Euler–Lagrange Equations for Lagrangians Containing Complex-order Fractional Derivatives

Abstract

Two variational problems of finding the Euler–Lagrange equations corresponding to Lagrangians containing fractional derivatives of real- and complex-order are considered. The first one is the unconstrained variational problem, while the second one is the fractional optimal control problem. The expansion formula for fractional derivatives of complex-order is derived in order to approximate the fractional derivative appearing in the Lagrangian. As a consequence, a sequence of approximated Euler–Lagrange equations is obtained. It is shown that the sequence of approximated Euler–Lagrange equations converges to the original one in the weak sense as well as that the sequence of the minimal values of approximated action integrals tends to the minimal value of the original one.

Appendix

We prove Proposition 2.1 by writing the left Riemann–Liouville fractional derivative of complex-order $\gamma =\alpha +\mathrm {i}\beta$, $\alpha \in ] 0,1[$, given by (2), as

$$\begin{aligned}&{}_{0}{\mathcal {D}}_{t}^{\gamma }y\left( t\right) \\&\quad =\frac{1}{\varGamma _{\mathrm {re}}^{2}+\varGamma _{\mathrm {im}}^{2}}\frac{\mathrm {d}}{\mathrm {d}t}\int _{0}^{t}\frac{ y\left( \tau \right) }{\left( t-\tau \right) ^{\alpha }}\left( \varGamma _{\mathrm {re}}\cos \left( \log \left( t-\tau \right) ^{\beta }\right) +\varGamma _{\mathrm {im}}\sin \left( \log \left( t-\tau \right) ^{\beta }\right) \right) \mathrm { d}\tau , \end{aligned}$$

where

$$\begin{aligned} \varGamma _{\mathrm {re}}=\int _{0}^{\infty }\mathrm {e}^{-x}x^{\alpha }\cos \left( \log x^{\beta }\right) \mathrm {d}x\;\;\text {and}\;\;\varGamma _{\mathrm {im}}=\int _{0}^{\infty }\mathrm {e}^{-x}x^{\alpha }\sin \left( \log x^{\beta }\right) \mathrm {d}x, \end{aligned}$$

are real and imaginary parts of $\varGamma \left( 1-\gamma \right)$, respectively. In order to obtain integration by parts formula (4), let $f,g\in {\mathrm {AC}}\left( \left[ 0,T\right] \right)$. Recall, then the first derivatives of f and g are integrable on [0, T]. We also use that the convolution of an absolutely continuous function on [0, T] and the integrable function on [0, t], $t<T$ is an absolutely continuous on [0, T] along with the fact that the partial integration holds for absolutely continuous functions, i.e., $\int _0^{t}f(\tau )\frac{\mathrm {d}}{\mathrm {d}\tau }g(\tau )d\tau = \left. f(\tau )g(\tau )\right| _{0}^{t} - \int _0^{t}g(\tau )\frac{\mathrm {d}}{\mathrm {d}\tau }f(\tau )d\tau$, for all $t\in [0,T]$, if $f,g \in {\mathrm {AC}}([0,T])$. Consider the integrals

$$\begin{aligned}&\int _{0}^{T}f\left( t\right) \,{}_{0}{\mathcal {D}}_{t}^{\gamma }g\left( t\right) \mathrm {d}t\nonumber \\&\quad =\frac{\varGamma _{\mathrm {re}}}{\varGamma _{\mathrm {re}}^{2}+\varGamma _{\mathrm {im}}^{2} }\int _{0}^{T}f\left( t\right) \left( \frac{\mathrm {d}}{\mathrm {d}t} \int _{0}^{t}g\left( \tau \right) \frac{\cos \left( \log \left( t-\tau \right) ^{\beta }\right) }{\left( t-\tau \right) ^{\alpha }}\mathrm {d}\tau \right) \mathrm {d}t \nonumber \\&\quad \quad +\,\frac{\varGamma _{\mathrm {im}}}{\varGamma _{\mathrm {re}}^{2}+\varGamma _{\mathrm {im}}^{2}}\int _{0}^{T}f\left( t\right) \left( \frac{\mathrm {d}}{\mathrm {d}t}\int _{0}^{t}g\left( \tau \right) \frac{\sin \left( \log \left( t-\tau \right) ^{\beta }\right) }{ \left( t-\tau \right) ^{\alpha }}\mathrm {d}\tau \right) \mathrm {d}t \end{aligned}$$

(49)

and

$$\begin{aligned}&\int _{0}^{T}g\left( t\right) \,{}_{t}{\mathcal {D}}_{T}^{\gamma }f\left( t\right) \mathrm {d}t\nonumber \\&\quad =-\frac{\varGamma _{\mathrm {re}}}{\varGamma _{\mathrm {re}}^{2}+\varGamma _{\mathrm {im}}^{2}}\int _{0}^{T}g\left( t\right) \left( \frac{\mathrm {d}}{\mathrm {d}t} \int _{t}^{T}f\left( \tau \right) \frac{\cos \left( \log \left( \tau -t\right) ^{\beta }\right) }{\left( \tau -t\right) ^{\alpha }}\mathrm {d}\tau \right) \mathrm {d}t \nonumber \\&\quad \quad -\,\frac{\varGamma _{\mathrm {im}}}{\varGamma _{\mathrm {re}}^{2}+\varGamma _{\mathrm {im}}^{2}}\int _{0}^{T}g\left( t\right) \left( \frac{\mathrm {d}}{\mathrm {d}t}\int _{t}^{T}f\left( \tau \right) \frac{\sin \left( \log \left( \tau -t\right) ^{\beta }\right) }{ \left( \tau -t\right) ^{\alpha }}\mathrm {d}\tau \right) \mathrm {d}t. \end{aligned}$$

