Open Journal of Mathematical Sciences
ISSN: 2523-0212 (Online) 2616-4906 (Print)
DOI: 10.30538/oms2019.0044
Nonlocal initial value problems for Katugampola-Caputo type fractional differential equations on time scales
Alagarsamy Nandhini, Devaraj Vivek\(^1\), Elsayed M. Elsayed
Department of Mathematics with Computer Applications, Sri Ramakrishna College of Arts and Science (Formerly SNR sons College), Coimbatore-641 006, India. (A.N & D.V)
Department of Mathematics, Faculty of Science, Mansoura University, Mansoura 35516, Egypt and Department of Mathematics, Faculty of Science, King Abdulaziz University,Jeddah 21589, Saudi Arabia. (E.M.E)
\(^{1}\)Corresponding Author: peppyvivek@gmail.com
Abstract
Keywords:
1. Introduction
The theory of FDEs has attracted attention of many researchers because of its wide applications in biology, medicine and in other applied fields, see for example [1, 2, 3, 4] and references therein. Throughout this paper \(\left(X,\left\|.\right\|\right)\) will be a Banach space, and \(I=[0,T],T>0,\) a compact interval in \(\mathbb{R}\). Let \(C = C\left([0,T],X\right)\) be the Banach space of all continuous functions \([0,T]\rightarrow X\) endowed with the topology of uniform convergence the norm in this space will be denoted by \(\left\|.\right\|_{c})\).
In this work we consider the following Cauchy problem for the FDEs with nonlocal conditions on time scales(H1)\(f :\mathbb{R}\times X \rightarrow X\) is jointly continuous.
(H2)\(\left\| f(t,x) - f(t,y)\right\|\leq L\left\|x - y\right\|\), \(\forall t \in \mathbb{R},x,y \in X\).
(H3)\(g : C \rightarrow X\) is continuous and \(\left\|g(x) - g(y)\right\| \leq b\left\|x - y\right\|\), \(\forall x,y \in C\).
2. Existence results
We are now ready to present our results.Theorem 1. Under assumptions (H1) and (H2), if \(b < \frac{1}{2}\) and \(L \leq \frac{\Gamma(q+1)}{2T^q}\), then the equations (1) and (2) has a unique solution .
Proof. Define \(F : C\rightarrow C\) by \begin{eqnarray*} (Fx)(t) = x_{0}-g(x)+ \frac{\rho^{(1-q)}}{\Gamma(q)} \int_0^t (t^{\rho}-s^{\rho})^{(q-1)} f(s,x(s))\Delta s \end{eqnarray*} choose \(r \geq 2(\left\|x_{0}\right\| + G + \frac{MT^q}{\Gamma(q+1)})\) ,and let \(sup_{t \in I}\left\|f(t,0)\right\| = M\). Then it is easy to see that \(FB_{r}\) \(\subset\) \(B_{r}\) where \(B_{r} = \{x \in C :\left\|x\right\| \leq r\}\). So let \(x \in B_{r}\) and set \(G= sup_{x \in C} \left\|g(x)\right\|\). Then we get \begin{align*} \left\|Fx(t)\right\| &\leq \left\|x_{0}\right\| + G + \frac{\rho^{(1-q)}}{\Gamma(q)} \int_{0}^{t}(t^{\rho}-s^{\rho})^{q-1}s^{\rho-1}\left\|f(s,x(s))\right\|\Delta s\\ &\leq \left\|x_{0}\right\| + G + \frac{\rho^{1-q}}{\Gamma(q)} \int_{0}^{t}(t^{\rho}-s^{\rho})^{q-1} s^{\rho-1}(\left\|f(s,x(s))-f(s,o)\right\| + \left\|f(s,0)\right\|)\Delta s\\ &\leq \left\|x_{0}\right\| + G + (Lr + M)\frac{1}{\Gamma(q)} \int_{0}^{t}(t^{\rho}-s^{\rho})^{q-1}\Delta s\\ &\leq \left\|x_{0}\right\| + G + (Lr + M)\frac{T^{q}}{\Gamma(q+1)} \leq r \end{align*} by the choice of \(L\) and \(r\). Now take \(x,y \in C\), then we get \begin{align*} \left\|(Fx)(t)-(Fx)(t)\right\| &\leq \left\|g(x)-g(y)\right\| + \frac{\rho^{1-q}}{\Gamma(q)}\int_{0}^{t}(t^{\rho}-s^{\rho})^{q-1}s^{\rho-1}\left\|f(s,x(s))-f(s,y(s))\right\|\Delta s\\ &\leq \Omega_{b,L,T,q,\rho}\left\|x-y\right\|, \end{align*} where \(\Omega_{b,L,T,q,\rho}= (b + \frac{LT^{\rho q}}{\rho^{q}\Gamma(q+1)})\) depends only on the parameters of the problem and since \(\Omega_{b,L,T,q,\rho} \leq 1\), the result follows in view of the contraction mapping principle.
