Existence result and approximate solutions for quadratic integro-differential equations of fractional order

Definition 1

A real function $f(x)\text{,}x>0$ is said to be in space $C_{\mu }\text{,}\mu \in R$ if there exists a real number $p>\mu$ , such that $f(t)=t^p{f}_1(t)$ , where $f_1(t)\in C(0\text{,}\infty )$ , and it is said to be in the space $C_{\mu }^n$ if and only if $f^n\in C_{\mu }\text{,}n\in N$ .

Definition 2

The Riemann-Liouville fractional integral operator of order $\alpha >0$ of a function $f\in C_{\mu }\text{,}\mu ⩾-1$ , is defined as

$$\begin{array}{cc}\hfill {\mathcal{J}}^{\alpha }f(t)& =\frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_0^t{(t-s)}^{\alpha -1}f(s)\mathit{ds}\text{,}\alpha >0\text{,}\hfill \\ \hfill {\mathcal{J}}^{0}f(t)& =f(t)\text{.}\hfill \end{array}$$ Some properties of the operator ${\mathcal{J}}^{\alpha }$ can be found in (Miller and Ross, 1993), which are needed here, as follows: for $f\in C_{\mu }\text{,}\mu ⩾-1\text{,}\alpha \text{,}\beta ⩾0$ and $\gamma ⩾-1$ :

1. ${\mathcal{J}}^{\alpha }{\mathcal{J}}^{\beta }f(t)={\mathcal{J}}^{\alpha +\beta }f(t)$ ,

2. ${\mathcal{J}}^{\alpha }{\mathcal{J}}^{\beta }f(t)={\mathcal{J}}^{\beta }{\mathcal{J}}^{\alpha }f(t)$ ,

3. ${\mathcal{J}}^{\alpha }{t}^{\gamma }=\frac{\mathrm{\Gamma }(\gamma +1)}{\mathrm{\Gamma }(\alpha +\gamma +1)}{t}^{\alpha +\gamma }$ .

Definition 3

The fractional derivative of $f(t)$ in the caputo sense is defined as

$$D^{\alpha }f(t)=\frac{1}{\mathrm{\Gamma }(m-\alpha )}{\int }_0^t{(t-s)}^{m-\alpha -1}{f}^m(s)\mathit{ds}\text{,}$$ for $m-1<\alpha ⩽m\text{,}m\in N\text{,}t>0\text{,}f\in C_{-1}^m$ .

Property 1

The Laplace transform of the Caputo derivative is defined as

$$\mathcal{L}[D^{\alpha }x(t)]=s^{\alpha }X(s)-{\displaystyle \sum _{k=0}^{n-1}}{s}^{\alpha -k-1}{x}^{(k)}(0)\text{,}$$ for $n-1<\alpha <n$ .The function $X(s)$ of the complex variable s defined by $X(s)=\mathcal{L}\{x(t)\text{;}s\}={\int }_0^{\infty }{e}^{-\mathit{st}}x(t)\mathit{dt}$ is called the Laplace transform of the function $x(t)$ .

Lemma 2.1

If $m-1<\alpha ⩽m\text{,}m\in N\text{,}f\in C_{\mu }^m\text{,}\mu ⩾-1$ , the following two properties hold:

1. $D^{\alpha }{\mathcal{J}}^{\alpha }f(t)=f(t)$ ,

2. $({\mathcal{J}}^{\alpha }{D}^{\alpha })f(t)=f(t)-{\sum }_{k=0}^{m-1}{f}^{(k)}(0)\frac{t^k}{k!}$ .

Theorem 3.1

Theorem 3.1 Schauder (Zeidler, 1995)

If E is a closed, bounded, convex subset of a Banach space B and $T:E\to E$ is completely continuous then T has a fixed point.

