*3.1. Representation of the Solution*

To derive the representation, we first focus on *t* ∈ [0, *t*1]. We can rewrite (3) as

$$
\partial\_t u + (\beta \* A u)\_t = f(\cdot, t), \mathfrak{u}(0) = u\_{0\prime} \tag{9}
$$

where *β*(*t*) = *<sup>t</sup> α*−1 <sup>Γ</sup>(*α*) and *Au* <sup>=</sup> −∇2*<sup>u</sup>* is a symmetric, self-adjoint, uniformly elliptic operator with domain *<sup>D</sup>*(*A*) = *<sup>H</sup>*2(Ω) <sup>∩</sup> *<sup>H</sup>*<sup>1</sup> <sup>0</sup> (Ω), the spectrum of *A* is entirely composed of a countable number of eigenvalues and we can set with finite multiplicities:

$$0 < \lambda\_1 \le \lambda\_2 \le \cdots \le \lambda\_n \le \cdots \dots$$

By *<sup>ϕ</sup><sup>n</sup>* <sup>∈</sup> *<sup>H</sup>*2(Ω) <sup>∩</sup> *<sup>H</sup>*<sup>1</sup> <sup>0</sup> (Ω), we denote the orthonormal eigenfunction corresponding to *λn*:

$$A\!\!\!\!p\_n = \lambda\_n \!\!p\_n, \quad n = 1, 2, \cdots \quad \text{.} $$

Then, the sequence {*ϕn*}*n*∈<sup>N</sup> is an orthonormal basis in *<sup>L</sup>*2(Ω). Since *<sup>u</sup>*(*t*) <sup>∈</sup> *<sup>L</sup>*2(Ω), we have

$$u(t) = \sum\_{j=1}^{\infty} u\_j(t) q\_{j'}$$

where *uj*(*t*)=(*u*(*t*), *ϕj*) is the *j*th Fourier coefficient. Taking an inner product between (9) and *ϕj*, we have an infinite number of linear integro-differential equations:

$$
\partial\_t \boldsymbol{u}\_j(t) + \lambda\_j (\boldsymbol{\beta} \* \boldsymbol{u}\_j)\_t = f\_j(\cdot, t), \tag{10}
$$

where *fj*(·, *t*)=(*f*(·, *t*), *ϕj*) and *uj*<sup>0</sup> = (*u*0, *ϕj*).

Taking Laplace Transform both sides of (10), we get

$$z\mathfrak{a}\_{\dot{j}}(z) - \mathfrak{u}\_{\dot{j}0} + \lambda\_{\dot{j}} z^{1-\kappa} \mathfrak{a}\_{\dot{j}}(z) = f\_{\dot{j}\prime}$$

where ˆ *hj*(*z*) = - ∞ <sup>0</sup> *<sup>e</sup>*−*ztuj*(*t*)*dt* is the Laplace Transform of *uj*. Simplifying, we get

$$\hat{u}\_{\hat{j}} = \left(\frac{u\_{\hat{j}0} + \hat{f}\_{\hat{j}}(z)}{z + \lambda\_{\hat{j}} z^{1-\alpha}}\right).$$

By taking the inverse Laplace Transform, we get

$$
\mathcal{L}^{-1}\left(\frac{1}{z + \lambda\_j z^{1-\mathfrak{a}}}\right) = E\_{\mathfrak{a}}(-\lambda\_j t^{\mathfrak{a}}).\tag{11}
$$

Now, the representation for *uj* of (10) is given by

$$
\mu\_j = E\_a(-\lambda\_j t^a) u\_{j0} + \int\_0^\infty E\_a(-\lambda\_j (t-s)^a) f\_j(\cdot, s) ds. \tag{12}
$$

Thus, a formal solution of (9) is given by

$$u(t) = \mathcal{E}(t)u\_0 + \int\_0^t \mathcal{E}(t-s)f(\cdot, s)ds,\tag{13}$$

where

$$\mathcal{E}(t)u\_0 = \sum\_{j=1}^{\infty} E\_{\mathfrak{a}} \left( -\lambda\_j t^{\mathfrak{a}} \right) \left( \mathfrak{u}\_{0\prime} \,\, \mathfrak{p}\_j \right) \mathfrak{p}\_{j\prime} \tag{14}$$

$$u(\cdot, t) = \begin{cases} \mathcal{E}(t)u\_0 + \int\_0^t \mathcal{E}(t - s)f(\cdot, s)ds, & t \in [0, t\_1], \\ \mathcal{E}(t)u\_0 + \sum\_{l\_i < l} \mathcal{E}(t - t\_l)I\_l(u(\cdot, t\_l)) + \int\_0^t \mathcal{E}(t - s)f(\cdot, s)ds, & t \in (t\_i, t\_{i+1}], \quad i = 1, \dots, P. \end{cases} \tag{15}$$
