(\alpha+1)} \\ & \hphantom{\Lambda_{2} =}{} +\Gamma(3-p) \biggl( 1 +\frac{1}{12}C'_{1} \biggr)\frac {1}{\Gamma(\alpha-p+1)} +\frac{1}{2} C'_{1} \frac{1}{\Gamma(\alpha+2)}, \end{aligned}$$
(2.7)
$$\begin{aligned}& \Lambda_{3} = \frac{1}{\Gamma(\alpha-1)} +C''_{1} \frac{1}{\Gamma(\alpha+1)} \\ & \hphantom{\Lambda_{3} =}{}+\Gamma(3-p) \biggl( 1+\frac{1}{12}C''_{1} \biggr)\frac {1}{\Gamma(\alpha-p+1)} +\frac{1}{2} C''_{1} \frac{1}{\Gamma(\alpha+2)} \end{aligned}$$
(2.8)
and
$$\begin{aligned} \Lambda_{4} =& \frac{1}{\Gamma(\alpha-p +1)}+D_{1} \frac{1}{\Gamma(\alpha+1)} \\ &{}+ \biggl(1+ \frac{\Gamma(3-p)}{12} D_{1} \biggr) \frac{1}{\Gamma(\alpha-p+1)}+\frac{1}{2} D_{1}\frac{1}{\Gamma(\alpha+2)}. \end{aligned}$$
(2.9)
Theorem 2.4
Suppose that:
- (H1):
-
\(F: J\times\mathbb{R}^{4} \to{\mathcal{P}}_{\mathrm{cp},\mathrm{c}}(\mathbb {R})\)
is a
\(L^{1}\)-Carathéodory multifunction.
- (H2):
-
There exist continuous nondecreasing functions
\(\psi_{i} : [0,\infty) \to (0,\infty)\)
and functions
\(p_{i} \in C(J,\mathbb{R}^{+})\), \(1\le i\le4\), such that
$$\bigl\Vert F(t, x_{1}, x_{2}, x_{3}, x_{4})\bigr\Vert :=\sup\bigl\{ |v|: v\in F(t, x_{1}, x_{2}, x_{3}, x_{4})\bigr\} \le\sum _{i=1}^{4} p_{i}(t)\psi_{i} \bigl(\vert x_{i}\vert \bigr) $$
for each
\((t,x_{i}) \in J\times\mathbb{R}\), \(1\le i\le4\).
- (H3):
-
There exists a constant
\(M>0\)
such that
$$\frac{M}{ \sum_{i=1}^{4}\Lambda_{i} \|p_{i}\|\psi_{i}(M)}> 1, $$
where
\(\Lambda_{i}\), \(1\le i\le4\)
are defined by (2.6)-(2.9).
Then the inclusion boundary value problem (1.1)-(1.2) has at least one solution.
Proof
To transform the problem (1.1)-(1.2) into a fixed point problem, we define an operator \({\mathcal{N}}:X\rightarrow{\mathcal{P}}(X)\) as
$${\mathcal{N}}(u)=\left \{ h \in X: h(t) = \left \{ \textstyle\begin{array}{l} I^{\alpha}v(t)+ (\frac{1}{2}-t )I^{\alpha-1}v(\xi)+ A(t) I^{\alpha }v(1) \\ \quad {}+ B(t)I^{\alpha-p}v(1) + C(t)I^{\alpha+1}v(1), t\in J, v\in S_{F,u} \end{array}\displaystyle \right \} \right \}. $$
We will show that \({\mathcal{N}}\) satisfies the assumptions of the nonlinear alternative of Leray-Schauder type. The proof consists of several steps. As a first step, we show that \({\mathcal{N}}\)
is convex for each
\(u \in X\). This step is obvious since \(S_{F,u}\) is convex (F has convex values), and therefore we omit the proof.