Theorem 3.2

Assume that

1. $f\in C[J\times {\mathbb{R}}^n]\text{,}K\in C[J\times J\times {\mathbb{R}}^n\text{,}{\mathbb{R}}^n]\text{,}{\int }_{t_0}^{\tau }\mid K(\tau \text{,}s\text{,}x(s))\mid \mathit{ds}⩽N$  for   $t_0⩽s\le t⩽t_0+a\text{,}x\in \mathrm{\Omega }=\left\{x\in C[J\text{,}{\mathbb{R}}^n]:x(t_0)=x_0\mathit{and}\mid x(t)-x_0\mid ⩽b\right\}$

2. $\Vert f(s\text{,}x(s))-f(s\text{,}y(s))\Vert <\frac{1}{2}\frac{∊\mathrm{\Gamma }(q+1)}{{\alpha }^q}$

3. $\Vert K(\sigma \text{,}s\text{,}x(s))-K(\sigma \text{,}s\text{,}y(s))\Vert ⩽\frac{1}{4}\frac{∊\mathrm{\Gamma }(q+1)}{\parallel x(t)\parallel {\alpha }^{q+1}}$

4. $\delta =\frac{1}{4}\frac{∊\mathrm{\Gamma }(q+1)}{N{\alpha }^q}$ then the (IVP) (3.1) has at least one solution $x(t)$ on $t_0⩽t⩽t_0+\alpha$ , for some $0<\alpha <a$ .

Proof

Consider the set $D=\left\{(t\text{,}x):t\in J\mathrm{and}\mid x-x_0\mid ⩽b\right\}$ and let $\mid f(t\text{,}x(t))\mid ⩽M$ on D. Choose $\alpha =\min \left\{a\text{,}{\left(\frac{b\mathrm{\Gamma }(q+1)}{M+N\Vert x\Vert }\right)}^{\frac{1}{q}}\right\}$ and let ${\mathrm{\Omega }}_0=\left\{x\in C[J_0\text{,}{\mathbb{R}}^n]:x(t_0)=x_0\mathrm{and}\mid x-x_0\mid ⩽b\right\}$ where $\Vert x\Vert ={\max }_{t_0⩽t⩽t_0+\alpha }\mid x(t)\mid$ and $J_0=[t_0\text{,}{t}_0+\alpha ]$ . Clearly the set ${\mathrm{\Omega }}_0$ is closed, convex and bounded. For any $x\in {\mathrm{\Omega }}_0$ define the operator Tx by

$$\mathit{Tx}(t)=x_0+\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-s)}^{q-1}f(s\text{,}x(s))\mathit{ds}+\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-\tau )}^{q-1}\left(x(\tau ){\int }_{t_0}^{\tau }K(\tau \text{,}s\text{,}x(s))\mathit{ds}\right)d\tau \text{,}$$ we may apply Schauder’s fixed point theorem to prove the existence of a fixed point of $T\in {\mathrm{\Omega }}_0$ which is equivalent to solving the IVP clearly $\mathit{Tx}(t_0)=x_0$ , and for $t\in J_0$ . $$\begin{array}{cc}\hfill |\mathit{Tx}(t)-x_0|& =\left|\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-s)}^{q-1}f(s\text{,}x(s))\mathit{ds}+\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-\tau )}^{q-1}\left(x(\tau ){\int }_{t_0}^{\tau }K(\tau \text{,}s\text{,}x(s))\mathit{ds}\right)d\tau \right|\text{,}\hfill \\ \hfill & ⩽\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-s)}^{q-1}\mid f(s\text{,}x(s))\mid \mathit{ds}+\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-\tau )}^{q-1}\left(x(\tau )\mid {\int }_{t_0}^{\tau }K(\tau \text{,}s\text{,}x(s))\mid \mathit{ds}\right)d\tau \text{,}\hfill \\ \hfill & ⩽\frac{M}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-s)}^{q-1}\mathit{ds}+\frac{N}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-\tau )}^{q-1}|x(\tau )|d\tau \hfill \\ \hfill & ⩽\left(\frac{M}{\mathrm{\Gamma }(q)}+\frac{N|x(t)|}{\mathrm{\Gamma }(q)}\right)\frac{{(t-t_0)}^q}{q}\text{,}\hfill \\ \hfill & =\frac{{\alpha }^q}{\mathrm{\Gamma }(q+1)}\left(M+N\parallel x\parallel \right)⩽b\text{.}\hfill \end{array}$$ Which implies that $T{\mathrm{\Omega }}_0\subset {\mathrm{\Omega }}_0$ . Furthermore, for any $t_1\text{,}{t}_2\in J_0$ , such that $t_2>t_1$ , we obtain

