Fundamentals of Fractional Transport

Abstract

In many physical systems, transport plays a central role. By transport we mean any macroscopic process that moves some physical quantity of interest across the system. For instance, transport processes are responsible for transporting mass and heat, including pollutants, throughout the atmosphere. Or for transporting water and energy, but also debris, across the ocean. Or for transporting plasma density and energy, but also impurities, out of a tokamak. Or for transporting angular momentum out of an accretion disk, but also mass and energy into the black hole at its center.

Appendices

Appendix 1: The Laplace Transform

The Laplace transform ⁴⁷ of a function f(t), t > 0, is defined as [46]:

$$\displaystyle \begin{aligned} {\mathrm{L}}[\,f(t)] := \skew2\tilde f(s) \equiv \int_{0}^\infty f(t)e^{-st} dt. \end{aligned} $$

(5.112)

The main requirement for f(t) to have a Laplace transform is that it cannot grow as t →∞ as fast as an exponential. Otherwise, the Laplace integral would not converge.⁴⁸ A list of common Laplace transforms is given in Table 5.1.

Table 5.1 Some useful Laplace transforms

The Laplace transform is a linear operation that has interesting properties. A first one has to do with the translation of the independent variable in Laplace space. That is,

$$\displaystyle \begin{aligned} \skew2\tilde f(s - a) = {\mathrm{L}}\left[ \exp(at) f(t) \right]. \end{aligned} $$

(5.113)

Another one, with the scaling of the independent variable,

$$\displaystyle \begin{aligned} \skew2\tilde f(s/a) = a {\mathrm{L}}\left[\,f(a t) \right]. \end{aligned} $$

(5.114)

The Laplace transform of the n-th order derivative of a function satisfies:

$$\displaystyle \begin{aligned} {\mathrm{L}}\left[ \frac{d^n f}{dt^n} \right] = s^n\skew2\tilde f(s) - s^{n-1}f(0) - s^{n-2}f'(0) -\cdots - f^{(n-1)}(0), \end{aligned} $$

(5.115)

whilst the Laplace transform of the integral of a function,

$$\displaystyle \begin{aligned} {\mathrm{L}}\left[ \int_0^t f(t')dt' \right] = \frac{\skew2\tilde f(s)}{s}. \end{aligned} $$

(5.116)

These properties are very useful to solve linear systems of ordinary differential equations [13]. Another interesting result is known as the convolution theorem. It refers to the temporal convolution of two functions [46],

$$\displaystyle \begin{aligned} h(t) = \int_0^t f(t') g(t - t') dt', \end{aligned} $$

(5.117)

whose Laplace transform is equal to the product of the Laplace transforms of the two functions,

$$\displaystyle \begin{aligned} \tilde h(s) = \skew2\tilde f(s)\cdot \tilde g(s). \end{aligned} $$

(5.118)

We will also discuss a property particularly useful in the context of self-similar functions. The Laplace transform of a power law, f(t) = t ^a, t ≥ 0 is given by $\skew 2\tilde f(s) = \varGamma (a+1)/s^{1+a}$, for a > −1, where Γ(x) is Euler’s gamma function [46]. This result, can also be extended to any function that asymptotically scales as (a > −1),

$$\displaystyle \begin{aligned} f(t) \sim t^a,~~~t\rightarrow \infty,~~~\Longleftrightarrow ~~~\skew2\tilde f(s) \sim s^{-(1+a)},~~s\rightarrow 0. \end{aligned} $$

(5.119)

Finally, we will mention that the inverse Laplace transform is given by [46]:

$$\displaystyle \begin{aligned} f(t) = \frac{1}{2\pi i} \int_{c -i\infty}^{c+i\infty} \skew2\tilde f(s) e^{st} ds, \end{aligned} $$

(5.120)

where s varies along an imaginary line, since c is real and must be chosen so that the integral converges.

Appendix 2: Riemann-Liouville Fractional Derivatives and Integrals

Fractional integrals and derivatives were introduced as interpolants between integrals and derivatives of integer order [21, 44, 45]. In this book, we will always use the Riemann-Lioville (RL) definition of fractional integrals and derivatives.

