1 Introduction
2 Preliminaries
3 The higher order fractional derivatives and integrals
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\(( {}^{ABR}_{a}\mathbf{D}^{\alpha}{}^{AB}_{a}\mathbf{I}^{\alpha}u)(t)=u(t)\).
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\(({}^{AB}_{a}\mathbf{I}^{\alpha}{}^{ABR}_{a}\mathbf{D}^{\alpha}u)(t)=u(t)-\sum_{k=0}^{n-1} \frac{u^{(k)}(a)}{k!} (t-a)^{k}\).
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\(({}^{AB}_{a}\mathbf{I}^{\alpha}{}^{ABC}_{a}\mathbf{D}^{\alpha}u)(t)=u(t)-\sum_{k=0}^{n} \frac{u^{(k)}(a)}{k!} (t-a)^{k}\).
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\(( {}^{ABR}\mathbf{D}_{b}^{\alpha}{}^{AB}\mathbf{I}_{b}^{\alpha}u)(t)=u(t)\).
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\(({}^{AB}\mathbf{I}_{b}^{\alpha}{}^{ABR}\mathbf{D}_{b}^{\alpha}u)(t)=u(t)-\sum_{k=0}^{n-1} \frac{(-1)^{k}u^{(k)}(b)}{k!} (b-t)^{k}\).
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\(({}^{AB}\mathbf{I}_{b}^{\alpha}{}^{ABC}\mathbf{D}_{b}^{\alpha}u)(t)=u(t)-\sum_{k=0}^{n} \frac{(-1)^{k}u^{(k)}(b)}{k!} (b-t)^{k}\).
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Assume \(0<\alpha\leq1\), \(y(0)=c\) and \(K(0)=0\). By applying \({}^{AB}_{0}\mathbf{I}^{\alpha}\) and making use of Proposition 3.1, we get the solutionNotice that the condition \(K(0)=0\) verifies that the initial condition \(y(0)=c\). Also notice that when \(\alpha\rightarrow1\) we reobtain the solution of the ordinary initial value problem \(y^{\prime}(t)=K(t)\), \(y(0)=c\).$$y(t)=c+\frac{1-\alpha}{B(\alpha)}K(t)+\frac{\alpha}{B(\alpha)}\bigl( {}_{0}I^{\alpha}K(\cdot)\bigr) (t). $$
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Assume \(1<\alpha\leq2\), \(K(0)=0 y(0)=c_{1}\), \(y^{\prime}(0)=c_{2}\): By applying \({}^{AB}_{0}\mathbf{I}^{\alpha}\) and making use of Proposition 3.1 and Definition 3.1 with \(\beta=\alpha -1\), we get the solutionNotice that the solution \(y(t)\) verifies \(y(0)=c_{1}\) without the use of \(K(0)=0\). However, it verifies \(y^{\prime}(0)=c_{2}\) under the assumption \(K(0)=0\). Also, note that when \(\alpha\rightarrow2\) we reobtain the solution of the second order ordinary initial value problem \(y^{\prime \prime}(t)=K(t)\).$$y(t)=c_{1}+c_{2}t+\frac{2-\alpha}{B(\alpha-1)} \int_{0}^{t}K(s)\,ds+\frac{\alpha -1}{B(\alpha-1)\Gamma(\alpha)} \int_{0}^{t} (t-s)^{\alpha-1}K(s)\,ds. $$
4 Existence and uniqueness theorems for the initial value problem types
5 The Lyapunov inequality for the ABR boundary value problem
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\(G(t,s)\geq0\) for all \(a\leq t,s \leq b\).
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\(\max_{t \in[a,b]} G(t,s)=G(s,s)\) for \(s \in[a,b]\).
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\(H(s,s)\) has a unique maximum, given by$$\max_{s \in[a,b]} G(s,s)= G\biggl(\frac{a+b}{2}, \frac{a+b}{2}\biggr)=\frac{(b-a)}{4}. $$