In this section, we briefly recall some basic known results which will be used in the sequel. Throughout this work, we set \(J = [0, b]\), where \(b >0\) is a constant. Let E be a Banach space with the norm \(\Vert \cdot \Vert \) and the pair \((A,B)\) generate a propagation family \(\{W(t)\}_{t\geq0}\) (see Definition 2.6). We denote by \(B(E)\) the Banach space of all bounded linear operators from E to E. For a closed and linear operator \(W:D(W)\subset E\rightarrow E\), where \(D(W)\) is the domain of T, we denote by \(\rho (W)\) its resolvent set. We also denote by \(C(J,E)\) the Banach space of all continuous E-value functions on the interval J with the norm \(\Vert u \Vert =\sup_{t\in J} \Vert u(t) \Vert \). Let
$$C_{1-\gamma}(J,E)= \bigl\{ u:J\rightarrow E|t^{1-\gamma}u(t)\in C(J,E) \bigr\} $$
with the norm \(\Vert \cdot \Vert _{C_{1-\gamma}}\) defined by
$$\Vert u \Vert _{C_{1-\gamma}}=\sup_{0\leq t\leq b} \bigl\vert t^{1-\gamma}u(t) \bigr\vert . $$
Evidently, \(C_{1-\gamma}(J,E)\) is a Banach space. Meanwhile, \(C_{1-\gamma}^{\gamma}(J,E)= \{f\in C_{1-\gamma }(J,E),^{c}D^{\alpha}_{0^{+}}f(t)\in C_{1-\gamma}(J,E) \}\).
For completeness we recall the following definitions from fractional calculus.
Definition 2.1
The Riemann–Liouville fractional integral of order α of a function \(f:[0,\infty)\rightarrow R\) is defined as
$$I_{0+}^{\alpha}f(t)=\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha -1}f(s)\,ds,\quad t>0, \alpha>0, $$
provided the right side is point-wise defined on \((0,\infty)\).
Definition 2.2
The Riemann–Liouville derivative of order α with the lower limit zero for a function \(f:[0,\infty)\rightarrow R\) can be written as
$$D^{\alpha}_{0^{+}}f(t)=\frac{1}{\Gamma(n-\alpha)}\frac {d^{n}}{dt^{n}} \int_{0}^{t}\frac{f(s)}{(t-s)^{\alpha +1-n}}\,ds,\quad t>0,n-1< \alpha< n. $$
Definition 2.3
The Caputo fractional derivative of order α for a function \(f:[0,\infty)\rightarrow R\) can be written as
$$^{c}D^{\alpha}_{0^{+}}f(t)=D^{\alpha}_{0^{+}} \Biggl[f(t)-\sum_{k=0}^{n-1} \frac{t^{k}}{k!}f^{(k)}(0) \Biggr],\quad t>0, n-1< \alpha< n, $$
where \(n = [\alpha] + 1\) and \([\alpha]\) denotes the integer part of α.
If u is an abstract function with values in E, then the integrals which appear in Definitions 2.2 and 2.3 are taken in Bochner’s sense.
Definition 2.4
(Hilfer fractional derivative; see [9]). The generalized Riemann–Liouville fractional derivative of order \(0\leq\nu\leq1\) and \(0<\mu<1\) with lower limit a is defined as
$$D^{\nu,\mu}_{a+}f(t)=I^{\nu(1-\mu)}_{a+}\, \frac{d}{dt}I^{(1-\nu )(1-\mu)}_{a+}f(t) $$
for functions such that the expression on the right hand side exists.
Recently (Hilfer et al. [32]), this definition for \(n-1 <\mu\leq n, n \in N, 0 \leq\nu\leq1\), was rewritten in a more general form:
$$D^{\nu,\mu}_{a+}f(t)=I^{\nu(n-\mu)}_{a+}\, \frac{d^{n}}{dt^{n}}I^{(1-\nu )(n-\mu)}_{a+}f(t)=I^{\nu(n-\mu)}_{a+}D_{a+}^{\mu+\nu n-\mu\nu}f(t), $$
where \(D_{a+}^{\mu+\nu n-\mu\nu}\) is the Riemann–Liouville fractional derivative and \(I_{a+}^{\nu(n-\mu)}\) is the Riemann–Liouville integral.
