Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Why must there be a unique solution to a linear constant-coefficient fractional differential equation of order $(n,q)$ with $\lceil\frac{n}{q}\rceil$ initial conditions? (All notation is as in Miller & Ross, An Introduction to the Fractional Calculus and Fractional DEs.)

More specifically, if $n$ and $q$ are positive integers and $v=\frac{1}{q}$ and $N=\lceil nv\rceil$ and $P(x)$ is a monic polynomial of degree $n$, consider the fractional DE $$P(_0D_t^v)y(t)=x(t)$$ $$y(0)=y'(0)=\dots=y^{(N-1)}(0)=0.$$ I can prove that $y(t)=\int_0^tK(t-\xi)x(\xi)d\xi$ is a solution, where $K$ is the fractional Green's function (inverse Laplace transform of $\frac{1}{P(s^v)}$), but Miller & Ross say it's the unique solution, and I can't see why!

I've been thinking of a proof along the lines of:

Assume there are 2 solutions, say $y_1(t)$ and $y_2(t)$. Then $y_1-y_2$ must satisfy the corresponding homogeneous DE, with the same initial conditions. But why does this mean $y_1-y_2\equiv0$? Is there some theorem from ODE theory about the dimension of the solution space which extends to cover fractional DEs - and if so, why does it extend?

Many thanks for any help with this!

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.