Riemann-Liouville Fractional Integrals

The left-sided RL fractional integral of order p > 0 of a function f(t) is defined as [21]:

$$\displaystyle \begin{aligned} _{a} D^{-p}_{t} f(t) \equiv \frac{1}{\varGamma(p)}\int_a^t (t-\tau)^{p-1} f(\tau) d\tau, \end{aligned} $$

(5.121)

A negative superscript, − p, is used to reveal that we are dealing with a RL fractional integral. a is known as the starting point of the integral.⁴⁹ As advertised, it can be shown that Eq. 5.121 reduces, for p = n > 0, to the usual integral of order n.⁵⁰ But they have some other interesting properties. For instance, they satisfy a commutative composition rule [21]. Indeed, one has that for p, q > 0:

(5.124)

The Laplace transform of the RL fractional integral, if the starting point is a = 0, is particularly simple [21]. It is given by:

$$\displaystyle \begin{aligned} L\left[{}_{0} D^{-p}_{t}\cdot f(t) \right] = s^{-p} \skew2\tilde f(s). \end{aligned} $$

(5.125)

This result naturally follows from the fact that, for a = 0, the RL fractional integral becomes a temporal convolution of f(t) with a power law, t ^p/Γ(p). Thus, Eq. 5.125 follows from applying the convolution theorem (Eq. 5.118).

A simple relation also provides the Fourier transform (see Appendix 1 of Chap. 2) of the RL fractional integral if the starting point is a = −∞. It is given by⁵¹:

$$\displaystyle \begin{aligned} {\mathrm{F}}\left[{}_{-\infty} D^{-p}_{t}\cdot f(t) \right] = (-i\omega)^{-p} \skew2\hat f(\omega). \end{aligned} $$

(5.127)

Riemann-Liouville Fractional Derivatives

The left-sided RL fractional derivative of order p > 0 of a function f(t) is defined as [21]:

$$\displaystyle \begin{aligned} {}_{a} D^{\,p}_{t} f(t) \equiv \frac{1}{\varGamma(k-p)} \frac{d^k}{dt^k}\int_a^t (t-\tau)^{k-p-1} f(\tau) d\tau, \end{aligned} $$

(5.128)

where the integer k satisfies that k − 1 ≤ p < k. Note that they are simply a combination of normal derivatives and RL fractional integrals: $_{a} D^{\,p}_{t} = (d/dt)^k \cdot _a D^{-(k - p)}_{t}$. Again, it turns out that, for p = n, the RL fractional derivative reduces to the standard derivative of order n.⁵²

RL fractional derivatives have interesting, but somewhat not intuitive properties. The most striking property is probably that the fractional derivative of a constant function is not zero. Indeed, using the fact that the derivative of a power law can be calculated to be [21]:

$$\displaystyle \begin{aligned} _{a} D^{\,p}_{t} \cdot (t-a)^\nu = \frac{\varGamma(1+\nu)}{\varGamma(1+\nu -p)} (t-a)^{\nu - p},~~~ p>0,~\nu > -1,~t>0, \end{aligned} $$

(5.130)

it is clear that choosing ν = 0 does not yield a constant, but (t − a)^−p/Γ(1 − p).

RL fractional derivatives can be combined with other derivatives (fractional or integer) and derivatives. But the combinations are not always simple. One of the simplest cases is when a fractional derivative of order p > 0 acts on a fractional integral of order q > 0:

$$\displaystyle \begin{aligned} _a D^{\,p}_{t} \left({}_{a} D^{-q}_{t} f(t)\right) = _a D^{\,p-q}_{t} f(t). \end{aligned} $$

(5.131)

However, the action of a RL fractional integral of order q > 0 on a RL fractional derivative of order p > 0 is given by a much more complicated expression [21].

If one sets p = q in Eq. 5.131, one finds that the inverse (from the left) of a RL fractional derivative of order p > 0 is the fractional integral of order p > 0:

$$\displaystyle \begin{aligned} _a D^{\,p}_{t} \left({}_a D^{-p}_{t} f(t)\right) = f(t). \end{aligned} $$

(5.132)

However, note that the RL fractional derivative of order p > 0 is not the inverse from the left of the RL fractional integral of order p. Instead, one has that [21]

$$\displaystyle \begin{aligned} _a D^{-p}_{t} \left({}_a D^{\,p}_{t} f(t)\right) = f(t) - \sum_{j=1}^k \left[-a D^{\,p-j}_{t} f(t)\right]_{t=a} \frac{(t-a)^{p-j}}{\varGamma(p-j+1)}. \end{aligned} $$