Remark 2.1
(i) When \(\nu=0,0<\mu<1\) and \(a=0\), the Hilfer fractional derivative corresponds to the classical Riemann–Liouville fractional derivative:
$$D_{0+}^{0,\mu}f(t)=\frac{d}{dt}I^{1-\mu}_{0+}f(t)=D^{\mu}_{0+}f(t). $$
(ii) When \(\nu=1, 0<\mu<1\) and \(a=0\), the Hilfer fractional derivative corresponds to the classical Caputo fractional derivative:
$$D^{1,\mu}_{0+}f(t)=I^{1-\mu}_{0+}\, \frac{d}{dt}f(t)=^{c}D^{\mu}_{0+}f(t). $$
Now, we recall the basic definitions and properties of the Kuratowski measure of noncompactness that will be used later.
Definition 2.5
([40])
Let E be a Banach space and \(\Omega_{E}\) be the bounded subsets of E. The Kuratowski measure of noncompactness is the map \(\alpha:\Omega_{E}\rightarrow[0,\infty)\) defined by (here \(B\in \Omega_{E}\))
$$\alpha(B)=\inf \Biggl\{ \varepsilon>0: B=\bigcup_{i=1}^{n} B_{i}\mbox{ and }\operatorname{diam}(B_{i})\leq\varepsilon \mbox{ for }i=1,\ldots,n \Biggr\} , $$
here \(\operatorname{diam}B_{i}=\sup\{ \vert x-y \vert :x,y\in B_{i}\}\).
Lemma 2.1
([41])
Let
S
and
T
be bounded sets of
E
and a be a real number. Then the noncompactness measure has the following properties:
-
(1)
\(\alpha(S)=0\)
if and only if S is a relatively compact set.
-
(2)
\(S\subset T\)
implies that
\(\alpha(S)\leq\alpha(T)\).
-
(3)
\(\alpha(\overline{S}) = \alpha(S)\).
-
(4)
\(\alpha(S\bigcup T) =\max\{\alpha(S),\alpha(T)\}\).
-
(5)
\(\alpha(aS) = |a|\alpha(S)\).
-
(6)
\(\alpha(S + T) \leq\alpha(S) +\alpha(T)\).
-
(7)
\(\alpha(\overline{co}S) = \alpha(S)\), where
\(\overline{co}S\)
is the convex closure of
S.
-
(8)
\(|\alpha(S)-\alpha(T)|\leq2d_{h}(S,T)\), where
\(d_{h}(S,T)\)
denotes the Hausdorff distance between the sets
S
and
T, that is,
$$d_{h}(S,T)=\max\Bigl\{ \sup_{x\in S}d(x,T),\sup _{x\in T}d(x,S)\Bigr\} , $$
where
\(d(\cdot,\cdot)\)
denotes the distance from an element of
E
to a set of
E.
Lemma 2.2
([42])
Let
E
be a Banach space, and let
\(D\subset E\)
be bounded. Then there exists a countable set
\(D_{0} \subset D\), such that
\(\alpha(D) \leq2\alpha(D_{0})\).
Lemma 2.3
([43])
Let
E
be a Banach space, and let
\(D\subset C(I,E)\)
be equicontinuous and bounded, then
\(\alpha(D(t))\)
is continuous on
I, and
\(\alpha(D)=\max_{t\in I} \alpha(D(t))\).
Lemma 2.4
([44])
Let E be a Banach space, and let
\(D = \{u_{n}\} \subset C(I, E)\)
be a bounded and countable set. Then
\(\alpha(D(t))\)
is the Lebesgue integral on
e, and
$$\alpha \biggl( \biggl\{ \int_{I}u_{n}(t)\,dt|n\in\mathbb{N} \biggr\} \biggr)\leq2 \int_{I}\alpha\bigl(D(t)\bigr)\,dt. $$
Lemma 2.5
([45])
Let
E
be a Banach space. Assume that
\(D \subset E\)
is a bounded closed and convex set on
E, \(Q : D \rightarrow D\)
is condensing. Then
Q
has at least one fixed point in
D.
