I know Richardson extrapolation can be used to estimate a parameter at a single point, but is there a 2D analogous of it where it estimates a parameter over a surface?
For example, I have a 1 m by 1 m square. I solve for the temperature on a mesh with spacing 0.5. I then run the simulation again on a mesh with spacing 0.25. Can I use these two temperature distribution with Richardson Extrapolation to estimate the what the continuum temperature distribution would be?
You've mixed Richardson extrapolation (which extrapolates single value obtained from two or more evaluations with different step sizes) and interpolation which is used to reconstruct continuum solutions from pointwise data.
Unless you're using some finite element method which gives continuum approximation whatever stepsize you take, you can use Richardson extrapolation only to refine solution on the coarse grid. If you want smooth data just interpolate the values using some spline, bicubic spline for example.
Consider the example. $$ -\Delta u = 4 \text{ in } D = [0,1] \times [0,1]\\ u = 0 \text{ on } \partial D $$ Let's approximate the problem with simple second order scheme $$ u_{m-1,n} + u_{m,n-1} -4u_{m,n} + u_{m+1,n} + u_{m,n+1} = 4h^2 \\ u_{0,n} = u_{M,n} = u_{m,0} = u_{m, N} = 0 $$ Solving that with $M = N = 2, h = 0.5$ gives $$ u\big|_{\substack{x = 0.5\\y = 0.5}} \approx u_{1,1}^{(h=0.5)} = 0.25 $$ and with $M = N = 4, h = 0.25$ $$ u\big|_{\substack{x = 0.5\\y = 0.5}} \approx u_{2,2}^{(h=0.25)} = 0.28125 $$ Since the method used is of second order ($p = 2$) $$ u\big|_{\substack{x = 0.5\\y = 0.5}} = u_{N/2, M/2}^{(h)} + \mathcal{O}(h^p) =u_{N/2, M/2}^{(h)} + Ch^p + o(h^p) $$ Now, by Richardson extrapolation, the value at center is $$ u\big|_{\substack{x = 0.5\\y = 0.5}} = \frac{2^p \cdot 0.28125 - 0.25}{2^p-1} + \mathcal{o}(h^2) = 0.291667 + o(h^2) $$ The actual errors for $u\big|_{\substack{x = 0.5\\y = 0.5}}$ are $$ \begin{array}{c|c} \text{method} & \text{error}\\\hline h = 0.5 & 0.044685\\ h = 0.25 & 0.013434\\ \text{extrapolated} & 0.0030175 \end{array} $$
Note, that you can not extrapolate value at $(0.25, 0.25)$, since you do not have a pair of values obtained from the same method but with different step sizes. Of course, you can use bilinear interpolation to interpolate value in $(0.25, 0.25)$ on the coarse grid $$ u\big|_{\substack{x = 0.25\\y = 0.25}} \approx \frac{1}{4}\left( u_{0,0}^{(h=0.5)}+ u_{0,1}^{(h=0.5)}+ u_{1,0}^{(h=0.5)}+ u_{1,1}^{(h=0.5)} \right) = 0.0625 $$ but you would introduce a new error of the same $h^2$ magnitude $$ u\big|_{\substack{x = 0.25\\y = 0.25}} = 0.0625 + Ch^2 + C_\text{interpolation} h^2 + o(h^2) $$ The error for the fine grid would have only the $Ch^2$ term, but not the $C_\text{interpolation} h^2$ $$ u\big|_{\substack{x = 0.25\\y = 0.25}} = u_{1,1}^{(h=0.25)} + Ch^2 + o(h^2) = 0.171875 + Ch^2 + o(h^2) $$ An attempt to blindly extrapolate that values result in $$ u\big|_{\substack{x = 0.25\\y = 0.25}} = \frac{2^p \cdot 0.171875 - 0.0625 }{2^p-1} + \mathcal{o}(h^2) = 0.208333 + o(h^2) $$ and similar error table for this point is $$ \begin{array}{c|c} \text{method} & \text{error}\\\hline h = 0.5 & 0.118643\\ h = 0.25 & 0.00926834\\ \text{extrapolated} & -0.0271897 \end{array} $$ which shows that extrapolated value is worse than $h = 0.25$ one.
Actually there's a way to make interpolation and Richardson extrapolation work together. You need to make the interpolation error even smaller, i.e. make it $o(h^2)$, for example by using biquadratic interpolation, and not the bilinear one. $$ u\big|_{\substack{x = 0.25\\y = 0.25}} \approx \begin{pmatrix} 0.375 & 0.75 & -0.125 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0\\ 0 & 0.25 & 0\\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0.375 \\ 0.75 \\ -0.125 \end{pmatrix} = 0.140625 $$ $$ u\big|_{\substack{x = 0.25\\y = 0.25}} = 0.140625 + Ch^2 + C_\text{interpolation} h^3 + o(h^2) = 0.0625 + Ch^2 + o(h^2) $$ Now the extrapolation performs better, but still not perfect $$ u\big|_{\substack{x = 0.25\\y = 0.25}} = \frac{2^p \cdot 0.171875 - 0.140625}{2^p-1} + \mathcal{o}(h^2) = 0.182292 + o(h^2) $$ and the error table is $$ \begin{array}{c|c} \text{method} & \text{error}\\\hline h = 0.5 & 0.0405183\\ h = 0.25 & 0.00926834\\ \text{extrapolated} & -0.00114766 \end{array} $$