A general polynomial solution to convection–dispersion equation using boundary layer theory

Abstract

A number of models have been established to simulate the behaviour of solute transport due to chemical pollution, both in croplands and groundwater systems. An approximate polynomial solution to convection–dispersion equation (CDE) based on boundary layer theory has been verified for the use to describe solute transport in semi-infinite systems such as soil column. However, previous studies have only proposed low order polynomial solutions such as parabolic and cubic polynomials. This paper presents a general polynomial boundary layer solution to CDE. Comparison with exact solution suggests the prediction accuracy of the boundary layer solution varies with the order of polynomial expression and soil transport parameters. The results show that prediction accuracy increases with increasing order up to parabolic or cubic polynomial function and with no distinct relationship between accuracy and order for higher order polynomials ($n\geqslant 3$). Comparison of two critical solute transport parameters (i.e., dispersion coefficient and retardation factor), estimated by the boundary layer solution and obtained by CXTFIT curve-fitting, shows a good agreement. The study shows that the general solution can determine the appropriate orders of polynomials for approximate CDE solutions that best describe solute concentration profiles and optimal solute transport parameters. Furthermore, the general polynomial solution to CDE provides a simple approach to solute transport problems, a criterion for choosing the right orders of polynomials for soils with different transport parameters. It is also a potential approach for estimating solute transport parameters of soils in the field.

Appendix

We assume the nth order polynomial approximate concentration profile for a boundary layer solution to CDE to express resident concentration as:

$$\begin{array}{@{}rcl@{}} &&C_{r} \left( {x,t} \right)= a_{0} \left( t \right)\,+\,a_{1} \left( t \right)x\,+\,a_{2} \left( t \right)x^{2}\,+\,\cdots\\ && +a_{n-2} \left( t \right)x^{n-2}\,+\,a_{n-1} \left( t \right)x^{n-1}\,+\,a_{n} \left( t \right)x^{n}. \end{array} $$

(A1)

The boundary-layer conditions assumed in this paper include

$$\begin{array}{@{}rcl@{}} C_{r} \left( {d\left( t \right),t} \right)\!\!\!\!&=&\!\!\!\!\frac{\partial C_{r} \left( {d\left( t \right),t} \right)}{\partial x}\,=\,\frac{\partial^{2}C_{r} \left( {d\left( t \right),t} \right)}{\partial x^{2}}\\ &=&\!\!\!\!\frac{\partial^{3}C_{r} \left( {d\left( t \right),t} \right)}{\partial x^{3}}\,=\,{\cdots} \\ &=&\!\!\!\!\frac{\partial^{n-3}C_{r} \left( {d\left( t \right),t} \right)}{\partial x^{n-3}}\,=\,\frac{\partial^{n-2}C_{r} \left( {d\left( t \right),t} \right)}{\partial x^{n-2}}\\ &=&\!\!\!\! \frac{\partial^{n-1}C_{r} \left( {d\left( t \right),t} \right)}{\partial x^{n-1}}\,=\,0. \end{array} $$

(A2)

Equation (A2) is corrected for n > 1. When n = 1, the boundary-layer condition is

$$ C_{r} \left( {d\left( t \right),t} \right)=0, $$

(A3)

and the (n − 1)-order derivative of equation (A1) is expressed as:

$$\begin{array}{@{}rcl@{}} \frac{\partial^{n-1}C_{r} \left( {d\left( t \right),t} \right)}{\partial x^{n-1}}\!\!\!\!&=&\!\!\!\!\left( \left( {n-1} \right)\left( {n-2} \right)\left( {n-3} \right)\cdots\right. \\ &&\!\!\!\! \left.\times\, 3\times 2\times 1 \right)a_{n-1} \left( t \right)\\ &&\!\!\!\! +\left( n\left( {n-1} \right)\left( {n-2} \right)\cdots\right.\\ &&\!\!\!\! \left. \times\, 4\times 3\times 2 \right)a_{n} \left( t \right)d\left( t \right)\\ &=&\!\!\!\!0. \end{array} $$

(A4)

Then, a _n (t) is expressed by a _n−1(t) as follows:

$$\begin{array}{@{}rcl@{}} a_{n} \left( t \right)=-\frac{a_{n-1} \left( t \right)}{nd\left( t \right)}. \end{array} $$

(A5)

