The one-dimensional GILTT solution

$$\frac{\stackrel{c }{{\partial }^{\alpha }}c(z,t)}{\partial {t}^{\alpha }}={k}_{z}^{{\prime }}\frac{\partial c(z,t)}{\partial z}+{k}_{z}\frac{{\partial }^{2}c(z,t)}{\partial {z}^{2}}+Q\delta (z-{h}_{s})\delta (t-{t}_{0}), 0<\alpha \le 1$$

7

Eq. (7) is solved together with the following boundary condition:

$${k}_{z}\frac{\partial c}{\partial z}=0 at z=0,z={z}_{i}$$

8

supposing the solution of Eq. (7) in the form:

$$c(z,t)=\sum _{n=0}^{N}\text{}\frac{{c}_{n}\left(t\right){\zeta }_{n}\left(z\right)}{{N}_{n}^{\frac{1}{2}}}$$

10

with \({N}_{n}\) given by

$${N}_{n}={\int }_{\upsilon }\text{}{\zeta }_{n}^{2}\left(z\right)d\upsilon$$

11

$${\zeta }_{n}\left(z\right)=\text{c}\text{o}\text{s}\left({\lambda }_{n}z\right) , {\lambda }_{n}=\frac{n\pi }{{z}_{i}}$$

12

where \({\zeta }_{n}\left(z\right)\) is the eigen function and \({\lambda }_{n}\) is the eigen value,

\(c(z,t)\) is the mean concentration in one dimension, \(c\) is the caputo derivative, \(\alpha\) is the order of the fractional, spatial operator \({k}_{z}\) is the vertical eddy diffusivity, \(Q\) is the emission rate and \(\delta (... )\) is the delta function

After introducing Eq. (10) in Eq. (7), multiplying with \(\frac{{\zeta }_{m}\left(z\right)}{{N}_{m}^{\frac{1}{2}}}\) and integrating with respect to \(z\) from \(0\to {z}_{i}\), one gets:

$$\begin{array}{l}\left.\begin{array}{c}\sum _{n=0}^{N}\text{}\frac{{d}^{\alpha }{c}_{n}\left(t\right)}{d{t}^{\alpha }}{\int }_{0}^{{z}_{i}}\text{}\frac{{\zeta }_{n}\left(z\right)}{{N}_{n}^{\frac{1}{2}}}\frac{{\zeta }_{m}\left(z\right)}{{N}_{m}^{\frac{1}{2}}}dz=\sum _{n=0}^{N}\text{}{c}_{n}\left(t\right){\int }_{0}^{{z}_{i}}\text{}\left(\frac{{\zeta }_{m}\left(z\right)}{{N}_{m}^{\frac{1}{2}}}{k}_{z}^{{\prime }}\frac{d\left(\frac{{\zeta }_{n}\left(z\right)}{{N}_{n}^{\frac{1}{2}}}\right)}{dz}\right.\left.+\frac{{\zeta }_{m}\left(z\right)}{{N}_{m}^{\frac{1}{2}}}{k}_{z}\frac{{d}^{2}\left(\frac{{\zeta }_{n}\left(z\right)}{{N}_{n}^{\frac{1}{2}}}\right)}{d{z}^{2}}\right)dz\\ +Q\delta (t-{t}_{0}){\int }_{0}^{{z}_{i}}\text{}\frac{{\zeta }_{m}\left(z\right)}{{N}_{m}^{\frac{1}{2}}}\delta (z-{h}_{s})dz\end{array}\right\}\end{array}$$

13

Upon, employing the properties of the eigenfunction \({\zeta }_{m}\left(z\right)\) and the eigenvalues \({\lambda }_{n}\) as given in Eq. (12), and defined the matrices **A, B** as:

$$A=(a{)}_{n,m}={\int }_{0}^{{z}_{i}}\text{}\frac{{\zeta }_{m}\left(z\right)}{{N}_{m}^{\frac{1}{2}}}\frac{{\zeta }_{n}\left(z\right)}{{N}_{n}^{\frac{1}{2}}}dz$$

