Initialized fractional calculus

Before discussing initialization of the differintegrals in fractional calculus, a certain oddity about the differintegral should be pointed out. Although:

$\mathbb{D}^q\mathbb{D}^{-q} = \mathbb{I}$

That is, D^q is the left inverse of D^−q. On first glance, the converse is not necessarily true.

$\mathbb{D}^{-q}\mathbb{D}^q \neq \mathbb{I}$

However, let's take a look at integral calculus, to get a better idea of what's happening. First, let's integrate, then differentiate, using the arbitrary function 3x² + 1:

$d\left[\int (3x^2+1)dx\right]/dx = d[x^3+x+c]/dx = 3x^2+1$

The process did work successfully. On exchanging the order of composition:

$\int [d(3x^2+1)/dx]dx = \int 6x \,dx = 3x^2+c$

The integration constant here is clear. Even if it wasn't obvious, we would simply use the initialization terms such as ƒ'(0) = c, ƒ''(0) = d, etc. If we neglected those initialization terms, the last equation would fail our test.

This is exactly the problem that we encountered with the differintegral. If the differintegral is initialized properly, then the hoped-for composition law holds. The problem is that in differentiation, we lose information, as we lost the c in the first equation. (see dynamical systems).

In fractional calculus, however, since the operator has been fractionalized and is thus continuous, an entire complementary function is needed, not just a constant or set of constants. We call this complementary function "Ψ".

$\mathbb{D}^q_t f(t) = \frac{1}{\Gamma(n-q)}\frac{d^n}{dt^n}\int_0^t (t-\tau)^{n-q-1}f(\tau)\,d\tau + \Psi(x)$

(Working with a properly initialized differintegral is the subject of initialized fractional calculus).

References

• Lorenzo, Carl F.; Hartley (2000), Initialized Fractional Calculus, NASA, doi:2060/20000031631  (technical report).