Analytical solutions of contaminant transport in homogeneous and isotropic aquifer in three-dimensional groundwater flow

Abstract

Modeling three-dimensional contaminant transport released from arbitrary shape source geometries is useful in hydrological and environmental sciences. This article produces several analytical solutions for three-dimensional contaminant transport in a homogeneous and isotropic aquifer by using Green’s function with the groundwater flow which is assumed to be in three directions. The solutions are obtained for both finite depth aquifer and semi-infinite depth aquifer. Various types of sources are discussed: point, line, plane, or cuboid sources. The continuous and instantaneous sources are also investigated. A MATLAB coding is developed to calculate the numerical integrals which occur at the solutions. Some solutions are verified with the solutions obtained in the literature. This study confirms the effect of groundwater velocities in all directions on the degree and the directions of contaminant spreading. Additionally, the results highlight the significant effect of the geometrical shape of the contaminant sources on contaminant concentrations for instantaneous and continuous sources. In particular, the cuboid source and the horizontal rectangular source provide the highest concentrations. The analytical solutions developed in this article can be applied for a wide range of contaminant transport.

Appendices

Appendix 1

To solve Eqs. (2)–(6), Green’s function approach is used. Initially, both the lower and upper boundaries along the z direction are assumed non-penetrable for a contaminant. The following transformations are used:

$${x}^{\ast }=x-\frac{U_xt}{R};\kern0.5em {y}^{\ast }=y-\frac{U_yt}{R},\kern0.5em {z}^{\ast }=z-{U}_zt/R,{\displaystyle \begin{array}{cc}\kern0.5em \overline{C}\left(x,y,t\right)=C\left(x,y,t\right)\mathit{\exp}\left(\nu t\right)& {\overline{C}}_s(t)={C}_s(t)\mathit{\exp}\left(\nu t\right)\end{array}}$$

(16)

with the following dimensionless representations:

$${\displaystyle \begin{array}{ccc}{t}_{\mathrm{D}}={D}_zt/\left({Rd}^2\right);& {C}_{\mathrm{D}}={D}_z\overline{C}/\left({RS}_0{d}^2\right);& {C}_{{\mathrm{s}}_{\mathrm{D}}}={\overline{C}}_{\mathrm{s}}/\left({S}_0R\right)\\ {}{x}_{\mathrm{D}}^{\ast }={x}^{\ast}\left(\sqrt{D_z/{D}_x}\right)/d;& {y}_D^{\ast }={y}^{\ast}\left(\sqrt{D_z/{D}_y}\right)/d;& {z}_D^{\ast }={z}^{\ast }/d\\ {}{U}_{x_D}={U}_xd/\sqrt{D_z{D}_x};& {U}_{y_D}={U}_yd/\sqrt{D_z{D}_y}& {U}_{z_D}={U}_zd/{D}_z\end{array}}$$

(17)

Thus, Eq. (1) can now be converted to another equation (see Appendix 2):

$${C}_D/\partial {t}_D-{\partial}^2{C}_D/\partial {x_D^{\ast}}^2-{\partial}^2{C}_D/\partial {y_D^{\ast}}^2-{\partial}^2{C}_D/\partial {z_D^{\ast}}^2=\left\{\begin{array}{l}{C}_{s_D}\left({t}_D\right)\kern0.5em \mathrm{inside}\kern0.17em \mathrm{the}\kern0.22em \mathrm{source}\ \\ {}0\kern3.5em \mathrm{outside}\kern0.17em \mathrm{the}\kern0.22em \mathrm{source}\end{array}\right.$$

