Paper 2, Section II, A

Numerical Analysis | Part II, 2017

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

Γ9[ui,j]:=103ui,j+23(ui+1,j+ui1,j+ui,j+1+ui,j1)+16(ui+1,j+1+ui+1,j1+ui1,j+1+ui1,j1)=h2fi,j\begin{aligned} \Gamma_{9}\left[u_{i, j}\right]:=&-\frac{10}{3} u_{i, j}+\frac{2}{3}\left(u_{i+1, j}+u_{i-1, j}+u_{i, j+1}+u_{i, j-1}\right) \\ &+\frac{1}{6}\left(u_{i+1, j+1}+u_{i+1, j-1}+u_{i-1, j+1}+u_{i-1, j-1}\right)=h^{2} f_{i, j} \end{aligned}

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 the grid points with h=1m+1h=\frac{1}{m+1}.

(i) Find the order of the local truncation error ηi,j\eta_{i, j} of the approximation.

(ii) Prove that the order of the truncation error is smaller if ff satisfies the Laplace equation 2f=0\nabla^{2} f=0.

(iii) Show that the modified nine-point scheme

Γ9[ui,j]=h2fi,j+112h2Γ5[fi,j]:=h2fi,j+112h2(fi+1,j+fi1,j+fi,j+1+fi,j14fi,j)\begin{aligned} \Gamma_{9}\left[u_{i, j}\right] &=h^{2} f_{i, j}+\frac{1}{12} h^{2} \Gamma_{5}\left[f_{i, j}\right] \\ &:=h^{2} f_{i, j}+\frac{1}{12} h^{2}\left(f_{i+1, j}+f_{i-1, j}+f_{i, j+1}+f_{i, j-1}-4 f_{i, j}\right) \end{aligned}

has a truncation error of the same order as in part (ii).

(iv) Let (ui,j)i,j=1m\left(u_{i, j}\right)_{i, j=1}^{m} be a solution to the m2×m2m^{2} \times m^{2} system of linear equations Au=bA \mathbf{u}=\mathbf{b} arising from the modified nine-point scheme in part (iii). Further, 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 approximation at grid points. Prove that there exists a constant cc such that

e2:=[i,j=1mei,j2]1/2<ch3,h0\|\mathbf{e}\|_{2}:=\left[\sum_{i, j=1}^{m}\left|e_{i, j}\right|^{2}\right]^{1 / 2}<c h^{3}, \quad h \rightarrow 0

[Hint: The nine-point discretization of 2u\nabla^{2} u can be written as

Γ9[u]=(Γ5+16Δx2Δy2)u=(Δx2+Δy2+16Δx2Δy2)u\Gamma_{9}[u]=\left(\Gamma_{5}+\frac{1}{6} \Delta_{x}^{2} \Delta_{y}^{2}\right) u=\left(\Delta_{x}^{2}+\Delta_{y}^{2}+\frac{1}{6} \Delta_{x}^{2} \Delta_{y}^{2}\right) u

where Γ5[u]=(Δx2+Δy2)u\Gamma_{5}[u]=\left(\Delta_{x}^{2}+\Delta_{y}^{2}\right) u is the five-point discretization and

Δx2u(x,y):=u(xh,y)2u(x,y)+u(x+h,y)Δy2u(x,y):=u(x,yh)2u(x,y)+u(x,y+h)]\left.\begin{array}{l} \Delta_{x}^{2} u(x, y):=u(x-h, y)-2 u(x, y)+u(x+h, y) \\ \Delta_{y}^{2} u(x, y):=u(x, y-h)-2 u(x, y)+u(x, y+h) \end{array}\right]

[Hint: The matrix A of the nine-point scheme is symmetric, with the eigenvalues

λk,=4sin2kπh24sin2πh2+83sin2kπh2sin2πh2,1k,m\lambda_{k, \ell}=-4 \sin ^{2} \frac{k \pi h}{2}-4 \sin ^{2} \frac{\ell \pi h}{2}+\frac{8}{3} \sin ^{2} \frac{k \pi h}{2} \sin ^{2} \frac{\ell \pi h}{2}, \quad 1 \leqslant k, \ell \leqslant m

Typos? Please submit corrections to this page on GitHub.