332 lines
12 KiB
Org Mode
332 lines
12 KiB
Org Mode
#+TITLE: Numerical solution of diffusion equation in 2D with ADI Scheme
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#+LaTeX_CLASS_OPTIONS: [a4paper,10pt]
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#+LATEX_HEADER: \usepackage{fullpage}
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#+LATEX_HEADER: \usepackage{amsmath}
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#+OPTIONS: toc:nil
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* Finite differences with nodes as cells' centres
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The 1D diffusion equation is:
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#+NAME: eqn:1
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\begin{align}
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\frac{\partial C }{\partial t} & = \frac{\partial}{\partial x} \left(\alpha \frac{\partial C }{\partial x} \right) \nonumber \\
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& = \alpha \frac{\partial^2 C}{\partial x^2}
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\end{align}
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We aim at numerically solving [[eqn:1]] on a spatial grid such as:
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[[./grid_pqc.pdf]]
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The left boundary is defined on $x=0$ while the center of the first
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cell - which are the points constituting the finite difference nodes -
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is in $x=dx/2$, with $dx=L/n$.
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** The explicit FTCS scheme (as in PHREEQC)
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We start by discretizing [[eqn:1]] following an explicit Euler scheme and
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specifically a Forward Time, Centered Space finite difference.
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For each cell index $i \in 1, \dots, n-1$ and assuming constant
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$\alpha$, we can write:
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#+NAME: eqn:2
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\begin{equation}\displaystyle
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\frac{C_i^{j+1} -C_i^{j}}{\Delta t} = \alpha\frac{\frac{C^j_{i+1}-C^j_{i}}{\Delta x}-\frac{C^j_{i}-C^j_{i-1}}{\Delta x}}{\Delta x}
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\end{equation}
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In practice, we evaluate the first derivatives of $C$ w.r.t. $x$ on
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the boundaries of each cell (i.e., $(C_{i+1}-C_i)/\Delta x$ on the
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right boundary of the i-th cell and $(C_{i}-C_{i-1})/\Delta x$ on its
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left cell boundary) and then repeat the differentiation to get the
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second derivative of $C$ on the the cell centre $i$.
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This discretization works for all internal cells, but not for the
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domain boundaries ($i=0$ and $i=n$). To properly treat them, we need
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to account for the discrepancy in the discretization.
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For the first (left) cell, whose center is at $x=dx/2$, we can
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evaluate the left gradient with the left boundary using such distance,
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calling $l$ the numerical value of a constant boundary condition:
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#+NAME: eqn:3
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\begin{equation}\displaystyle
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\frac{C_0^{j+1} -C_0^{j}}{\Delta t} = \alpha\frac{\frac{C^j_{1}-C^j_{0}}{\Delta x}-
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\frac{C^j_{0}-l}{\frac{\Delta x}{2}}}{\Delta x}
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\end{equation}
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This expression, once developed, yields:
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#+NAME: eqn:4
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\begin{align}\displaystyle
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C_0^{j+1} & = C_0^{j} + \frac{\alpha \cdot \Delta t}{\Delta x^2} \cdot \left( C^j_{1}-C^j_{0}- 2 C^j_{0}+2l \right) \nonumber \\
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& = C_0^{j} + \frac{\alpha \cdot \Delta t}{\Delta x^2} \cdot \left( C^j_{1}- 3 C^j_{0} +2l \right)
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\end{align}
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In case of constant right boundary, the finite difference of point
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$C_n$ - calling $r$ the right boundary value - is:
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#+NAME: eqn:5
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\begin{equation}\displaystyle
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\frac{C_n^{j+1} -C_n^j}{\Delta t} = \alpha\frac{\frac{r - C^j_{n}}{\frac{\Delta x}{2}}-
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\frac{C^j_{n}-C^j_{n-1}}{\Delta x}}{\Delta x}
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\end{equation}
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Which, developed, gives
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#+NAME: eqn:6
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\begin{align}\displaystyle
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C_n^{j+1} & = C_n^{j} + \frac{\alpha \cdot \Delta t}{\Delta x^2} \cdot \left( 2 r - 2 C^j_{n} -C^j_{n} + C^j_{n-1} \right) \nonumber \\
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& = C_n^{j} + \frac{\alpha \cdot \Delta t}{\Delta x^2} \cdot \left( 2 r - 3 C^j_{n} + C^j_{n-1} \right)
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\end{align}
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If on the right boundary we have closed or Neumann condition, the left derivative in eq. [[eqn:5]]
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becomes zero and we are left with:
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#+NAME: eqn:7
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\begin{equation}\displaystyle
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C_n^{j+1} = C_n^{j} + \frac{\alpha \cdot \Delta t}{\Delta x^2} \cdot (C^j_{n-1} - C^j_n)
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\end{equation}
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A similar treatment can be applied to the BTCS implicit scheme.
