Flux Characterization in Heterogeneous Transport Problems by the Boundary Integral Method

Abstract

Potential flow problems with spatially varying transport coefficients are commonplace. While numerical methods are often implemented for such problems, coupled analytic solutions to heterogeneous, anisotropic problems enable below-grid, nonlinear flow effects to be accurately captured, even for systems with complex source functions. Coupling occurs through construction of boundary integrals to capture material transport between homogeneous domains. Various nonparametric methods are examined to compute the unknown flux distribution, including Gauss node distribution schemes with point source functions and patch-wise, semi-analytic integration. The merits of a posed parametric method that asserts a hybrid boundary condition and avoids numerical integration altogether within an overall analytic structure are further examined here. Reduction of heterogeneous problems to equivalent homogeneous transport property problems yields an additional term with linkage to uniform flux and uniform pressure contributions at increasingly distance interfaces in prolonged systems.

Appendix: Partial Integration of the Neumann Function on a Boundary

In patch-wise uniform flux boundary elements, assuming $\frac{a^{2}} {k_{x}} =\max (\frac{a^{2}} {k_{x}}, \frac{b^{2}} {k_{y}}, \frac{h^{2}} {k_{z}})$, we examine partial integration of the point source Neumann function on the boundary as t $\rightarrow \infty$. Focusing on the triple infinite series term requiring computational advantage, we obtain the following:

$$ \displaystyle\begin{array}{rcl} & & \int _{z_{1}}^{z_{2} }\int _{x_{1}}^{x_{2} }\sum _{l,m,n=1}^{\infty }\frac{C_{lmn}} {\pi ^{2}} \cdot \frac{1} {D_{lmn}^{2}} {}\\ & & \quad \cdot cos(\frac{\pi lx} {a} )cos(\frac{\pi my} {b} )cos(\frac{\pi nz} {h} )cos(\frac{\pi lx_{o}} {a} )cos(\frac{\pi my_{o}} {b} )cos(\frac{\pi mz_{o}} {b} )dx_{o}dz_{o} = {}\\ & & \qquad \frac{ah \cdot H(x - x_{1})H(x_{2} - x)} {4\pi } \cdot \sum _{j=1}^{4}\sum _{ m,n=1}^{\infty }\frac{sin(\frac{\pi nZ_{j}} {h} )cos(\frac{\pi my} {b} )cos(\frac{\pi my_{o}} {b} )} {nD_{mn}^{2}} {}\\ & & -\frac{ah} {8\pi } \cdot sign(X_{i}) \cdot \sum _{m,n=1}^{\infty }\frac{sin(\frac{\pi nZ_{j}} {h} )cos(\frac{\pi my} {b} )cos(\frac{\pi my_{o}} {b} )} {nD_{mn}^{2}} \cdot \frac{sinh[ \frac{\pi a} {\sqrt{k_{x}}}D_{mn}(1 -\frac{\vert X_{i}\vert } {a} )} {sinh[ \frac{\pi a} {\sqrt{k_{x}}}D_{mn})} {}\\ \end{array} $$

where

$$ \displaystyle\begin{array}{rcl} & D_{mn}^{2} \equiv \frac{k_{y}m^{2}} {b^{2}} + \frac{k_{z}n^{2}} {h^{2}} & {}\\ & X_{i} \equiv [(x_{2} + x),(x_{2} - x),-(x + x_{1}),(x - x_{1})]& {}\\ & Z_{j} \equiv [(z_{2} + z),(z_{2} - z),-(z + z_{1}),(z - z_{1})] & {}\\ \end{array} $$

The first term on the RHS can be reduced to a single series with exponential damping, whereas the hyperbolic functions are replaced with exponentials and reformulated in terms of a rapidly convergent double infinite series with exponential damping.

If $\frac{h^{2}} {k_{z}} \geq \frac{b^{2}} {k_{y}}$, then

$$\displaystyle\begin{array}{rcl} & & \sum _{j=1}^{4}\sum _{ m,n=1}^{\infty }\frac{sin(\frac{\pi nZ_{j}} {h} )cos(\frac{\pi my} {b} )cos(\frac{\pi my_{o}} {b} )} {nD_{mn}^{2}} = {}\\ & & \qquad \qquad \qquad \qquad \qquad \qquad \frac{1} {2} \cdot \sum _{m=1}^{\infty }\frac{h^{2}} {k_{z}}cos(\frac{\pi m(y \pm y_{o})} {b} ) \cdot [\sum _{n=1}^{\infty } \frac{\sum _{k=1}^{4}sin(\frac{\pi nZ_{k}} {h} )} {n(n^{2} + \frac{k_{y}} {k_{z}} \frac{h^{2}} {b^{2}} m^{2})}] {}\\ \end{array}$$

The portion in parentheses can be further reduced. Otherwise,

$$\displaystyle\begin{array}{rcl} & & \sum _{j=1}^{4}\sum _{ m,n=1}^{\infty }\frac{sin(\frac{\pi nZ_{j}} {h} )cos(\frac{\pi my} {b} )cos(\frac{\pi my_{o}} {b} )} {nD_{mn}^{2}} = {}\\ & & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \frac{1} {2} \cdot \sum _{n=1}^{\infty } \frac{b^{2}} {nk_{y}}\sum _{k=1}^{4}sin(\frac{\pi nZ_{k})} {h} \cdot \sum _{m=1}^{\infty } \frac{cos(\frac{\pi m(y\pm y_{o})} {b} )} {(m^{2} + \frac{k_{z}} {k_{y}} \frac{b} {h}n^{2})} {}\\ \end{array}$$

Concerning the second term in the integration,

$$\displaystyle\begin{array}{rcl} & & sign(X_{i}) \cdot \sum _{m,n=1}^{\infty }\frac{sin(\frac{\pi nZ_{j}} {h} )cos(\frac{\pi my} {b} )cos(\frac{\pi my_{o}} {b} )} {nD_{mn}^{2}} \cdot \frac{sinh[ \frac{\pi a} {\sqrt{k_{x}}}D_{mn}(1 -\frac{\vert X_{i}\vert } {a} )} {sinh[ \frac{\pi a} {\sqrt{k_{x}}}D_{mn})} = {}\\ & & \qquad \qquad 4 \cdot \sum _{m=1}^{\infty }\sum _{ n=1}^{\infty }\frac{cos(\frac{\pi my} {b} )cos(\frac{\pi my_{o}} {b} )\sum _{k=1}^{4}sin(\frac{\pi nZ_{k}} {h} )} {n(m^{2}\frac{k_{y}} {b^{2}} + \frac{k_{z}} {h^{2}} n^{2})} {}\\ & & \qquad \qquad \qquad \qquad \qquad \qquad \cdot \frac{\sum _{k=1}^{4}sign(X_{k}) \cdot (e^{-\pi D_{mn}\vert \frac{X_{k}} {a} \vert }- e^{-\pi D_{mn}(2-\vert \frac{X_{k}} {a} \vert )})} {1 - e^{-2\pi D_{mn}}} {}\\ \end{array}$$

This last term can be efficiently coded with termination of infinite summations to desired accuracy.