On a grid I am having the values of a physical quantity say for example Temperature, at the E,W,N,S and P node all of them being calculated using a second order discretization scheme. I want a second order accurate estimate of the value of Temperature at point X. How to get that.
Note- I can get a second order accurate value of temperature at point IP1 (using second order polynomial fitting with point E, W and P) and at point IP2 (using second order polynomial fitting with point N, S and P). Both IP1 and IP2 are at distance $\Delta x/4$ from point P.
Kindly refer to the stencil for details.
It seems to me that you need to wipe out the first and second order partials of $T$ using a linear combination of values of $T$ at the other $5$ grid points. $$\begin{bmatrix}hT_x&0&0&0&0\\0&hT_y&0&0&0\\0&0&h^2T_{xx}&0&0\\0&0&0&h^2T_{xy}&0\\0&0&0&0&h^2T_{yy}\end{bmatrix}\begin{bmatrix}-3/4 &1/4 &1/4 &5/4 &1/4\\ 1/4 &-3/4 &5/4 &1/4 &1/4\\ 9/32 &1/32 &1/32 &25/32 &1/32\\ -3/16 &-3/16 &5/16 &5/16 &1/16\\ 1/32 &9/32 &25/32 &1/32 &1/32\end{bmatrix}\begin{bmatrix}x_W\\x_S\\x_N\\x_E\\x_P\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\\0\end{bmatrix}$$ But since the determinant of that matrix is $-\frac1{16}$, it follows that you can't get a nontrivial linear combination of the coefficients that wipes out the first and second order terms. Even a condition such as $\nabla^2T=0$ wouldn't help here because as it turns out that would not add a linearly independent eqution in this case.
EDIT: You could add another point to the formula. The closest is $T_{SW}$, and if you did so, you would find $$T_{x}=\frac1{32}(3T_W+3T_S-3T_N-3T_E+30T_P+2T_{SW})$$