(18)

and the boundary conditions are converted to

$${\displaystyle \begin{array}{l}{C}_D\left(\pm \infty, {y}_D^{\ast },{z}_D^{\ast },{t}_D\right)=0\\ {}{C}_D\left({x}_D^{\ast },\pm \infty, {z}_D^{\ast },{t}_D\right)=0\\ {}\partial {C}_D\left({x}_D^{\ast },\kern0.5em {y}_D^{\ast }-{U}_{z_D}{t}_D,{t}_D\right)/\partial {z}_d=\partial {C}_D\left({x}_D^{\ast },{y}_D^{\ast },1-{U}_{z_D}{t}_D,{t}_D\right)/\partial {z}_d=0\end{array}}\kern1.25em {\displaystyle \begin{array}{l}--\infty \le {x}_D^{\ast}\le \infty, \kern0.5em -{U}_{z_D}{t}_D\le {z}_D^{\ast}\le 1-{U}_{z_D}{t}_D\\ {}-\infty \le {x}_D^{\ast}\le \infty \kern0.75em ,\kern0.75em -{U}_{z_D}{t}_D\le {z}_D^{\ast}\le 1-{U}_{z_D}{t}_D\\ {}-\infty \le {y}_D^{\ast}\le \infty, \kern2.25em -\infty \le {x}_D^{\ast}\le \infty \end{array}}\kern1.25em {\displaystyle \begin{array}{l}\mathrm{and}\ {t}_D\ge 0\\ {}\mathrm{and}\ {t}_D\ge 0\\ {}\mathrm{and}\ {t}_D\ge 0\end{array}}$$

(19)

While the initial condition is converted to

$${\displaystyle \begin{array}{ccc}{C}_D\left({x}_D^{\ast },{y}_D^{\ast },{z}_D^{\ast },0\right)=0,\kern0.5em & -\infty <{x}_D^{\ast }<\infty & -{U}_{z_D}{t}_D\le {z}_D^{\ast}\le 1-{U}_{z_D}{t}_D\end{array}}$$

(20)

Green’s function approach is implemented to solve Eqs. (18)–(20). Here, Green’s function is used to represent the pollutants’ concentration at the point (x, y, z, t) in the aquifer generated by an instantaneous point source (x^′, y^′, z^′, τ) of strength unity. The concentration is assumed equal to zero at the aquifer and boundary surface. Initially, Green’s function (G) will be obtained by solving the following equation:

$${\partial}^2G/\partial {x_D^{\ast}}^2+{\partial}^2G/\partial {x_D^{\ast}}^2+{\partial}^2G/\partial {z_D^{\ast}}^2-\partial G/\partial {t}_D=\delta \left({x}_D^{\ast }-{x}_D^{\prime}\right)\delta \left({y}_D^{\ast }-{y}_D^{\prime}\right)\delta \left({z}_D^{\ast }-{z}_D^{\prime}\right)\delta \left({t}_D-{\tau}_D\right)$$

(21)

where δ is the Dirac delta function. Consequently, Eq. (36) can be solved by calculating the following integral:

$${\displaystyle \begin{array}{c}{C}_D\left({x}_D^{\ast },{y}_D^{\ast },{z}_D^{\ast },{t}_D\right)=\int_0^{t_D}{C}_{s_D}\left({\tau}_D\right){\int}_{\varDelta }G\left({x}_D^{\ast },{y}_D^{\ast },{z}_D^{\ast },{t}_D-{\tau}_D\right) d\varDelta d{\tau}_D\\ {}\kern5em =\int_0^{t_D}{C}_{s_D}\left({\tau}_D\right) SF\left({x}_D^{\ast },{y}_D^{\ast },{z}_D^{\ast },{t}_D-{\tau}_D\right)d{\tau}_D\end{array}}$$

(22)

where ∆ is the source domain and SF is the source function that is equal to the integration of Green’s function over the area or volume of the source. To obtain the three-dimensional solution function (G) for Eq. (21), one-dimensional Green’s functions in x, y, and z directions will be calculated and multiplied as follows (Carslaw and Jaeger 1986; Park and Zhan 2001):

$$G\left({x}_D^{\ast },{y}_D^{\ast },{z}_D^{\ast },{t}_D-{\tau}_D\right)=G\left({x}_D^{\ast },{t}_D-{\tau}_D\right)G\left({y}_D^{\ast },{t}_D-{\tau}_D\right)G\left({z}_D^{\ast },{t}_D-{\tau}_D\right)$$

(23)

