## 3 Numerical Algorithm of Solving the Problem
\tag{21}\]  
Conditions (15) are also reduced to a grid form as follows  
\[\left(c^{(i)}\right)_{k}^{0}=\left(\rho_{a}^{(i)}\right)_{k}^{0}=\left(\rho_{ n}^{(i)}\right)_{k}^{0}=0,\ k=0,\ 1,\ 2,\...,\left(c^{(i)}\right)_{0}^{j}=c_{0}^{(i)},j=\overline{0,\ J},\ \left(c^{(i)}\right)_{K}^{j}=0,j= \overline{0,\ J}, \tag{22}\]  
where \(K\) is a sufficiently large number for which \(\left(c^{(i)}\right)_{K}^{j}=0\) approximately. After simple transformations, the schemes (18) and (19) take the form  
\[\left(\rho_{a}^{(i)}\right)_{k}^{j+1}=\left(\rho_{a}^{(i)}\right)_{k}^{j}+ \tau\beta_{a}^{(i)}\left(\frac{\left(c^{(i)}\right)_{k}^{j}}{1+\gamma^{(i)} \left|\
abla p\right|_{k}^{j+1}}-\frac{\left(\rho_{a}^{(i)}\right)_{k}^{j} \left(1+\omega^{(i)}\left|\
abla p\right|_{k}^{j+1}\right)}{\rho_{a0}^{(i)}}c _{0}^{(i)}\right), \tag{23}\]  
\[\left(\rho_{p}^{(i)}\right)_{k}^{j+1}=\left(\rho_{p}^{(i)}\right)_{k}^{j}+ \tau\beta_{p}^{(i)}\phi_{i}\left(\left(\rho_{p}^{(1)}\right)_{k}^{j},\ \left(\rho_{p}^{(2)}\right)_{k}^{j}\right)\left(c^{(i)}\right)_{k}^{j},k= \overline{0,\ K},j=\overline{0,\ J-1}. \tag{24}\]  
We transform the scheme (16) to obtain the following system of linear equations  
\[A^{(i)}\left(c^{(i)}\right)_{k-1}^{j+1}-B^{(i)}\left(c^{(i)}\right)_{k}^{j+1}+E ^{(i)}\left(c^{(i)}\right)_{k+1}^{j+1}=-\left(F^{(i)}\right)_{k}^{j},\ k= \overline{1,K-1},\ j=\overline{0,J-1}, \tag{25}\]  
where  
\[A^{(i)}=D^{(i)}\frac{\tau}{h^{2}}+v\frac{\tau}{h},\left(B^{(i)}\right)_{j}^{k} =2D^{(i)}\frac{\tau}{h^{2}}+v\frac{\tau}{h}+m_{j^{\prime}}^{k},E^{(i)}=D^{(i)} \frac{\tau}{h^{2}},\]  
\[\left(F^{(i)}\right)_{k}^{j}=m_{k}^{j}\left(c^{(i)}\right)_{k}^{j}-\left(\left( \rho_{a}^{(i)}\right)_{k}^{j+1}-\left(\rho_{a}^{(i)}\right)_{k}^{j}\right)- \left(\left(\rho_{p}^{(i)}\right)_{k}^{j+1}-\left(\rho_{p}^{(i)}\right)_{k}^{j }\right).\]  
Using the values \(\left(\rho_{a}^{(i)}\right)_{k}^{j+1}\) and \(\left(\rho_{p}^{(i)}\right)_{k}^{j+1}\) we solve (25) by the tridiagonal matrix algorithm (Thomas' algorithm) [47, 48]  
\[\left(c^{(i)}\right)_{k}^{j+1}=\alpha_{k+1}^{(i)}\left(c^{(i)}\right)_{k+1}^{j+1 }+\beta_{k+1}^{(i)},k=\overline{0,K-1},\ \ \ j=\overline{0,J-1}, \tag{26}\]  
where  
\[\alpha_{k+1}^{(i)}=\frac{E^{(i)}}{B^{(i)}-A^{(i)}\alpha_{k}^{(i)}},\beta_{k+1}^{ (i)}=\frac{A^{(i)}\beta_{k}^{(i)}+\left(F^{(i)}\right)_{k}^{j}}{B^{(i)}-A^{(i)} \alpha_{k}^{(i)}}.\]  
From boundary conditions we have  
\[\left(c^{(i)}\right)_{0}^{j+1}=\alpha_{1}^{(i)}\left(c^{(i)}\right)_{1}^{j+1}+ \beta_{1}^{(i)}=c_{0}^{(i)},\,so,\,\alpha_{1}^{(i)}=0,\ \ \ \beta_{1}^{(i)}=c_{0}^{(i)}. \tag{27}\]Numerical algorithm is follows. First by using the initial conditions we find values \(\left(c^{(i)}\right)_{k^{\prime}}^{j}\left(\rho_{a}^{(i)}\right)_{k^{\prime}}^{j}\), \(\left(\rho_{p}^{(i)}\right)_{k}^{j}\) in zero time layer. Then, use (17) to find \(\left|\
abla p\right|_{k}^{j+1}\). Next, substitute the values found into (23) to find \(\left(\rho_{a}^{(i)}\right)_{k}^{j+1}\). Then, according to (24), for each component of the suspension, the concentration of the irreversible deposition in passive zone \(\left(\rho_{p}^{(i)}\right)_{k}^{j+1}\) is determined. Finally, substituting \(\left(\rho_{a}^{(i)}\right)_{k}^{j+1}\) and \(\left(\rho_{p}^{(i)}\right)_{k}^{j+1}\) into (25), by using [PERSON] algorithm we determine \(\left(c^{(i)}\right)_{k}^{j+1}\) on the \((j+1)\) th layer of the grid \(\omega_{hr}\).