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** Implicit BTCS scheme
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First, we define the Backward Time difference:
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\begin{equation}
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\frac{\partial C^{j+1} }{\partial t} = \frac{C^{j+1}_i - C^{j}_i}{\Delta t}
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\end{equation}
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Second the spatial derivative approximation, evaluated at time level $j+1$:
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#+NAME: eqn:secondderivative
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\begin{equation}
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\frac{\partial^2 C^{j+1} }{\partial x^2} = \frac{\frac{C^{j+1}_{i+1}-C^{j+1}_{i}}{\Delta x}-\frac{C^{j+1}_{i}-C^{j+1}_{i-1}}{\Delta x}}{\Delta x}
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\end{equation}
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Taking the 1D diffusion equation from [[eqn:1]] and substituting each term by the
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equations given above leads to the following equation:
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# \begin{equation}\displaystyle
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# \frac{C_i^{j+1} -C_i^{j}}{\Delta t} = \alpha\frac{\frac{C^{j+1}_{i+1}-C^{j+1}_{i}}{\Delta x}-\frac{C^{j+1}_{i}-C^{j+1}_{i-1}}{\Delta x}}{\Delta x}
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# \end{equation}
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# Since we are not able to solve this system w.r.t unknown values in $C^{j-1}$ we
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# are shifting each j by 1 to $j \to (j+1)$ and $(j-1) \to j$ which leads to:
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#+NAME: eqn:1DBTCS
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\begin{align}\displaystyle
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\frac{C_i^{j+1} - C_i^{j}}{\Delta t} & = \alpha\frac{\frac{C^{j+1}_{i+1}-C^{j+1}_{i}}{\Delta x}-\frac{C^{j+1}_{i}-C^{j+1}_{i-1}}{\Delta x}}{\Delta x} \nonumber \\
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& = \alpha\frac{C^{j+1}_{i-1} - 2C^{j+1}_{i} + C^{j+1}_{i+1}}{\Delta x^2}
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\end{align}
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This only applies to inlet cells with no ghost node as neighbor. For the left
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cell with its center at $\frac{dx}{2}$ and the constant concentration on the
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left ghost node called $l$ the equation goes as followed:
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\begin{equation}\displaystyle
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\frac{C_0^{j+1} -C_0^{j}}{\Delta t} = \alpha\frac{\frac{C^{j+1}_{1}-C^{j+1}_{0}}{\Delta x}-
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\frac{C^{j+1}_{0}-l}{\frac{\Delta x}{2}}}{\Delta x}
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\end{equation}
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This expression, once developed, yields:
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\begin{align}\displaystyle
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C_0^{j+1} & = C_0^{j} + \frac{\alpha \cdot \Delta t}{\Delta x^2} \cdot \left( C^{j+1}_{1}-C^{j+1}_{0}- 2 C^{j+1}_{0}+2l \right) \nonumber \\
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& = C_0^{j} + \frac{\alpha \cdot \Delta t}{\Delta x^2} \cdot \left( C^{j+1}_{1}- 3 C^{j+1}_{0} +2l \right)
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\end{align}
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Now we define variable $s_x$ as followed:
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\begin{equation}
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s_x = \frac{\alpha \cdot \Delta t}{\Delta x^2}
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\end{equation}
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Substituting with the new variable $s_x$ and reordering of terms leads to the equation applicable to our model:
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\begin{equation}\displaystyle
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-C^j_0 = (2s_x) \cdot l + (-1 - 3s_x) \cdot C^{j+1}_0 + s_x \cdot C^{j+1}_1
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\end{equation}
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The right boundary follows the same scheme. We now want to show the equation for the rightmost inlet cell $C_n$ with right boundary value $r$:
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\begin{equation}\displaystyle
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\frac{C_n^{j+1} -C_n^{j}}{\Delta t} = \alpha\frac{\frac{r-C^{j+1}_{n}}{\frac{\Delta x}{2}}-
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\frac{C^{j+1}_{n}-C^{j+1}_{n-1}}{\Delta x}}{\Delta x}
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\end{equation}
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This expression, once developed, yields:
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\begin{align}\displaystyle