Consequently,

$$SF\left({x}_D^{\ast },{y}_D^{\ast },{z}_D^{\ast },{t}_D-{\tau}_D\right)= SF\left({x}_D^{\ast },{t}_D-{\tau}_D\right) SF\left({y}_D^{\ast },{t}_D-{\tau}_D\right) SF\left({z}_D^{\ast },{t}_D-{\tau}_D\right)$$

(24)

The SF function in the x direction is the integration of Green’s function over the source domain along the x-axis (Gringarten and Ramey Jr. 1973; Carslaw and Jaeger 1986; Park and Zhan 2001).

$${\displaystyle \begin{array}{l} SF\left({x}_D^{\ast },{t}_D-{\tau}_D\right)=1/\left(2\sqrt{\pi}\right){\int}_{-{U_x}_D{\tau}_D}^{x_{0_D}-{U}_{x_D}{\tau}_D}1/\left(\sqrt{t_D-{\tau}_D}\right)\exp \left(-{\left({x}_D^{\ast }-{\psi}_D\right)}^2/\left(4\left({t}_D-{\tau}_D\right)\right)\right)d{\psi}_D\\ {}=\frac{1}{2}\left(\operatorname{erfc}\ \left(\left({x}_D^{\ast }-{x}_{0_D}+{U}_{x_D}{\tau}_D\right)/\left(2\sqrt{t_D-{\tau}_D}\right)\right)-\operatorname{erfc}\left(\left({x}_D^{\ast }+{U}_{x_D}{\tau}_D\right)/\left(2\sqrt{t_D-{\tau}_D}\right)\right)\right)\end{array}}$$

(25)

In the same way the dimensionless function can be calculated along y-axis:

$${\displaystyle \begin{array}{l} SF\left({y}_D^{\ast },{t}_D-{\tau}_D\right)=1/\left(2\sqrt{\pi}\right){\int}_{-{y}_{0_D}-{U_y}_D{\tau}_D}^{y_{0_D}-{U}_{y_D}{\tau}_D}1/\left(\sqrt{t_D-{\tau}_D}\right)\exp \left(-{\left({y}_D^{\ast }-{\zeta}_D\right)}^2/\left(4\left({t}_D-{\tau}_D\right)\right)\right)d{\zeta}_D\\ {}=\frac{1}{2}\left(\operatorname{erfc}\ \left(\left({y}_D^{\ast }-{y}_{0_D}+{U}_{y_D}{\tau}_D\right)/\left(2\sqrt{t_D-{\tau}_D}\right)\right)-\operatorname{erfc}\left(\left({y}_D^{\ast }+{y}_{0_D}+{U}_{y_D}{\tau}_D\right)/\left(2\sqrt{t_D-{\tau}_D}\right)\right)\right)\end{array}}$$

(26)

Since the domain is finite in the vertical direction, the SF function along the z-axis is equal to the following (Gringarten and Ramey Jr. 1973; Carslaw and Jaeger 1986; Park and Zhan 2001):

$${\displaystyle \begin{array}{l}\begin{array}{l} SF\left({z}_D^{\ast },{t}_D-{\tau}_D\right)={\int}_{z_{0_D}-{U}_{z_D}{\tau}_D}^{z_{1_D}-{U}_{z_D}{\tau}_D}\Big[1+2\sum_{n=1}^{\infty}\cos n\pi {\eta}_D\cos n\pi {z}_D^{\ast}\exp \left(-{n}^2{\pi}^2\left({t}_D-{\tau}_D\right)\right)d{\eta}_D\\ {}={z}_{1_D}-{z}_{0_D}+\frac{2}{\pi }{\Sigma_{n=1}^{\infty}}^{\frac{1}{n}\left[\sin \left( n\pi \left({z}_{1_D}-{U}_{z_D}{\tau}_D\right)\right)-\sin \left( n\pi \left({z}_{0_D}-{U}_{z_D}{\tau}_D\right)\right)\right]\cos n\pi {z}_D^{\ast }}\end{array}\\ {}\kern14.5em \exp \left(-{n}^2{\pi}^2\left({t}_D-{\tau}_D\right)\right)\end{array}}$$