We recall the abstract degenerate Cauchy problem as follows [35]:
$$ \textstyle\begin{cases} \frac{d}{dt}Bu(t)=Au(t),&t\in J,\\ Bu(0)=Bu_{0}. \end{cases} $$
(2.1)
Definition 2.6
(See [19, Definition 1.4])
A strongly continuous operator family \(\{W(t)\}_{t\geq0}\) of \(D(B)\) to a Banach space E, satisfying the requirement that \(\{W(t)\}_{t\geq0}\) is exponentially bounded, which means that for any \(u\in D(B)\) there exist \(a>0\), \(M>0\) such that
$$\bigl\Vert W(t)u \bigr\Vert \leq Me^{at} \Vert u \Vert ,\quad t\geq0, $$
is called an exponentially bounded propagation family for (2.1) if for \(\lambda>a\),
$$ (\lambda B-A)^{-1}Bu= \int_{0}^{\infty}e^{-\lambda t}W(t)u\,dt,\quad u\in D(B). $$
In this case, we also say that (2.1) has an exponentially bounded propagation family \(\{W(t)\}_{t\geq0}\).
Motivated by the above definition, we can give the following definition.
Definition 2.7
Let \(A:D(A)\subseteq E\rightarrow E\), \(B:D(B)\subseteq E\rightarrow E\) be closed linear operators defined on a Banach space E satisfying \(D(A)\cap D(B)\neq\{0\}\). Let \(\mu>0\). We say the pair \((A,B)\) is the generator of an μ-resolvent family, if there exist \(a\geq0\) and a strongly continuous function \(K_{\mu}:[0,\infty)\rightarrow B(E)\) such that \(K_{\mu}(t)\) is exponentially bounded, \(\{\lambda^{\mu}: \operatorname{Re}\lambda > a\}\subset\rho(A)\), and for all \(x\in E\),
$$ \bigl(\lambda^{\mu}B-A\bigr)^{-1}Bu= \int_{0}^{\infty}e^{-\lambda t}K_{\mu}(t)u \,dt,\quad \operatorname{Re}\lambda>a. $$
(2.2)
In this case, \(\{K_{\mu}(t)\}_{t\geq0}\) is called the μ-resolvent family generated by the pair \((A,B)\).
Lemma 2.6
([34])
Let
\(f:J\times E\times E\rightarrow E\)
be a function such that
\(f\in C_{1-\gamma}(J)\)
for any
\(u\in C_{1-\gamma}(J)\). A function
\(u\in C_{1-\gamma}(J)\)
is a solution of the fractional initial value problem
$$ \textstyle\begin{cases} D^{\nu,\mu}_{0+}(Bu(t))=Au(t)+Bf (t,u(t),\int_{0}^{t}\rho (t,s)h(t,s,u(s))\,ds ),& t\in J,\\ I^{1-\gamma}_{0+}Bu(0)=Bu_{0}, \end{cases} $$
if
u
satisfies
$$\begin{aligned} u(t)=S_{\nu,\mu}(t)u_{0}+ \int_{0}^{t}K_{\mu}(t-s)f \biggl(s,u(s), \int _{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds, \end{aligned}$$
(2.3)
where
$$S_{\nu,\mu}(t)=I_{0+}^{\nu(1-\mu)}K_{\mu}(t),\qquad K_{\mu}(t)= t^{\mu-1}P_{\mu}(t),\qquad P_{\mu}(t)=\mu \int_{0}^{\infty}\sigma\xi_{\mu}(\sigma )W \bigl(t^{\mu}\sigma\bigr)\,d\sigma, $$
the function
\(\xi_{\mu}\)
is a probability density function defined on
\((0,\infty)\)
such that
$$\xi_{\mu}(\sigma)=\frac{1}{\mu}\sigma^{-1-\frac{1}{\mu}} \varpi_{\mu}\bigl(\sigma^{-\frac{1}{\mu}}\bigr)\geq0 $$
and the one sided stable probability density in [24] is as follows:
$$\varpi_{\mu}(\sigma)=\frac{1}{\pi}\sum _{n=1}^{\infty}(-1)^{n-1}\sigma^{-\mu n-1} \frac{\Gamma(n\mu+1)}{n!}\sin(n\pi\mu),\quad \sigma\in(0,\infty). $$
Lemma 2.7
Suppose
A
and
B
are closed (unbounded) linear operator with domains contained in
E, \(0\leq\nu\leq1\), \(0<\mu<1\), then
$$D^{\nu,\mu}_{0+}\bigl[BS_{\nu,\mu}(t)u_{0} \bigr]=A\bigl[S_{\nu,\mu}(t)u_{0}\bigr] $$