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C_n^{j+1} & = C_n^{j} + \frac{\alpha \cdot \Delta t}{\Delta x^2} \cdot \left( 2r - 2C^{j+1}_{n} - C^{j+1}_{n} + C^{j+1}_{n-1} \right) \nonumber \\
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& = C_0^{j} + \frac{\alpha \cdot \Delta t}{\Delta x^2} \cdot \left( 2r - 3C^{j+1}_{n} + C^{j+1}_{n-1} \right)
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\end{align}
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Now rearrange terms and substituting with $s_x$ leads to:
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\begin{equation}\displaystyle
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-C^j_n = s_x \cdot C^{j+1}_{n-1} + (-1 - 3s_x) \cdot C^{j+1}_n + (2s_x) \cdot r
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\end{equation}
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*TODO*
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- Tridiagonal matrix filling
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** 2D ADI using BTCS scheme
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In the previous sections we described the usage of FTCS and BTCS on 1D grids. To
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make usage of the BTCS scheme we are following the alternating-direction
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implicit method. Therefore we make use of second difference operator defined in
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equation [[eqn:secondderivative]] in both $x$ and $y$ direction for half the time
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step $\Delta t$. We are denoting the numerator of equation [[eqn:1DBTCS]] as
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$\delta^2_x C^n_{ij}$ and $\delta^2_y C^n_{ij}$ respectively with $i$ the
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position in $x$, $j$ the position in $y$ direction and $n$ as the current time
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step:
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\begin{align}\displaystyle
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\delta^2_x C^n_{ij} &= C^{n}_{i-1,j} - 2C^{n}_{i,j} + C^{n}_{i+1,j} \nonumber \\
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\delta^2_y C^n_{ij} &= C^{n}_{i,j-1} - 2C^{n}_{i,j} + C^{n}_{i,j+1}
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\end{align}
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Assuming a constant $\alpha_x$ and $\alpha_y$ in each direction the equation can
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be defined:
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#+NAME: eqn:genADI
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\begin{align}\displaystyle
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\frac{C^{n+1/2}_{ij}-C^n_{ij}}{\frac{\Delta t}{2}} &= \alpha_x \frac{\left( \delta^2_x C^{n+1/2}_{ij} + \delta^2_y C^{n}_{ij}\right)}{\Delta x^2} \nonumber \\
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\frac{C^{n+1}_{ij}-C^{n+1/2}_{ij}}{\frac{\Delta t}{2}} &= \alpha_y \frac{\left( \delta^2_x C^{n+1/2}_{ij} + \delta^2_y C^{n+1}_{ij}\right)}{\Delta y^2}
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\end{align}
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Now we will define $s_x$ and $s_y$ respectively as followed:
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\begin{align}\displaystyle
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s_x &= \frac{\alpha_x \cdot \frac{\Delta t}{2}}{\Delta x^2} \nonumber \\
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s_y &= \frac{\alpha_y \cdot \frac{\Delta t}{2}}{\Delta y^2}
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\end{align}
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Equation [[eqn:genADI]], once developed in the $x$ direction, yields:
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\begin{align}\displaystyle
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\frac{C^{n+1/2}_{ij}-C^n_{ij}}{\frac{\Delta t}{2}} &= \alpha_x \frac{\left( \delta^2_x C^{n+1/2}_{ij} + \delta^2_y C^{n}_{ij}\right)}{\Delta x^2} \nonumber \\
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\frac{C^{n+1/2}_{ij}-C^n_{ij}}{\frac{\Delta t}{2}} &= \alpha_x \frac{\left( \delta^2_x C^{n+1/2}_{ij} \right)}{\Delta x^2} + \alpha_x \frac{\left(\delta^2_y C^{n}_{ij}\right)}{\Delta x^2} \nonumber \\
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C^{n+1/2}_{ij}-C^n_{ij} &= s_x \delta^2_x C^{n+1/2}_{ij} + s_x \delta^2_y C^{n}_{ij} \nonumber \\
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-C^n_{ij} - s_x \delta^2_y C^{n}_{ij} &= C^{n+1/2}_{ij} + s_x \delta^2_x C^{n+1/2}_{ij} \nonumber \\
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-C^n_{ij} - s_x \left(C^n_{i,j-1}-2C^n_{i,j}+C^n_{i,j+1} \right)&= C^{n+1/2}_{ij} + s_x \left( C^{n+1/2}_{i-1,j} - 2C^{n+1/2}_{i,j} + C^{n+1/2}_{i+1,j} \right)
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\end{align}
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All values on the left side of the equation are known and we are solving for
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$C^{1/2}$. The calculation in $y$ direction for another $\frac{\Delta t}{2}$ is
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similar and can also be easily derived. Therefore we do not show it here.