(27)

By substituting Eqs. (24), (25), and (26) into Eq. (23), we get

$${\displaystyle \begin{array}{l}\begin{array}{l}\begin{array}{l} SF\left({x}_D^{\ast },{y}_D^{\ast },{z}_D^{\ast },\kern0.5em {t}_D-{\tau}_D\right)=\left[\frac{1}{2}\left(\operatorname{erfc}\ \left(\left({x}_D^{\ast }-{x}_{0_D}+{U}_{x_D}{\tau}_D\right)/\left(2\sqrt{t_D-{\tau}_D}\right)\right)-\operatorname{erfc}\left(\left({x}_D^{\ast }+{U}_{x_D}{\tau}_D\right)/\left(2\sqrt{t_D-{\tau}_D}\right)\right)\right)\right]\\ {}\times \left[\frac{1}{2}\left(\operatorname{erfc}\ \left(\left({y}_D^{\ast }-{y}_{0_D}+{U}_{y_D}{\tau}_D\right)/\left(2\sqrt{t_D-{\tau}_D}\right)\right)-\operatorname{erfc}\left(\left({y}_D^{\ast }+{y}_{0_D}+{U}_{y_D}{\tau}_D\right)/\left(2\sqrt{t_D-{\tau}_D}\right)\right)\right)\right]\end{array}\\ {}\times \left[{z}_{1_D}-{z}_{0_D}+\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{1}{n}\left[\sin \left( n\pi \left({z}_{1_D}-{U}_{z_D}{\tau}_D\right)\right)-\sin \left( n\pi \left({z}_{0_D}-{U}_{z_D}{\tau}_D\right)\right)\right]\cos n\pi {z}_D^{\ast}\exp \left(-{n}^2{\pi}^2\left({t}_D-{\tau}_D\right)\right)\right]\end{array}\\ {}\kern14.5em \end{array}}$$

(28)

Thus, from Eq. (22), the solution is

$$\begin{array}{l}\begin{array}{l}\begin{array}{l}\begin{array}{l}\begin{array}{l}C_D\left(x_D^\ast,y_D^\ast,z_D^\ast,t_D\right)=\frac14\int_0^{t_D}C_{s_D}\left(\tau_D\right)\\\times\left[\left(\operatorname{erfc}\;\left(\left(x_D^\ast-x_{0_D}+U_{x_D}\tau_D\right)/\left(2\sqrt{t_D-\tau_D}\right)\right)-\operatorname{erfc}\left(\left(x_D^\ast+U_{x_D}\tau_D\right)/\left(2\sqrt{t_D-\tau_D}\right)\right)\right)\right]\end{array}\\\times\left[\left(\operatorname{erfc}\;\left(\left(y_D^\ast-y_{0_D}+U_{y_D}\tau_D\right)/\left(2\sqrt{t_D-\tau_D}\right)\right)-\operatorname{erfc}\left(\left(y_D^\ast+y_{0_D}U_{x_D}\tau_D\right)/\left(2\sqrt{t_D-\tau_D}\right)\right)\right)\right]\end{array}\\\times\left[z_{1_D}-z_{0_D}+\frac2\pi\sum\limits_{n=1}^\infty\frac1n\left[\sin\left(n\pi\left(z_{1_D}-U_{z_D}\tau_D\right)\right)-\sin\left(n\pi\left(z_{0_D}-U_{z_D}\tau_D\right)\right)\right]\cos n\pi z_D^\ast\right.\end{array}\\\left.\;\;\;\;\;\;\;\;\;\;\;\;\;\exp\left(-n^2\pi^2\left(t_D-\tau_D\right)\right)\right]d\tau_{D,}\end{array}\\\end{array}$$

(29)