(50)

The integral in the first term of (49) is transformed as

$$\begin{aligned}&\int _{0}^{T}f\left( t\right) \left( \frac{\mathrm {d}}{\mathrm {d}t} \int _{0}^{t}g\left( \tau \right) \frac{\cos \left( \log \left( t-\tau \right) ^{\beta }\right) }{\left( t-\tau \right) ^{\alpha }}\mathrm {d}\tau \right) \mathrm {d}t \nonumber \\&\quad =\left[ f(t)\int _{0}^{t}g\left( \tau \right) \frac{\cos \left( \log \left( t-\tau \right) ^{\beta }\right) }{\left( t-\tau \right) ^{\alpha }} \mathrm {d}\tau \right] _{0}^{T}\nonumber \\&\qquad -\int _{0}^{T}f^{\prime }(t)\left( \int _{0}^{t}g\left( \tau \right) \frac{\cos \left( \log \left( t-\tau \right) ^{\beta }\right) }{\left( t-\tau \right) ^{\alpha }}\mathrm {d}\tau \right) \mathrm {d}t \nonumber \\&\quad =f\left( T\right) \int _{0}^{T}g\left( \tau \right) \frac{\cos \left( \log \left( T-\tau \right) ^{\beta }\right) }{\left( T-\tau \right) ^{\alpha }}\mathrm {d}\tau \nonumber \\&\qquad -\int _{0}^{T}g\left( \tau \right) \left( \int _{\tau }^{T}f^{\prime }\left( t\right) \frac{\cos \left( \log \left( t-\tau \right) ^{\beta }\right) }{\left( t-\tau \right) ^{\alpha }}\mathrm {d}t\right) \mathrm {d}\tau \nonumber \\&\quad =\int _{0}^{T}g\left( \tau \right) \left( f\left( T\right) \frac{\cos \left( \log \left( T-\tau \right) ^{\beta }\right) }{\left( T-\tau \right) ^{\alpha }}-\int _{\tau }^{T}f^{\prime }\left( t\right) \frac{\cos \left( \log \left( t-\tau \right) ^{\beta }\right) }{\left( t-\tau \right) ^{\alpha }}\mathrm {d}t\right) \mathrm {d}\tau . \end{aligned}$$

(51)

Similarly, transforming the integral in the second term of (49) we obtain

$$\begin{aligned}&\int _{0}^{T}f\left( t\right) \left( \frac{\mathrm {d}}{\mathrm {d}t} \int _{0}^{t}g\left( \tau \right) \frac{\sin \left( \log \left( t-\tau \right) ^{\beta }\right) }{\left( t-\tau \right) ^{\alpha }}\mathrm {d}\tau \right) \mathrm {d}t \nonumber \\&\quad =\int _{0}^{T}g\left( \tau \right) \left( f\left( T\right) \frac{\sin \left( \log \left( T-\tau \right) ^{\beta }\right) }{\left( T-\tau \right) ^{\alpha }}-\int _{\tau }^{T}f^{\prime }\left( t\right) \frac{\sin \left( \log \left( t-\tau \right) ^{\beta }\right) }{\left( t-\tau \right) ^{\alpha }}\mathrm {d}t\right) \mathrm {d}\tau . \end{aligned}$$

(52)

The integral in the first term of (50) yields

$$\begin{aligned}&\frac{\mathrm {d}}{\mathrm {d}t}\int _{t}^{T}f\left( \tau \right) \frac{\cos \left( \log \left( \tau -t\right) ^{\beta }\right) }{\left( \tau -t\right) ^{\alpha }}\mathrm {d}\tau \nonumber \\&\quad =\frac{\mathrm {d}}{\mathrm {d}t}\int _{0}^{T-t}f\left( u+t\right) \frac{\cos u^{\beta }}{u^{\alpha }}\mathrm {d}u \nonumber \\&\quad =-f\left( T\right) \frac{\cos \left( T-t\right) ^{\beta }}{\left( T-t\right) ^{\alpha }}+\int _{0}^{T-t}f^{\prime }\left( u+t\right) \frac{\cos \left( \log u^{\beta }\right) }{u^{\alpha }}\mathrm {d}u \nonumber \\&\quad =-f\left( T\right) \frac{\cos \left( T-t\right) ^{\beta }}{\left( T-t\right) ^{\alpha }}+\int _{t}^{T}f^{\prime }\left( \tau \right) \frac{\cos \left( \log \left( \tau -t\right) ^{\beta }\right) }{\left( \tau -t\right) ^{\alpha }}\mathrm {d}\tau , \end{aligned}$$

(53)

while, similarly, the integral in the second term of (50) gives

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}t}\int _{t}^{T}f\left( \tau \right) \frac{\sin \left( \log \left( \tau -t\right) ^{\beta }\right) }{\left( \tau -t\right) ^{\alpha }}\mathrm {d}\tau= & {} -f\left( T\right) \frac{\sin \left( T-t\right) ^{\beta }}{\left( T-t\right) ^{\alpha }}\nonumber \\&\quad +\int _{t}^{T}f^{\prime }\left( \tau \right) \frac{\sin \left( \log \left( \tau -t\right) ^{\beta }\right) }{\left( \tau -t\right) ^{\alpha }}\mathrm {d}\tau . \end{aligned}$$

(54)

By plugging (51) and (52) into (49), as well as by plugging (53) and (54) into (50), we obtain integration by parts formula (4).