and
$$\begin{aligned} &D^{\nu,\mu}_{0+} \biggl( \int_{0}^{t}K_{\mu}(t-s)Bf \biggl(s,u(s), \int _{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds \biggr) \\ &\quad =A \int_{0}^{t}K_{\mu}(t-s)f \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau ,u(\tau)\bigr) \,d\tau \biggr)\,ds\\ &\qquad {}+Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau ,u(\tau)\bigr) \,d\tau \biggr). \end{aligned}$$
Proof
By Definition 2.7, let \(\lambda>a\) be fixed, then we have
$$ \bigl(\lambda^{\mu}B-A\bigr)^{-1}Bu= \int_{0}^{\infty}e^{-\lambda t}K_{\mu}(t)u \,dt,\quad u\in D(B). $$
(2.4)
It follows from (2.4) and the Laplace transform; it is obvious that
$$\begin{aligned} \mathcal{L}\bigl(BS_{\nu,\mu}(t)u_{0}\bigr)&= \mathcal{L}\bigl(I_{0+}^{\nu(1-\mu )}BK_{\mu}(t)u_{0} \bigr) \\ &=\lambda^{\nu(\mu-1)}B\bigl(\lambda^{\mu}B-A\bigr)^{-1}Bu_{0}. \end{aligned}$$
(2.5)
The difference between fractional derivatives of different types becomes apparent from the Laplace transformation. In [33] it is found for \(0<\mu<1\) that
$$\mathcal{L}\bigl[D^{\mu,\nu}_{0+}f(x)\bigr](\lambda)= \lambda^{\mu}\mathcal {L}\bigl[f(x)\bigr](\lambda)-\lambda^{\nu(\mu-1)} \bigl(I_{0+}^{(1-\nu)(1-\mu)}f\bigr) (0+), $$
where \((I_{0+}^{(1-\nu)(1-\mu)}f)(0+)\) is the Riemann–Liouville fractional integral of order \((1-\nu)(1-\mu)\) evaluated in the limits as \(t\rightarrow0+\), it being understood that
$$\mathcal{L}\bigl[f(x)\bigr](\lambda)= \int_{0}^{\infty}e^{-\lambda x}f(x)\,dx. $$
Therefore, we obtain
$$\begin{aligned} \mathcal{L}\bigl( D^{\nu,\mu}_{0+} \bigl[BS_{\nu,\mu}(t)u_{0}\bigr]\bigr)&=\lambda^{\mu}\mathcal{L}\bigl(BS_{\nu,\mu}(t)u_{0}\bigr)-\lambda^{\nu(\mu-1)}Bu_{0} \\ &=\lambda^{\mu}B \bigl[\lambda^{\nu(\mu-1)}\bigl( \lambda^{\mu}B-A\bigr)^{-1}B \bigr]u_{0}- \lambda^{\nu(\mu-1)}Bu_{0} \\ &=\lambda^{\nu(\mu-1)}\bigl(\lambda^{\mu}B-A\bigr)^{-1}B \bigl[\lambda^{\mu}B-\bigl(\lambda^{\mu}B-A\bigr) \bigr]u_{0} \\ &=\lambda^{\nu(\mu-1)}\bigl(\lambda^{\mu}B-A\bigr)^{-1}B \bigl[\lambda^{\mu}B-\lambda^{\mu}B+A \bigr]u_{0} \\ &=\lambda^{\nu(\mu-1)}\bigl(\lambda^{\mu}B-A\bigr)^{-1}BAu_{0} \\ &=A\lambda^{\nu(\mu-1)}\bigl(\lambda^{\mu}B-A\bigr)^{-1}Bu_{0}. \end{aligned}$$
(2.6)
Combining (2.4) and (2.6) yields
$$D^{\nu,\mu}_{0+}\bigl[BS_{\nu,\mu}(t)u_{0} \bigr]=A\bigl[S_{\nu,\mu}(t)u_{0}\bigr]. $$
Similarly, we have
$$\begin{aligned} &\mathcal{L} \biggl( \int_{0}^{t}K_{\mu}(t-s)Bf \biggl(s,u(s), \int_{0}^{s}\rho (s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds \biggr) \\ &\quad=\mathcal{L}\bigl(K_{\mu}(t)\bigr)\cdot\mathcal{L} \biggl(Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr) \biggr) \end{aligned}$$
(2.7)
and
