In this paper, we study Katugampola fractional differential equations (FDEs) with nonlocal conditions on time scales. By means of standard fixed point theorems, some new sufficient conditions for the existence of solutions are established.
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 \(\left\|g(x) – g(y)\right\| \leq b\left\|x – y\right\|\), \(\forall x,y \in C\).
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
Theorem 3. Assume the conditions (H1)and (H3) holds, \(b < 1\) and
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 \(q 1\), \((t_{1}^{\rho}-s^{\rho})^{q-1}\) \(\leq\) \((t_{2}^{\rho}-s^{\rho})^{q-1}\), 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.