(5.133)

The action of normal derivatives on RL fractional derivatives is also simple to express⁵³:

$$\displaystyle \begin{aligned} \frac{d^m }{dt^m} \cdot~ _a D^{\,p}_{t} f(t) = _a D^{\,p+m}_{t} f(t). \end{aligned} $$

(5.135)

But again, the action of the RL fractional derivative on a normal derivative is much more complicated, and given by the expression:

$$\displaystyle \begin{aligned} _a D^{\,p}_{t}\cdot \frac{d^m }{dt^m} f(t) =_a D^{\,p+m}_{t} f(t) - \sum_{j=0}^{m-1} \frac{f^{(j)}(a)(t-a)^{\,j-p-m}} {\varGamma(1+ j -p-m)}. \end{aligned} $$

(5.136)

Finally, the composition of RL fractional derivatives of different orders is given by a rather complex expression, that is never equal to a fractional derivative of a higher order, except in very special cases [21],

Relatively simple expressions also exist for the Laplace transform of the left-sided RL fractional derivative of order p if the starting point is a = 0:

$$\displaystyle \begin{aligned} L\left[{}_0 D^{\,p}_{ t}\cdot f(t) \right] = s^{p} \skew2\tilde f(s) - \sum_{j=0}^{k-1} s^{\,j} \left[{}_0 D^{\,p-j-1}_{t} \cdot f(t)\right]_{t=0}. \end{aligned} $$

(5.137)

This expression is very reminiscent of the one obtained for normal derivatives (Eq. 5.115). Similarly, the Fourier transform of the left-sided RL fractional derivative satisfies a very simple relation,⁵⁴ but only for starting point a = −∞:

$$\displaystyle \begin{aligned} {\mathrm{F}}\left[{}_{-\infty} D^{\,p}_{t} \cdot f(t) \right] = (i\omega)^{p} \skew2\hat f(\omega). \end{aligned} $$

(5.139)

Appendix 3: The Riesz-Feller Fractional Derivative

The Riesz fractional derivative of order α is defined by the integral⁵⁵:

$$\displaystyle \begin{aligned} \frac{\partial^\alpha f}{\partial |x|{}^\alpha}:= -\frac{1}{2\varGamma(\alpha)\cos{}(\alpha\pi/2)}\int_{-\infty}^\infty dx' \frac{f(x')}{|x-x'|{}^{\alpha + 1}} dx'. \end{aligned} $$

(5.140)

The most remarkable property of this derivative, and one of the reasons why it appears so often in the context of transport (see Sect. 5.3.1), has to do with its Fourier transform, which is given by [47]:

$$\displaystyle \begin{aligned} {\mathrm{F}}\left[\frac{\partial^\alpha f}{\partial |x|{}^\alpha}\right] = -|k|{}^\alpha \skew2\hat f(k). \end{aligned} $$

(5.141)

Using now the complex identity ($\i = \sqrt {-1}$),

$$\displaystyle \begin{aligned} (\i k)^\alpha + (-\i k)^\alpha = 2 \cos\left(\frac{\pi\alpha}{2}\right) |k|{}^\alpha, \end{aligned} $$

(5.142)

it is very easy to prove that the Riesz derivative can also be expressed as a symmetrized sum of two (one left-sided, one right-sided) RL fractional derivatives of order α [21],

$$\displaystyle \begin{aligned} \frac{\partial^\alpha f}{\partial |x|{}^\alpha} = -\frac{1}{2\varGamma(\alpha)\cos{}(\alpha\pi/2)} \left( \mbox{}_{-\infty}D^{\alpha}_{x} +\mbox{}^\infty D^{\alpha}_{x} \right) \end{aligned} $$

(5.143)

It is also possible to define an asymmetric version of the Riesz-Feller derivative, often known as the Riesz-Feller fractional derivative of order α and asymmetry parameter λ [48]. This can be done through its Fourier transform, that is given by:

$$\displaystyle \begin{aligned} {\mathrm{F}}\left[\frac{\partial^{\alpha,\lambda} f}{\partial |x|{}^{\alpha,\lambda}}\right] = -|k|{}^\alpha\left[1- \i \lambda {\mathrm{sgn}}(k) \tan{}(\pi\alpha/2) \right] \skew2\hat f(k). \end{aligned} $$

