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Given the following 9 point Laplacian
\begin{align} -\nabla^2u_{i,j} = \frac{2}{3h^2}\left[5u_{i,j} - u_{i-1,j} - u_{i+1,j} - u_{i,j-1} - u_{i,j+1} - u_{i-1,j-1} - u_{i-1,j+1} - u_{i+1,j-1} - u_{i+1,j+1}\right] \quad\quad (1) \end{align} Show using Taylor series that
\begin{align} -\nabla^2u_{i,j} = -\nabla^2u - \frac{h^2}{12}(u_{xxxx} + 2u_{xxyy} + u_{yyyy}) + O(h^4) \quad\quad (2) \end{align} $h$ is the distance between two adjacent points and is uniform along both axes. That is,
\begin{align} u_{i,j} - u_{i-1,j} = u_{i,j} - u_{i,j-1} = h \end{align} So I just took the Taylor series for the terms in (1):
\begin{align} u_{i-1,j} = u - hu_x + \frac{h^2}{2}u_{xx} - \frac{h^3}{6}u_{xxx} + \frac{h^4}{24}u_{xxxx} + O(h^5)\\ u_{i+1,j} = u + hu_x + \frac{h^2}{2}u_{xx} + \frac{h^3}{6}u_{xxx} + \frac{h^4}{24}u_{xxxx} + O(h^5)\\ u_{i,j-1} = u - hu_y + \frac{h^2}{2}u_{yy} - \frac{h^3}{6}u_{yyy} + \frac{h^4}{24}u_{yyyy} + O(h^5)\\ u_{i,j+1} = u + hu_y + \frac{h^2}{2}u_{yy} + \frac{h^3}{6}u_{yyy} + \frac{h^4}{24}u_{yyyy} + O(h^5)\\ u_{i-1,j-1} = u - h(u_x + u_y) + \frac{h^2}{2}(u_{xx} + 2u_{xy} + u_{yy}) - \frac{h^3}{6}(u_{xxx} + 3u_{xxy} + 3u_{xyy} + u_{yyy}) + \frac{h^4}{24}(u_{xxxx} + 4u_{xxxy} + 6u_{xxyy} + 4u_{xyyy} + u_{yyyy}) + O(h^5)\\ u_{i-1,j+1} = u - h(u_x - u_y) + \frac{h^2}{2}(u_{xx} - 2u_{xy} + u_{yy}) - \frac{h^3}{6}(u_{xxx} - 3u_{xxy} + 3u_{xyy} - u_{yyy}) + \frac{h^4}{24}(u_{xxxx} - 4u_{xxxy} + 6u_{xxyy} - 4u_{xyyy} + u_{yyyy}) + O(h^5)\\ u_{i+1,j-1} = u + h(u_x - u_y) + \frac{h^2}{2}(u_{xx} - 2u_{xy} + u_{yy}) + \frac{h^3}{6}(u_{xxx} - 3u_{xxy} + 3u_{xyy} - u_{yyy}) + \frac{h^4}{24}(u_{xxxx} - 4u_{xxxy} + 6u_{xxyy} - 4u_{xyyy} + u_{yyyy}) + O(h^5)\\ u_{i+1,j+1} = u + h(u_x + u_y) + \frac{h^2}{2}(u_{xx} + 2u_{xy} + u_{yy}) + \frac{h^3}{6}(u_{xxx} + 3u_{xxy} + 3u_{xyy} + u_{yyy}) + \frac{h^4}{24}(u_{xxxx} + 4u_{xxxy} + 6u_{xxyy} + 4u_{xyyy} + u_{yyyy}) + O(h^5) \end{align} Assuming I got all the Taylor series terms right, adding the first 4 and the last 4 terms gives \begin{align} u_{i-1,j} + u_{i+1,j} + u_{i,j-1} + u_{i,j+1} = 4u + h^2(u_{xx} + u_{yy}) + \frac{h^4}{12}(u_{xxxx} + u_{yyyy}) + O(h^6)\\ u_{i-1,j-1} + u_{i-1,j+1} + u_{i+1,j-1} + u_{i+1,j+1} = 4u + 2h^2(u_{xx} + u_{yy}) + \frac{h^4}{6}(u_{xxxx} + 6u_{xxyy} + u_{yyyy}) + O(h^6) \end{align} Substituting in eqn (1), \begin{align} -\nabla^2u_{i,j} &= \frac{2}{3h^2}\left[-3u_{i,j} - 3h^2(u_{xx} + u_{yy}) - \frac{h^4}{4}(u_{xxxx} + 4u_{xxyy} + u_{yyyy}) + O(h^6)\right]\\ \Rightarrow -\nabla^2u_{i,j} &=\left[\frac{-2u_{i,j}}{h^2} -2(u_{xx} + u_{yy}) - \frac{h^4}{6}(u_{xxxx} + 4u_{xxyy} + u_{yyyy}) + O(h^4)\right] \end{align} That's what I've got so far. Clearly I have some additional terms (compared to eqn (2)), but I'm not really sure what else I can do here. Any help would be much appreciated!

James
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  • is this question related to a specific PDE problem? if yes, what's the original PDE you're trying to solve? – etothepitimesi May 04 '14 at 00:58
  • This is related to a generic Poisson equation: $-\nabla^2u = f$. For this stencil, we are asked to show that the truncation term is $O(h^4)$ as in equation (2). – James May 04 '14 at 01:08

1 Answers1

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The approximation with unit weights on the outside nodes should be (method A) $$ \nabla^2 u \approx \frac{-8u_{i,j}+u_{i+1,j}+u_{i-1,j}+u_{i,j+1}+u_{i,j-1}+u_{i+1,j+1}+u_{i-1,j-1}+u_{i+1,j-1}+u_{i-1,j+1}}{3h^2} $$ you can get this formula simply by summing up all of your Taylor expansions (they look pretty good) - this will also give the $O(h^4)$ accuracy.

the coefficients in the numerator must always sum to zero - yours in equation (1) sum to 8-5 = 3. This form in equation (1: method Q) is probably a typo!

figure demonstrating simple function, exact Laplacian and estimates

the first plot is a simple function that we can calculate the Laplacian exacly, the second is the exact Laplacian, the third is using the formula that I have provided, and the last is using the one from your equation (1).

eigenjohnson
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