On the cumulant expansion up scaling of ground water contaminant transport equation with nonequilibrium sorption

Abstract

The laboratory-scale ground water transport equation with nonequilibrium sorption reaction subjected to unsteady, nondivergence-free, and nonstationary velocity fields is up-scaled to the field-scale by using the ensemble-averaged equations obtained from the cumulant expansion ensemble-averaging method. It is found that existing ensemble-averaged equations obtained with the help of the cumulant expansion method for the system of linear partial differential equations are not second-order exact. Although the cumulant expansion methodology is designed for noncommuting operators, it is found that there are still commudativity requirements that need to be satisfied by the functions and constants exist in the coefficient matrix of the system of ordinary/partial differential equations. A reversibility requirement, which covers the commudativity requirements, is also proposed when applying the cumulant expansion method to a system of partial differential equations/a partial differential equation. The significance of the new velocity correction obtained in this study due to the applied second-order exact cumulant expansion is investigated on a numerical example with a linear trend in the distribution coefficient. It is found that the effect of the new velocity correction can be significant enough to affect the maximum concentration values and the plume center of mass in the case of a trending distribution coefficient in a physically heterogeneous environment.

Appendices

Appendix 1

Following van Kampen (1981) the terms inside the ordering symbol may be considered as if they commute time wise so the Eq. 17 can be written as

$$ {\left\lceil{{\sum\limits_{n = 0}^\infty\frac{1}{n!}\Bigg({\int\limits_0^t{d \tau A_{0} (\tau) +{\int\limits_0^t{d \tau \alpha A_{1} (\tau)\Bigg)^{n}}}}}}} \right\rceil}a ={\left\lceil{\exp\Bigg [{\int\limits_0^t{d \tau A_{0} (\tau)\Bigg]}}} \right\rceil} {\left\lceil{\exp \Bigg[{\int\limits_0^t{d\tau \alpha A_{1} (\tau)}}\Bigg]} \right\rceil}a $$

(36)

If ${\int\nolimits_0^t{d \tau A_{0} (\tau)}}$ and ${\int\nolimits_0^t{d \tau \alpha A_{1} (\tau)}}$ commute Eq. 36 becomes

$$ {\left\lceil {{\sum\limits_{n = 0}^\infty {\frac{1} {{n!}}{\sum\limits_{k = 0}^n {{\left( {\begin{array}{*{20}c} {n} \\ {k} \\ \end{array} } \right)}} }\Bigg({\int\limits_0^t {d\tau A_{0} (\tau )\Bigg)^{{n - k}}\Bigg ({\int\limits_0^t {d\tau \alpha A_{1} (\tau )\Bigg)^{k} } }} }} }} \right\rceil }a = {\left\lceil {\exp\Bigg[{\int\limits_0^t {d\tau A_{0} (\tau )\Bigg]} }} \right\rceil }\,{\left\lceil {\exp\Bigg[{\int\limits_0^t {d\tau \alpha A_{1} (\tau )} }\Bigg]} \right\rceil }a $$

(37)

which is

$$ {\left\lceil {{\sum\limits_{n = 0}^\infty {{\sum\limits_{k = 0}^n {\frac{{\Bigg({\int\limits_0^t {d\tau A_{0} (\tau )\Bigg)^{{n - k}} \Bigg({\int\limits_0^t {d\tau \alpha A_{1} (\tau )\Bigg)^{k} } }} }}} {{(n - k)!k!}}} }} }} \right\rceil }a = {\left\lceil {\exp \Bigg[{\int\limits_0^t {d\tau A_{0} (\tau )\Bigg]} }} \right\rceil }\,{\left\lceil {\exp\Bigg[{\int\limits_0^t {d\tau \,\alpha A_{1} (\tau )}}\Bigg]} \right\rceil }a $$

(38)

Eq. 38 is also equal to

$$ {\left\lceil{\exp\Bigg[{\int\limits_0^t{d \tau A_{0} (\tau)\Bigg]\exp\Bigg[{\int\limits_0^t{d\tau \alpha A_{1} (\tau)}}\Bigg]}}} \right\rceil}a ={\left\lceil{\exp\Bigg[{\int\limits_0^t{d \tau A_{0} (\tau)\Bigg]}}} \right\rceil} {\left\lceil{\exp\Bigg[{\int\limits_0^t{d\tau \alpha A_{1} (\tau)}}\Bigg]} \right\rceil}a $$

(39)

Since the time ordered and regular exponentials of a constant matrix is the same Eq. 39 is automatically satisfied by A ₀(τ) and α A ₁(τ) matrices made up of constants. When A ₀(τ) and α A ₁(τ) contain time dependent functions at least one of those matrices should be a diagonal matrix in order to have the Eq. 39 satisfied since the time ordered exponential and regular exponential of a diagonal matrix with time dependent functions is the same.