In the second step, we show that \({\mathcal{N}}\)
maps bounded sets (balls) into bounded sets in
X. For a positive number r, let \(B_{r} = \{u \in X: \|u\| \le r \}\) be a bounded ball in X. Then, for each \(h \in{\mathcal{N}} (u)\) and \(u \in B_{r}\), there exists \(v \in S_{F,u}\) such that
$$h(t) = I^{\alpha}v(t)+ \biggl(\frac{1}{2}-t \biggr)I^{\alpha -1}v( \xi)+ A(t) I^{\alpha}v(1) + B(t)I^{\alpha-p}v(1) + C(t)I^{\alpha+1}v(1), \quad t\in J. $$
Then we have
$$\begin{aligned} \bigl\vert h(t)\bigr\vert \le& I^{\alpha}\bigl\vert v(t)\bigr\vert +\biggl\vert \frac{1}{2}-t\biggr\vert I^{\alpha -2}\bigl\vert v( \xi)\bigr\vert + \bigl\vert A(t)\bigr\vert I^{\alpha}\bigl\vert v(1) \bigr\vert \\ &{}+ \bigl\vert B(t)\bigr\vert I^{\alpha-p}\bigl\vert v(1) \bigr\vert + \bigl\vert C(t)\bigr\vert I^{\alpha+1}\bigl\vert v(1) \bigr\vert \\ \le& I^{\alpha}\bigl[p_{1}(t)\psi_{1}\bigl(\bigl\vert u(t)\bigr\vert \bigr)+p_{2}(t)\psi _{2}\bigl(\bigl\vert u'(t)\bigr\vert \bigr) \\ &{}+p_{3}(t)\psi_{3} \bigl(\bigl\vert u''(t)\bigr\vert \bigr) +p_{4}(t)\psi_{4}\bigl(\bigl\vert {}^{\mathrm{c}}D^{p}u(t) \bigr\vert \bigr)\bigr] \\ &{} + \frac{1}{2} I^{\alpha-1}\bigl[p_{1}(\xi) \psi_{1}\bigl(\bigl\vert u(\xi)\bigr\vert \bigr)+p_{2}( \xi)\psi _{2}\bigl(\bigl\vert u'(\xi)\bigr\vert \bigr) \\ &{}+p_{3}(\xi)\psi_{3}\bigl(\bigl\vert u''(\xi)\bigr\vert \bigr) +p_{4}(\xi) \psi_{4}\bigl(\bigl\vert {}^{\mathrm{c}}D^{p}u(\xi )\bigr\vert \bigr)\bigr] \\ &{}+ \frac{1}{2}(1+C_{1}) I^{\alpha}\bigl[p_{1}(1) \psi_{1}\bigl(\bigl\vert u(1)\bigr\vert \bigr)+p_{2}(1) \psi _{2}\bigl(\bigl\vert u'(1)\bigr\vert \bigr) \\ &{}+p_{3}(1)\psi_{3}\bigl(\bigl\vert u''(1) \bigr\vert \bigr) +p_{4}(1)\psi_{4}\bigl(\bigl\vert {}^{\mathrm{c}}D^{p}u(1)\bigr\vert \bigr)\bigr] \\ &{} +\Gamma(3-p) \biggl(\frac{7}{4}+\frac{1}{12}C_{1} \biggr)I^{\alpha-p} \bigl[p_{1}(1)\psi_{1}\bigl(\bigl\vert u(1)\bigr\vert \bigr)+p_{2}(1)\psi_{2}\bigl(\bigl\vert u'(1)\bigr\vert \bigr) \\ &{}+p_{3}(1) \psi_{3}\bigl(\bigl\vert u''(1)\bigr\vert \bigr)+p_{4}(1)\psi_{4}\bigl(\bigl\vert {}^{\mathrm{c}}D^{p}u(1) \bigr\vert \bigr)\bigr] \\ &{} + C_{1}I^{\alpha+1}\bigl[p_{1}(1)\psi_{1} \bigl(\bigl\vert u(1)\bigr\vert \bigr)+p_{2}(1)\psi _{2} \bigl(\bigl\vert u'(1)\bigr\vert \bigr) \\ &{}+p_{3}(1) \psi_{3}\bigl(\bigl\vert u''(1)\bigr\vert \bigr) +p_{4}(1)\psi_{4}\bigl(\bigl\vert {}^{\mathrm{c}}D^{p}u(1)\bigr\vert \bigr)\bigr] \\ \le& \bigl[\|p_{1}\|\psi_{1}(r)+\|p_{2}\| \psi_{2}(r)+\|p_{3}\|\psi_{3}(r)+\|p_{4}\| \psi_{4}(r) \bigr] \biggl\{ \frac{1}{\Gamma(\alpha+1)}+\frac{1}{2} \frac{\xi ^{\alpha-2}}{\Gamma(\alpha-1)} \\ &{}+\frac{1}{2}(1+C_{1})\frac{1}{\Gamma(\alpha+1)}+\Gamma(3-p) \biggl( \frac {7}{4}+\frac{1}{12}C_{1} \biggr)\frac{1}{\Gamma(\alpha-p+1)} + \frac{1}{2} C_{1}\frac{1}{\Gamma(\alpha+2)} \biggr\} \\ =&\Lambda_{1}\sum_{i=1}^{4} \|p_{i}\|\psi_{i}(r) \end{aligned}$$
for all \(t\in J\). In a similar manner we obtain
