The Linear Boltzmann Equation in Column Experiments of Porous Media

Abstract

The use of the linear Boltzmann equation is proposed for transport in porous media in a column. By column experiments, we show that the breakthrough curve is reproduced by the linear Boltzmann equation. The advection–diffusion equation is derived from the linear Boltzmann equation in the asymptotic limit of large propagation distance and long time.

Appendices

Appendix 1: Eigenvalues

Let us write the homogeneous equation as

$$\begin{aligned} \begin{aligned} \left( \mu _tI-\frac{1}{\nu }\Xi _+\right) \Phi _+&= \frac{\mu _s}{2}W\left( \Phi _++\Phi _-\right) ,\\ \left( \mu _tI-\frac{1}{\nu }\Xi _-\right) \Phi _-&= \frac{\mu _s}{2}W\left( \Phi _++\Phi _-\right) , \end{aligned} \end{aligned}$$

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where I is the identity, $\Xi _{\pm }$ and W are matrices, and $\Phi _{\pm }$ are vectors defined as

$$\begin{aligned} \Xi _{\pm }=\begin{pmatrix} \pm \mu _1+\eta &{}&{}&{} \\ &{} \pm \mu _2+\eta &{}&{} \\ &{}&{} \ddots &{} \\ &{}&{}&{} \pm \mu _N+\eta \end{pmatrix}, \quad \Phi _{\pm }=\begin{pmatrix} \phi (\nu ,\pm \mu _1) \\ \phi (\nu ,\pm \mu _2) \\ \vdots \\ \phi (\nu ,\pm \mu _N) \end{pmatrix}, \quad \left\{ W\right\} _{ij}=w_j. \end{aligned}$$

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We obtain

$$\begin{aligned} SM\begin{pmatrix}\Phi _+ \\ \Phi _-\end{pmatrix}= \frac{1}{\nu }\begin{pmatrix}\Phi _+ \\ \Phi _-\end{pmatrix}, \end{aligned}$$

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where

$$\begin{aligned} S=\begin{pmatrix}\Xi _+^{-1} &{} \\ &{} \Xi _-^{-1}\end{pmatrix}, \quad M=\mu _t(s)\begin{pmatrix}I &{} \\ &{} I \end{pmatrix}-\frac{\mu _s}{2} \begin{pmatrix}W &{} W \\ W &{} W\end{pmatrix}. \end{aligned}$$

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We label the eigenvalues as $\nu _n$ ($n=1,2,\dots ,2N$).

By deforming the Bromwich contour, we obtain a contour $(-\infty ,\gamma )$, $\gamma >0$, on which s is real. In this case, we have

$$\begin{aligned} SMw_0=\lambda _0w_0, \end{aligned}$$

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where $\lambda _0$ is real and $w_0$ a 2N-dimensional real vector. We note that $M=\mathfrak {R}{M}$ in this case and moreover we can write

$$\begin{aligned} \mu _t(s)=\frac{\sigma _a+\sigma _s+\mathfrak {R}{s}}{v_0}. \end{aligned}$$

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Since M is a symmetric real matrix, by the Cholesky decomposition we can write $M=U^TU$, where U is a real triangular matrix with positive diagonal entries. Hence,

$$\begin{aligned} USU^T\widetilde{w_0}=\lambda \widetilde{w_0},\quad \widetilde{w_0}=Uw_0. \end{aligned}$$

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Since U is nonsingular, S and $USU^T$ have the same inertia, i.e., the same number of positive, negative, and zero eigenvalues according to Sylvester’s law of inertia. Clearly, S has $N_{\eta }$ positive eigenvalues. Therefore, there are $N_{\eta }$ eigenvalues $\lambda _0$.

