*6.2. FDS Transients and the Mittag–Leffler Function*

First, we consider the elementary FDE initial-value problem:

$$D^n(\mathbf{x}(t)) = a\mathbf{x}(t) \quad 0 < n < 1\tag{46}$$

with the distributed initial condition *z*(ω, 0) ∀ω ∈ [0, ∞).

System (46) can be transformed into the distributed frequency one:

$$\begin{cases} \frac{\partial z(\boldsymbol{\omega},t)}{\partial t} = -\omega z(\boldsymbol{\omega},t) + a\boldsymbol{x}(t) \\ \boldsymbol{x}(t) = \int \boldsymbol{\mu}\_{n}(\boldsymbol{\omega}) z(\boldsymbol{\omega},t) d\boldsymbol{w} \end{cases} \tag{47}$$

with the initial condition *z*(ω, 0).

Using the Laplace transform, we can write

$$sZ(\omega, \mathbf{s}) - z(\omega, \mathbf{0}) = -\omega Z(\omega, \mathbf{s}) + aX(\mathbf{s})\_{\prime\prime}$$

$$\text{i.e.}\,\chi(s) = \frac{s^{\pi}}{s^{\pi}-a} \int\_{0}^{\infty} \mu\_{n}(\omega) \frac{z(\omega,0)}{s+\omega} d\omega$$

$$\text{for } \lambda(s) = s \frac{s^{n-1}}{s^n - a} X\_0(s), \tag{48}$$

with *X*0(*s*) = # ∞ 0 *<sup>μ</sup>n*(*ω*) *<sup>z</sup>*(*ω*,0) *<sup>s</sup>*+*<sup>ω</sup> dω* So,

$$\mathbf{x}\_{0}(t) = L^{-1} \left\{ \int\_{0}^{\infty} \mu\_{n}(\omega) \frac{z(\omega, 0)}{s + \omega} d\omega \right\} = \int\_{0}^{\infty} \mu\_{n}(\omega) z(\omega, 0) e^{-\omega t} d\omega \tag{49}$$

Let us remember that

$$L\{E\_{n,1}(at^n)\} = \frac{s^{n-1}}{s^n - a} \tag{50}$$

where *En*,1(*atn*) is the Mittag–Leffler function:

$$E\_{n,1}(at^n) = \sum\_{k=0}^{\infty} \frac{(at^n)^k}{\Gamma(nk+1)},\tag{51}$$

which is the generalization of the exponential function, since for *n* = 1 we obtain *E*1,1 *at*1 <sup>=</sup> <sup>∞</sup> ∑ (*at*<sup>1</sup>) *k <sup>k</sup>*! = *<sup>e</sup>at*.

*k*=0 Consequently, we obtain

$$\mathbf{x}(t) = L^{-1}\{X(s)\} = \frac{d}{dt}\{E\_{n,1}(at^n) \* \mathbf{x}\_0(t)\}.\tag{52}$$

We can easily generalize this result to any linear FDS initial value problem:

$$D^{\mathfrak{u}}(\underline{\mathfrak{x}}(t)) = A\underline{\mathfrak{x}}(t) + \underline{\mathfrak{B}}\mathfrak{u}(t) \; 0 < \mathfrak{u} < 1,\tag{53}$$

with the distributed initial condition *z*(*ω*, 0) ∀*ω* ∈ [0, ∞).

Defining the matrix Mittag–Leffler function:

$$E\_{n,1}(At^n) = \sum\_{k=0}^{\infty} \frac{(At^n)^k}{\Gamma(nk+1)},\tag{54}$$

we can write

$$\underline{\mathbf{x}}(t) = \frac{d}{dt} \{ E\_{n,1}(At^n) \* \underline{\mathbf{x}}\_0(t) \} + \int\_0^t E\_{n,1}\left( A(t-\tau)^n \right) \underline{B}\tilde{u}(\tau)d\tau,\tag{55}$$

where *<sup>u</sup>*(τ) = *<sup>D</sup>*1−*n*(*u*(*τ*)) and *<sup>x</sup>*0(*t*) = # ∞ 0 *μn*(*ω*)*z*(*ω*, 0)*e*−*ω<sup>t</sup> dω*.

Obviously, the free response *<sup>d</sup> dt* {*En*,1(*Atn*) <sup>∗</sup> *<sup>x</sup>*0(*t*)} of *<sup>x</sup>*(*t*) is more complex than the wrong usual one derived from (8), i.e., *x*(*t*) = *En*,1(*Atn*)*x*(0).

Moreover, there is a major difficulty, i.e., the convolution between the Mittag–Leffler function and *x*0(*t*), which is not a straightforward operation. Thus, the interest of this expression is essentially theoretical.

Notice also that the matrix Mittag–Leffler function does not verify the semi-group properties of the matrix exponential function [53,54].

**Remark 4**: The use of the Caputo derivative is based on the assumption *x*0(*t*) = *x*(0) = *cte* ∀*t*. Is this requirement always wrong? In addition, if it is correct, what does it mean?

Consider the following system

$$D^n(\mathfrak{x}(t)) = a\mathfrak{x}(t) + b\mathfrak{u}(t) \text{ } 0 < n < 1.$$

and suppose that it is at rest at *t* = 0 and that *u*(*t*) = *UH*(*t*).