$$\begin{aligned}& \bigl\vert h'(t)\bigr\vert \le \bigl[\|p_{1}\| \psi_{1}(r)+\|p_{2}\|\psi_{2}(r)+\|p_{3}\| \psi_{3}(r)+\|p_{4}\| \psi_{4}(r) \bigr] \biggl\{ \frac{1}{\Gamma(\alpha)}+ \frac{\xi^{\alpha -2}}{\Gamma(\alpha-1)} \\& \hphantom{\bigl\vert h'(t)\bigr\vert \le}{} +\frac{1}{2}C'_{1} \frac{1}{\Gamma(\alpha+1)}+\Gamma(3-p) \biggl( 1 +\frac {1}{12}C'_{1} \biggr)\frac{1}{\Gamma(\alpha-p+1)} +\frac{1}{2} C'_{1} \frac{1}{\Gamma(\alpha+2)} \biggr\} \\& \hphantom{\bigl\vert h'(t)\bigr\vert } = \Lambda_{2}\sum_{i=1}^{4} \|p_{i}\|\psi_{i}(r), \\& \bigl\vert h''(t)\bigr\vert \le \bigl[ \|p_{1}\|\psi_{1}(r)+\|p_{2}\|\psi_{2}(r)+ \|p_{3}\|\psi_{3}(r)+\|p_{4}\| \psi_{4}(r) \bigr] \biggl\{ \frac{1}{\Gamma(\alpha-1)} +C''_{1} \frac{1}{\Gamma (\alpha+1)} \\& \hphantom{\bigl\vert h''(t)\bigr\vert \le}{} +\Gamma(3-p) \biggl( 1 +\frac{1}{12}C''_{1} \biggr)\frac{1}{\Gamma(\alpha-p+1)} +\frac{1}{2} C''_{1} \frac{1}{\Gamma(\alpha+2)} \biggr\} \\& \hphantom{\bigl\vert h''(t)\bigr\vert } = \Lambda_{3}\sum_{i=1}^{4} \|p_{i}\|\psi_{i}(r) \end{aligned}$$
and
$$\begin{aligned} \bigl\vert {}^{\mathrm{c}}D^{p}h(t)\bigr\vert \le& \bigl[ \|p_{1}\|\psi_{1}(r)+\|p_{2}\|\psi_{2}(r)+ \|p_{3}\|\psi_{3}(r)+\|p_{4}\| \psi_{4}(r) \bigr] \biggl\{ \frac{1}{\Gamma(\alpha-p +1)} \\ &{}+D_{1}\frac{1}{\Gamma (\alpha+1)}+\biggl(1+\frac{\Gamma(3-p)}{12} D_{1}\biggr)\frac{1}{\Gamma(\alpha-p+1)} + \frac{1}{2} D_{1}\frac{1}{\Gamma(\alpha+2)} \biggr\} \\ =&\Lambda_{4}\sum_{i=1}^{4} \|p_{i}\|\psi_{i}(r) \end{aligned}$$
for all \(t\in J\). Thus we get
$$\|h\|\le(\Lambda_{1}+\Lambda_{2}+\Lambda_{3}+ \Lambda_{4}) \sum_{i=1}^{4} \|p_{i}\|\psi_{i}(r)=\sum_{i=1}^{4} \Lambda_{i}\|p_{i}\|\psi_{i}(r), $$
which implies that \({\mathcal{N}}\) maps bounded sets into bounded sets in X.
Now, we prove that \({\mathcal{N}}\) maps bounded sets into equi-continuous subsets of X. Suppose that \(u\in B_{r}\) and \(t_{1}, t_{2}\in J\) with \(t_{1}< t_{2}\). Then we have
$$\begin{aligned}& \bigl\vert h(t_{2})-h(t_{1})\bigr\vert \\& \quad \le \biggl\vert \frac{1}{\Gamma(\alpha)}\int_{0}^{t_{2}} (t_{2}-s)^{\alpha -1}v(\tau)\, d\tau-\frac{1}{\Gamma(\alpha)}\int _{0}^{t_{1}} (t_{1}-s)^{\alpha -1}v(\tau) \, d\tau\biggr\vert \\& \qquad {} +\vert t_{2}-t_{1}\vert I^{\alpha-2}\bigl\vert v(\xi)\bigr\vert + \bigl\vert A(t_{2})-A(t_{1}) \bigr\vert I^{\alpha}\bigl\vert v(1)\bigr\vert + \bigl\vert B(t_{2})-B(t_{1})\bigr\vert I^{\alpha-p}\bigl\vert v(1)\bigr\vert \\& \qquad {}+ \bigl\vert C(t_{2})-C(t_{1})\bigr\vert I^{\alpha+1}\bigl\vert v(1)\bigr\vert \\& \quad \le \frac{1}{\Gamma(\alpha)}\int_{0}^{t_{1}} \bigl[(t_{2}-s)^{\alpha -1}-(t_{1}-s)^{\alpha-1}\bigr] \bigl\vert v( \tau)\bigr\vert \, d\tau+ \frac{1}{\Gamma(\alpha)}\int _{t_{1}}^{t_{2}}(t_{2}-s)^{\alpha-1}\bigl\vert v( \tau)\bigr\vert \, d\tau \\& \qquad {} +\vert t_{2}-t_{1}\vert I^{\alpha-2}\bigl\vert v(\xi)\bigr\vert + \bigl\vert