(3.4)

$$\begin{array}{cc}\hfill |\mathit{Tx}(t_2)-\mathit{Tx}(t_1)\mid & =\left|\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^{t_2}{(t_2-s)}^{q-1}f(s\text{,}x(s))\mathit{ds}+\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^{t_2}{(t_2-\tau )}^{q-1}\left(x(\tau ){\int }_{t_0}^{\tau }K(\tau \text{,}s\text{,}x(s))\mathit{ds}\right)d\tau \right.\hfill \\ \hfill & \left.-\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^{t_1}{(t_1-s)}^{q-1}f(s\text{,}x(s))\mathit{ds}-\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^{t_1}{(t_1-\tau )}^{q-1}\left(x(\tau ){\int }_{t_0}^{\tau }K(\tau \text{,}s\text{,}x(s))\mathit{ds}\right)d\tau \right|\hfill \\ \hfill & =\left|\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^{t_1}{(t_2-s)}^{q-1}f(s\text{,}x(s))\mathit{ds}+\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^{t_1}{(t_2-\tau )}^{q-1}\left(x(\tau ){\int }_{t_0}^{\tau }K(\tau \text{,}s\text{,}x(s))\mathit{ds}\right)d\tau \right.\hfill \\ \hfill & +\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_1}^{t_2}{(t_2-s)}^{q-1}f(s\text{,}x(s))\mathit{ds}+\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_1}^{t_2}{(t_2-\tau )}^{q-1}\left(x(\tau ){\int }_{t_0}^{\tau }K(\tau \text{,}s\text{,}x(s))\mathit{ds}\right)d\tau \hfill \\ \hfill & \left.-\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^{t_1}{(t_1-s)}^{q-1}f(s\text{,}x(s))\mathit{ds}-\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^{t_1}{(t_1-\tau )}^{q-1}\left(x(\tau ){\int }_{t_0}^{\tau }K(\tau \text{,}s\text{,}x(s))\mathit{ds}\right)d\tau \right|\hfill \\ \hfill & ⩽\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^{t_1}\left[{(t_2-s)}^{q-1}-{(t_1-s)}^{q-1}\right]\mid f(s\text{,}x(s))\mid \mathit{ds}+\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_1}^{t_2}{(t_2-s)}^{q-1}\mid f(s\text{,}x(s))\mid \mathit{ds}\hfill \\ \hfill & +\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^{t_1}\left[{(t_2-\tau )}^{q-1}-{(t_1-\tau )}^{q-1}\right]\left(\mid x(\tau )\mid {\int }_{t_0}^{\tau }\mid K(\tau \text{,}s\text{,}x(s))\mid \mathit{ds}\right)d\tau \hfill \\ \hfill & +\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_1}^{t_2}{(t_2-\tau )}^{q-1}\left(\mid x(\tau )\mid {\int }_{t_0}^{\tau }\mid K(\tau \text{,}s\text{,}x(s))\mid \mathit{ds}\right)d\tau \hfill \\ \hfill & ⩽\frac{M}{\mathrm{\Gamma }(q)}{\int }_{t_0}^{t_1}\left[{(t_2-s)}^{q-1}-{(t_1-s)}^{q-1}\right]\mathit{ds}+\frac{M}{\mathrm{\Gamma }(q)}{\int }_{t_1}^{t_2}{(t_2-s)}^{q-1}\mathit{ds}\hfill \\ \hfill & +\frac{N\parallel x\parallel }{\mathrm{\Gamma }(q)}{\int }_{t_0}^{t_1}\left[{(t_2-\tau )}^{q-1}-{(t_1-\tau )}^{q-1}\right]d\tau +\frac{N\parallel x\parallel }{\mathrm{\Gamma }(q)}{\int }_{t_1}^{t_2}{(t_2-\tau )}^{q-1}d\tau \hfill \\ \hfill & ⩽\left(\frac{M+N\parallel x\parallel }{\mathrm{\Gamma }(q+1)}\right)\left[{(t_2-t_1)}^q-{(t_2-t_0)}^q+{(t_1-t_0)}^q+{(t_2-t_1)}^q\right]\hfill \\ \hfill & ⩽\left(\frac{M+N\parallel x\parallel }{\mathrm{\Gamma }(q+1)}\right)\left[2{(t_2-t_1)}^q\right]\text{,}\hfill \end{array}$$ as $t_1\to t_2$ the right hand side of inequality (3.4) tends to zero. Therefore the operator $T:{\mathrm{\Omega }}_0\to {\mathrm{\Omega }}_0$ is equicontinuous, and consequently the closure of $T({\mathrm{\Omega }}_0)$ is compact.

