*1.4. The Problem under Consideration*

Let Ω be a bounded domain of R*<sup>d</sup>* with *C*<sup>2</sup> boundary Γ = *∂*Ω. We consider the following initial value/boundary value problem of an impulsive sub-diffusion equation of order *α* ∈ (0, 1):

$$\begin{cases} \partial\_t u + \partial\_t^{1-\alpha} A u = f(\mathbf{x}, t) & \text{in } \quad \Omega \times (0, T), \\ \Delta u(\cdot, t\_i) = l\_i(u(\cdot, t\_i)), i = 1, 2, 3, \dots, P, \\ u = 0 & \text{on } \quad \Gamma \times (0, T), \\ u(\cdot, 0) = u\_0 & \text{in } \quad \Omega. \end{cases} \tag{3}$$

In Equation (3), *u* = *u*(*x*, *t*) is the state to be controlled and *f* = *f*(*x*, *t*) is the control which is localized in a subdomain *ω* of Ω. We will act by *f* to drive the initial state *u*<sup>0</sup> = *u*0(*x*) to some target function *u*<sup>1</sup> = *u*1(*x*). The operator *A* is a symmetric and uniformly elliptic operator. The details will be specified later; *T* > 0 is also a constant. Several problems in applications can be modeled by the above equation. Some of them are: thermal diffusion in media with fractional geometry, underground environmental problems, highly heterogeneous aquifer, etc. [18]. In this paper, we study approximate controllability for fractional partial differential equations with impulses. We say that Equation (3) is approximately controllable if, for any *<sup>u</sup>*<sup>1</sup> <sup>∈</sup> *<sup>L</sup>*2(Ω) and *<sup>ε</sup>* <sup>&</sup>gt; 0, there exists a control *<sup>f</sup>* such that the solution *u* of (3) satisfies

$$\|\mu(\cdot, T) - \mu\_1\|\_{L^2(\Omega)} \le \varepsilon. \tag{4}$$

This paper is divided into four sections. In Section 2, we study requisite function spaces and some important basic results. In Section 3, we analyse the mild solutions of the Equation (3) by eigenfunction expansion. Section 4 is devoted to the study of a dual system of (3) and to establish a unique continuation property. In the last section, we establish the proof of approximate controllability.

#### **2. Preliminaries**

In this section, we state a few function spaces, notations and results in order to establish our main results. For the smooth reading of the manuscript, we first define the following class of spaces (for more details, we refer to Adams [19], Mahto [12]):

*<sup>L</sup>p*[*a*, *<sup>b</sup>*] = *<sup>f</sup>* : [*a*, *<sup>b</sup>*] <sup>→</sup> <sup>R</sup><sup>|</sup> *<sup>f</sup>* is Lebesgue measurable and *<sup>b</sup> a* | *f*(*t*)| *pdt* < ∞ , *AC*[*a*, *<sup>b</sup>*] = *<sup>f</sup>* : [0, *<sup>T</sup>*] <sup>→</sup> <sup>R</sup><sup>|</sup> *<sup>f</sup>* is absolutely continuous on [*a*, *<sup>b</sup>*] , *<sup>C</sup>*[*a*, *<sup>b</sup>*] = *<sup>f</sup>* : [0, *<sup>T</sup>*] <sup>→</sup> <sup>R</sup><sup>|</sup> *<sup>f</sup>* is continuous on [*a*, *<sup>b</sup>*] , *<sup>L</sup>p*(Ω) = *<sup>f</sup>* : <sup>Ω</sup> <sup>→</sup> <sup>R</sup><sup>|</sup> *<sup>f</sup>* is Lebesgue measurable and Ω | *f*(*x*)| *pdx* < ∞ , *<sup>H</sup>*1(Ω) = *<sup>f</sup>* : *<sup>f</sup>* , *<sup>∂</sup> <sup>f</sup> ∂x*<sup>1</sup> , ··· , *<sup>∂</sup> <sup>f</sup> ∂xd* <sup>∈</sup> *<sup>L</sup>*2(Ω) , *H*1 <sup>0</sup> (Ω) = *<sup>f</sup>* : *<sup>f</sup>* <sup>∈</sup> *<sup>H</sup>*1(Ω) and *<sup>f</sup>* <sup>=</sup> 0 on <sup>Γ</sup> , *AC*(0, *<sup>T</sup>*; *<sup>L</sup>*2(Ω)) = *<sup>f</sup>* : [0, *<sup>T</sup>*] <sup>→</sup> *<sup>L</sup>*2(Ω)<sup>|</sup> *<sup>f</sup>* <sup>∈</sup> *AC*([*t*0, *<sup>t</sup>*1]; *<sup>L</sup>*2(Ω)) <sup>∪</sup> *AC*((*ti*, *ti*+1], *<sup>L</sup>*2(Ω), *<sup>i</sup>* <sup>=</sup> 1, 2, ··· , *P*, *x*(*t* + *<sup>i</sup>* ), *x*(*t* − *<sup>i</sup>* ) exist and *x*(*ti*) = *x*(*t* − *i* ) , *PC*(0, *<sup>T</sup>*; *<sup>L</sup>*2(Ω)) = *<sup>f</sup>* : [0, *<sup>T</sup>*] <sup>→</sup> *<sup>L</sup>*2(Ω)<sup>|</sup> *<sup>f</sup>* <sup>∈</sup> *<sup>C</sup>*([*t*0, *<sup>t</sup>*1]; *<sup>L</sup>*2(Ω)) <sup>∪</sup> *<sup>C</sup>*((*ti*, *ti*+1], *<sup>L</sup>*2(Ω), *<sup>i</sup>* <sup>=</sup> 1, 2, ··· , *P*, *x*(*t* + *<sup>i</sup>* ), *x*(*t* − *<sup>i</sup>* ) exist and*x*(*ti*) = *x*(*t* − *i* ) .