A(t_{2})-A(t_{1}) \bigr\vert I^{\alpha}\bigl\vert v(1)\bigr\vert + \bigl\vert B(t_{2})-B(t_{1})\bigr\vert I^{\alpha-p}\bigl\vert v(1)\bigr\vert \\& \qquad {}+ \bigl\vert C(t_{2})-C(t_{1})\bigr\vert I^{\alpha+1}\bigl\vert v(1)\bigr\vert \\& \quad \le \sum_{i=1}^{4}\|p_{i} \|\psi_{i}(r) \biggl\{ \bigl(t_{2}^{\alpha}-t_{1}^{\alpha } \bigr)\frac{1}{\Gamma(\alpha+1)}+\vert t_{2}-t_{1}\vert \frac{\xi^{\alpha-2}}{\Gamma (\alpha-1)}+ \bigl\vert A(t_{2})-A(t_{1})\bigr\vert \frac{1}{\Gamma(\alpha+1)} \\& \qquad {}+\bigl\vert B(t_{2})-B(t_{1})\bigr\vert \frac{1}{\Gamma(\alpha-p+1)}+ \bigl\vert C(t_{2})-C(t_{1})\bigr\vert \frac {1}{\Gamma(\alpha+2)} \biggr\} . \end{aligned}$$
Proceeding as above we have
$$\begin{aligned}& \bigl\vert h'(t_{2})-h'(t_{1}) \bigr\vert \\& \quad \le \biggl\vert \frac{1}{\Gamma(\alpha-1)}\int_{0}^{t_{2}} (t_{2}-s)^{\alpha -2}v(\tau)\, d\tau-\frac{1}{\Gamma(\alpha-1)}\int _{0}^{t_{1}} (t_{1}-s)^{\alpha -2}v(\tau) \, d\tau\biggr\vert \\& \qquad {}+ \bigl\vert A'(t_{2})-A'(t_{1}) \bigr\vert I^{\alpha}\bigl\vert v(1)\bigr\vert + \bigl\vert B'(t_{2})-B'(t_{1})\bigr\vert I^{\alpha-p}\bigl\vert v(1)\bigr\vert \\& \qquad {}+ \bigl\vert C'(t_{2})-C'(t_{1}) \bigr\vert I^{\alpha+1}\bigl\vert v(1)\bigr\vert \\& \quad \le \sum_{i=1}^{4}\|p_{i} \|\psi_{i}(r) \biggl\{ \bigl(t_{2}^{\alpha-1}-t_{1}^{\alpha -1} \bigr)\frac{1}{\Gamma(\alpha)} + \bigl\vert A'(t_{2})-A'(t_{1}) \bigr\vert \frac{1}{\Gamma(\alpha +1)} \\& \qquad {}+\bigl\vert B'(t_{2})-B'(t_{1}) \bigr\vert \frac{1}{\Gamma(\alpha-p+1)}+ \bigl\vert C'(t_{2})-C'(t_{1}) \bigr\vert \frac{1}{\Gamma(\alpha+2)} \biggr\} , \\& \bigl\vert h''(t_{2})-h''(t_{1}) \bigr\vert \\& \quad \le\sum_{i=1}^{4}\|p_{i}\| \psi_{i}(r) \biggl\{ \bigl(t_{2}^{\alpha-1}-t_{1}^{\alpha -1} \bigr)\frac{1}{\Gamma(\alpha-1)} + \bigl\vert A''(t_{2})-A''(t_{1}) \bigr\vert \frac{1}{\Gamma (\alpha+1)} \\& \qquad {} +\bigl\vert B''(t_{2})-B''(t_{1}) \bigr\vert \frac{1}{\Gamma(\alpha-p+1)}+ \bigl\vert C''(t_{2})-C''(t_{1}) \bigr\vert \frac{1}{\Gamma(\alpha+2)} \biggr\} \end{aligned}$$
and
$$\begin{aligned}& \bigl\vert {}^{\mathrm{c}}D^{p}h(t_{2})-{}^{\mathrm{c}}D^{p}h(t_{1}) \bigr\vert \\& \quad \le \sum_{i=1}^{4}\|p_{i} \|\psi_{i}(r) \biggl\{ \bigl(t_{2}^{\alpha-p}-t_{1}^{\alpha -p} \bigr)\frac{1}{\Gamma(\alpha-p+1)} + |t_{2}-t_{1}|\frac{12\gamma}{12+\gamma \Gamma(3-p)} \frac{1}{\Gamma(\alpha+1)} \\& \qquad {} +|t_{2}-t_{1}| \biggl(1+ \frac{12\gamma}{12+\gamma\Gamma(3-p)} \biggr) \frac {1}{\Gamma(\alpha-p+1)} \\& \qquad {} +|t_{2}-t_{1}| \frac{1}{2} \frac{\Gamma(3-p) \gamma}{12+\gamma\Gamma (3-p)} \frac{1}{\Gamma(\alpha+2)} \biggr\} . \end{aligned}$$
Obviously the right-hand side of the above inequalities tends to zero independently of \(u \in B_{r}\) as \(t_{2}- t_{1} \to0\). Therefore it follows by the Ascoli-Arzelá theorem that \({\mathcal{N}}: X \to {\mathcal{P}}(X)\) is completely continuous.