To show that T is a continuous map, let us take an $∊>0$ and $x\text{,}y$ in ${\mathrm{\Omega }}_0$ , it follows, using uniform continuity of f and K that for any $∊>0$ there exists $\delta >0$ such that $$\begin{array}{cc}\hfill \mid \mathit{Tx}(t)-\mathit{Ty}(t)\mid =& \left|\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-s)}^{q-1}\left(f(s\text{,}x(s))-f(s\text{,}y(s))\right)\mathit{ds}\right.\hfill \\ \hfill & +\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-\tau )}^{q-1}x(\tau )\left({\int }_{t_0}^{\tau }K(\tau \text{,}s\text{,}x(s))\mathit{ds}\right)d\tau \hfill \\ \hfill & -\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-\tau )}^{q-1}y(\tau )\left({\int }_{t_0}^{\tau }K(\tau \text{,}s\text{,}y(s))\mathit{ds}\right)d\tau \hfill \\ \hfill & +\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-\tau )}^{q-1}x(\tau )\left({\int }_{t_0}^{\tau }K(\tau \text{,}s\text{,}y(s))\mathit{ds}\right)d\tau \hfill \\ \hfill & \left.-\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-\tau )}^{q-1}x(\tau )\left({\int }_{t_0}^{\tau }K(\tau \text{,}s\text{,}y(s))\mathit{ds}\right)d\tau \right|\hfill \\ \hfill =& \left|\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-s)}^{q-1}\left(f(s\text{,}x(s))-f(s\text{,}y(s))\right)\mathit{ds}\right.\hfill \\ \hfill & +\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-\tau )}^{q-1}x(\tau )\left({\int }_{t_0}^{\tau }(K(\tau \text{,}s\text{,}x(s))-K(\tau \text{,}s\text{,}y(s)))\mathit{ds}\right)d\tau \hfill \\ \hfill & +\left.\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-\tau )}^{q-1}(x(\tau )-y(\tau ))\left({\int }_{t_0}^{\tau }K(\tau \text{,}s\text{,}y(s))\mathit{ds}\right)d\tau \right|\hfill \\ \hfill ⩽& \frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-s)}^{q-1}\left(\mid f(s\text{,}x(s))-f(s\text{,}y(s))\mid \right)\mathit{ds}\hfill \\ \hfill & +\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-\tau )}^{q-1}\mid x(\tau )\mid \left({\int }_{t_0}^{\tau }(\mid K(\tau \text{,}s\text{,}x(s))-K(\tau \text{,}s\text{,}y(s)))\mid \mathit{ds}\right)d\tau \hfill \\ \hfill & +\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-\tau )}^{q-1}\mid x(\tau )-y(\tau )\mid \left({\int }_{t_0}^{\tau }\mid K(\tau \text{,}s\text{,}x(s))\mid \mathit{ds}\right)d\tau \text{.}\hfill \end{array}$$ Since f and K is uniformly continuous for the above $∊>0$ , there exist $\delta >0$ such that $\mid x(t)-y(t)\mid <\delta$ , by using (ii)-(iv) we have $$\begin{array}{cc}\hfill ⩽& \frac{1}{\mathrm{\Gamma }(q)}\frac{∊\mathrm{\Gamma }(q+1)}{2{\alpha }^q}{\int }_{t_0}^t{(t-s)}^{q-1}\mathit{ds}+\frac{\parallel x\parallel }{\mathrm{\Gamma }(q)}\frac{∊\mathrm{\Gamma }(q+1)}{4\parallel x\parallel {\alpha }^{q+1}}{\int }_{t_0}^t{(t-\tau )}^{q-1}\left({\int }_{t_0}^{\tau }\mathit{ds}\right)d\tau \hfill \\ \hfill & +\frac{\delta N}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-\tau )}^{q-1}d\tau \hfill \\ \hfill ⩽& \frac{1}{2}∊+\frac{\parallel x\parallel }{q\mathrm{\Gamma }(q)}\frac{1}{4}\frac{∊\mathrm{\Gamma }(q+1)\alpha }{{\alpha }^{q+1}\parallel x\parallel }{(t-t_0)}^q+\frac{\delta N}{\mathrm{\Gamma }(q+1)}{(t-t_0)}^q⩽∊\text{.}\hfill \end{array}$$