Also, the (n − 2)-order derivative of equation (A5) is given as:

$$\begin{array}{@{}rcl@{}}\frac{\partial^{n-2}C_{r} \left( {d\left( t \right),t} \right)}{\partial x^{n-2}}\!\!\!\!&=&\!\!\!\! \left( \left( {n\,-\,2} \right)\left( {n\,-\,3} \right)\left( {n\,-\,4} \right)\cdots\right.\\ &&\!\!\!\! \left. \times\, 3\!\times\! 2\!\times\! 1 \right)a_{n-2} \left( t \right) \\ &&\!\!\!\! + \left( \left( {n\,-\,1} \right)\left( {n\,-\,2} \right)\left( {n\,-\,3} \right)\cdots\right.\\ &&\!\!\!\! \left. \times\, 4\!\times\! 3\!\times\! 2 \right)a_{n-1} \left( t \right)d\left( t \right) \\ &&\!\!\!\! + \left( n\left( {n\,-\,1} \right)\left( {n\,-\,2} \right){\cdots} \!\times\! 5\right.\\ &&\!\!\!\! \left.\times\, 4\!\times\! 3 \right)a_{n} \left( t \right)d^{2}\left( t \right)\,=\,0. \end{array} $$

(A6)

Substituting equation (A5) to equation (A6), the relationship of a _n−1(t) and a _n−2(t) is obtained by

$$\begin{array}{@{}rcl@{}} a_{n-1} \left( t \right)=-\frac{2a_{n-2} \left( t \right)}{\left( {n-1} \right)d\left( t \right)}. \end{array} $$

(A7)

Then, integrating equation (A5) with equation (A7), a _n (t) is expressed in terms of a _n−2(t) as follows:

$$ a_{n} \left( t \right)=\frac{2a_{n-2} \left( t \right)}{n\left( {n-1} \right)d^{2}\left( t \right)}. $$

(A8)

In the same way, the k-order derivative of equation (A1) is obtained as follows:

$$\begin{array}{@{}rcl@{}} \frac{\partial^{k}C_{r} ({d(t),t})}{\partial x^{k}}\!\!\!\! &=&\!\!\!\! (k({k\,-\,1})({k\,-\,2})\!\times\! {\cdots} \!\times\! 3\!\times\! 2\! \times\! 1)a_{k} (t) \\ &&\!\!\!\! +\, (({k\,+\,1})k({k\,-\,1})\!\cdots\\ &&\!\!\!\! \times\, 4\!\times\! 3 \!\times\! 2)a_{k+1} (t)d (t) \,+\, \cdots\\ &&\!\!\!\! +\, (({n\,-\,1}) ({n\,-\,2}) ({n\,-\,3}) \!\times\! \cdots\\ &&\!\!\!\! \times\, ({n\,-\,k})a_{n-1} (t) d^{n-1-k} (t))\\ &&\!\!\!\! +\, (n({n\,-\,1})({n\,-\,2})\times \cdots\\ &&\!\!\!\! \times\, ({n\,-\,k\,+\,1})a_{n} (t)d^{n-k} (t)). \end{array} $$

(A9)

Then, equation (A9) is correct when 1 ≤ k ≤ n − 1. By using boundary layer conditions, equations (A2) and (A9) are simplified as follows:

$$\begin{array}{@{}rcl@{}} ({n-k})!a_{k} (t)\!\!\!\!\!&+&\!\!\!\!\frac{({k\,+\,1})!}{k!}\frac{({n\,-\,k})!}{1!}a_{k+1} (t)d(t)\,+\,{\cdots} \\ &+&\!\!\!\! \frac{({n\,-\,1})!}{k!}\frac{({n\,-\,k})!}{({n\,-\,1\,-\,k})!}a_{n-1} (t)d^{n-1-k}(t)\\ &+&\!\!\!\! \frac{n!}{k!}\frac{({n\,-\,k})!}{({n\,-\,k})!}a_{n} (t)d^{n-k}(t)\,=\,0 . \end{array} $$

(A10)

Each term in equation (A10) can be expressed in terms of a _k (t). From this analogy, the nth time coefficients in equation (A1) are reduced to a single coefficient. The coefficient of the kth term of equation (A1) then becomes

$$\begin{array}{@{}rcl@{}} a_{k}(t)\!\!\!&=&\!\!\!({-1})^{k}\frac{n!}{k!({n\,-\,k})!d^{k}(t)}a_{0}(t)\\ \!\!&=&\!\!\!({-1})^{k}{C}_{n}^{k} a_{0} (t) . \end{array} $$

(A11)

Equation (A11) is valid for k ≤ n. Therefore, the alternative expression of equation (A1) with only one coefficient is given as

$$ C_{r} \left( {x,t} \right)=\frac{vd\left( t \right)C_{0}} {vd\left( t \right)+nD}\left( {1-\frac{x}{d\left( t \right)}} \right)^{n}\!\!\!. $$

(A12)

Then,using the same deduction procedure discussed in the main text of the paper, d(t) is obtained as

$$ d(t)\,=\,\frac{({n\,+\,1})vt}{2R}\,+\,\sqrt{\left( {\frac{({n\,+\,1})vt}{2R}}\right)^{2}\,+\,\frac{n({n\,+\,1})Dt}{R}} . $$

(A13)