Since, in this case,

$$Z(\omega, s) = \frac{aX(s) + b\frac{\mathcal{U}}{s}}{s + \omega} \text{ and } \mathcal{X}(s) = \int\_0^\infty \mu\_\mathbb{R}(\omega) Z(\omega, s) d\omega \text{, we obtain } Z(\omega, s) = b\frac{s^\pi}{s^\pi - a} \frac{1}{s + \omega} \frac{\mathcal{U}}{s}.$$

Thus

$$z(\omega, \infty) = \limsup\_{s \to 0} Z(\omega, s) = 0 \quad \forall \omega \neq 0 \text{ and } z(0, \infty) = \lim\_{s \to 0} \frac{bL}{s^{1-n}} = \infty \tag{56}$$

On the other hand, we can also write *<sup>x</sup>*(∞) = <sup>−</sup>*<sup>b</sup> <sup>a</sup>U* for a stable system, i.e., *a* < 0 in our case.

Assume that we have applied the same step input *UH*(*t*), but, at *t* = −∞, this means that at *<sup>t</sup>* = 0, we obtained *<sup>x</sup>*(0) = <sup>−</sup>*<sup>b</sup> <sup>a</sup>U* for a long time in the past (*t* < 0).

Consequently, the condition *x*0(*t*) = *x*(0) = *cte* ∀*t* requires that the system has been at rest for a very long time.

Then

$$\mu\_0(t) = \int\_0^\infty \mu\_\pi(\omega) z(\omega, 0) e^{-\omega \upsilon} d\omega = \int\_0^\infty \mu\_\pi(\omega) z(0, 0) e^{-0t} d\omega = x(0) = cte. \tag{57}$$

Since *z*(ω, ∞) = 0∀ω = 0.

Theoretically, the condition *x*0(*t*) = *x*(0) = *cte* ∀*t* can be achieved, but it is completely unrealistic. Moreover, it would require infinite energy [34].

**Remark 5**: Consider again the revisited definition of the Caputo derivative (see Section 5.6):

$$\prescript{\mathbb{C}}{}{D}\_{t}^{\mathbb{H}}(\mathbf{x}(t)) = \prescript{}{l}{l}^{1-n} \left(\frac{d\mathbf{x}(t)}{dt}\right) \Rightarrow \begin{cases} \frac{\partial z\_{\mathbb{C}}(\omega,t)}{\partial t} = -\omega z\_{\mathbb{C}}(\omega,t) + \frac{d\mathbf{x}(t)}{dt} & \omega \in [0,\infty) \\\ \prescript{\mathbb{C}}{}{D}\_{t}^{n}(\mathbf{x}(t)) = \int \mu\_{1-n}(\omega) z\_{\mathbb{C}}(\omega,t) d\omega \\\ \mu\_{1-n}(\omega) = \frac{\sin((1-n)\pi)}{\pi} \omega^{-(1-n)} \text{ and } 0 < n < 1 \end{cases}$$

with *zC*(*ω*, *t*) = *zC*(*ω*, 0) at *t* = 0.

We have demonstrated (41) that:

$$L\left\{{}^{\mathbb{C}}D\_{t}^{\boldsymbol{n}}(\boldsymbol{x}(t))\right\} = {}^{\boldsymbol{n}}\boldsymbol{X}(\boldsymbol{s}) - \frac{\boldsymbol{x}(0)}{\operatorname{s}^{1-\boldsymbol{n}}} + \int\_{0}^{\infty} \frac{\mu\_{1-\boldsymbol{n}}(\boldsymbol{\omega})}{\boldsymbol{s}+\boldsymbol{\omega}} \boldsymbol{z}\_{\mathbb{C}}(\boldsymbol{\omega},0)d\boldsymbol{\omega}.$$

If *x*(*t*) = *cte*, for a long time in the past (*t* < 0), then *dx*(*t*) *dt* = 0. This means that *zC*(ω, *t*) = 0 ∀ ω for a long time in the past (*t* < 0), and consequently *zC*(ω, 0) = 0 ∀ ω.

Then, we can write *L CDn <sup>t</sup>* (*x*(*t*)) <sup>=</sup> *<sup>s</sup>nX*(*s*) <sup>−</sup> *<sup>x</sup>*(0) *<sup>s</sup>*1−*<sup>n</sup>* .

Obviously, the previous conditions are very restrictive, and the usual initial condition is wrong as soon as there is a variation of *x*(*t*).

**Remark 6**: We have demonstrated (52) that:

$$\mathbf{x}(t) = \frac{d}{dt} \{ E\_{n,1}(at^n) \* \mathbf{x}\_0(t) \} = E\_{n,1}(at^n) \* \frac{d}{dt} \mathbf{x}\_0(t).$$

Since *<sup>x</sup>*0(*t*) = *<sup>x</sup>*(0) = *cte* <sup>∀</sup>*t*, we can write *<sup>x</sup>*0(*t*) = *<sup>x</sup>*(0)*H*(*t*), so *<sup>d</sup> dt x*0(*t*) = *x*(0)*δ*(*t*).

Then,

$$\mathbf{x}(t) = E\_{n,1}(at^n) \* \mathbf{x}(0)\delta(t) = E\_{n,1}(at^n)\mathbf{x}(0) \tag{58}$$

is the usual result corresponding to the Caputo derivative assumption.

The conclusion is that many results obtained with the Caputo derivative approach are not necessarily wrong, but they require the previous very restrictive assumptions related to *x*(0).