Appendix 2

By using Eqs. 25 and 29 together with the separation given in Eqs. 23 and 24 the right hand side of Eq. 16 can be written as

$$ \begin{aligned} {\rm RHS} &={\left\lceil{\exp\Bigg[{\int\limits_0^t{d\tau \Bigg(- \langle v_{x} (\tau) \rangle \frac{{\partial }}{{\partial x}}\Bigg)}}\Bigg]} \right\rceil} \\ & \exp\Bigg[{\int\limits_0^t{d\tau \Bigg (- \frac{{\partial \langle v_{x} (x + \varsigma ^{*}, \tau) \rangle}}{{\partial x}} - k_{r} K_{d} (x + \varsigma ^{*}, \tau) - k_{r}\Bigg)}}\Bigg] {\hbox{I}} \\ & {\left\lceil{{\hbox{exp}}igg[{\int\limits_{{0}}^{{\rm t}} \begin{aligned} d\tau \exp\Bigg[ -{\int\limits_0^\tau{d\eta (- \frac{{\partial \langle v_{x} (x + \xi ^{*}, \eta) \rangle}}{{\partial x}} - k_{r} K_{d} (x + \xi ^{*}, \eta) - k_{r})}}\Bigg] {\hbox{I}}\\ \alpha{\hbox{A}}_{{{1}}} (x + \varsigma ^{*}, \tau) \\ \exp \Bigg[{\int\limits_0^\tau{d\eta (- \frac{{\partial \langle v_{x} (x + \xi ^{*}, \eta) \rangle}}{{\partial x}} - k_{r} K_{d} (x + \xi ^{*}, \eta) - k_{r})}}\Bigg] {\hbox{I}} \Bigg] \\ \end{aligned}}} \right\rceil}a \\ \end{aligned} $$

(40)

in which

$$ \xi ^{*} ={\int\limits_0^\eta{d\mu \langle v_{x} (x,\mu) \rangle }} \quad \varsigma ^{*} ={\int\limits_0^\tau{d\eta \langle v_{x} (x,\eta) \rangle }} $$

Applying functions and operators in Eq. 40 one obtains

$$ \begin{aligned} {\rm RHS} &={\left\lceil{\exp \Bigg[{\int\limits_0^t{d\tau \Bigg(- \langle v_{x} (\tau) \rangle \frac{{\partial }}{{\partial x}}\Bigg)}}\Bigg]} \right\rceil} \\ & \exp \Bigg[{\int\limits_0^t{d\tau (- \frac{{\partial \langle v_{x} (x + \varsigma ^{*}, \tau) \rangle}}{{\partial x}} - k_{r} K_{d} (x + \varsigma ^{*}, \tau) - k_{r})}}\Bigg] {\hbox{I}} \\ & {\left\lceil{ \Bigg[{\hbox{exp}}{\int\limits_{{0}}^{t}{d\tau \alpha{\hbox{A}}^+_{{{1}}} (x + \varsigma ^{*}, \tau)\Bigg ]}}} \right\rceil}a \\\end{aligned} $$

(41)

in which

$$ \alpha A^{ + }_{1} (x + \varsigma ^{*} ,\tau ) = {\left[ {\begin{array}{*{20}c} {\begin{aligned} & ( - v_{x} (x + \varsigma ^{*} ,\tau ) + \langle v_{x} (x + \varsigma ^{*} ,\tau ) \rangle )\frac{\partial } {{\partial x}} \\ & + ( - v_{x} (x + \varsigma ^{*} ,\tau ) + \langle v_{x} (x + \varsigma ^{*} ,\tau ) \rangle )\frac{{\partial E}} {{\partial x}} \\ & - \frac{{\partial ( - v_{x} (x + \varsigma ^{*} ,\tau ) + \langle v_{x} (x + \varsigma ^{*} ,\tau ) \rangle )}} {{\partial x}} \\ & + D_{x} \frac{{\partial ^{2} \,}} {{\partial x^{2} }} + D_{x} \frac{{\partial ^{2} E\,}} {{\partial x^{2} }} + k_{r} \\ \end{aligned} } & {{k_{r} }} \\ {{k_{r} K_{d} (x + \varsigma ^{*} ,\tau )}} & {\begin{aligned} & \langle v_{x} (x + \varsigma ^{*} ,\tau ) \rangle \frac{\partial } {{\partial x}} \\ & + \langle v_{x} (x + \varsigma ^{*} ,\tau ) \rangle \frac{{\partial E}} {{\partial x}} \\ & + \frac{{\partial \langle v_{x} (x + \varsigma ^{*} ,\tau ) \rangle }} {{\partial x}} \\ & + k_{r} K_{d} (x + \varsigma ^{*} ,\tau ) \\ \end{aligned} } \\ \end{array} } \right]} $$

$$ \begin{aligned} E &={\int\limits_0^\tau{d\eta \Bigg(- \frac{{\partial \langle v_{x} (x + \xi ^{*}, \eta) \rangle}}{{\partial x}} - k_{r} K_{d} (x + \xi ^{*}, \eta) - k_{r}\Bigg)}}\\ \xi ^{*} &={\int\limits_0^\eta{d\mu \langle v_{x} (x,\mu) \rangle }}\quad \varsigma ^{*} ={\int\limits_0^\tau{d\eta \langle v_{x} (x,\eta) \rangle }}\\ \end{aligned} $$