In our next step, we show that \({\mathcal{N}}\) is upper semicontinuous. It is well known by Lemma 1.1 that \({\mathcal{N}}\) will be upper semicontinuous if we prove that it has a closed graph, since \({\mathcal{N}}\) is already shown to be completely continuous. Thus we will prove that \({\mathcal{N}}\)
has a closed graph. Let \(u_{n} \to u_{*}\), \(h_{n} \in{\mathcal{N}} (u_{n})\), and \(h_{n} \to h_{*}\). Then we need to show that \(h_{*} \in {\mathcal{N}} (u_{*})\). Associated with \(h_{n} \in{\mathcal{N}} (u_{n})\), there exists \(v_{n} \in S_{F,u_{n}}\) such that, for each \(t \in J\),
$$ h_{n}(t) = I^{\alpha}v_{n}(t)+ \biggl( \frac{1}{2}-t \biggr)I^{\alpha-2}v_{n}(\xi)+ A(t) I^{\alpha}v_{n}(1) + B(t)I^{\alpha-p}v_{n}(1) + C(t)I^{\alpha+1}v_{n}(1). $$
Thus it suffices to show that there exists \(v_{*} \in S_{F,u_{*}}\) such that, for each \(t \in J\),
$$ h_{*}(t) = I^{\alpha}v_{*}(t)+ \biggl(\frac{1}{2}-t \biggr)I^{\alpha-2}v_{*}( \xi)+ A(t) I^{\alpha}v_{*}(1) + B(t)I^{\alpha-p}v_{*}(1) + C(t)I^{\alpha+1}v_{*}(1). $$
Let us consider the linear operator \(\Theta: L^{1}(J, \mathbb{R}) \to C(J, \mathbb{R})\) given by
$$ f \mapsto\Theta(v) (t) =I^{\alpha}v(t)+ \biggl(\frac{1}{2}-t \biggr)I^{\alpha-1}v(\xi)+ A(t) I^{\alpha}v(1) + B(t)I^{\alpha-p}v(1) + C(t)I^{\alpha+1}v(1). $$
Observe that
$$\begin{aligned} \bigl\| h_{n}(t)-h_{*}(t)\bigr\| =& \biggl\| I^{\alpha}\bigl(v_{n}(t)-v_{*}(t) \bigr) + \biggl(\frac {1}{2}-t \biggr)I^{\alpha-2}\bigl(v_{n}( \xi)-v_{*}(\xi)\bigr) \\ &{} + A(t) I^{\alpha}\bigl(v_{n}(1)-v_{*}(1)\bigr)+ B(t)I^{\alpha-p} \bigl(v_{n}(1)-v_{*}(1)\bigr) \\ & {}+ C(t)I^{\alpha+1}\bigl(v_{n}(1)-v_{*}(1)\bigr) \biggr\| \to 0,\quad \mbox{as } n\to\infty. \end{aligned}$$
Thus, it follows by Lemma 1.2 that \(\Theta\circ S_{F}\) is a closed graph operator. Further, we have \(h_{n}(t) \in \Theta(S_{F,u_{n}})\). Since \(u_{n} \to u_{*}\), therefore, we have
$$ h_{*}(t) = I^{\alpha}v_{*}(t)+ \biggl(\frac{1}{2}-t \biggr)I^{\alpha-2}v_{*}( \xi)+ A(t) I^{\alpha}v_{*}(1) + B(t)I^{\alpha-p}v_{*}(1) + C(t)I^{\alpha+1}v_{*}(1) $$
for some \(v_{*} \in S_{F,u_{*}}\).
Finally, we show that there exists an open set \(U\subseteq X\) with \(u\notin{\mathcal{N}} (u)\) for any \(\lambda\in (0,1)\) and all \(u\in\partial U\). Let \(\lambda\in(0,1)\) and \(u\in \lambda{\mathcal{N}} (u)\). Then there exists \(v \in L^{1}(J, \mathbb{R})\) with \(v \in S_{F,u}\) such that, for \(t \in J\), we have
$$ u(t) = \lambda I^{\alpha}v(t)+\lambda \biggl(\frac {1}{2}-t \biggr)I^{\alpha-1}v(\xi)+ \lambda A(t) I^{\alpha}v(1) +\lambda B(t)I^{\alpha-p}v(1) + \lambda C(t)I^{\alpha+1}v(1). $$
Using the computations of the second step above we have
$$\begin{aligned} \|u\| \le& (\Lambda_{1}+\Lambda_{2}+\Lambda_{3}+ \Lambda_{4}) \bigl[\|p_{1}\|\psi _{1}\bigl(\Vert u \Vert \bigr)+\|p_{2}\|\psi_{2}\bigl(\Vert u\Vert \bigr) \\ &{}+ \|p_{3}\|\psi_{3}\bigl(\Vert u\Vert \bigr)+ \|p_{4}\|\psi_{4}\bigl(\Vert u\Vert \bigr) \bigr], \end{aligned}$$
which implies that
$$\frac{\|u\|}{ \sum_{i=1}^{4}\Lambda_{i}\|p_{i}\|\psi_{i}(\|u\| )}\le1. $$
In view of (H3), there exists M such that \(\|u\| \ne M\). Let us set
$$U = \bigl\{ u \in X : \Vert x\Vert < M\bigr\} . $$
Note that the operator \({\mathcal{N}} :\overline{U} \to\mathcal{P}(X)\) is upper semicontinuous and completely continuous. From the choice of U, there is no \(u \in\partial U\) such that \(u \in\lambda{\mathcal{N}} (u)\) for some \(\lambda\in(0,1)\). Consequently, by the nonlinear alternative of Leray-Schauder type (Lemma 1.3), we deduce that \({\mathcal{N}}\) has a fixed point \(u \in\overline{U}\) which is a solution of the problem (1.1)-(1.2). This completes the proof. □
In the next theorem, we prove the existence of solution for the inclusion boundary value problem (1.1)-(1.2) when the multifunction F is non-convex valued.
Theorem 2.5
Assume that:
- (H4):
-
\(F : J \times\mathbb{R}^{4} \to{\mathcal{P}}_{\mathrm{cp}}(\mathbb{R})\)
is such that
\(F(\cdot,u,v,z,w) : J \to{\mathcal{P}}_{\mathrm{cp}}(\mathbb{R})\)
is measurable for each
\(u,v,z,w \in\mathbb{R}\).
