Paper 3, Section II, E

Numerical Analysis | Part II, 2018

The diffusion equation for u(x,t)u(x, t) :

ut=2ux2,xR,t0\frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial x^{2}}, \quad x \in \mathbb{R}, \quad t \geqslant 0

is solved numerically by the difference scheme

umn+1=umn+32μ(um1n2umn+um+1n)12μ(um1n12umn1+um+1n1).u_{m}^{n+1}=u_{m}^{n}+\frac{3}{2} \mu\left(u_{m-1}^{n}-2 u_{m}^{n}+u_{m+1}^{n}\right)-\frac{1}{2} \mu\left(u_{m-1}^{n-1}-2 u_{m}^{n-1}+u_{m+1}^{n-1}\right) .

Here μ=kh2\mu=\frac{k}{h^{2}} is the Courant number, with k=Δt,h=Δxk=\Delta t, h=\Delta x, and umnu(mh,nk)u_{m}^{n} \approx u(m h, n k).

(a) Prove that, as k0k \rightarrow 0 with constant μ\mu, the local error of the method is O(k2)\mathcal{O}\left(k^{2}\right).

(b) Applying the Fourier stability analysis, show that the method is stable if and only if μ14\mu \leqslant \frac{1}{4}. [Hint: If a polynomial p(x)=x22αx+βp(x)=x^{2}-2 \alpha x+\beta has real roots, then those roots lie in [a,b][a, b] if and only if p(a)p(b)0p(a) p(b) \geqslant 0 and α[a,b]\alpha \in[a, b].]

(c) Prove that, for the same equation, the leapfrog scheme

umn+1=umn1+2μ(um1n2umn+um+1n)u_{m}^{n+1}=u_{m}^{n-1}+2 \mu\left(u_{m-1}^{n}-2 u_{m}^{n}+u_{m+1}^{n}\right)

is unstable for any choice of μ>0\mu>0.

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