$$\begin{aligned} &\mathcal{L} \biggl(D^{\nu,\mu}_{0+} \biggl[ \int_{0}^{t}K_{\mu}(t-s)Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds \biggr] \biggr) \\ &\quad=\lambda^{\mu}\mathcal{L} \biggl( \int_{0}^{t}K_{\mu}(t-s)Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds \biggr)-\lambda^{\nu(\mu-1)}\cdot0 \\ &\quad=\lambda^{\mu}\mathcal{L}\bigl(K_{\mu}(t)\bigr)\cdot \mathcal{L} \biggl(Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr) \biggr) \\ &\quad=\lambda^{\mu}\bigl(\lambda^{\mu}B-A \bigr)^{-1}B\cdot\mathcal{L} \biggl(Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr) \biggr) \\ &\quad=\bigl(\lambda^{\mu}B-A+A\bigr) \bigl(\lambda^{\mu}B-A \bigr)^{-1}B\cdot\mathcal {L} \biggl(f \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr) \biggr) \\ &\quad=A\bigl(\lambda^{\mu}B-A\bigr)^{-1}B\cdot\mathcal{L} \biggl(f \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr) \biggr) \\ &\qquad{}+B\cdot\mathcal{L} \biggl(f \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau )h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr) \biggr). \end{aligned}$$
(2.8)
Thus, it follows from (2.7) and (2.8) that
$$\begin{aligned} &D^{\nu,\mu}_{0+} \biggl[ \int_{0}^{t}K_{\mu}(t-s)Bf \biggl(s,u(s), \int _{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds \biggr] \\ &\quad =A \int_{0}^{t}K_{\mu}(t-s)f \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau ,u(\tau)\bigr) \,d\tau \biggr)\,ds \\ &\qquad {}+Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr). \end{aligned}$$
According to the above fundamental result, a new and important equivalent mixed type integral equation for the problem can be established. □
Lemma 2.8
Let
f
be a functions such that
\(f(\cdot,v,w)\in C_{1-\gamma}(J)\)
for any
\(u\in C_{1-\gamma}(J)\). A function
\(u\in C_{1-\gamma}^{\gamma }(J)\)
is a solution of Eq. (1.1) if and only if
u
satisfies the mixed type integral equation
$$\begin{aligned} u(t)={}&S_{\nu,\mu}(t)d\sum_{i=1}^{m} \lambda_{i}I_{0+}^{\gamma} \int _{0}^{\tau_{i}}K_{\mu}(\tau_{i}-s)f \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau )h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds \\ &{}+ \int_{0}^{t}K_{\mu}(t-s)f \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau ,u(\tau)\bigr) \,d\tau \biggr)\,ds, \end{aligned}$$
(2.9)
where
$$d=\frac{1}{1-\sum_{i=1}^{m}\lambda_{i}I_{0+}^{\gamma}S_{\nu,\mu}(\tau_{i})}. $$
Proof
According to Lemma 2.6, a solution of Eq. (1.1) can be expressed by
$$ u(t)=S_{\nu,\mu}(t) I^{1-\gamma}_{0+}u(0)+ \int_{0}^{t}K_{\mu}(t-s)f \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds. $$
(2.10)
Next, we substitute \(t=\tau_{i}\) into the above equation, we have
$$\begin{aligned} \lambda_{i}u(\tau_{i})={}&\lambda_{i}S_{\nu,\mu}( \tau_{i})u_{0} \\ &{}+\lambda_{i} \int _{0}^{\tau_{i}}K_{\mu}(\tau_{i}-s)f \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau )h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds. \end{aligned}$$
(2.11)
Thus, we have
$$\begin{aligned} I^{1-\gamma}_{0+}Bu(0)={}&\sum_{i=1}^{m} \lambda_{i}I^{\gamma}_{0+}Bu(\tau _{i})\\ ={}&\sum _{i=1}^{m}I^{\gamma}_{0+} \lambda_{i}Bu(\tau_{i}) \\ ={}&\sum_{i=1}^{m}\lambda_{i}I^{\gamma}_{0+}S_{\nu,\mu}( \tau _{i})Bu_{0}\\ &{}+\sum_{i=1}^{m} \lambda_{i}I^{\gamma}_{0+} \int_{0}^{\tau_{i}}K_{\mu}(\tau_{i}-s)Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau )\bigr) \,d\tau \biggr)\,ds \\ ={}&I^{1-\gamma}_{0+}Bu(0)\sum_{i=1}^{m} \lambda_{i}I^{\gamma}_{0+}S_{\nu ,\mu}( \tau_{i})\\ &{}+\sum_{i=1}^{m} \lambda_{i}I^{\gamma}_{0+} \int_{0}^{\tau _{i}}K_{\mu}(\tau_{i}-s)Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau ,u(\tau)\bigr) \,d\tau \biggr)\,ds, \end{aligned}$$