- (H5):
-
For almost all
\(t \in J\)
and
\(u_{1},u_{2}, u_{3}, u_{4}, w_{1},w_{2}, w_{3}, w_{4}\in \mathbb{R}\)
we have
$$\begin{aligned}& H_{d}\bigl(F(t,u_{1},u_{2},u_{3},u_{4}), F(t,w_{1},w_{2},w_{3},w_{4})\bigr) \\& \quad \leq m(t) \bigl(|u_{1}-w_{1}|+|u_{2}-w_{2}|+|u_{3}-w_{3}|+|u_{4}-w_{4}|\bigr) \end{aligned}$$
with
\(m \in C(J, \mathbb{R}^{+})\)
and
\(d(0,F(t,0,0,0,0))\le m(t)\), for almost all
\(t \in J\).
Then the boundary value problem (1.1)-(1.2) has at least one solution on
J
if
\(\|m\|\sum_{i=1}^{4} \Lambda_{i} <1\).
Proof
Observe that the set \(S_{F,u}\) is nonempty for each \(u \in X\) by the assumption (H4), so F has a measurable selection (see Theorem III.6 in [36]). Now we show that the operator \({\mathcal{N}}\), defined in the beginning of proof of Theorem 2.4, satisfies the assumptions of Lemma 1.4. To show that \({\mathcal{N}}(u) \in{\mathcal{P}}_{\mathrm{cl}}(X)\) for each \(u \in X\), let \(\{u_{n}\}_{n \ge0} \in {\mathcal{N}}(u)\) be such that \(u_{n} \to u\) (\(n \to\infty\)) in X. Then \(u \in X\) and there exists \(v_{n} \in S_{F,u_{n}}\) such that, for each \(t \in J\),
$$ u_{n}(t) = I^{\alpha}v_{n}(t)+ \biggl( \frac{1}{2}-t \biggr)I^{\alpha-2}v_{n}(\xi)+ A(t) I^{\alpha}v_{n}(1) + B(t)I^{\alpha-p}v_{n}(1) + C(t)I^{\alpha+1}v_{n}(1). $$
As F has compact values, we pass onto a subsequence (if necessary) to find that \(v_{n}\) converges to v in \(L^{1} (J,\mathbb{R})\). Thus, \(v \in S_{F,u}\) and for each \(t \in J\), we have
$$ u_{n}(t)\to u(t) = I^{\alpha}v(t)+ \biggl(\frac{1}{2}-t \biggr)I^{\alpha-1}v(\xi )+ A(t) I^{\alpha}v(1) + B(t)I^{\alpha-p}v(1) + C(t)I^{\alpha+1}v(1). $$
Hence, \(u \in{\mathcal{N}}(u)\).
Next we show that \({\mathcal{N}}\) is a contractive multifunction with constant \(\delta=\|m\|\sum_{i=1}^{4}\Lambda_{i} <1\). Let \(u,w \in X\) and \(h_{1} \in{\mathcal{N}}(u)\). Then there exists \(v_{1} \in S_{F,u}\) such that, for each \(t \in J\),
$$ h_{1}(t) = I^{\alpha}v_{1}(t)+ \biggl( \frac{1}{2}-t \biggr) I^{\alpha-2}v_{1}(\xi) + A(t) I^{\alpha}v_{1}(1) + B(t)I^{\alpha-p} v_{1}(1) + C(t)I^{\alpha+1}v_{1}(1). $$
By (H5), we have
$$\begin{aligned}& H_{d}\bigl(F\bigl(t, u(t), u'(t), u''(t), {}^{\mathrm{c}}D^{p} u(t)\bigr), F\bigl(t, w(t), w'(t), w''(t), {}^{\mathrm{c}}D^{p} w(t)\bigr) \bigr) \\& \quad \leq m(t) \bigl(\bigl\vert u(t)-w(t)\bigr\vert +\bigl\vert u'(t)-w'(t)\bigr\vert + \bigl\vert u''(t)-w''(t)\bigr\vert + \bigl\vert {}^{\mathrm{c}}D^{p} u(t)-{}^{\mathrm{c}}D^{p} w(t)\bigr\vert \bigr), \end{aligned}$$
so there exists \(z\in F(t, u(t), u'(t), u''(t), {}^{\mathrm{c}}D^{p} u(t)) \) such that
$$\bigl\vert v_{1}(t)-z\bigr\vert \leq m(t) \bigl(\bigl\vert u(t)-w(t)\bigr\vert +\bigl\vert u'(t)-w'(t)\bigr\vert + \bigl\vert u''(t)-w''(t) \bigr\vert + \bigl\vert ^{\mathrm{c}}D^{p} u(t)-{}^{\mathrm{c}}D^{p} w(t)\bigr\vert \bigr) $$
for almost all \(t\in J\). Define the multifunction \(U: J \rightarrow {\mathcal{P}}(\mathbb{R})\) by
$$\begin{aligned} U(t) =&\bigl\{ z\in\mathbb{R} : \bigl\vert v_{1}(t)-z\bigr\vert \leq m(t) \bigl( \bigl\vert u(t)-w(t)\bigr\vert +\bigl\vert u'(t)-w'(t) \bigr\vert + \bigl\vert u''(t)-w''(t) \bigr\vert \\ &{}+ \bigl\vert {}^{\mathrm{c}}D^{p} u(t)-{}^{\mathrm{c}}D^{p} w(t)\bigr\vert \bigr) \mbox{ for almost all } t\in J\bigr\} . \end{aligned}$$