14

$$=-\frac{1}{{N}_{n}^{\frac{1}{2}}{N}_{m}^{\frac{1}{2}}}\left[{\int }_{0}^{{z}_{i}}\text{}{k}_{z}^{{\prime }}{\zeta }_{m}\left(z\right){\zeta }_{n}^{{\prime }}\left(z\right)dz+{\int }_{0}^{{z}_{i}}\text{}{k}_{z}{\zeta }_{m}\left(z\right){\zeta }_{n}^{{\prime }{\prime }}\left(z\right)dz\right]$$

$$=-\frac{1}{{N}_{n}^{\frac{1}{2}}{N}_{m}^{\frac{1}{2}}}\left[{\int }_{0}^{{z}_{i}}\text{}{k}_{z}^{{\prime }}{\zeta }_{m}\left(z\right){\zeta }_{n}^{{\prime }}\left(z\right)dz-{\lambda }_{n}^{2}{\int }_{0}^{{z}_{i}}\text{}{k}_{z}{\zeta }_{m}\left(z\right){\zeta }_{n}\left(z\right)dz\right]$$

15

Eq. (13) becomes:

$$\sum _{n=0}^{N}\text{}(a{)}_{n,m}\frac{{d}^{\alpha }{c}_{n}\left(t\right)}{d{t}^{\alpha }}=\sum _{n=0}^{N}\text{}-(b{)}_{n,m}{c}_{n}\left(t\right)+Q\frac{{\zeta }_{m}\left({h}_{s}\right)}{{N}_{m}^{\frac{1}{2}}}M\delta (t-{t}_{0})$$

16

where \(M\) is the unit column matrix, stand for:

$$M=\left(\begin{array}{l}1\\ 1\\ ⋮\\ 1\\ \end{array}\right)$$

Eq. (16) can be written in matrix form as:

$$\frac{{\partial }^{\alpha }}{\partial {t}^{\alpha }}Y\left(t\right)+FY\left(t\right)=\eta \delta (t-{t}_{0})$$

17

For \(t>0\), where \(Y\left(t\right)\) is the column matrix, stand for:

$$Y\left(t\right)=\left(\begin{array}{l}{c}_{0}\left(t\right)\\ {c}_{1}\left(t\right)\\ .\\ .\\ {c}_{N}\left(t\right)\\ \end{array}\right)$$

18

and

$$F={A}^{-1}B , \eta ={A}^{-1}Q\frac{{\zeta }_{m}\left({h}_{s}\right)}{{N}_{m}^{\frac{1}{2}}}M$$

19

Eq. (17) is one dimensional matrix differential equation in \(t\). upon, using Laplace transform on Eq. (17), which transform \(t\) to \(s\) and \(Y\left(t\right)\) to \(\stackrel{\sim}{Y}\left(s\right)\) namely:

$$\stackrel{\sim}{Y}\left(s\right)={\int }_{0}^{\infty }\text{}{e}^{-st}Y\left(t\right)dt$$

20

and by virtue of definition of Laplace transform of a fractional derivative of \(\alpha\) order, which given by Caputo’s formula [20], namely:

$$\mathcal{L}\left(\frac{{\partial }^{\alpha }}{\partial {t}^{\alpha }}Y\left(t\right)\right)={s}^{\alpha }\stackrel{\sim}{Y}\left(s\right)-\sum _{k=0}^{n-1}\text{}{s}^{\alpha -k-1}{Y}^{k}\left(0\right), n-1<\alpha \le n$$

21

which in our case is:

$$\mathcal{L}\left(\frac{{\partial }^{\alpha }}{\partial {t}^{\alpha }}Y\left(t\right)\right)={s}^{\alpha }\stackrel{\sim}{Y}\left(s\right)-{s}^{\alpha -1}Y\left(0\right), 0<\alpha \le 1$$