In a dimensional format, the solution becomes

$$\begin{array}{l}\begin{array}{l}\begin{array}{l}\begin{array}{l}\begin{array}{l}\mathrm C\left(x,y,z,t\right)=1/\left(4dR\right)\int_0^tC_s\left(\tau\right)\;\exp\left(-\nu\left(t-\tau\right)\right)\\\times\left[\operatorname{erfc}\left[\sqrt R\left(x-x_0-U_x\left(t-\tau\right)/R\right)/\left(2\sqrt{D_x\left(t-\tau\right)}\right)\right]-\operatorname{erfc}\left[\sqrt R\left(x-U_x\left(t-\tau\right)/R\right)/\left(2\sqrt{D_x\left(t-\tau\right)}\right)\right]\right]\end{array}\\\times\left[\operatorname{erfc}\left[\sqrt R\left(y-y_0-U_y\left(t-\tau\right)/R\right)/\left(2\sqrt{D_y\left(t-\tau\right)}\right)\right]-\operatorname{erfc}\left[\sqrt R\left(y+y_0-U_y\left(t-\tau\right)/R\right)/\left(2\sqrt{D_y\left(t-\tau\right)}\right)\right]\right]\end{array}\\\times\left[z_1-z_0+\frac2\pi\sum\limits_{n=1}^\infty\frac1n\left[\sin\left(\frac{n\pi\left(z_1-U_z\tau/R\right)}d\right)-\sin\left(\frac{n\pi\left(z_0-U_z\tau/R\right)}d\right)\right]\cos\left(\frac{n\pi\left(z-U_zt/R\right)}d\right)\right.\end{array}\\\left.\;\exp\left(\frac{-n^2\pi^2D_z\left(t-\tau\right)}{Rd^2}\right)\right]d\tau\end{array}\\\end{array}$$

(30)

By changing the integration parameter from τ to τ^′ = t − τ we will get Eq. (8)

$${\displaystyle \begin{array}{l}C\left(x,y,z,t\right)=1/\left(4 dR\right)\int_0^t{C}_s\left(t-{\tau}^{\prime}\right)\;\exp \left(-\nu {\tau}^{\prime}\right)\\ {}\times \left[\operatorname{erfc}\ \left[\sqrt{R}\left(x-{x}_0-{U}_x{\tau}^{\prime }/R\right)/\left(2\sqrt{D_x{\tau}^{\prime }}\right)\right]-\operatorname{erfc}\left[\sqrt{R}\left(x-{U}_x{\tau}^{\prime }/R\right)/\left(2\sqrt{D_x{\tau}^{\prime }}\right)\right]\right]\\ {}\begin{array}{l}\times \left[\operatorname{erfc}\ \left[\sqrt{R}\left(y-{y}_0-{U}_y{\tau}^{\prime }/R\right)/\left(2\sqrt{D_y{\tau}^{\prime }}\right)\right]-\operatorname{erfc}\left[\sqrt{R}\left(y+{y}_0-{U}_y{\tau}^{\prime }/R\right)/\left(2\sqrt{D_y{\tau}^{\prime }}\right)\right]\right]\\ {}\begin{array}{c}\times \left[{z}_1-{z}_0+\frac{2}{\pi }\sum\limits_{n=1}^{\infty}\left[\sin \left(\frac{n\pi \left({z}_1-{U}_z\left(t-{\tau}^{\prime}\right)/R\right)}{d}\right)-\sin \left(\frac{n\pi \left({z}_0-{U}_z\left(t-{\tau}^{\prime}\right)/R\right)}{d}\right)\right]\cos \left(\frac{n\pi \left(z-{U}_zt/R\right)}{d}\right)\right.\\ {}\left.\exp \left(\frac{-{n}^2{\pi}^2{D}_z{\tau}^{\prime }}{R{d}^2}\right)\right]d{\tau}^{\prime}\end{array}\end{array}\end{array}}$$

