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The aim is to construct symmetric Laplacian matrix $L$ for 2D square domain.

I have the following discretization of second derivatives on non-uniform grid (will skip the steps of derivation, any textbook works). velocity vectors $u$ and $v$ are defined on cell faces, $x_i$ corresponds to $u_{i+mk}$, where m is amount of unknown $u$'s and k is an integer.

$$ \left(\frac{\partial^2 u}{\partial x^2}\right)_i \approx \frac{u_{i+1}\left(x_i-x_{i-1}\right)-u_i\left(x_{i+1}-x_{i-1}\right)+u_{i-1}\left(x_{i+1}-x_i\right)}{\left(\frac{x_{i+1}-x_{i-1}}{2}\right)\left(x_i-x_{i-1}\right)\left(x_{i+1}-x_i\right)}+O(\Delta x). $$ This scheme becomes second order when $\Delta x_i\approx\Delta x_{i-1}$. The above discretization leads to matrix $\hat{L}$.

Now, consider the symmetrizer matrix $$ \hat{M} =\left[\begin{array}{cc} \frac{1}{2}\left(x_{i+1}+x_{i-1}\right) & 0 \\ 0 & \frac{1}{2}\left(y_{j+1}+ y_{j-1}\right) \end{array}\right], $$ which is a diagonal.

The product $\hat{M}\hat{L}=L$ makes L symmetric for Dirichlet BC and Neumann (1st order) BC.

Now, if I want to use the second order scheme for Neumann BC, discretization becomes 3 point. Therefore, it starts affecting the off diagonal elements in $\hat{L}$ (first oder boundary condition scheme changes only the diagonal element, which does not kill the symmetricity). Hence, $L$ is not symmetric anymore.

Is there a way to modify the diagonal symmetrizer $\hat{M}$ without inducing a lot of computational time? $\hat{M}$ is multiplied by the rhs vector at every time step. I tried the simple $\hat{M}=(\frac{\hat{L}+\hat{L}'}{2})\hat{L}^{-1}$ but it is too dense (see pic below).

M becomes this

Edit: I figured it out. Will post the solution when finish typesetting it in LaTeX.

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1 Answer 1

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I finally figured it out.

We need to move to the new variables.

(Idk why the typeset result formulas look different than the preview.)

Let us have $m$ intervals in $x$ direction and $n$ intervals in $y$ direction. The number of unknowns is equal to the number of face velocities in the interior of the domain excluding the boundaries. Rewrite $\vec{u}$ as a vector of unknowns

$$\vec{u} = [\underbrace{u_1;u_2;\dotsc;u_{m-1};u_1;u_2;\dotsc;u_{m-1};\dotsc;u_1;u_2;\dotsc;u_{m-1}}_{n\text{ times for each }1...m-1\text{ block}}, \underbrace{v_1;v_2;\dotsc;v_{m};v_1;v_2;\dotsc;v_{m};\dotsc;v_1;v_2;\dotsc;v_{m}}_{n-1 \text{ of each }1,\dotsc,m\text{ block}}].$$

To maintain consistency - indexation goes left to right from bottom to top.

Define matrices

$\Delta y_j=diag([\underbrace{\Delta y_1; \Delta y_1; \dotsc;\Delta y_1}_{m-1\text{ times}}; \underbrace{\Delta y_2; \Delta y_2; \dotsc;\Delta y_2}_{m-1\text{ times}};\dotsc;\underbrace{\Delta y_n; \Delta y_n; \dotsc;\Delta y_n}_{m-1\text{ times}}]),$

$ \Delta x_i=diag([\underbrace{\Delta x_1; \Delta x_2;\dotsc\Delta x_m;\Delta x_1; \Delta x_2;\dotsc\Delta x_m; \dotsc;\Delta x_1; \Delta x_2;\dotsc;\Delta x_m}_{n\text{ times}}])$

\begin{equation} R =\left[\begin{array}{cc} \Delta y_j & 0 \\ 0 & \Delta x_i \end{array}\right], \end{equation}

Let the new vector variable of unknowns (something like mass flux) be defined as

$$\vec{q} = R\vec{u},$$

where $R$ is a diagonal matrix of size $(m-1)n\times m(n-1)$ defined above.

Next, we need to define the difference matrices to cancel out the $\frac{x_{i+1}−x_{i-1}}{2}$ central term in diffusion discretization.

Let

${\Delta x_i+\Delta x_{i-1}} = diag[\underbrace{\Delta x_2+\Delta x_1; \Delta x_3+\Delta x_2; \dotsc \Delta x_m-\Delta x_{m-1};...;\Delta x_2+\Delta x_1; \Delta x_3+\Delta x_2; \dotsc \Delta x_m-\Delta x_{m-1}}_{n\text{ times for each block of } \Delta x_2+\Delta x_1\text{ to }\Delta x_m-\Delta x_{m-1}}]$

${\Delta y_j+\Delta y_{j-1}} = diag[\underbrace{\Delta y_2+\Delta y_1; \dotsc \Delta y_2+\Delta y_1;}_{m\text{ times}} \underbrace{\Delta y_3+\Delta y_2; \dotsc \Delta y_3+\Delta y_2;}_{m\text{ times}} \underbrace{\Delta y_n+\Delta y_{n-1}\dotsc \Delta y_n+\Delta y_{n-1}}_{m\text{ times}}],$

\begin{equation} \hat{M} =\left[\begin{array}{cc} \frac{1}{2}\left(\Delta x_i+\Delta x_{i-1}\right) & 0 \\ 0 & \frac{1}{2}\left(\Delta y_j+\Delta y_{j-1}\right) \end{array}\right]. \end{equation}

Now, since $\vec{q}=R\vec{u}$ then $\vec{u}=R^{-1}\vec{q}$, Going back to Laplacian matrix

$$L\vec{u}= LR^{-1}\vec{q}.$$

Premultiply by $\hat M$, the expression will become

$$\hat MLR^{-1}\vec{q},$$

where $\hat M L R^{-1}$ is symmetric by construction.

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