Theorem 2. Let \(M\) be a nonempty convex subset of the Banach space \(X.\) Let \(A,B\) be two operators such that
- \(Ax + By \in M\) whenever \(x,y \in M\),
- \(A\) is compact and continuous,
- \(B\) is a contraction mapping,
Theorem 3. Assume the conditions (H1)and (H3) holds, \(b < 1\) and
- [(H4)] \(\left\|f(t,x)\right\|\leq\mu(t),\quad \forall(t,x)\in I\times X,\) where \(\mu \in L^{1}(I,\mathbb{R}^{+})\),
Proof. Choose \(r \geq \left\|x_{0}\right\| + G + \frac{T^{\rho q}\left\|\mu\right\|_{L^{1}}}{\rho^{q}\Gamma(q+1)}\) and consider \(B_{r}:{x \in C :\left\|x\right\| \leq r}\). Now define on \(B_{r}\) the operators \(A\) and \(B\) as \begin{align*} (Ax)(t) &= \frac{\rho^{1-q}}{\Gamma(q)} \int_{0}^{t}(t^{\rho}-s^{\rho})^{q-1}s^{\rho-1}f(s,x(s))\Delta s, and \\ (Bx)(t) &= x_{0} - g(x) \end{align*} If \(x,y \in B_{r}\) then \(Ax + By \in B_{r}\). Indeed \begin{eqnarray*} \left\|Ax + By\right\| \leq \left\|x_{0}\right\| + G + \frac{T^{\rho q}\left\|\mu\right\|_{L^{1}}}{\rho^{q}\Gamma(q+1)} \leq r. \end{eqnarray*} By \((H3)\), it is also clear that \(B\) is a contraction mapping for \(b< 1\). Since \(x\) is continuous, then \((Ax)(t)\) is continuous in view of (H1). Note that \(A\) is uniformly bounded on \(B_{r}\). Hence we have \begin{eqnarray*} \left\|(Ax)(t)\right\| \leq \frac{T^{\rho q}\left\|\mu\right\|_{L^{1}}}{\rho^{q}\Gamma{(q+1)}}. \end{eqnarray*} Now, we prove that \((Ax)(t)\) is equicontinuous. Let \(t_{1},t_{2} \in I\) and \(x \in B_{r}\). Using the fact that \(f\) is bounded on the compact set \(I \times B_{r}\) (thus \(sup_{(t,x)\in I\times B_{r}}\left\|f(t,x)\right\| = c_{0} < \infty)\), we get \begin{align*} &\left\|Ax(t_{1})-Ax(t_{2})\right\|\\ &= \left\|\frac{\rho^{1-q}}{\Gamma(q)} \int_{0}^{t_{1}}(t_{1}^{\rho}-s^{\rho})^{q-1}s^{\rho-1}f(s,x(s))\Delta s - \frac{\rho^{1-q}}{\Gamma(q)} \int_{0}^{t_{2}}(t_{2}^{\rho}-s^{\rho})^{q-1}s^{\rho-1}f(s,x(s))\Delta s\right\|\\ &\leq \frac{\rho^{1-q}}{\Gamma(q)}\left\|\int_{0}^{t_{1}}[(t_{1}^{\rho}-s^{\rho})^{q-1}-(t_{2}^{\rho}-s^{\rho})^{q-1}]s^{\rho-1}f(s,x(s))\Delta