which implies
$$ I^{1-\gamma}_{0+}Bu(0)=d\sum _{i=1}^{m}\lambda_{i}I^{\gamma}_{0+} \int _{0}^{\tau_{i}}K_{\mu}(\tau_{i}-s)Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau )h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds, $$
(2.12)
i.e.
$$I^{1-\gamma}_{0+}u(0)=u_{0}=d\sum _{i=1}^{m}\lambda_{i}I^{\gamma}_{0+} \int _{0}^{\tau_{i}}K_{\mu}(\tau_{i}-s)f \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau )h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds. $$
Submitting (2.12) to (2.10), we derive that (2.9). It is probative that u is also a solution of the integral of Eq. (2.2) when u is a solution of Eq. (1.1).
The necessity has been already proved, next, we are read to prove its sufficiency. Applying \(I^{1-\gamma}_{0+}\) to both sides of (2.9), we have
$$\begin{aligned} &I^{1-\gamma}_{0+}Bu(t)\\ &\quad =I^{1-\gamma}_{0+}S_{\nu,\mu}(t)d \sum_{i=1}^{m}\lambda_{i}I_{0+}^{\gamma} \int_{0}^{\tau_{i}}K_{\mu}(\tau _{i}-s)Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds \\ &\qquad {}+I^{1-\gamma}_{0+} \int_{0}^{t}K_{\mu}(t-s)Bf \biggl(s,u(s), \int_{0}^{s}\rho (s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds. \end{aligned}$$
Substituting \(t=\tau_{i}\) into (2.9), we have
$$\begin{aligned} u(\tau_{i})={}&S_{\nu,\mu}(\tau_{i})d\sum _{i=1}^{m}\lambda _{i}I_{0+}^{\gamma} \int_{0}^{\tau_{i}}K_{\mu}(\tau_{i}-s)f \biggl(s,u(s), \int _{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds \\ &{}+ \int_{0}^{\tau_{i}}K_{\mu}(\tau_{i}-s)f \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau )h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds. \end{aligned}$$
Then we derive
$$\begin{aligned} &\sum_{i=1}^{m} \lambda_{i}I^{\gamma}_{0+}Bu(\tau_{i}) \\ &\quad = \sum_{i=1}^{m}\lambda_{i}I^{\gamma}_{0+}S_{\nu,\mu}( \tau_{i})d\sum_{i=1}^{m} \lambda_{i}I_{0+}^{\gamma} \int_{0}^{\tau_{i}}K_{\mu}(\tau _{i}-s)Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds \\ &\qquad {}+\sum_{i=1}^{m}\lambda_{i}I^{\gamma}_{0+} \int_{0}^{\tau_{i}}K_{\mu}(\tau _{i}-s)Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds \\ &\quad =\frac{\sum_{i=1}^{m}\lambda_{i}I^{\gamma}_{0+}S_{\nu,\mu}(\tau _{i})}{1-\sum_{i=1}^{m}\lambda_{i}I^{\gamma}_{0+}S_{\nu,\mu}(\tau _{i})} \\ &\qquad {}\times\sum_{i=1}^{m} \lambda_{i}I_{0+}^{\gamma} \int_{0}^{\tau_{i}}K_{\mu}(\tau _{i}-s)Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds \\ &\qquad {}+\sum_{i=1}^{m}\lambda_{i}I_{0+}^{\gamma} \int_{0}^{\tau_{i}}K_{\mu}(\tau _{i}-s)Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds \\ &\quad =(d-1)\sum_{i=1}^{m} \lambda_{i}I_{0+}^{\gamma} \int_{0}^{\tau_{i}}K_{\mu}(\tau_{i}-s)Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau )\bigr) \,d\tau \biggr)\,ds \\ &\qquad {}+\sum_{i=1}^{m}\lambda_{i}I_{0+}^{\gamma} \int_{0}^{\tau_{i}}K_{\mu}(\tau _{i}-s)Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds \\ &\quad =d\sum_{i=1}^{m}\lambda_{i}I_{0+}^{\gamma} \int_{0}^{\tau_{i}}K_{\mu}(\tau_{i}-s)Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau )\bigr) \,d\tau \biggr)\,ds. \end{aligned}$$