(5.144)

The indices are restricted, in this case, to α ∈ (0, 2) and λ ∈ [−1, 1]. For λ = 0, the standard symmetric Riesz derivative is recovered. It can also be shown that the Riesz-Feller derivative can also be expressed as an asymmetric sum of the same two RL fractional derivatives of order α [49]:

$$\displaystyle \begin{aligned} \frac{\partial^{\alpha,\lambda} f}{\partial |x|{}^{\alpha,\lambda}} = -\frac{1}{2\varGamma(\alpha)\cos{}(\alpha\pi/2)} \left( c_-(\alpha,\lambda) \mbox{}_{-\infty}D^{\alpha}_{x} + c_+(\alpha,\lambda) \mbox{}^\infty D^{\alpha}_{x} \right) \end{aligned} $$

(5.145)

with the coefficients defined as:

$$\displaystyle \begin{aligned} c_{\pm}(\alpha,\lambda) := \frac{1 \mp \lambda}{1 + \lambda\tan{}(\pi\alpha/2)} \end{aligned} $$

(5.146)

Thus, in the limit of λ = 1, only the left-sided α-RL derivative $\mbox{ }_{-\infty } D^{\alpha }_{x}$ remains, whilst for λ = −1, only the right-side one $\mbox{ }^\infty D^{\alpha }_{x}$ does.

Appendix 4: Discrete Approximations for Fractional Derivatives

In order to be useful for actual applications, discrete representations of fractional derivatives are needed that could be easily implemented in a computer [44, 50,51,52]. We discuss here one possible way to do this [51]. The main difficulty has to do with the singularities of the Riemann-Lioville definition, since it turns out that it becomes singular at the starting point, t = a, whenever a is finite. This will certainly be the case in practical cases, since a computer must always deal with finite intervals.

Case 1 < p < 2

One can made these singularities explicit by rewriting the fractional derivative $\mbox{ }_aD_t^p f$, 1 < p < 2, as the infinite series [52]:

$$\displaystyle \begin{aligned} \frac{1}{\varGamma(1-p)}\frac{f(a)}{(t-a)^p} + \sum_{k=1}^\infty \frac{f^{(k)}(a) (t-a)^{k-p}}{\varGamma(k+1-p)} \end{aligned} $$

(5.147)

simply by Taylor expanding f(t) around t = a and using Eq. 5.130 to derive the different powers of x. Clearly, the first two terms of Eq. 5.147 are singular. At least, unless one sets f(a) = 0, and f′(a) = 0 as well for 1 < p < 2, which puts too much of a restriction in some cases.

One way to circumvent this problem is to replace in practice the Riemann-Liouville derivative by the so-called Caputo fractional derivative [21]:

$$\displaystyle \begin{aligned} \mbox{}^{\mathrm{C}}_aD_t^p f := \mbox{}_aD_t^p \left[\,f - \sum_{i=0}^{{\mathrm{int}}(p)} f^{(i)}(a)(t-a)^{i} \right], \end{aligned} $$

(5.148)

where int(p) is the integer part of p.⁵⁶ The Caputo definition removes the singularities so that the expression is now well defined and can be easily discretized. To do it, we give its analytical expression [21],

$$\displaystyle \begin{aligned} \mbox{}_a^{\mathrm{C}} D^{\,p}_t f(t) \equiv \frac{1}{\varGamma(k-p)} \int_a^t (t-\tau)^{k-p-1} f^{(k)}(\tau) d\tau, \end{aligned} $$

(5.150)

with k − 1 ≤ p < k, which is very similar to the RL definition (Eq. 5.128) but with the integer derivatives acting inside of the integral, instead of outside. It can be shown that RL and Caputo derivatives are identical, for most functions for a →−∞ and t →∞. However, they are different for finite starting points. For starters, it should be noted that the Caputo derivative of a constant is now zero!

