Optimal Control of Diffusion Equation with Fractional Time Derivative with Nonlocal and Nonsingular Mittag-Leffler Kernel

Abstract

In this paper, we consider a diffusion equation with fractional time derivative with nonsingular Mittag-Leffler kernel in Hilbert spaces. We first prove the existence and uniqueness of solution by means of a spectral argument. Then, we consider a distributed controlled fractional diffusion problem. We show that there exists a unique optimal control, which can act on the system in order to approach the state of the system by a given state at minimal cost. Finally, using the Euler–Lagrange first-order optimality condition, we obtain an optimality system, which characterizes the optimal control.

Appendix: Proof of Proposition 2.1

Let $\phi , y \in {\mathcal {C}}^{\infty }({\bar{Q}})$ be given. Then,

$$\begin{aligned}&\displaystyle \int _{\varOmega }\int _{0}^{T} \bigg ({}^{abc}_{0} \mathbf D ^{\alpha }_{t}~y(x,t) - \Delta y(x,t) \bigg ) \phi (x,t) \hbox {d}x~\hbox {d}t \\&\quad =\displaystyle \int _{\varOmega }\int _{0}^{T} {}^{abc}_{0} \mathbf D ^{\alpha }_{t}~y(x,t)\phi (x,t)\hbox {d}x~\hbox {d}t \\&\qquad - \int _{\varOmega }\int _{0}^{T} \Delta y(x,t)\phi (x,t) \hbox {d}x~\hbox {d}t = (A) + (B) \end{aligned}$$

Let us compute each part separately.

• Computation of (A): By using the definition of the Atangana–Baleanu fractional derivative (1) at the base point $a=0$ we have

  $$\begin{aligned} \int _{0}^{T} {}^{abc}_{0} \mathbf D ^{\alpha }_{t}~y(x,t)\phi (x,t)\hbox {d}t = \int _{0}^{T} \left[ \frac{B(\alpha )}{1-\alpha } \int _{0}^{t} y'(\tau ) E_{\alpha } \left[ -\gamma (t-\tau )^{\alpha } \right] \hbox {d}\tau \right] \phi (t)\hbox {d}t.\nonumber \\ \end{aligned}$$

  (29)

Let us define $A_{1} := \int _{0}^{t} y'(\tau ) E_{\alpha } \left[ -\gamma (t-\tau )^{\alpha } \right] \hbox {d}\tau$. By using integration by parts it yields

$$\begin{aligned} A_{1} = y(t) - y(0)E_{\alpha }[-\alpha t^{\alpha }] - \int _{0}^{t}\gamma (t-\tau )^{\alpha -1}E_{\alpha ,\alpha } \left[ -\gamma (t-\tau )^{\alpha } \right] y(\tau )\hbox {d}\tau . \end{aligned}$$

Replacing $A_{1}$ by its value in (29) we get that

$$\begin{aligned} \displaystyle \int _{0}^{T} {}^{abc}_{0} \mathbf D ^{\alpha }_{t}~y(x,t)\phi (x,t)\hbox {d}t= & {} \displaystyle \frac{B(\alpha )}{1-\alpha } \int _{0}^{T} y(x,t) \phi (x,t) \hbox {d}t \nonumber \\&- \frac{B(\alpha )}{1-\alpha } \int _{0}^{T} y(x,0)E_{\alpha }[-\alpha t^{\alpha }]\phi (x,t)\hbox {d}t \nonumber \\&\displaystyle - \frac{B(\alpha )}{1-\alpha } \int _{0}^{T} \left[ \int _{0}^{t} \gamma (t-\tau )^{\alpha -1}E_{\alpha ,\alpha } \left[ -\gamma (t-\tau )^{\alpha } \right] y(x,\tau )\hbox {d}\tau \right] \phi (x,t)\hbox {d}t \nonumber \\&=: A_{2} + A_{3} + A_{4}, \end{aligned}$$

(30)

with

$$\begin{aligned} A_{2}:= & {} \displaystyle \frac{B(\alpha )}{1-\alpha } \int _{0}^{T} y(x,t) \phi (x,t) \hbox {d}t;\\ A_{3}:= & {} \displaystyle \frac{B(\alpha )}{1-\alpha } \int _{0}^{T} y(x,0)E_{\alpha }[-\alpha t^{\alpha }]\phi (x,t)\hbox {d}t; \\ A_{4}:= & {} \displaystyle -\frac{B(\alpha )}{1-\alpha } \int _{0}^{T} \left[ \int _{0}^{t} \gamma (t-\tau )^{\alpha -1}E_{\alpha ,\alpha } \left[ -\gamma (t-\tau )^{\alpha } \right] y(x,\tau )\hbox {d}\tau \right] \phi (x,t)\hbox {d}t. \end{aligned}$$