(2.13)
It follows (2.12) and (2.13) that
$$I^{1-\gamma}_{0+}Bu(0)=\sum_{i=1}^{m} \lambda_{i}I_{0+}^{1-\gamma }Bu(\tau_{i}). $$
Next, by applying \(D^{\nu,\mu}_{0+}\) to both sides of (2.9) and using Lemma 2.7, we can obtain
$$\begin{aligned} D^{\nu,\mu}_{0+}(Bu)={}&D^{\nu,\mu}_{0+} \Biggl[BS_{\nu,\mu}(t)d\sum_{i=1}^{m} \lambda_{i}I_{0+}^{\gamma}\\ &{}\times \int_{0}^{\tau_{i}}K_{\mu}(\tau _{i}-s)f \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds \\ &{}+ \int_{0}^{t}K_{\mu}(t-s)Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau ,u(\tau)\bigr) \,d\tau \biggr)\,ds \Biggr] \\ ={}&D^{\nu,\mu}_{0+} \Biggl[BS_{\nu,\mu}(t)d\sum _{i=1}^{m}\lambda _{i}I_{0+}^{\gamma}\\ &{}\times \int_{0}^{\tau_{i}}K_{\mu}(\tau_{i}-s)f \biggl(s,u(s), \int _{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds \Biggr] \\ &{}+D^{\nu,\mu}_{0+} \biggl[ \int_{0}^{t}K_{\mu}(t-s)Bf \biggl(s,u(s), \int _{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds \biggr] \\ ={}& \Biggl[d\sum_{i=1}^{m} \lambda_{i}I_{0+}^{\gamma} \int_{0}^{\tau_{i}}K_{\mu}(\tau_{i}-s)f \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau )\bigr) \,d\tau \biggr)\,ds \Biggr]\\ &{}\times D^{\nu,\mu}_{0+} \bigl[BS_{\nu,\mu}(t) \bigr] \\ &{}+D^{\nu,\mu}_{0+} \biggl[ \int_{0}^{t}K_{\mu}(t-s)Bf \biggl(s,u(s), \int _{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds \biggr] \\ ={}& \Biggl[d\sum_{i=1}^{m} \lambda_{i}I_{0+}^{\gamma} \int_{0}^{\tau_{i}}K_{\mu}(\tau_{i}-s)f \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau )\bigr) \,d\tau \biggr)\,ds \Biggr]AS_{\nu,\mu}(t) \\ &{}+A \int_{0}^{t}K_{\mu}(t-s)f \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau ,u(\tau)\bigr) \,d\tau \biggr)\,ds\\ &{}+Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau ,u(\tau)\bigr) \,d\tau \biggr) \\ ={}&A \Biggl(S_{\nu,\mu}(t)d\sum_{i=1}^{m} \lambda_{i}I_{0+}^{\gamma} \int _{0}^{\tau_{i}}K_{\mu}(\tau_{i}-s)f \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau )h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr)\,ds \\ &{}+ \int_{0}^{t}K_{\mu}(t-s)f \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau ,u(\tau)\bigr) \,d\tau \biggr)\,ds \Biggr)\\ &{}+Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau )h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr) \\ ={}&Au(t)+Bf \biggl(s,u(s), \int_{0}^{s}\rho(s,\tau)h\bigl(s,\tau,u(\tau)\bigr) \,d\tau \biggr). \end{aligned}$$
Hence, it reduces to \(D^{\nu,\mu}_{0+}(Bu)=Au(t)+Bf (s,u(s),\int _{0}^{s}\rho(s,\tau)h(s,\tau,u(\tau))\,d\tau )\). The results are proved completely. □
Lemma 2.9
(See [46, Property 2.1])
If
α> and
\(\beta>0\), then
$$\bigl[I^{\alpha}_{a+}(t-a)^{\beta-1} \bigr](x)= \frac{\Gamma(\beta )}{\Gamma(\alpha+\beta)}(x-a)^{\alpha+\beta-1}. $$
To end this section, we give the following lemma.