Next, we assume that s has a small imaginary part. Then, we can write

$$\begin{aligned} S\left( \mathfrak {R}{M}+i\frac{\mathfrak {I}{s}}{v_0}\begin{pmatrix}I&{}\\ {} &{}I\end{pmatrix}\right) w= \lambda w. \end{aligned}$$

(57)

We treat the imaginary part as perturbation and express the matrix-vector equation as

$$\begin{aligned} S\left( \mathfrak {R}{M}+i\epsilon \frac{\mathfrak {I}{s}}{v_0} \begin{pmatrix}I&{}\\ {} &{}I\end{pmatrix}\right) \left( w_0+\epsilon w_1+\cdots \right) = \left( \lambda _0+\lambda _1+\cdots \right) \left( w_0+\epsilon w_1+\cdots \right) . \end{aligned}$$

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By collecting terms of order $O(\epsilon ^0)$, we have $S(\mathfrak {R}{M})w_0=\lambda _0w_0$. From terms of $O(\epsilon ^1)$, we have

$$\begin{aligned} S(\mathfrak {R}{M})w_1+i\frac{\mathfrak {I}{s}}{v_0}Sw_0=\lambda _0w_1+\lambda _1w_0. \end{aligned}$$

(59)

Let us multiply $w_0^T$ on both sides of the above equation. We obtain

$$\begin{aligned} w_0^TS(\mathfrak {R}{M})w_1+i\frac{\mathfrak {I}{s}}{v_0}w_0^TSw_0=\lambda _0w_0^Tw_1+\lambda _1w_0^Tw_0. \end{aligned}$$

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Using $w_0^T(S(\mathfrak {R}{M}))^T=w_0^TS\mathfrak {R}{M}=\lambda _0w_0^T$, we obtain

$$\begin{aligned} \lambda _1=i\frac{(\mathfrak {I}{s})w_0^TSw_0}{v_0w_0^Tw_0}. \end{aligned}$$

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That is, $\lambda _1$ is pure imaginary and the number of positive $\mathfrak {R}{\lambda }$ does not change. This fact implies that there are $N_{\eta }$ eigenvalues $\nu _n$ ($n=1,\dots ,2N$) such that $\mathfrak {R}{\nu _n}>0$.

Appendix 2: Advection–Diffusion Equation with Absorption

Let us consider the following advection–diffusion equation with the absorption term:

$$\begin{aligned} \left( \frac{\partial }{\partial t}-D\frac{\partial ^2}{\partial x^2}+u\frac{\partial }{\partial x}+\sigma _a \right) C(x,t)=0, \end{aligned}$$

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$$\begin{aligned} C(x,0)=0, \end{aligned}$$

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$$\begin{aligned} C(0,t)=C_0. \end{aligned}$$

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Let us introduce $\Gamma (x,t)$ as

$$\begin{aligned} C(x,t)=\Gamma (x,t)\exp \left( \frac{ux}{2D}-\frac{u^2t}{4D}-\sigma _at\right) . \end{aligned}$$

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We have

$$\begin{aligned} \frac{\partial }{\partial t}\Gamma (x,t)-D\frac{\partial ^2}{\partial x^2}\Gamma (x,t)=0, \end{aligned}$$

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$$\begin{aligned} \Gamma (x,0)=0, \end{aligned}$$

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$$\begin{aligned} \Gamma (0,t)=C_0\exp \left( \frac{u^2t}{4D}+\sigma _at\right) . \end{aligned}$$

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Let us consider the Laplace transform:

$$\begin{aligned} \hat{\Gamma }(x,{p})=\int _0^{\infty }\mathrm {e}^{-{p}t}\Gamma (x,t)\,\mathrm {d}t, \quad \mathfrak {R}{{p}}>\frac{u^2}{4D}+\sigma _a. \end{aligned}$$

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Then, we obtain

$$\begin{aligned} \frac{\partial ^2}{\partial x^2}\hat{\Gamma }(x,{p})=\frac{{p}}{D}\hat{\Gamma }(x,{p}),\quad \hat{\Gamma }(0,{p})=\frac{C_0}{{p}-\frac{u^2}{4D}-\sigma _a}. \end{aligned}$$

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We obtain

$$\begin{aligned} \hat{\Gamma }(x,{p})=\frac{C_0}{{p}-\frac{u^2}{4D}-\sigma _a} \mathrm {e}^{-\sqrt{\frac{{p}}{D}}x}. \end{aligned}$$

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Thus,

$$\begin{aligned} \Gamma (x,t)=C_0\int _0^tf(t-\tau )g(\tau )\,d\tau , \end{aligned}$$

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where

$$\begin{aligned} \hat{f}({p})=\frac{1}{{p}-\frac{u^2}{4D}-\sigma _a},\quad \hat{g}({p})=\mathrm {e}^{-\sqrt{\frac{{p}}{D}}x}. \end{aligned}$$

