Help in proof of theorem about Riemann-Liouville Fractional Calculus

Question

Theorem:

Let, $$\left[D^{nv}+a_{1}D^{\left(n-1\right)v}+\dots+a_{n}D^{0}\right]\left(y\right)=0$$ be a fractional differential equation of order $\left(n,q\right)$, where $v=\frac{n}{q}$, and let $$P\left(x\right)=x^{n}+a_{1}x^{n-1}+\dots+a_{n}$$ be the corresponding indicial polynomial.

Let $$y_{1}\left(t\right)=\mathscr{L}^{-1}\left\{ P^{-1}\left(s^{v}\right)\right\} $$ Then if $N$ is the smallest integer with the property $N\geq n\cdot v$, then $$y_{1}\left(t\right), y_{2}\left(t\right)=D^{1}y_{1}\left(t\right),...,y_{N}\left(t\right)=D^{N-1}y_{1}\left(t\right)$$ are $N$ independent solutions.

This theorem is from the book "An Introduction to Fractional Calculus and Fractional Differential Equations" by Miller and Ross. I don't manage to understand the proof provided in the book.

Proof:

We take the Laplace transform of the equation:

$\mathscr{L}\left\{ P\left(D^{v}y\left(t\right)\right)\right\} =0$,

and also $$\mathscr{L}\left\{ P\left(D^{v}y\left(t\right)\right)\right\} =P\left(s^{v}\right)Y\left(s\right)-\sum\limits_{r=0}^{N-1}B_{r}\left(y\right)s^{r}$$ where $B_{r}\left(y\right)$ is a linear combination of terms of the form:

$$D^{kv-\left(r+1\right)}y\left(0\right)$$
$k=rq+1,\dots,n$ $r=0,1,\dots,N-1$

In particular $B_{0}\left(y\right)=P\left(D^{v}\right)D^{-1}y\left(0\right)-a_{n}D^{-1}y\left(0\right)$.

We have,

$$Y\left(s\right)=\frac{\sum\limits_{r=0}^{N-1}B_{r}\left(y\right)s^{r}}{P\left(s^{v}\right)}$$ and $$y\left(t\right)=\mathscr{L}^{-1}\left\{ Y\left(s\right)\right\} $$ is a solution of the equation.

Now let,

$y_{1}\left(t\right)=\mathscr{L}^{-1}\left\{ P^{-1}\left(s^{v}\right)\right\} $, then

$\mathscr{L}\left\{ P\left(D^{v}\right)y_{1}\left(t\right)\right\} =P\left(s^{v}\right)-\sum\limits_{r=0}^{N-1}B_{r}\left(y_{1}\right)s^{r}$.

The initial value theorem for the Laplace transform says that, $\underset{s\rightarrow+\infty}{\lim}s^{v+1}\mathscr{L}\left\{ f\left(t\right)\right\} =D^{v}f\left(0\right)$ for all $v\in\mathbb{R}$. Thus, $B_{0}\left(y_{1}\right)=1$, $B_{r}\left(y_{1}\right)=0$ for $r>1$

Is this last deduction in italics where I'm stuck. It is not clear to me.

Answer 1

@Janus
You've probably already worked out the answer by now, but here's my explanation.

Putting $f=y_1$ in the initial value theorem gives $\underset{s\rightarrow+\infty}{\lim}\frac{s^{v+1}}{P(s^v)}=D^vy_1(0)$ for all $v\in\mathbb{R}$. And $P$ is a monic polynomial of degree $n$, so this gives $D^uy_1(0)=0$ for all $u<nv-1$ and $D^{nv-1}y_1(0)=0$.

And we already know that every $B_{r}\left(y\right)$ is a linear combination of terms of the form $D^{kv-\left(r+1\right)}y\left(0\right)$ for $k=rq+1,\dots,n$, $r=0,1,\dots,N-1$ and in particular $B_{0}\left(y\right)=P\left(D^{v}\right)D^{-1}y\left(0\right)-a_{n}D^{-1}y\left(0\right)$. Plugging in the above expressions gives the result.

PS. Nice to see someone else studying FC! :-) Are you working through Miller & Ross right now?

Answer 2

As you might know, we have Laplace's transform of various sense of fractional derivatives. So it is not clear to me as to whether you really refer to Riemann Liouville, Caputo or something like that, please just clarify a little and may be I can help.

Answer 3

By definition, $(D^vf)(x)=\dfrac{1}{\Gamma(1-v)}\dfrac{d}{dx}\displaystyle\int^x_0\dfrac{f(t)dt}{(x-t)^{v-[\Re(v)]}}$ ......$(1)$ and $\mathscr{L}(D^k\phi(t))(s)=s^k(\mathscr{L}\phi)(s)-\displaystyle\sum_{j=0}^{k-1}s^{k-j-1}(D^j\phi)(0)$ ......$(2)$, $k\in{\mathbb{N}}$. Then with these two, a more general relation is drived: $\mathscr{L}(D^vf(t))(s)=s^v(\mathscr{L}f)(s)-\displaystyle\sum_{r=0}^{N-1}s^{N-r-1}(D^v(I^{N-v}f)(0)$ ......$(3)$, where $(\Re(s)>q_0)$. Study these three equations by passing $\lim_{s\rightarrow{\infty}}$ throughout equation $(3)$ and continue with your work.

But remember, this is only true on the half axis and I omitted the $+0$ subscript sign that indicates Riemann Liouville on the half axis.