The functions and operators defined below are very standard in the fractional calculus. For more details, we refer to [20]:

1. Mittag-Leffler function by

$$E\_{\alpha, \mathfrak{E}}(z) := \sum\_{k=0}^{\infty} \frac{z^k}{\Gamma(\alpha k + \beta)}, \quad z \in \mathbb{C}\_{\omega}$$

where *<sup>α</sup>* <sup>&</sup>gt; 0 and *<sup>β</sup>* <sup>∈</sup> <sup>R</sup> are arbitrary constants. We can directly verify that *<sup>E</sup>α*,*β*(*z*) is an entire function of *<sup>z</sup>* <sup>∈</sup> <sup>C</sup>. As for the Mittag–Leffler functions, we have the following lemma.

**Lemma 1.** *Let* <sup>0</sup> <sup>&</sup>lt; *<sup>α</sup>* <sup>&</sup>lt; <sup>2</sup> *and <sup>β</sup>* <sup>∈</sup> <sup>R</sup> *be arbitrary and <sup>μ</sup> satisfy πα*/2 <sup>&</sup>lt; *<sup>μ</sup>* <sup>&</sup>lt; min{*π*, *πα*}*. Then, there exists a constant C* = *C*(*α*, *β*, *μ*) > 0 *such that*

$$|E\_{n, \theta}(z)| \le \frac{C}{1 + |z|}, \quad \mu \le |\arg(z)| \le \pi. \tag{5}$$

2. Reimann-Liouville integrals: For *<sup>α</sup>* <sup>&</sup>gt; 0 and *<sup>f</sup>* <sup>∈</sup> *<sup>L</sup>*1(0, *<sup>T</sup>*), we define *<sup>α</sup>*-th order forward and backward integrals of *f* by

$$\begin{aligned} I\_{0+}^{\alpha}f(t) &:= \frac{1}{\Gamma(\alpha)} \int\_0^t (t-\tau)^{\alpha-1} f(\tau)d\tau, \\\ I\_{T-}^{\alpha}f(t) &:= \frac{1}{\Gamma(\alpha)} \int\_t^T (\tau-t)^{\alpha-1} f(\tau)d\tau. \end{aligned}$$

In other words, the forward integral operators of *α*-th order is the convolution with *t <sup>α</sup>*−1/Γ(*α*) and consequently *I<sup>α</sup>* <sup>0</sup><sup>+</sup> *<sup>f</sup>* also belongs to *<sup>L</sup>*1(0, *<sup>T</sup>*). The same argument is also valid for the backward integrals.

3. The Riemann-Liouvill fractional derivatives: For *α* ∈ (0, 1), we define the forward and backward fractional derivatives of *f* ∈ *AC*[0, *T*] by

$$
\partial\_t^a f(t) := \frac{d}{dt} I^{1-a} h(t) = \frac{1}{\Gamma(1-a)} \frac{d}{dt} \int\_0^t (t-\tau)^{-a} h(\tau) d\tau,\tag{6}
$$

$$D\_t^\kappa f(t) := \frac{1}{\Gamma(1-a)} \left(-\frac{d}{dt}\right) \int\_t^T (\tau - t)^{-a} h(\tau) d\tau. \tag{7}$$

We also have the following lemmas for fractional integration by parts.

