I am solving the following PDE;
$$
\nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \rho,
$$
where $\rho(0.5,0.5) = 2$ (zero elsewhere), $0\leq x,y\leq1$ and the boundaries will satisfy the condition $u = 0$.
I am solving by finite differences using both a 5 point second order stencil and a 9 point fourth order stencil. It is my understanding that the solution to this PDE is the zero function and this is what I am using to calculate rates of convergence. I have used various error measures including the average difference, max difference and the value at the point $(0.5,0.5)$.
For each error measure I am getting convergence rates of approximately order 2 for both stencils, is there a reason why the fourth order stencil would not converge with order 4?
Best Answer
Ian made two excellent points in comments:
I will elaborate on the second point. The discrete values of $\rho$ you feed into any finite difference method do not represent pointwise values of $\rho$, but rather its averages on scale $h$. So, feeding in $\rho=2$ at $(0.5,0.5)$ (which I assume is one of the grid points) corresponds to prescribing the Laplacian of $2$ on the square with vertices $(0.5\pm h/2, 0.5\pm h/2)$. The analytic solution to the latter problem is of order $h^2$. Indeed, it is given by the integral of Green's function $G$: $$ u(x,y) = \int_{0.5-h/2}^{0.5+h/2}\int_{0.5-h/2}^{0.5+h/2} 2G(x,y;x',y')\,dx'\,dy' $$ which, at a point $(x,y)\ne (0.5,0.5)$, is asymptotic to $h^2 G(x,y,0.5,0.5)$.
At $(x,y)=(0.5,0.5)$ the analytic solution is actually slightly larger: $h^2\log(1/h)$, but this is probably not apparent in the numerical results.