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*TODO*:
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- ADI at boundaries
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#+LATEX: \clearpage
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* Old stuff
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** Input
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- =c= $\rightarrow c$
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- containing current concentrations at each grid cell for species
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- size: $N \times M$
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- row-major
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- =alpha= $\rightarrow \alpha$
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- diffusion coefficient for both directions (x and y)
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- size: $N \times M$
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- row-major
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- =boundary_condition= $\rightarrow bc$
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- Defines closed or constant boundary condition for each grid cell
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- size: $N \times M$
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- row-major
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** Internals
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- =A_matrix= $\rightarrow A$
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- coefficient matrix for linear equation system implemented as sparse matrix
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- size: $((N+2)\cdot M) \times ((N+2)\cdot M)$ (including ghost zones in x direction)
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- column-major (not relevant)
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- =b_vector= $\rightarrow b$
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- right hand side of the linear equation system
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- size: $(N+2) \cdot M$
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- column-major (not relevant)
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- =x_vector= $\rightarrow x$
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- solutions of the linear equation system
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- size: $(N+2) \cdot M$
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- column-major (not relevant)
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** Calculation for $\frac{1}{2}$ timestep
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** Symbolic addressing of grid cells
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[[./grid.png]]
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** Filling of matrix $A$
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- row-wise iterating with $i$ over =c= and =\alpha= matrix respectively
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- addressing each element of a row with $j$
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- matrix $A$ also containing $+2$ ghost nodes for each row of input matrix $\alpha$
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- $\rightarrow offset = N+2$
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- addressing each object $(i,j)$ in matrix $A$ with $(offset \cdot i + j, offset \cdot i + j)$
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*** Rules
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$s_x(i,j) = \frac{\alpha(i,j)*\frac{t}{2}}{\Delta x^2}$ where $x$ defining the domain size in x direction.
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For the sake of simplicity we assume that each row of the $A$ matrix is addressed correctly with the given offset.
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**** Ghost nodes
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$A(i,-1) = 1$
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$A(i,N) = 1$
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**** Inlet
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$A(i,j) = \begin{cases}
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1 & \text{if } bc(i,j) = \text{constant} \\
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-1-2*s_x(i,j) & \text{else}
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\end{cases}$
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$A(i,j\pm 1) = \begin{cases}
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0 & \text{if } bc(i,j) = \text{constant} \\
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s_x(i,j) & \text{else}
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\end{cases}$
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** Filling of vector $b$
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- each elements assign a concrete value to the according value of the row of matrix $A$
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- Adressing would look like this: $(i,j) = b(i \cdot (N+2) + j)$
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- $\rightarrow$ for simplicity we will write $b(i,j)$
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*** Rules
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**** Ghost nodes
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$b(i,-1) = \begin{cases}
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0 & \text{if } bc(i,0) = \text{constant} \\
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c(i,0) & \text{else}
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\end{cases}$
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$b(i,N) = \begin{cases}
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0 & \text{if } bc(i,N-1) = \text{constant} \\
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c(i,N-1) & \text{else}
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\end{cases}$
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*** Inlet
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$p(i,j) = \frac{\Delta t}{2}\alpha(i,j)\frac{c(i-1,j) - 2\cdot c(i,j) + c(i+1,j)}{\Delta x^2}$
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\noindent $p$ is called =t0_c= inside code
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$b(i,j) = \begin{cases}
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bc(i,j).\text{value} & \text{if } bc(i,N-1) = \text{constant} \\
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-c(i,j)-p(i,j) & \text{else}
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\end{cases}$
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