It is easy to check that the multifunction \(U( \cdot) \cap F(\cdot ,u(\cdot),u'(\cdot), u''(\cdot),{}^{\mathrm{c}}D^{p} u(\cdot )) \) is measurable. Hence, we can choose \(v_{2}\in S_{F, u}\) such that
$$\bigl\vert v_{1}(t)-v_{2}(t)\bigr\vert \leq m(t) \bigl( \bigl\vert u(t)-w(t)\bigr\vert +\bigl\vert u'(t)-w'(t) \bigr\vert + \bigl\vert u''(t)-w''(t) \bigr\vert + \bigl\vert {}^{\mathrm{c}}D^{p} u(t)-{}^{\mathrm{c}}D^{p} w(t)\bigr\vert \bigr) $$
for almost all \(t\in J\). Consider \(h_{2}\in{\mathcal{N}}(u)\) which is defined by
$$ h_{2}(t) = I^{\alpha}v_{2}(t)+ \biggl( \frac{1}{2}-t \biggr)I^{\alpha-2}v_{2}(\xi)+ A(t) I^{\alpha}v_{2}(1) + B(t)I^{\alpha-p} v_{2}(1) + C(t)I^{\alpha+1}v_{2}(1). $$
Thus,
$$\begin{aligned} \bigl\vert h_{1}(t)-h_{2}(t)\bigr\vert =& I^{\alpha}\bigl\vert v_{1}(t)-v_{2}(t)\bigr\vert + \biggl\vert \frac{1}{2}-t \biggr\vert I^{\alpha-2}\bigl\vert v_{1}(1)-v_{2}(1) \bigr\vert + \bigl\vert A(t)\bigr\vert I^{\alpha}\bigl\vert v_{1}(1)-v_{2}(1)\bigr\vert \\ &{} + \bigl\vert B(t) \bigr\vert I^{\alpha-p} \bigl\vert v_{1}(1)-v_{2}(1)\bigr\vert + \bigl\vert C(t)\bigr\vert I^{\alpha+1}\bigl\vert v_{1}(1)-v_{2}(1)\bigr\vert \\ \le& \|m\| \biggl\{ \frac{1}{\Gamma(\alpha+1)}+\frac{1}{2}\frac{\xi ^{\alpha-2}}{\Gamma(\alpha-1)}+ \frac{1}{2}(1+C_{1})\frac{1}{\Gamma(\alpha +1)} \\ &{}+\Gamma(3-p) \biggl(\frac{7}{4}+\frac{1}{12}C_{1} \biggr)\frac{1}{\Gamma (\alpha-p+1)} +\frac{1}{2} C_{1}\frac{1}{\Gamma(\alpha+2)} \biggr\} \|u-w\| \\ =&\|m\|\Lambda_{1}\|u-w\|. \end{aligned}$$
In a similar manner we obtain
$$\begin{aligned}& \bigl\vert h'_{1}(t)-h'_{2}(t) \bigr\vert \le \|m\| \biggl\{ \frac{1}{\Gamma(\alpha)}+ \frac{\xi^{\alpha -2}}{\Gamma(\alpha-1)}+ \frac{1}{2}C'_{1}\frac{1}{\Gamma(\alpha+1)} \\& \hphantom{\bigl\vert h'_{1}(t)-h'_{2}(t) \bigr\vert \le}{} +\Gamma(3-p) \biggl(1+\frac{1}{12}C'_{1} \biggr)\frac{1}{\Gamma(\alpha-p+1)} +\frac{1}{2} C'_{1} \frac{1}{\Gamma(\alpha+2)} \biggr\} \|u-w\| \\& \hphantom{\bigl\vert h'_{1}(t)-h'_{2}(t) \bigr\vert } = \|m\|\Lambda_{2}\|u-w\|, \\& \bigl\vert h''_{1}(t)-h''_{2}(t) \bigr\vert \le \|m\| \biggl\{ \frac{1}{\Gamma(\alpha-1)} +C''_{1} \frac{1}{\Gamma (\alpha+1)}+\Gamma(3-p) \biggl(1+\frac{1}{12}C''_{1} \biggr)\frac{1}{\Gamma (\alpha-p+1)} \\& \hphantom{\bigl\vert h''_{1}(t)-h''_{2}(t) \bigr\vert \le }{}+\frac{1}{2} C''_{1} \frac{1}{\Gamma(\alpha+2)} \biggr\} \|u-w\| \\& \hphantom{\bigl\vert h''_{1}(t)-h''_{2}(t) \bigr\vert } = \|m\|\Lambda_{3}\|u-w\| \end{aligned}$$
and
$$\begin{aligned} \bigl\vert {}^{\mathrm{c}}D^{p}h_{1}(t)-{}^{\mathrm{c}}D^{p}h_{2}(t) \bigr\vert \le& \|m\| \biggl\{ \frac{1}{\Gamma(\alpha-p + 1)}+D_{1} \frac{1}{\Gamma (\alpha+1)} \\ &{}+ \biggl(1+ \frac{\Gamma(3-p)}{12}D_{1} \biggr)\frac{1}{\Gamma(\alpha-p+1)} + \frac{1}{2} D_{1}\frac{1}{\Gamma(\alpha+2)} \biggr\} \|u-w\| \\ =&\|m\|\Lambda_{4}\|u-w\|. \end{aligned}$$
Hence,
$$\| h_{1}-h_{2}\| \le \|m\| \sum _{i=1}^{4}\Lambda_{i}\|u-w\|. $$
Analogously, interchanging the roles of u and w, we obtain
$$H_{d}\bigl({\mathcal{N}}(u), {\mathcal{N}}(w)\bigr) \le \|m\| \sum _{i=1}^{4}\Lambda_{i}\|u-w\|. $$
Since \(\delta= \|m\| \sum_{i=1}^{4}\Lambda_{i}<1\), \({\mathcal{N}}\) is a contraction, it follows by Lemma 1.4 that \({\mathcal{N}}\) has a fixed point u which is a solution of (1.1)-(1.2). This completes the proof. □
Now, we give an illustrative example.