Theorem 3.3

Theorem 3.3 Tychonoff (Zeidler, 1995)

Let B be a complete, locally convex, linear space and $B_0$ a closed convex subset of B. Let the mapping $T:B\to B$ be a continuous and $T(B_0)\subset B_0$ if the closure of $T(B_0)$ is compact then Thas a fixed point in $B_0$ .

Theorem 3.4

Assume that

• (i) $f\in C[{\mathbb{R}}_{+}\times {\mathbb{R}}^n\text{,}{\mathbb{R}}^n]\text{,}g\in C[{\mathbb{R}}_{+}^2\text{,}{\mathbb{R}}_{+}]\text{,}g(t\text{,}u)$ is monotone nondecreasing in u for each $T\in J$ and $$\mid f(t\text{,}x)\mid ⩽g(t\text{,}\mid x\mid )\text{,}(t\text{,}x)\in {\mathbb{R}}_{+}\times {\mathbb{R}}^n\text{;}$$

• (ii) $K\in C[{\mathbb{R}}_{+}^2\times {\mathbb{R}}^n\text{,}{\mathbb{R}}^n]\text{,}G\in C[{\mathbb{R}}_{+}^3\text{,}{\mathbb{R}}_{+}]\text{,}G(t\text{,}s\text{,}u)$ is monotone nondecreasing in u for each $(t\text{,}s)\in {\mathbb{R}}_{+}^2$ and $$\mid K(t\text{,}s\text{,}x)\mid ⩽G(t\text{,}s\text{,}\mid x\mid )\text{,}(t\text{,}s\text{,}x)\in {\mathbb{R}}_{+}^2\times {\mathbb{R}}^n\text{;}$$

• (iii) ${\int }_s^t\left|K(\sigma \text{,}s\text{,}x(s))\right|d\sigma ⩽N\text{,}\mathit{fort}\text{,}s\in {\mathbb{R}}_{+}\text{,}x\in C[{\mathbb{R}}_{+}\text{,}{\mathbb{R}}^n]$

Then the fractional quadratic integro differential equation

(3.5)

$$D^{q}u(t)=g(t\text{,}u(t))+u(t){\int }_{t_0}^{t}G(t\text{,}s\text{,}u(s))\mathit{ds}\text{,}u(t_0)=u_0$$ has a solution $u(t)$ existing for for every $u_0>0\text{,}t⩾t_0$ , also, then for every $x_0\in {\mathbb{R}}^n$ such that $\mid x_0\mid ⩽u_0$ , there exists a solution $x(t)$ of (3.1) for $t⩾t_0$ satisfying $\mid x(t)\mid ⩽u(t)\text{,}t⩾t_0$ .

Proof

Let us consider the real vector space B of all continuous functions from $[t_0\text{,}\infty ]$ into ${\mathbb{R}}^n$ , the topology on B being that induced by the family of pseudo-norms ${\left\{V_n(x)\right\}}_{n=1}^{\infty }$ where for $x\in B\text{,}{V}_n(x)={\sup }_{t_0⩽t⩽n}\mid x(t)\mid$ . A fundamental system of neighborhoods is the given by ${\left\{S_n\right\}}_{n=1}^{\infty }$ , where $S_n=\left\{x\in B:V_n(x)⩽1\right\}$ under this topology, B is a complete, locally convex linear space.