Separating $(\frac{{\partial \langle v_{x} (x + \varsigma ^{*}, \tau) \rangle}}{{\partial x}} + k_{r} K_{d} (x + \varsigma ^{*}, \tau) + k_{r}) {\hbox{I}}$ from $\alpha A^{+}_{1} (x + \varsigma ^{*}, \tau)$ with the help of Eq. 29, one obtains

$$ \begin{aligned} {\rm RHS} =&{\left\lceil{\exp \Bigg[{\int\limits_0^t{d\tau \Bigg(- \langle v_{x} (\tau) \rangle \frac{{\partial }}{{\partial x}}\Bigg)}}\Bigg]} \right\rceil} \\ & \exp \Bigg[{\int\limits_0^t{d\tau \Bigg(- \frac{{\partial \langle v_{x} (x + \varsigma ^{*}, \tau) \rangle}}{{\partial x}} - k_{r} K_{d} (x + \varsigma ^{*}, \tau) - k_{r}\Bigg)}}\Bigg] {\hbox{I}} \\ & \exp \Bigg[{\int\limits_0^t{d\tau \Bigg(\frac{{\partial \langle v_{x} (x + \varsigma ^{*}, \tau) \rangle}}{{\partial x}} + k_{r} K_{d} (x + \varsigma ^{*}, \tau) + k_{r}\Bigg)}}\Bigg] {\hbox{I}} \\ & {\left\lceil{ \Bigg[{\hbox{exp}}{\int\limits_{{0}}^{{\rm t}}{d\tau \alpha{\hbox{A}}^{{{{+ +}}}}_{{{1}}} (x + \varsigma ^{*}, \tau) \Bigg]}}} \right\rceil}a \\ \end{aligned} $$

(42)

in which

$$ \alpha A^{{ + + }}_{1} (x + \varsigma ^{*} ,\tau ) = {\left[ {\begin{array}{*{20}c} {\begin{aligned} & ( - v_{x} (x + \varsigma ^{*} ,\tau ) + \langle v_{x} (x + \varsigma ^{*} ,\tau ) \rangle )\frac{\partial } {{\partial x}} \\ & - \frac{{\partial ( - v_{x} (x + \varsigma ^{*} ,\tau ))}} {{\partial x}} + D_{x} \frac{{\partial ^{2} }} {{\partial x^{2} }} - k_{r} K_{d} (x + \varsigma ^{*} ,\tau ) \\ \end{aligned} } & {{k_{r} } } \\ {{k_{r} K_{d} (x + \varsigma ^{*} ,\tau )} } & {{ \langle v_{x} (x + \varsigma ^{*} ,\tau ) \rangle \frac{\partial } {{\partial x}} - k_{r} } } \\ \end{array} } \right]} $$

$$ \varsigma ^{*} ={\int\limits_0^\tau{d\eta \langle v_{x} (x,\eta) \rangle }} $$

Since the second and the third exponentials cancel each other, we are left only with the first and the fourth exponentials in Eq. 42. If the equivalence of the fourth exponential according to Eq. 29 is used in Eq. 42, one can write

$$ \begin{aligned} {\rm RHS}=& {\left\lceil{\exp \Bigg[{\int\limits_0^t{d\tau \Bigg(- \langle v_{x} (\tau) \rangle \frac{{\partial }}{{\partial x}}\Bigg)}}\Bigg]} \right\rceil}{\left\lfloor {\exp \Bigg[{\int\limits_0^t {d\tau \Bigg(\langle v_{x} (\tau) \rangle \frac{{\partial }}{{\partial x}}\Bigg)}}\Bigg]} \right\rfloor} \\ & {\left\lceil { {\hbox{exp}} \Bigg[ {\int\limits_{{0}}^{{\rm t}} {d\tau (A_{0} (\tau) + \alpha A_{1} (\tau)) \Bigg]}}} \right\rceil}a \\ \end{aligned} $$

(43)

Eq. 43 is equal to

$$ {\rm RHS} = {\left\lceil {\Bigg[{\hbox {exp}}{\int\limits_{{0}}^{{\rm t}} {d\tau (A_{0} (\tau) + \alpha A_{1} (\tau)) \Bigg]}}} \right\rceil}a $$

(44)

which is the left hand side of Eq. 16.