Paper 3, Section II, A

Numerical Analysis | Part II, 2010

The Poisson equation 2u=f\nabla^{2} u=f in the unit square Ω=[0,1]×[0,1],u=0\Omega=[0,1] \times[0,1], u=0 on Ω\partial \Omega, is discretized with the five-point formula

ui,j1+ui,j+1+ui+1,j+ui1,j4ui,j=h2fi,ju_{i, j-1}+u_{i, j+1}+u_{i+1, j}+u_{i-1, j}-4 u_{i, j}=h^{2} f_{i, j}

where 1i,jM,ui,ju(ih,jh)1 \leqslant i, j \leqslant M, u_{i, j} \approx u(i h, j h) and (ih,jh)(i h, j h) are grid points.

Let u(x,y)u(x, y) be the exact solution, and let ei,j=ui,ju(ih,jh)e_{i, j}=u_{i, j}-u(i h, j h) be the error of the five-point formula at the (i,j)(i, j) th grid point. Justifying each step, prove that

e=[i,j=1Mei,j2]1/2ch for sufficiently small h>0,\|\mathbf{e}\|=\left[\sum_{i, j=1}^{M}\left|e_{i, j}\right|^{2}\right]^{1 / 2} \leqslant c h \quad \text { for sufficiently small } h>0,

where cc is some constant independent of hh.

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