Now define a subset $B_0$ of B as follows: $$B_0=\left\{x\in B:\mid x(t)\mid ⩽u(t)\text{,}t⩾t_0\right\}\text{,}$$ where $u(t)$ is a solution of (3.5) existing for $t⩾t_0$ . It is clear that in the topology of $B\text{,}{B}_0$ is closed convex and bounded. Consider the integral operator defined by (3.3) whose fixed point corresponds to a solution of (3.1), evidently, the operator T is compact in the topology of B, and hence the closure of $T(B_0)$ is compact. In view of the boundedness of $B_0$ . Now to prove $T(B_0)\subset B_0$ , observe that for any $x\in B_0$ , $$\begin{array}{cc}\hfill \Vert \mathit{Tx}(t)\Vert ⩽& \Vert x_0\Vert +\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-s)}^{q-1}\parallel f(s\text{,}x(s))\parallel \mathit{ds}\hfill \\ \hfill & +\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-\tau )}^{q-1}\parallel x(\tau )\parallel \left({\int }_{t_0}^{\tau }\parallel K(\tau \text{,}s\text{,}x(s))\parallel \mathit{ds}\right)d\tau \hfill \end{array}$$ $$\begin{array}{cc}\hfill ⩽& \Vert x_0\Vert +\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-s)}^{q-1}g(s\text{,}\parallel x(s)\parallel )\mathit{ds}\hfill \\ \hfill & +\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-\tau )}^{q-1}\parallel x(\tau )\parallel \left({\int }_{t_0}^{\tau }G(\tau \text{,}s\text{,}\parallel x(s)\parallel )\mathit{ds}\right)d\tau \hfill \\ \hfill ⩽& u_0+\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-s)}^{q-1}g(s\text{,}u(s))\mathit{ds}\hfill \\ \hfill & +\frac{1}{\mathrm{\Gamma }(q)}{\int }_{t_0}^t{(t-\tau )}^{q-1}u(\tau )\left({\int }_{t_0}^{\tau }G(\tau \text{,}s\text{,}u(s))\mathit{ds}\right)d\tau \hfill \\ \hfill =& u(t)\text{.}\hfill \end{array}$$ Using the monotonicity of g and G, the definition of $B_0$ , and the fact $u(t)$ is a solution of (3.5), therefore $\parallel \mathit{Tx}(t)\parallel ⩽u(t)$ , which implies $T(B_0)\subset B_0$ .

Hence by Tychonoff’s fixed point theorem, T has a fixed point in $B_0$ , which completes the proof of the theorem.

Example 5.1

$$D^{\alpha }u(t)=\frac{1}{\mathrm{\Gamma }(\frac{1}{2})}\left(\frac{8}{3}{t}^3-2t^{\frac{1}{2}}\right)u(t)+\frac{t}{1260}+u(t){\int }_0^{t}t\sin (s)u(s)\mathit{ds}\text{,}$$ with the initial condition is $u(0)=0$ and $0<\alpha <1$ . First, we apply the Laplace transform to both sides of (5.1) $$\mathcal{L}\left[D^{\alpha }u(t)\right]=\mathcal{L}\left[\frac{1}{\mathrm{\Gamma }(\frac{1}{2})}\left(\frac{8}{3}{t}^3-2t^{\frac{1}{2}}\right)u(t)\right]+\mathcal{L}\left[\frac{t}{1260}\right]+\mathcal{L}\left[u(t){\int }_0^{t}t\sin (s)u(s)\mathit{ds}\right]\text{,}$$ using the property of Laplace transform and the initial condition we get $$s^{\alpha }\mathcal{L}\left[u(t)\right]=\mathcal{L}\left[\frac{1}{\mathrm{\Gamma }(\frac{1}{2})}\left(\frac{8}{3}{t}^3-2t^{\frac{1}{2}}\right)u(t)\right]+\mathcal{L}\left[\frac{t}{1260}\right]+\mathcal{L}\left[u(t){\int }_0^{t}t\sin (s)u(s)\mathit{ds}\right]\text{,}$$ and $$\mathcal{L}\left[u(t)\right]=\frac{1}{s^{\alpha }}\left\{\mathcal{L}\left[\frac{1}{\mathrm{\Gamma }(\frac{1}{2})}\left(\frac{8}{3}{t}^3-2t^{\frac{1}{2}}\right)u(t)\right]+\mathcal{L}\left[\frac{t}{1260}\right]+\mathcal{L}\left[u(t){\int }_0^{t}t\sin (s)u(s)\mathit{ds}\right]\right\}\text{,}$$ substituting (4.4), (4.5) and (4.6) into above equation, we have

