*5.6. The Caputo Derivative Definition Revisited*

We have demonstrated with the previous elementary counter example that the integration technique based on the Caputo derivative definition is unable to provide a correct expression of the free response of an elementary initial-value problem. Of course, we have pointed out the reason of this failure, i.e., *x*(0) does not represent the initial condition of the fractional integrator. Basically, what is the origin of this error?

In order to understand why so many researchers have been misled by the so-called "initial condition" *x*(0), we have to once again consider the definition (12) of the Caputo derivative:

$${}^{C}D\_{t}^{n}(x(t)) = I\_{t}^{1-n} \left(\frac{d\mathbf{x}(t)}{dt}\right).$$

This derivative relies on a fractional integrator <sup>1</sup> *<sup>s</sup>*1−*<sup>n</sup>* , so we have to take into account its internal state variables *zC*(*ω*, *t*) at *t* = 0 (notice that *zC*(*ω*, *t*) = *z*(*ω*, *t*)) [34].

Thus, the distributed-frequency model of the Caputo derivative corresponds to that of the fractional integrator *I* <sup>1</sup>−*<sup>n</sup> <sup>t</sup>* (.), where, in this case, the input and the output are *dx*(*t*) *dt* and *CDn <sup>t</sup>* (*x*(*t*)), respectively:

$$\begin{cases} \begin{array}{c} \frac{\partial z\_{\mathbb{C}}(\omega,t)}{\partial t} = -\omega z\_{\mathbb{C}}(\omega,t) + \frac{dx(t)}{dt} \,\omega \, [0,\infty) \\ \prescript{\mathbb{C}}{}{D}\_{t}^{n}(x(t)) = \int \mu\_{1-n}(\omega) z\_{\mathbb{C}}(\omega,t) d\omega \\ \mu\_{1-n}(\omega) = \frac{\sin((1-n)\pi)}{\pi} \omega^{-(1-n)} \text{ and } 0 < n < 1 \end{array} \end{cases} \tag{40}$$

with the initial condition *zC*(*ω*, 0) and *ω* ∈ [0, ∞).

By using the Laplace transform, we can write

$$\begin{cases} Z\_{\mathbb{C}}(\omega, \mathbf{s}) = \frac{z\_{\mathbb{C}}(\omega, \mathbf{0})}{s + \omega} + \frac{L\left\{\frac{dx(t)}{dt}\right\}}{s + \omega} \,\,\omega \in [0, \infty) \\ \qquad \text{with } L\left\{\frac{dx(t)}{dt}\right\} = \mathbf{s}X(\mathbf{s}) - \mathbf{x}(0) \end{cases}$$

Thus, we obtain

$$L\left\{{}^{\mathbb{C}}D\_{t}^{\boldsymbol{u}}(\mathbf{x}(t))\right\} = \int\_{0}^{\infty} \mu\_{1-\boldsymbol{u}}(\omega) Z\_{\mathbb{C}}(\omega,\mathbf{s}) d\omega = \mathbf{s}^{\boldsymbol{u}}X(\boldsymbol{s}) - \frac{\mathbf{x}(0)}{\mathbf{s}^{1-\boldsymbol{u}}} + \int\_{0}^{\infty} \frac{\mu\_{1-\boldsymbol{u}}(\omega)}{\mathbf{s}+\omega} z\_{\mathbb{C}}(\omega,0) d\omega \tag{41}$$

We can conclude that the usual initial condition of the Caputo derivative is wrong because it does not take into account the transients of its associated integrator <sup>1</sup> *<sup>s</sup>*1−*<sup>n</sup>* , i.e., its distributed initial conditions *zC*(*ω*, 0).

Notice that the same conclusions apply to the Riemann–Liouville derivative [34].

Consequently, the Caputo derivative approach to the integration of FDE/FDS must be rejected because it provides wrong solutions to fractional initial-value problems. This approach is wrong for two main reasons:


Numerical simulations of the Caputo and Riemann–Liouville derivatives exhibiting the role of their initial conditions are available in [38] and in Volume 1 of [34].
