I would like to solve a system of differential equations $u+\nabla(\nabla\cdot u)=f$ or in more detail$a+\partial_t^2a+\partial_t\partial_xb+\partial_t\partial_yc=f$$b+\partial_x^2b+\partial_x\partial_ta+\partial_x\partial_yc=g$$c+\partial_y^2c+\partial_y\partial_tb+\partial_y\partial_xc=h$with periodic boundary conditions on a regular grid by the method of Finite Differences. $a=a(t,x)$, $b=b(t,x)$ and $c=c(t,x)$ are my unknowns and $f=f(t,x)$, $g=g(t,x)$ and $h=h(t,x)$ are known.I discretized my domain in time ($t$) and space ($x,y$). What I already have now is a big equation system. The matrix has full rank. For a grid with only 3 nodes (=27 nodes overall, =81 equations) in each direction it looks like thisWhen increasing the number of nodes the system gets to big very soon for my computers memory. That's why I actually want the system matrix to be symmetric in order to use Cholesky factorization (hopefully) or something else. But I am not sure if there exists a numbering of nodes which would make the system matrix symmetric.In literature the commonly used example is the Laplace equation, where they just use a symmetric stencil in order to approximate $\Delta$. In contrast to that I have to use one dimensional finite difference for the approximation of e.g. $\partial_x^2$ in a 3 dimensional space. Furthermore I am using one-sided differences for the mixed derivatives at boundaries, which might destroy symmetric. However I am not sure if this really destroys symmetry. Only because I found this thread I've got the hope that there is a numbering of my nodes, which would make my system matrix symmetric. But actually I cannot imagine that, because a non-symmetric matrix cannot become a symmetric one by only permuting rows and columns in the same way.So is there a way to get a symmetric system matrix for my problem? If not, what would you do instead in order to solve the system?
Get symmetric Finite Difference matrix in non Laplacian settings
linear algebra;finite difference;boundary conditions;matrix;symmetry