Lemma 2.10
Let
\(\mu>0,k>0,z\in R\)
and
\(f\in C(J)\), then
$$I^{k}_{0+} \int_{0}^{z}(z-t)^{\mu-1}P_{\mu}(t-s)f(s) \,ds=\frac{B(\mu ,k)}{\Gamma(k)} \int_{0}^{z}(z-t)^{\mu+k-1}P_{\mu}(t-s)f(s) \,ds. $$
Proof
According to Lemma 2.9, we have
$$\begin{aligned} &I^{k}_{0+} \int_{0}^{z}(z-t)^{\mu-1}P_{\mu}(t-s)f(s) \,ds \\ &\quad =\frac{1}{\Gamma(k)} \int_{0}^{z}(z-u)^{k-1} \int_{0}^{u}(u-t)^{\mu-1}P_{\mu}(t-s)f(s)\,ds\,du \\ &\quad =\frac{1}{\Gamma(k)} \int_{0}^{z} \int_{t}^{z}(z-u)^{k-1}(u-t)^{\mu-1}P_{\mu}(t-s)f(s)\,ds\,du \\ &\quad =\frac{1}{\Gamma(k)} \int_{0}^{z}P_{\mu}(t-s)f(s)\,ds \int _{t}^{z}(z-u)^{k-1}(u-t)^{\mu-1}du \\ &\quad =\frac{1}{\Gamma(k)} \int_{0}^{z}B(\mu,k) (z-t)^{\mu+k-1}P_{\mu}(t-s)f(s)\,ds \\ &\quad =\frac{B(\mu,k)}{\Gamma(k)} \int_{0}^{z}(z-t)^{\mu+k-1}P_{\mu}(t-s)f(s) \,ds. \end{aligned}$$
The desired result is obtained. □
Lemma 2.11
([34])
Assume that
\(\{W(t)\}_{t\geq0}\)
is a norm continuous family for
\(t>0\)
and
\(\Vert W(t) \Vert \leq M\), for any fixed
\(t>0\), \(\{K_{\mu}(t)\} _{t>0}\), and
\(\{S_{\nu,\mu}(t)\}_{t>0}\)
are linear operators, and for any
\(u\in E\)
$$\bigl\Vert K_{\mu}(t) \bigr\Vert \leq\frac{Mt^{\mu-1}}{\Gamma(\mu)},\qquad \bigl\Vert S_{\nu,\mu }(t) \bigr\Vert \leq\frac{Mt^{(\nu-1)(1-\mu)}}{\Gamma(\nu(1-\mu)+\mu)}. $$
Lemma 2.12
([34])
Assume that
\(\{W(t)\}_{t\geq0}\)
is a norm continuous family for
\(t>0\)
and
\(\Vert W(t) \Vert \leq M\), \(\{K_{\mu}(t)\}_{t>0}\)
and
\(\{S_{\nu ,\mu}(t)\}_{t>0}\)
are strongly continuous for
\(t>0\).