(5.2)

$$\mathcal{L}\left[{\displaystyle \sum _{n=0}^{\infty }}{u}_n(t)\right]=\frac{1}{s^{\alpha }}\left\{\mathcal{L}\left[\frac{1}{\mathrm{\Gamma }(\frac{1}{2})}\left(\frac{8}{3}{t}^3-2t^{\frac{1}{2}}\right){\displaystyle \sum _{n=0}^{\infty }}{u}_n(t)\right]+\mathcal{L}\left[\frac{t}{1260}\right]\right.=1\left.+\mathcal{L}\left[{\displaystyle \sum _{n=0}^{\infty }}{u}_n(t){\int }_0^{t}t\sin (s){\displaystyle \sum _{n=0}^{\infty }}{u}_n(s)\mathit{ds}\right]\right\}\text{.}$$ Match both side of (5.2), we have the following relation $$\begin{array}{cc}\hfill & \mathcal{L}[u_0(t)]=\frac{1}{s^{\alpha }}\mathcal{L}\left[\frac{t}{1260}\right]\text{,}\hfill \\ \hfill & \mathcal{L}[u_{n+1}(t)]=\frac{1}{s^{\alpha }}\left\{\mathcal{L}\left[\frac{1}{\mathrm{\Gamma }(\frac{1}{2})}\left(\frac{8}{3}{t}^3-2t^{\frac{1}{2}}\right)u_n(t)\right]+\mathcal{L}\left[u_n(t){\int }_0^{t}t\sin (s)u_n(s)\mathit{ds}\right]\right\}\text{.}\hfill \end{array}$$ Applying inverse Laplace transform to above equation we get $$\begin{array}{cc}\hfill & u_0(t)={\mathcal{L}}^{-1}\left\{\frac{1}{s^{\alpha }}\mathcal{L}\left[\frac{t}{1260}\right]\right\}\text{,}\hfill \\ \hfill & u_1(t)={\mathcal{L}}^{-1}\left\{\frac{1}{s^{\alpha }}\left\{\mathcal{L}\left[\frac{1}{\mathrm{\Gamma }(\frac{1}{2})}\left(\frac{8}{3}{t}^3-2t^{\frac{1}{2}}\right)u_n(t)\right]+\mathcal{L}\left[u_n(t){\int }_0^{t}t\sin (s)u_n(s)\mathit{ds}\right]\right\}\right\}\text{.}\hfill \end{array}$$ Therefor, the solution is obtained to be

(5.3)

$$u(t)=\frac{1}{\mathrm{\Gamma }(\alpha +5)}\frac{1}{1260}{t}^{\alpha +1}(2+\alpha )(3+\alpha )(4+\alpha )+\dots \text{,}$$ According to ADM, the recursive ADM is $$\begin{array}{cc}\hfill & u_0(t)=J^{\alpha }\left(\frac{t}{1260}\right)\text{,}\hfill \\ \hfill & u_{n+1}(t)=J^{\alpha }\left(\frac{1}{\mathrm{\Gamma }(\frac{1}{2})}\left(\frac{8}{3}{t}^3-2t^{\frac{1}{2}}\right)u_n(t)\right)+J^{\alpha }\left(u_n(t){\int }_{t_0}^{t}t\sin (s)u_n(s)\mathit{ds}\right)\text{.}\hfill \end{array}$$ Therefor, the solution is obtained to be

(5.4)

$$u(t)=\frac{1}{3175200}\left(1680t^{2\alpha +4}\mathrm{\Gamma }\left(2\alpha +\frac{5}{2}\right)\sqrt{\pi }\mathrm{\Gamma }(3\alpha +6){\alpha }^7{4}^{-\alpha }\right)+\dots \text{,}$$

Tables 1, 2 presents the approximate solution for different values of $\alpha$ , we have noticed that the accuracy is improving by computing more terms of the approximate solutions (see Fig. 1).