We proceed now to find discrete expressions for the Caputo derivative for 1 < p < 2 on the discrete regular mesh, t _i  = (i − 1)Δt, with i = 0, 2, ⋯ , N _t . The integral from t = t ₁ up to t = t _i is then discretized⁵⁷ following the scheme [44, 51],

$$\displaystyle \begin{aligned} \begin{array}{rcl} \mbox{}^{\mathrm{C}}_aD_t^p f &\displaystyle = &\displaystyle \frac{1}{\varGamma(2-p)} \int_a^t (t-\tau)^{1-p} f^{(2)}(\tau) d\tau \\ &\displaystyle \simeq &\displaystyle ~\frac{1}{\varGamma(2-p)} \sum_{j=0}^{i-1}\int_{t_j}^{t_{j+1}}\frac{f''(t-s)}{s^{p-1}}ds \\ &\displaystyle \simeq &\displaystyle \frac{1}{\varGamma(2-p)}\sum_{j=0}^{i-1}\left[ \frac{f(t_i - t_{j+1}) + f(t_i-t_{j-1}) - 2f(t_i-t_j)}{(\varDelta t)^2} \int_{t_j}^{t_{j+1}}\frac{ds}{s^{p-1}} \right] \\ &\displaystyle \simeq &\displaystyle \sum_{j=0}^{i-1 }\left[ \frac{(f(t_i - t_{j+1}) + f(t_i-t_{j-1}) - 2f(t_i-t_j))\left[ (j+1)^{2-p} - j^{2-p}\right]}{\varGamma(3-p)(\varDelta t)^p} \right]. \end{array} \end{aligned} $$

(5.151)

This expression is more conveniently expressed as,

$$\displaystyle \begin{aligned} \mbox{}^{\mathrm{C}}_aD_t^p f = \sum_{j=-1}^i \frac{W_j^{p>1}}{\varGamma(3-p)(\varDelta t)^p}f(t_i - t_j), \end{aligned} $$

(5.152)

in terms of the weights, $W_j^{p>1}$,

$$\displaystyle \begin{aligned} W_j^{p>1} = \left\lbrace \begin{array}{ll} \displaystyle 1, & j = -1\\ \displaystyle 2^{2-p}-3, & j = 0\\ \displaystyle (j+2)^{2-p} - 3(j+1)^{2-p} + 3j^{2-p} - (j-1)^{2-p},~~~ & 0 < j < i - i\\ \displaystyle -2i^{2-p} + 3(i-1)^{2-p}- (i-2)^{2-p}, & j = i-1 \\ \displaystyle i^{2-p} - (i-1)^{2-p}, & j = i \end{array} \right. \end{aligned} $$

(5.153)

Case 0 < α < 1

When p < 1 there is only one singularity at the starting point a of the fractional derivative. We can make it explicit by expressing again the fractional derivative $\mbox{ }_aD_t^p f$, as the infinite series:

$$\displaystyle \begin{aligned} \frac{1}{\varGamma(1-p)}\frac{f(a)}{(t-a)^p} + \sum_{k=1}^\infty \frac{f^{(k)}(a) (t-a)^{k-p}}{\varGamma(k+1-p)}, \end{aligned} $$

(5.154)

done by Taylor expanding f(t) around t = a and using Eq. 5.130 to derive the different powers of x. Clearly, the first term of Eq. 5.154 is singular. At least, unless one sets f(a) = 0.

The solution is again to use the Caputo derivative (Eq. 5.150) instead. Now, int(p) = 0 and the integral to discretize is:

$$\displaystyle \begin{aligned} \mbox{}_a^{\mathrm{C}} D^{\,p}_t f(t) \equiv \frac{1}{\varGamma(1-p)} \int_a^t (t-\tau)^{-p} f^{(k)}(\tau) d\tau, \end{aligned} $$

(5.155)

For 0 < p < 1, on the other hand, we discretize the integral from t = t ₀ up to t = t _i as,

$$\displaystyle \begin{aligned} \begin{array}{rcl} \mbox{}^{\mathrm{C}}_aD_t^p f &\displaystyle = &\displaystyle \frac{1}{\varGamma(1-p)} \int_a^t (t-\tau)^{-p} f^{(1)}(\tau) d\tau \\ &\displaystyle \simeq &\displaystyle ~\frac{1}{\varGamma(1-p)} \sum_{j=0}^{i-1}\int_{t_j}^{t_{j+1}}\frac{f'(t-s)}{s^{p}}ds \\ &\displaystyle \simeq &\displaystyle \frac{1}{\varGamma(1-p)}\sum_{j=0}^{i-1}\left[ \frac{f(t_i - t_{j+1}) - f(t_i-t_{j-1})}{2\varDelta t} \int_{t_j}^{t_{j+1}}\frac{ds}{s^{p}} \right] \\ &\displaystyle \simeq &\displaystyle \sum_{j=0}^{i-1 }\left[ \frac{(f(t_i - t_{j+1}) - f(t_i-t_{j-1}))\left[ (j+1)^{1-p} - j^{1-p}\right]}{2\varGamma(2-p)(\varDelta t)^p} \right]. \end{array} \end{aligned} $$