Appendix 2

Here, we will show the derivation of Eq. (8) from Eq. (2). By using transformations and dimensionless variables described in Eqs. (16) and (17) in Appendix 1, we get

$${\displaystyle \begin{array}{l}\frac{\partial {C}_D}{\partial {t}_D}=\frac{\partial {C}_D}{\partial t}.\frac{\partial t}{\partial {t}_D}=\frac{\partial {C}_D}{\partial t}.\frac{R{d}^2}{D_z}\\ {}\kern2em =\frac{1}{S_0}\frac{\partial \overline{C}}{\partial t}=\frac{\exp \left(\nu t\right)}{S_0}\left(\frac{\partial C}{\partial t}+\frac{\partial C}{\partial t}.\frac{\partial t}{\partial x}+\frac{\partial C}{\partial t}.\frac{\partial t}{\partial y}+\frac{\partial C}{\partial t}.\frac{\partial t}{\partial x}z\right)\\ {}\kern1.75em =\frac{\exp \left(\nu t\right)}{S_0}\frac{\partial C}{\partial t}\left(1+\frac{R}{U_x}+\frac{R}{U_y}+\frac{R}{U_z}\right)\end{array}}$$

(31)

$${\displaystyle \begin{array}{l}\frac{\partial {C}_D}{\partial {x}_D^{\ast }}=\frac{\partial {C}_D}{\partial x}.\frac{\partial x}{\partial {x}^{\ast }}.\frac{\partial {x}^{\ast }}{\partial {x}_D^{\ast }}=\frac{\partial {C}_D}{\partial x}.\frac{d}{\sqrt{D_z/{D}_x}}\\ {}\kern1.75em =\frac{\sqrt{D_z{D}_x}}{R{S}_0d}.\frac{\partial \overline{C}}{\partial x}=\exp \left(\nu t\right)\frac{\sqrt{D_z{D}_x}}{R{S}_0d}.\frac{\partial C}{\partial x}\end{array}}$$

(32)

$${\displaystyle \begin{array}{l}\frac{\partial^2{C}_D}{\partial {x}_D^{\ast 2}}=\frac{\partial }{\partial {x}_D^{\ast }}\left(\frac{\partial {C}_D}{\partial {x}_D^{\ast }}\right)=\frac{\partial }{\partial x}.\frac{\partial x}{\partial {x}^{\ast }}.\frac{\partial {x}^{\ast }}{\partial {x}_D^{\ast }}\left(\frac{\partial {C}_D}{\partial {x}_D^{\ast }}\right)=\frac{d}{\sqrt{D_z/{D}_x}}.\left(\frac{\partial {C}_D}{\partial {x}_D^{\ast }}\right)\\ {}\kern2.25em =\frac{d}{\sqrt{D_z/{D}_x}}.\exp \left(\nu t\right)\frac{\sqrt{D_z{D}_x}}{R{S}_0d}.\frac{\partial C}{\partial x}=\frac{D_x}{R{S}_0}\exp \left(\nu t\right)\frac{\partial C}{\partial x}\end{array}}$$

(33)

$${\displaystyle \begin{array}{l}\frac{\partial {C}_D}{\partial {y}_D^{\ast }}=\frac{\partial {C}_D}{\partial y}.\frac{\partial y}{\partial {y}^{\ast }}.\frac{\partial {y}^{\ast }}{\partial {y}_D^{\ast }}=\frac{\partial {C}_D}{\partial y}.\frac{d}{\sqrt{D_z/{D}_y}}\\ {}\kern2em =\frac{\sqrt{D_z{D}_y}}{R{S}_0d}.\frac{\partial \overline{C}}{\partial y}=\exp \left(\nu t\right)\frac{\sqrt{D_z{D}_y}}{R{S}_0d}.\frac{\partial C}{\partial y}\end{array}}$$

(34)

$${\displaystyle \begin{array}{l}\frac{\partial^2{C}_D}{\partial {y}_D^{\ast 2}}=\frac{\partial }{\partial {y}_D^{\ast }}\left(\frac{\partial {C}_D}{\partial {y}_D^{\ast }}\right)=\frac{\partial }{\partial y}.\frac{\partial y}{\partial {y}^{\ast }}.\frac{\partial {y}^{\ast }}{\partial {y}_D^{\ast }}\left(\frac{\partial {C}_D}{\partial {y}_D^{\ast }}\right)\frac{d}{\sqrt{D_z/{D}_y}}.\left(\frac{\partial {C}_D}{\partial {y}_D^{\ast }}\right)\\ {}\kern2em =\frac{d}{\sqrt{D_z/{D}_y}}.\exp \left(\nu t\right)\frac{\sqrt{D_z{D}_y}}{R{S}_0d}.\frac{\partial C}{\partial y}=\frac{D_y}{R{S}_0}\exp \left(\nu t\right)\frac{\partial C}{\partial y}\end{array}}$$

