We study the time-fractional advection-diffusion equation in a line segment \(0< x< L\),

$$ \frac{\partial^{\alpha} c}{\partial t^{\alpha} } = a \frac{\partial^{2}c}{\partial x ^{2}} - v \frac{\partial c}{\partial x } , \quad 0< \alpha\leq1. $$

(8)

As usually \(a>0\), \(v>0\), \(0< t< \infty\).

The advection diffusion equation (8) is considered under the zero initial condition

$$ t=0:\quad c(x,t) =0, $$

(9)

and the Dirichlet boundary conditions at the ends of the segment:

$$\begin{aligned}& x=0:\quad c(x,t) = g_{0} \delta(t), \end{aligned}$$

(10)

$$\begin{aligned}& x=L:\quad c(x,t) =0, \end{aligned}$$

(11)

where \(\delta(t)\) is the Dirac delta function. The constant multiplier \(g_{0}\) has been introduced to obtain the nondimensional quantity displayed in the figures.

The new sought-for function

$$ c(x,t) = \exp \biggl( \frac{vx}{2a} \biggr) u(x,t) $$

(12)

reduces the considered initial-boundary-value problem to the following one:

$$\begin{aligned}& \frac{\partial^{\alpha} u}{\partial t^{\alpha} } = a \frac{\partial^{2}u}{\partial x ^{2}} - \frac{v^{2}}{4a} u, \end{aligned}$$

(13)

$$\begin{aligned}& t=0:\quad u(x,t) =0, \end{aligned}$$

(14)

$$\begin{aligned}& x=0:\quad u(x,t) = g_{0} \delta(t), \end{aligned}$$

(15)

$$\begin{aligned}& x=L:\quad u(x,t) =0 . \end{aligned}$$

(16)

To solve the Dirichlet problem under examination, we use the finite sin-Fourier transform with respect to the spatial coordinate *x*. Such a transform is the convenient reformulation of the sin-Fourier series in the domain \(0\leq x \leq L\):

$$\begin{aligned}& \mathcal{F} \bigl\{ u(x) \bigr\} = \widetilde{u} ( \xi_{k} ) = \int _{0}^{L} u(x) \sin ( \xi_{k} x ) \,\mathrm{d} x, \end{aligned}$$

(17)

$$\begin{aligned}& \mathcal{F}^{-1} \bigl\{ \widetilde{u} ( \xi_{k} ) \bigr\} = u(x) = \frac{2}{L} \sum_{k=1}^{\infty} \widetilde{u} ( \xi _{k} ) \sin ( \xi_{k} x ) , \end{aligned}$$

(18)

with

$$ \xi_{k}= \frac{k\pi}{L},\quad k=1,2,3,\ldots. $$

(19)

The finite sin-Fourier transform of the second order derivative of a function is calculated according to the relation

$$ \mathcal{F} \biggl\{ \frac{\mathrm{d}^{2}u(x)}{\mathrm{d}x^{2}} \biggr\} = - \xi^{2}_{k} \widetilde{u} ( \xi_{k} ) + \xi_{k} \bigl[u(0) - (-1 )^{k} u(L) \bigr] . $$

(20)

Application of the finite sin-Fourier transform (17) to equation (13) using (15), (16), and (20) leads to

$$\begin{aligned}& \frac{\partial^{\alpha} \widetilde{u} (\xi_{k},t )}{\partial t^{\alpha}} = \biggl( - a \xi_{k}^{2} - \frac{v^{2}}{4a} \biggr) \widetilde{u} (\xi_{k},t ) + a \xi_{k}g_{0} \delta(t) , \end{aligned}$$

(21)

$$\begin{aligned}& t=0: \quad\widetilde{u} (\xi_{k},t ) =0. \end{aligned}$$

(22)

Next, we use the Laplace transform with respect to the time *t*. Recall that the Caputo fractional derivative for the Laplace transform requires the knowledge of the initial values of the function and its integer derivatives of the order \(k=1,2, \dots, n-1\) [20–22]:

$$ \mathcal{L} \biggl\{ \frac{\mathrm{d}^{\alpha} u(t)}{\mathrm{d}t^{\alpha}} \biggr\} = s^{\alpha} u^{*} (s) - \sum_{k=0}^{n-1} u^{(k)}\bigl(0^{+}\bigr)s^{\alpha-1-k},\quad 1 < \alpha< n, $$

(23)

where *s* is the transform variable.

Applying the Laplace transform to equation (21) under the initial condition (22) gives

$$ \widetilde{u}^{ *} (\xi_{k},s ) =ag_{0} \frac{\xi_{k}}{s^{\alpha} + a\xi_{k}^{2} + \frac {v^{2}}{4a}} . $$

(24)

Inversion of the integral transforms results in the solution

$$ u(x,t) = \frac{2ag_{0}t^{\alpha-1}}{L} \sum_{k=1}^{\infty} \xi_{k} \sin ( \xi_{k}x ) E_{\alpha,\alpha} \biggl[- \biggl(a\xi_{k}^{2} + \frac{v^{2}}{4a} \biggr)t^{\alpha} \biggr] , $$

(25)

where the formula [20–22]

$$ \mathcal{L}^{-1} \biggl\{ \frac{s^{\alpha- \beta}}{s^{\alpha}+b} \biggr\} = t^{\beta-1}E_{\alpha, \beta} \bigl(-bt^{\alpha} \bigr) $$

(26)

has been used with \(E_{\alpha, \alpha}\) being the Mittag-Leffler function in two parameters *α* and *β*:

$$ E_{\alpha, \beta}(z) = \sum_{n=0}^{\infty} \frac{z^{n}}{\Gamma(\alpha n + \beta)},\quad \alpha, \beta>0. $$

(27)

Returning to the quantity \(c(x,t)\) according to (12), we finally get the fundamental solution to the Dirichlet problem:

$$ c(x,t) = \frac{2ag_{0}t^{\alpha-1}}{L} \exp \biggl(\frac{vx}{2a} \biggr) \sum _{k=1}^{\infty} \xi_{k} \sin ( \xi_{k}x ) E_{\alpha,\alpha} \biggl[- \biggl(a\xi_{k}^{2} + \frac{v^{2}}{4a} \biggr)t^{\alpha} \biggr] . $$

(28)

Using the nondimensional quantities

$$ \bar{x} = \frac{x}{L},\qquad \bar{\xi_{k}}=L\xi_{k}=k\pi,\qquad \bar{v}= \frac{vL}{a},\qquad \kappa= \frac{\sqrt{a}t^{\alpha/2}}{L} ,\qquad \bar{c} = \frac{L^{2}c}{ag_{0}t^{\alpha-1}} , $$

(29)

we obtain

$$ \bar{c}(\bar{x},\kappa) = 2 \exp \biggl(\frac{\bar{v}\bar{x}}{2} \biggr) \sum _{k=1}^{\infty} \bar{\xi}_{k} \sin ( \bar{\xi} _{k}\bar{x} ) E_{\alpha,\alpha} \biggl[-\kappa^{2} \biggl( \bar{\xi} _{k}^{2} + \frac{\bar{v}^{2}}{4} \biggr)t^{\alpha} \biggr] . $$

(30)

The fundamental solution (30) is shown in Figures 1-4 for different values of the order of fractional derivative *α*, the time parameter *κ*, and the drift parameter *v̄*.