Example 2.6
We consider the following fractional differential inclusion:
$$\begin{aligned} \begin{aligned}[b] {}^{\mathrm{c}}D^{\frac{5}{2}} u(t)\in{}& \biggl[ 0, \frac{\frac{t}{200} \vert\sin \frac{\pi}{2}t \vert\vert u(t)\vert}{1+\vert u(t)\vert} + \frac {t\vert\sin u'(t) \vert^{3}}{200 + 200 \vert\sin u'(t) \vert^{3} } + \frac {\frac{t}{200} \vert\cos u''(t) \vert^{4}}{1+ \vert\cos u''(t)\vert^{4} } \\ &{}+ \frac{t \vert{}^{\mathrm{c}}D^{\frac{5}{3}}u(t)\vert}{200 \vert\cos\pi t \vert( 1+ \vert{}^{\mathrm{c}}D^{\frac{5}{3}}u(t)\vert)} \biggr], \end{aligned} \end{aligned}$$
(2.10)
via the boundary value conditions
$$\begin{aligned}& u(0)+u(1)=0, \qquad {}^{\mathrm{c}}D^{\frac{5}{3}}u( 1)= \frac{1}{1\text{,}000} \int_{0}^{1} u(\tau)\, d\tau, \\& u'(0)+u'' \biggl( \frac{1}{100} \biggr)= \frac{1}{1\text{,}000} \int_{0}^{1} u(\tau)\, d\tau, \end{aligned}$$
where \(t\in J=[0,1]\). By the above inclusion problem, we have \(\alpha= 5/2\), \(p=5/3\), \(\xi= 1/100\), and \(\gamma= \eta= 1/1\text{,}000\). Now, we define an operator \(F: J \times\mathbb{R} \times\mathbb{R} \times \mathbb{R} \times\mathbb{R} \to P_{\mathrm{cp}}(\mathbb{R})\) by
$$\begin{aligned} F(t, x_{1},x_{2},x_{3},x_{4}) =& \biggl[0, \frac{\frac{t}{200} \vert\sin\frac {\pi}{2}t \vert\vert x_{1}\vert}{1+\vert x_{1}\vert} + \frac{t\vert \sin x_{2} \vert^{3}}{200 + 200 \vert\sin x_{2} \vert^{3} } + \frac{\frac {t}{200} \vert\cos x_{3} \vert^{4}}{1+ \vert\cos x_{3}\vert^{4} } \\ &{} + \frac{t \vert x_{4}\vert}{200 \vert\cos\pi t \vert( 1+ \vert x_{4}\vert)} \biggr]. \end{aligned}$$
Also, we define the function \(m: J \to\mathbb{R}^{+}\) by \(m(t)=t/200\). It is clear that m is continuous on J and \(\|m\| = 1/200\). Finally, one can write
$$\begin{aligned}& H_{d}\bigl(F(t,x_{1},x_{2},x_{3},x_{4}), F(t,y_{1},y_{2},y_{3},y_{4})\bigr) \\& \quad \leq m(t) \bigl(\vert x_{1}-y_{1}\vert +\vert x_{2}-y_{2}\vert +\vert x_{3}-y_{3} \vert +\vert x_{4}-y_{4}\vert \bigr), \end{aligned}$$
where \(t\in J\) and \(x_{i}, y_{i} \in\mathbb{R}\) (\(i=1,2, 3, 4\)). On the other hand, there exist the following values:
$$\Lambda_{1} = 2.13208,\qquad \Lambda_{2} = 1.79407, \qquad \Lambda_{3} = 2.057974, \qquad \Lambda_{4} = 2.08023. $$
Then \(\Vert m\Vert(\Lambda_{1} + \Lambda_{2} + \Lambda_{3} + \Lambda_{4} ) = \frac{1}{200}\times8.064354 = 0.04 < 1 \). Consequently, all assumptions and conditions of Theorem 2.5 are satisfied. Hence, Theorem 2.5 implies that the fractional differential inclusion problem (2.10) has at least one solution.