Added 2D ADI scheme for boundaries

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Max Luebke 2022-04-26 10:07:51 +02:00
parent ffefb6ab50
commit 57cfa682c5

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@ -186,12 +186,12 @@ implicit method. Therefore we make use of second difference operator defined in
equation [[eqn:secondderivative]] in both $x$ and $y$ direction for half the time
step $\Delta t$. We are denoting the numerator of equation [[eqn:1DBTCS]] as
$\delta^2_x C^n_{ij}$ and $\delta^2_y C^n_{ij}$ respectively with $i$ the
position in $x$, $j$ the position in $y$ direction and $n$ as the current time
position in $x$, $j$ the position in $y$ direction and $T$ as the current time
step:
\begin{align}\displaystyle
\delta^2_x C^n_{ij} &= C^{n}_{i-1,j} - 2C^{n}_{i,j} + C^{n}_{i+1,j} \nonumber \\
\delta^2_y C^n_{ij} &= C^{n}_{i,j-1} - 2C^{n}_{i,j} + C^{n}_{i,j+1}
\delta^2_x C^T_{ij} &= C^{T}_{i-1,j} - 2C^{T}_{i,j} + C^{T}_{i+1,j} \nonumber \\
\delta^2_y C^T_{ij} &= C^{T}_{i,j-1} - 2C^{T}_{i,j} + C^{T}_{i,j+1}
\end{align}
Assuming a constant $\alpha_x$ and $\alpha_y$ in each direction the equation can
@ -199,8 +199,8 @@ be defined:
#+NAME: eqn:genADI
\begin{align}\displaystyle
\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 \\
\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}
\frac{C^{T+1/2}_{ij}-C^T_{ij}}{\frac{\Delta t}{2}} &= \alpha_x \frac{\left( \delta^2_x C^{T+1/2}_{ij} + \delta^2_y C^{T}_{ij}\right)}{\Delta x^2} \nonumber \\
\frac{C^{T+1}_{ij}-C^{T+1/2}_{ij}}{\frac{\Delta t}{2}} &= \alpha_y \frac{\left( \delta^2_x C^{T+1/2}_{ij} + \delta^2_y C^{T+1}_{ij}\right)}{\Delta y^2}
\end{align}
Now we will define $s_x$ and $s_y$ respectively as followed:
@ -213,19 +213,34 @@ s_y &= \frac{\alpha_y \cdot \frac{\Delta t}{2}}{\Delta y^2}
Equation [[eqn:genADI]], once developed in the $x$ direction, yields:
\begin{align}\displaystyle
\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 \\
\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 \\
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 \\
-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 \\
-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)
\frac{C^{T+1/2}_{ij}-C^T_{ij}}{\frac{\Delta t}{2}} &= \alpha_x \frac{\left( \delta^2_x C^{T+1/2}_{ij} + \delta^2_y C^{T}_{ij}\right)}{\Delta x^2} \nonumber \\
\frac{C^{T+1/2}_{ij}-C^T_{ij}}{\frac{\Delta t}{2}} &= \alpha_x \frac{\left( \delta^2_x C^{T+1/2}_{ij} \right)}{\Delta x^2} + \alpha_x \frac{\left(\delta^2_y C^{T}_{ij}\right)}{\Delta x^2} \nonumber \\
C^{T+1/2}_{ij}-C^T_{ij} &= s_x \delta^2_x C^{T+1/2}_{ij} + s_x \delta^2_y C^{T}_{ij} \nonumber \\
-C^T_{ij} - s_x \delta^2_y C^{T}_{ij} &= C^{T+1/2}_{ij} + s_x \delta^2_x C^{T+1/2}_{ij}
\end{align}
All values on the left side of the equation are known and we are solving for
$C^{1/2}$. The calculation in $y$ direction for another $\frac{\Delta t}{2}$ is
similar and can also be easily derived. Therefore we do not show it here.
Substituting $\delta$ in the equation leads to following epxression:
*TODO*:
- ADI at boundaries
\begin{equation}
-C^T_{ij} - s_x \left(C^T_{i,j-1}-2C^T_{i,j}+C^T_{i,j+1} \right) = C^{T+1/2}_{ij} + s_x \left( C^{T+1/2}_{i-1,j} - 2C^{T+1/2}_{i,j} + C^{T+1/2}_{i+1,j} \right)
\end{equation}
This scheme also only applies to inlet cells without a relation to boundaries.
Fortunately we already derived both cases of outer left and right inlet cell
respectively. Hence we are able to redefine each $\delta^2$ case in x and y
direction, assuming $l_x$ and $l_y$ the be the left boundary value and $r_x$ and
$r_y$ the right one for each direction $x$ and $y$. The equations are exemplary
for timestep $T+1/2$:
\begin{align}\displaystyle
\delta^2_d C^{T+1/2}_{0,j} &= 2l_x - 3C^{T+1/2}_{0,j} + C^{T+1/2}_{1,j} \nonumber \\
\delta^2_d C^{T+1/2}_{n,j} &= 2r_x - 3C^{T+1/2}_{n,j} + C^{T+1/2}_{n-1,j} \nonumber \\
\delta^2_d C^{T+1/2}_{i,0} &= 2l_y - 3C^{T+1/2}_{i,0} + C^{T+1/2}_{i,1} \nonumber \\
\delta^2_d C^{T+1/2}_{i,n} &= 2r_y - 3C^{T+1/2}_{i,n} + C^{T+1/2}_{i,n-1}
\end{align}
#+LATEX: \clearpage