**Lemma 2.** *Let <sup>α</sup>* <sup>&</sup>gt; <sup>0</sup>*. If f* , *<sup>g</sup>* <sup>∈</sup> *PC*([0, *<sup>T</sup>*], *<sup>L</sup>*2(Ω))*, then*

$$\int\_0^T I\_{0+}^{\alpha} f(t) \mathcal{g}(t) dt = \int\_0^T f(t) I\_{T-}^{\alpha} \mathcal{g}(t) dt.$$

**Proof.**

$$\begin{aligned} \int\_0^T g(t) I\_{0+}^a f(t) dt &= \int\_0^T g(t) \int\_0^t \frac{(t-s)^{a-1}}{\Gamma(a)} f(s) ds dt \\ &= \int\_0^T f(t) \int\_t^T \frac{(s-t)^{a-1}}{\Gamma(a)} g(s) ds dt \\ \text{ (using Eubini theorem for chain)} \end{aligned}$$

(using Fubini theorem for change of order of integration.)

$$=\int\_0^T f(t)I^a\_T \, g(t)dt.$$

**Lemma 3.** *Let f* <sup>∈</sup> *PC*(0, *<sup>T</sup>*), *<sup>g</sup>* <sup>∈</sup> *<sup>C</sup>*<sup>∞</sup> <sup>0</sup> (0, *T*)*. Then, we have the following identity:*

$$\int\_{0}^{T} \mathcal{g}(t) \partial\_{t}^{a} f(t) dt = \int\_{0}^{T} f(t) D\_{t}^{a} \mathcal{g}(t) dt. \tag{8}$$

**Proof.** By substituing the value of R-L fractional derivative, we obtain

$$\begin{split} &\quad \int\_{0}^{T} g(t) \partial\_{t}^{a} f(t) dt \\ &= \int\_{0}^{T} g(t) \frac{d}{dt} \int\_{0}^{t} \frac{(t-s)^{-a}}{\Gamma(1-a)} f(s) ds dt \\ &= \left(\frac{g(t)}{\Gamma(1-a)} \int\_{0}^{t} (t-s)^{-a} f(s) ds\right)\_{t=0}^{t=T} - \int\_{0}^{T} g'(t) \int\_{0}^{t} \frac{(t-s)^{-a}}{\Gamma(1-a)} f(s) ds dt \\ &\quad \text{(using integration by parts.)} \\ &= \frac{g(T)}{\Gamma(1-a)} \int\_{0}^{T} (t-s)^{-a} f(s) ds - \int\_{0}^{T} g'(t) \int\_{0}^{t} \frac{(t-s)^{-a}}{\Gamma(1-a)} f(s) ds dt \\ &= -\int\_{0}^{T} g'(t) \int\_{0}^{t} \frac{(t-s)^{-a}}{\Gamma(1-a)} f(s) ds dt \quad (\because \ g(T) = 0.) \\ &= -\int\_{0}^{T} f(t) \int\_{t}^{T} \frac{(s-t)^{-a}}{\Gamma(1-a)} g'(s) ds dt \\ \text{(using Euler's theorem for change of order of integration)} \end{split}$$

(using Fubini theorem for change of order of integration.)

$$=-\int\_{0}^{T} f(t) \frac{d}{dt} \int\_{t}^{T} \frac{(s-t)^{1-\kappa}}{\Gamma(2-\alpha)} g'(s) ds dt$$

(using Leibnitz theorem for differentiation under integration.)

$$\begin{split} &= \int\_{0}^{T} f(t) \left( \frac{\mathbf{g}(T)(T-t)^{1-\alpha}}{\Gamma(1-\alpha)} - \frac{d}{dt} \int\_{t}^{T} \frac{(s-t)^{1-\alpha}}{\Gamma(2-\alpha)} \mathbf{g}'(s) ds \right) dt \\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \int\_{0}^{T} \frac{\mathbf{g}(s)}{\Gamma(2-\alpha)} \frac{\mathbf{g}'(s)}{\Gamma(2-\alpha)} \mathbf{g}'(s) ds \right) dt \\ &= \int\_{0}^{T} f(t) \left( \frac{d}{dt} \left( \frac{\mathbf{g}(s)(s-t)^{1-\alpha}}{\Gamma(2-\alpha)} \right)\_{s=t}^{s=T} - \frac{d}{dt} \int\_{t}^{T} \frac{(s-t)^{1-\alpha}}{\Gamma(2-\alpha)} \mathbf{g}'(s) ds \right) dt \\ &= - \int\_{0}^{T} f(t) \frac{d}{dt} \int\_{t}^{T} \frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} \mathbf{g}(s) ds dt \\ &\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad$$