s + \int_{t_{1}}^{t_{2}}(t_{2}^{\rho}-s^{\rho})^{q-1}s^{\rho-1}f(s,x(s))\Delta s\right\|\\ &\leq \frac{C_{0}\rho^{1-q}}{\Gamma(q)}\left(\int_{0}^{t_{1}}\left|(t_{1}^{\rho}-s^{\rho})^{q-1}-(t_{2}^{\rho}-s^{\rho})^{q-1}\right|s^{\rho-1}\Delta s + \int_{t_{1}}^{t_{2}}(t_{2}^{\rho}-s^{\rho})^{q-1}s^{\rho-1}\Delta s\right). \end{align*} For \(q < 1\), \((t_{1}^{\rho}-s^{\rho})^{q-1}\) \(\geq\) \((t_{2}^{\rho}-s^{\rho})^{q-1}\), we have \begin{align*} \int_{t_{0}}^{t_{1}}\left|(t_{1}^{\rho}-s^{\rho})^{q-1} - (t_{2}^{\rho}-s^{\rho})^{q-1}\right|s^{\rho-1}\Delta s &=\int_{t_{0}}^{t_{1}}[(t_{1}^{\rho}-s^{\rho})^{q-1}-(t_{2}^{\rho}-s^{\rho})^{q-1}]s^{q-1}\Delta s\\ &=\frac{1}{\rho q}(t_{1}^{\rho} q - t_{2}^{\rho} q) + \frac{1}{\rho q}(t_{2}^{\rho}-t_{1}^{\rho})^{q}\\ &=\frac{1}{\rho q}(t_{2}^{\rho}-t_{1}^{\rho})^{q}. \end{align*} If \(q > 1\), \((t_{1}^{\rho}-s^{\rho})^{q-1}\) \(\leq\) \((t_{2}^{\rho}-s^{\rho})^{q-1}\), we have \begin{align*} \int_{t_{0}}^{t_{1}}\left|(t_{1}^{\rho}-s^{\rho})^{q-1} - (t_{2}^{\rho}-s^{\rho})^{q-1}\right|s^{\rho-1}\Delta s &= \int_{t_{0}}^{t_{1}}[(t_{1}^{\rho}-s^{\rho})^{q-1}-(t_{1}^{\rho}-s^{\rho})^{q-1}]s^{\rho-1}\Delta s\\ &= \frac{1}{\rho q}(t_{2}^{\rho} q - t_{1}^{\rho} q) - \frac{1}{\rho q}(t_{2}^{\rho}-t_{1}^{\rho})^{q}\\ &= \frac{1}{\rho q}(t_{2}^{\rho} q - t_{1}^{\rho} q) \end{align*} \begin{align*} \left\|(Ax)(t_{1}) - (Ax)(t_{2})\right\| &\leq \frac{2C_{0}}{\rho^{q}\Gamma(q+1)}(t_{2}^{\rho}-t_{1}^{\rho})^{q} , \quad q \leq 1 \\ &\leq \frac{C_{0}}{\rho^{q}\Gamma(q+1)}\left[(t_{2}^{\rho}-t_{1}^{\rho})^{q} + (t_{2}^{\rho q}-t_{1}^{\rho q})\right] , \quad q > 1, \end{align*} which does not depend on \(x\), so \(A(B_{r})\) is relatively compact. By the Arzela-Ascoli Theorem, \(A\) is compact. We now conclude the result of the theorem based on the Krasnoselkii's theorem above.
Competing Interests
The author(s) do not have any competing interests in the manuscript.References
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