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Noting that

$$\begin{aligned} \int _0^{\infty }\mathrm {e}^{-{p}t}\mathrm{e}^{at}\,\mathrm {d}t= \frac{1}{{p}-a}\;(\mathfrak {R}{{p}}>a),\quad \int _0^{\infty }\mathrm {e}^{-{p}t}\frac{b}{2\sqrt{\pi }t^{3/2}} \mathrm{e}^{-\frac{b^2}{4t}}\,\mathrm {d}t= \mathrm {e}^{-b\sqrt{{p}}}\;(\mathfrak {R}{{p}}>0), \end{aligned}$$

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we obtain

$$\begin{aligned} \Gamma (x,t)&= C_0\int _0^t\exp \left( \left( \frac{u^2}{4D}+\sigma _a\right) (t-\tau )\right) \frac{x/\sqrt{D}}{2\sqrt{\pi }\tau ^{3/2}}\mathrm {e}^{-\frac{x^2}{4D\tau }} \,\mathrm {d}\tau \nonumber \\&= \frac{x/\sqrt{D}}{2\sqrt{\pi }}\exp \left( \left( \frac{u^2}{4D}+\sigma _a\right) t\right) \exp \left( -\sqrt{\frac{u^2}{4D}+\sigma _a}\frac{x}{\sqrt{D}}\right) \nonumber \\&\quad \times \int _0^t\frac{1}{\tau ^{3/2}}\exp \left( -\left( \frac{x/\sqrt{D}}{2\sqrt{\tau }}- \sqrt{\left( \frac{u^2}{4D}+\sigma _a\right) \tau }\right) ^2\right) \,\mathrm {d}\tau \nonumber \\&= \frac{C_0}{2}\exp \left( \left( \frac{u^2}{4D}+\sigma _a\right) t\right) \exp \left( -\sqrt{\frac{u^2}{4D}+\sigma _a}\frac{x}{\sqrt{D}}\right) \nonumber \\&\quad \times \Biggl [ \mathop {\mathrm {erfc}}\left( \frac{x}{\sqrt{4Dt}}-\sqrt{\left( \frac{u^2}{4D}+\sigma _a\right) t}\right) \nonumber \\&\quad + \exp \left( \sqrt{\frac{u^2}{4D}+\sigma _a}\frac{2x}{\sqrt{D}}\right) \mathop {\mathrm {erfc}}\left( \frac{x}{\sqrt{4Dt}}+\sqrt{\left( \frac{u^2}{4D}+\sigma _a\right) t}\right) \Biggr ], \end{aligned}$$

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where the complementary error function is given by $\mathop {\mathrm {erfc}}(x)=(2/\sqrt{\pi })\int _x^{\infty }\mathrm {e}^{-t^2}\,\mathrm {d}t$. Finally,

$$\begin{aligned} \begin{aligned} C(x,t)&= \frac{C_0}{2}\mathrm {e}^{\frac{ux}{2D}} \exp \left( -\sqrt{\frac{u^2}{4D}+\sigma _a}\frac{x}{\sqrt{D}}\right) \\&\quad \times \Biggl [ \mathop {\mathrm {erfc}}\left( \frac{x}{\sqrt{4Dt}}-\sqrt{\left( \frac{u^2}{4D}+\sigma _a\right) t}\right) \\&\quad + \exp \left( \sqrt{\frac{u^2}{4D}+\sigma _a}\frac{2x}{\sqrt{D}}\right) \mathop {\mathrm {erfc}}\left( \frac{x}{\sqrt{4Dt}}+\sqrt{\left( \frac{u^2}{4D}+\sigma _a\right) t}\right) \Biggr ]. \end{aligned} \end{aligned}$$

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The above solution reduces to the Ogata–Banks solution (Ogata and Banks 1961) when $\sigma _a=0$. In particular, we have $C(x,t)\rightarrow C_0$ as $t\rightarrow \infty$ if $\sigma _a=0$. When $\sigma _a=0$, we have

$$\begin{aligned} C(x,t)= \frac{C_0}{2}\left[ \mathop {\mathrm {erfc}}\left( \frac{x-ut}{\sqrt{4Dt}}\right) + \mathrm {e}^{ux/D} \mathop {\mathrm {erfc}}\left( \frac{x+ut}{\sqrt{4Dt}}\right) \right] . \end{aligned}$$

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