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.