Fractional calculus I have this exercise :
"Consider the Cauchy problem's :
$$ 
^C D^{\alpha}y(t)=f(t,y(t),y'(t)), t\in [0,T] ....(1)
$$
$$
y(0)=y_0, y'(0)=y_1 ....                              (2)
$$
Where $f:[0,T]\times\mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R} $is a function , $y_0,y_1\in \mathbb{R}$ and $\alpha \in ]1,2]$
give a result of existence and uniqueness for the problem by using the contraction principle of Banach "
Can someone help me to solve this exercise ?
Thank you.
 A: We can assume $1<\alpha<2$ since the case $\alpha=2$ is the usual ODE case. So the Cauchy problem is
$$
\gamma\int_0^t(t-s)^{1-\alpha}y''(s) \, \mathrm{d}s = f(t,y(t),y'(t)),
\qquad\qquad(*)
$$
with $y(0)=y_0$ and $y'(0)=y_1$, where $\gamma=1/\Gamma(2-\alpha)$. Note that the integral in the left hand side is well defined if for instance we assume that $y\in C^2([0,t])$. In order to mimic the proof for the ODE case, we need to write the problem in a form like
$$
y'(T) = \int_0^T \big[\textrm{some function of }t\textrm{ and }f(t,y(t),y'(t))\big]  \,\mathrm{d}t.
$$
To do so, we multiply $(*)$ by $(T-t)^{\alpha-2}$ and integrate over $t$ from $0$ to $T$.
$$
\gamma \int_0^T\int_0^t(T-t)^{\alpha-2}(t-s)^{1-\alpha}y''(s) \, \mathrm{d}s \,\mathrm{d}t
= \int_0^T (T-t)^{\alpha-2} f(t,y(t),y'(t))\,\mathrm{d}t.
\qquad\qquad(**)
$$
Interchanging the order of integration in the left hand side, which is justified under the assumption $y\in C^2([0,T])$, we get
$$
\begin{split}
\int_0^T\int_0^t(T-t)^{\alpha-2}(t-s)^{1-\alpha}y''(s) \, \mathrm{d}s \,\mathrm{d}t
&=
\int_0^T\int_s^T(T-t)^{\alpha-2}(t-s)^{1-\alpha}y''(s) \,\mathrm{d}t\, \mathrm{d}s\\
&=
\Gamma(\alpha-1)\Gamma(2-\alpha)\int_0^T y''(s)\, \mathrm{d}s \\
&=\Gamma(\alpha-1)\Gamma(2-\alpha)\big( y'(T)-y'(0) \big),
\end{split}
$$
where we have used the fact that
$$
\int_0^1x^{a-1}(1-x)^{b-1}\, \mathrm{d}x = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)},
$$
for $a>0$ and $b>0$.
To conclude, we have
$$
y'(T) = y_1 + c\int_0^T (T-t)^{\alpha-2} f(t,y(t),y'(t))\,\mathrm{d}t,
\qquad\qquad(\dagger)
$$
where the integration is justified if $f$ is continuous and $y\in C^1([0,T])$.
In the converse direction, if $y\in C^2([0,T])$ then $(\dagger)$ implies $(*)$.
Since now it is in a standard form and the question is a homework question, I leave the rest as an exercise.