22

Eq. (17) is transformed to:

$${s}^{\alpha }\stackrel{\sim}{Y}\left(s\right)-{s}^{\alpha -1}Y\left(0\right)+F\stackrel{\sim}{Y}\left(s\right)=\eta {e}^{-s{t}_{0}}$$

23

and after putting the matrix \(F\) as:

Eq. (23) can be rewritten as:

$$\stackrel{\sim}{Y}\left(s\right)=X\frac{1}{{s}^{\alpha }I+D}{X}^{-1}\left({s}^{\alpha -1}Y\left(0\right)+\eta {e}^{-s{t}_{0}}\right)$$

25

$$Y\left(t\right)={\mathcal{L}}^{-1}\stackrel{\sim}{Y}\left(s\right)$$

$$=X{\mathcal{L}}^{-1}\left(\frac{{s}^{\alpha -1}}{{s}^{\alpha }I+D}\right){X}^{-1}Y\left(0\right)+X{\mathcal{L}}^{-1}\left(\frac{{e}^{-s{t}_{0}}}{{s}^{\alpha }I+D}\right){X}^{-1}\eta$$

26

the inverse Laplace transform of the first term in the RHS (Right Hand Side) of the pervious Equation can be evaluated as:

$${\mathcal{L}}^{-1}\left(\frac{{s}^{\alpha -1}}{{s}^{\alpha }I+D}\right)=\left(\begin{array}{llllll}{E}_{\alpha }(-{d}_{1}{t}^{\alpha })& 0& & & & \\ 0& {E}_{\alpha }(-{d}_{2}{t}^{\alpha })& & & & \\ .& 0& {E}_{\alpha }(-{d}_{3}{t}^{\alpha })& & & \\ .& .& .& & & \\ .& .& & & & \\ 0& 0& & & {E}_{\alpha }(-{d}_{N}{t}^{\alpha })& \\ & & & & & \end{array}\right)$$

27

in which \({E}_{\alpha }(-{d}_{1}{t}^{\alpha })\) is Mittage-Leffer function defined as:

$${E}_{\alpha }(-{d}_{i}{t}^{\alpha })=\sum _{n=0}^{\infty }\text{}\frac{(-{d}_{i}{t}^{\alpha })}{{\Gamma }(n\alpha +1)} ,\alpha >0$$

28

\(Y\left(0\right)\) in Eq. (26) can be evaluated by introducing Eq. (10) in Eq. (3), one gets:

Then Eq. (26) becomes:

$$Y\left(t\right)=X{\mathcal{L}}^{-1}\left(\frac{{e}^{-s{t}_{0}}}{{s}^{\alpha }I+D}\right){X}^{-1}\eta$$

$$=X\left(\sum _{n=0}^{\infty }\text{}\frac{(-{t}_{0}{)}^{n}}{n!}{\mathcal{L}}^{-1}\left(\frac{{s}^{n}}{{s}^{\alpha }I+D}\right)\right){X}^{-1}\eta$$

$$=XG\left(t\right){X}^{-1}\eta$$

30

where \(G\left(t\right)\) is a diagonal matrix which elements are series in the form:

$$\sum _{n=0}^{\infty }\text{}\frac{(-{t}_{0}{)}^{n}}{n!}{t}^{\alpha -n}{E}_{\alpha ,\alpha -n}(-{\lambda }_{i}{t}^{\alpha })$$

$${E}_{\alpha ,\beta }\left(x\right)=\sum _{k=0}^{\infty }\text{}\frac{{x}^{k}}{{\Gamma }\left(k\alpha +\beta \right)} \alpha ,\beta >0$$

Then we get the solution \(c(z,t)\) by substituting Eq. (12) and Eq. (??) in Eq. (10). Similarly, we will find \(c(x,t)\) and \(c(y,t)\) as previously in section (3).