Let us compute $A_{4}$

$$\begin{aligned} A_{4}:= & {} \displaystyle -\frac{B(\alpha )}{1-\alpha } \int _{0}^{T} \left[ \int _{0}^{t} \gamma (t-\tau )^{\alpha -1}E_{\alpha ,\alpha } \left[ -\gamma (t-\tau )^{\alpha } \right] y(x,\tau )\hbox {d}\tau \right] \phi (x,t)\hbox {d}t \\= & {} \displaystyle \frac{B(\alpha )}{1-\alpha } \int _{0}^{T} y(x,\tau )\left[ -\int _{\tau }^{T} \gamma (t-\tau )^{\alpha -1}E_{\alpha ,\alpha } \left[ -\gamma (t-\tau )^{\alpha } \right] \phi (x,t) \hbox {d}t \right] \hbox {d}\tau \\= & {} \displaystyle \frac{B(\alpha )}{1-\alpha } \phi (x,T)\int _{0}^{T} y(x,t) E_{\alpha ,\alpha } \left[ -\gamma (T-t)^{\alpha } \right] \hbox {d}t -\frac{B(\alpha )}{1-\alpha } \int _{0}^{T} y(x,t) \phi (x,t) \hbox {d}t \\&\displaystyle + \int _{0}^{T} y(x,t) \left[ -\frac{B(\alpha )}{1-\alpha } \int _{t}^{T} E_{\alpha ,\alpha } \left[ -\gamma (t-\tau )^{\alpha } \right] \phi '(x,\tau ) \hbox {d}\tau \right] \hbox {d}t \\= & {} \displaystyle \frac{B(\alpha )}{1-\alpha } \phi (x,T)\int _{0}^{T} y(x,t) E_{\alpha ,\alpha } \left[ -\gamma (T-t)^{\alpha } \right] \hbox {d}t -\frac{B(\alpha )}{1-\alpha } \int _{0}^{T} y(x,t) \phi (x,t) \hbox {d}t \\&\displaystyle +\int _{0}^{T}y(x,t) {}^{abc}_{T} \mathbf D ^{\alpha }_{t}~y(x,t)\phi (x,t)\hbox {d}t. \end{aligned}$$

Hence, replacing $A_{4}$ by its value in (30) we finally have that

$$\begin{aligned} \displaystyle \int _{0}^{T} {}^{abc}_{0} \mathbf D ^{\alpha }_{t}~y(x,t)\phi (x,t)\hbox {d}t= & {} -\int _{0}^{T}y(x,t) {}^{abc}_{T} \mathbf D ^{\alpha }_{t}~y(x,t)\phi (x,t)\hbox {d}t \nonumber \\&+ \frac{B(\alpha )}{1-\alpha } \phi (x,T)\int _{0}^{T} y(x,t) E_{\alpha ,\alpha } \left[ -\gamma (T-t)^{\alpha } \right] \hbox {d}t\nonumber \\&-\frac{B(\alpha )}{1-\alpha } y(x,0)\int _{0}^{T}E_{\alpha }[-\alpha t^{\alpha }]\phi (x,t)\hbox {d}t. \end{aligned}$$

(31)

• Computation of (B) By a using integration by parts we have that

  $$\begin{aligned} (B):= & {} \displaystyle -\int _{\varOmega }\int _{0}^{T}\phi (x,t)\Delta u(x,t) \hbox {d}x~\hbox {d}t = \int _{0}^{T} \int _{\partial \varOmega } u \frac{\partial \phi }{\partial \sigma } \phi \hbox {d}\sigma ~\hbox {d}t \nonumber \\&- \int _{0}^{T} \int _{\partial \varOmega } \phi \frac{\partial u}{\partial \sigma } \hbox {d}\sigma ~\hbox {d}t \nonumber \\&\displaystyle - \int _{\varOmega }\int _{0}^{T}u(x,t) \Delta \phi (x,t)\hbox {d}t~\hbox {d}x. \end{aligned}$$

  (32)

Hence, by adding (31) to (32), we obtain the desired result (5).