(35)

$${\displaystyle \begin{array}{c}\frac{\partial {C}_D}{\partial {z}_D^{\ast }}=\frac{\partial {C}_D}{\partial z}.\frac{\partial z}{\partial {z}^{\ast }}.\frac{\partial {z}^{\ast }}{\partial {z}_D^{\ast }}=\frac{\partial {C}_D}{\partial z}.d\\ {}=\frac{D_z}{R{S}_0d}.\frac{\partial \overline{C}}{\partial z}=\exp \left(\nu t\right)\frac{D_z}{R{S}_0d}.\frac{\partial C}{\partial z}\end{array}}$$

(36)

$${\displaystyle \begin{array}{l}\frac{\partial^2{C}_D}{\partial {z}_D^{\ast 2}}=\frac{\partial }{\partial {z}_D^{\ast }}\left(\frac{\partial {C}_D}{\partial {z}_D^{\ast }}\right)=\frac{\partial }{\partial z}.\frac{\partial z}{\partial {z}^{\ast }}.\frac{\partial {z}^{\ast }}{\partial {z}_D^{\ast }}\left(\frac{\partial {C}_D}{\partial {z}_D^{\ast }}\right)=d.\left(\frac{\partial {C}_D}{\partial {z}_D^{\ast }}\right)\\ {}\kern2.25em =\frac{D_y}{R{S}_0}\exp \left(\nu t\right)\frac{\partial C}{\partial z}\end{array}}$$

(37)

$${C}_{s_D}=\left\{\begin{array}{l}\frac{1}{R{S}_0}{\overline{C}}_s=\frac{1}{R{S}_0}\exp \left(\nu t\right){C}_s(t)\kern0.75em \mathrm{inside}\kern0.17em \mathrm{the}\kern0.22em \mathrm{source}\\ {}0\kern12em \mathrm{outside}\kern0.17em \mathrm{the}\kern0.22em \mathrm{source}\end{array}\right.$$

(38)

Multiply Eq. (2) by $\frac{1}{R{S}_0}\exp \left(\nu t\right)$, then

$${\displaystyle \begin{array}{l}\frac{1}{R{S}_0}\exp \left(\nu t\right)\left(R\frac{\partial C}{\partial t}-{D}_x\frac{\partial^2C}{\partial {x}^2}-{D}_y\frac{\partial^2C}{\partial {y}^2}-{D}_z\frac{\partial^2C}{\partial {z}^2}+{U}_x\frac{\partial C}{\partial x}+{U}_y\frac{\partial C}{\partial y}\right)\\ {}+{U}_z\frac{\partial C}{\partial z}+\nu RC=\frac{1}{R{S}_0}\exp \left(\nu t\right).{C}_s(t)\end{array}}$$

(39)

Which implies that

$${C}_D/\partial {t}_D-{\partial}^2{C}_D/\partial {x_D^{\ast}}^2-{\partial}^2{C}_D/\partial {y_D^{\ast}}^2-{\partial}^2{C}_D/\partial {z_D^{\ast}}^2=\left\{\begin{array}{l}{C}_{s_D}\left({t}_D\right)\kern1.25em \mathrm{inside}\kern0.17em \mathrm{the}\kern0.22em \mathrm{source}\ \\ {}0\kern4.25em \mathrm{outside}\kern0.17em \mathrm{the}\kern0.22em \mathrm{source}\end{array}\right.$$

Appendix 3

Table 5 Directional Green’s functions for the instantaneous point sources

Table 6 Analytical solutions for different continuous source types including point source, the line sources with different positions, and area sources with different positions