(5.156)

Again, we can express this formula more conveniently by introducing weights, $W_j^{p<1}$,

$$\displaystyle \begin{aligned} \mbox{}^{\mathrm{C}}_aD_t^p f = \sum_{j=-1}^i \frac{W_j^{p<1}}{2\varGamma(2-p)(\varDelta t)^p}f(t_i - t_j), \end{aligned} $$

(5.157)

which are now given by the expressions,

$$\displaystyle \begin{aligned} W_j^{p<1} = \left\lbrace \begin{array}{ll} \displaystyle -1, & j = -1\\ \displaystyle -(2^{1-p}-1), & j = 0\\ \displaystyle j^{1-p} -(j-1)^{1-p} - (j+2)^{1-p} +(j+1)^{1-p},~~~ & 0 < j < i - i\\ \displaystyle (i-1)^{1-p}- (i-2)^{1-p}, & j = i-1 \\ \displaystyle i^{1-p} - (i-1)^{1-p}, & j = i \end{array} \right. \end{aligned} $$

(5.158)

Equations 5.152 and 5.157 are not however, the only possible discrete representation of the Caputo fractional derivative of order p. These formulas are based on central differencing. But other discretizations also exist, that may sometimes offer either better accuracy over some ranges of p or higher orders of discretization [44, 50,51,52]. Equations 5.152 and 5.157 are however sufficient for the purposes of this book.

Problems

5.1

CTRW: Fluid Limit

Write a code that evolves in time a one-dimensional CTRW that has a Gaussian jump size distribution, $p(\varDelta x) = N_{[0,\sigma ^2]}(\varDelta x)$, and an exponential waiting time pdf, $\psi (\varDelta t) = E_{\tau _0}(\varDelta t)$. In order to generate values for the jumps and waiting times at run-time, use the algorithms described in Appendix 1 of Chap. 2. Use the code to calculate numerically the CTRW propagator for σ ² = 2 and τ ₀ = 1, using N = 100, 000 particles. Compare the results with its fluid limit: G(x         −     x ₀, t)                         =                         N _[0,2t] (x − x ₀).

5.2

Langevin Equation: Propagator

Write a code that evolves in time a collection of N particles according to the Langevin equation (Eq. 5.26). Use a uniform noise with autocorrelation given by Eq. 5.27 with D = 2. Use N = 100, 000 particles to calculate the propagator of the Langevin equation and compare it with its analytical solution: G(x − x ₀, t) = N _[0,2t](x − x ₀).

5.3

Langevin Equation: Moments of the Propagator

Compute all moments of the propagator of the Langevin equation and show that all odd moments vanish and all even moments n > 2 are given by m _n  = (Dt)ⁿ(n − 1)!!.

5.4

Fractional Transport Equation: Propagator for α < 2, β = 1

Calculate the propagator of the fractional transport equation (Eq. 5.66) in the case in which β = 1.

5.5

Scale-Invariant Generalization of Fick’s Law

Find the expression that gives the local particle flux for the fractional transport equation. To do it, recast Eq. 5.66 into the form ∂n/∂t + ∇⋅ Γ = S, using the properties of fractional derivatives discussed in Appendices 2 and 3.

5.6

Advanced Problem: Propagator of the Running Sandpile

Write a code that uses the rules discussed in Sect. 5.5 to advances an arbitrary number of tracers on the height profile evolution calculated by the sandpile code previously developed (see Prob 1.5). Use the code to estimate the numerical propagator of the running sandpile for a sandpile with L = 2000, Z _c  = 200, N _f  = 20, N _b  = 10 and p ₀ = 10⁻⁴.

5.7

Advanced Problem: Numerical Integration of the fTe

Write a code that integrates the fractional transport equation (Eq. 5.66 ) for β = 1 and arbitrary α, arbitrary initial condition, n ₀(x), and external source, S(x, t). Use, for that purpose, the discrete expressions of the Caputo fractional derivative (Eqs. 5.152 and 5.157) given in Appendix 4.