Kernel of Fractional Differential Operator Suppose we have a fractional differential equation:
$$\left[D^{nv}+a_{1}D^{\left(n-1\right)v}+\dots+a_{n}D^{0}\right]y(t)=0$$
where $\nu=\frac{1}{q}$ and $q\in\mathbb{N}$ and y is an analytic function.
How can we prove that the kernel of the differential operator
$$\left[D^{nv}+a_{1}D^{\left(n-1\right)v}+\dots+a_{n}D^{0}\right]$$
has dimension $N=\lceil{n\nu}\rceil$?
Thanks a lot.
 A: This is exactly Theorem 1 in K.S. Miller & B. Ross, "An introduction to the fractional calculus and fractional differential equations," John Wiley and Sons, Inc., 1993.
Let's rewrite the given FDE as follows:
$$
P(D)y(t)=0, \qquad (\spadesuit)
$$
where $P$ is the "fractional polynomial" operator $P(D^v)=D^{nv}+a_1 D^{(n-1)v}+\ldots+a_n D^{0}$ generated by the function $p(x)=x^n+a_1 x^{n-1}+\ldots+a_n$ replacing $x$ by $D^v$. Let us take the Laplace transform on () and let us denote by $Y(s)$ the Laplace transform of $y(t)$:
$$
\mathscr{L}\{P(D^v)y(t)\}(s)=0\Leftrightarrow 
P(s^v)Y(s)-\sum_{i=0}^{N-1}Q_i(y)s^i=0,\qquad(\clubsuit) 
$$
where $Q_i$ are functions of the form
$$
Q_i(y) = \sum_{i=0}^{N-1}\sum_{k=rq+1}^{n}(D^{kv-(r+1)}y)(0).
$$
The term $Q_0$ is given by
$$
Q_0(y) = P(D^v)D^{-1}y(0) - a_n D^{-1}y(0).
$$
From ($\clubsuit$) it follows that
$$
Y(s)=\frac{\sum_{i=0}^{N-1}Q_i(y)s^r}{P(s^v)},
$$
and $y(t)=\mathscr{L}^{-1}\{Y(s)\}(t)$ is a solution of ($\spadesuit$).
Define $y_1(t)=\mathscr{L}^{-1}\{\frac{1}{P(s^v)}\}$; then it can be shown that $y_1$ is a solution of (). Recursively, define
$$
y_{j+1}(t) = D^j y_{1}(t),\ j=0,1,\ldots, N-1.
$$
It can again be shown that all $y_{j}$ are solutions of ($\spadesuit$). Finally, it is easy to show that these are all linearly independent. 
