Question 5.68: A diffusion-transport problem is described by the differenti......

(a) Replacing the derivatives y^{\prime \prime} and y^{\prime} in the differential equation by their central difference approximations, we obtain

\frac{p}{h^{2}}\left(y_{n-1}-2 y_{n}+y_{n+1}\right)+\frac{V}{2 h}\left(y_{n+1}-y_{n-1}\right)+O\left(h^{2}\right)=0 .

Neglecting the truncation error, we get

\left(1+\frac{V h}{2 p}\right) y_{n+1}-2 y_{n}+\left(1-\frac{V h}{2 p}\right) y_{n-1}=0.

The characteristic equation is given by

\left(1+\frac{V h}{2 p}\right) \xi^{2}-2 \xi+\left(1-\frac{V h}{2 p}\right)=0,

or (1+R e) \xi^{2}-2 \xi+(1-R e)=0

where R e=V h /(2 p) is called the cell Reynold number.

The roots are given by \xi=1 and \xi=(1-R e) /(1+R e).

The solution of the difference equation is given by

y_{n}=C_{1}+C_{2}\left(\frac{1-R e}{1+R e}\right)^{n}

when (p / h) is too small, that is

Hence, move if R e>>1, then the solution will contain slowly damped oscillations.

(b) Let now the forward difference approximation to y^{\prime} be used. Neglecting the truncation error we get the difference equation

\left(1+\frac{V h}{p}\right) y_{n+1}-2\left(1+\frac{V h}{2 p}\right) y_{n}+y_{n-1}=0,

or (1+2 R e) y_{n+1}-2(1+R e) y_{n}+y_{n-1}=0 .

The characteristic equation is given by

(1+2 R e) \xi^{2}-2(1+R e) \xi+1=0,

whose roots are \xi=1 and 1 /(1+2 R e). The solution of the difference equation is

y_{n}=A+\frac{B}{(1+2 R e)^{n}} .

Hence, for R e>1, the solution does not have any oscillations.

(c) The truncation error of the difference scheme in (b) is defined by

T_{n}=\left(1+\frac{V h}{p}\right) y\left(x_{n+1}\right)-2\left(1+\frac{V h}{2 p}\right) y\left(x_{n}\right)+y\left(x_{n-1}\right) .

Expanding each term in Taylor’s series, we get

T_{n}=\left[\frac{V}{p} y^{\prime}\left(x_{n}\right)+y^{\prime \prime}\left(x_{n}\right)\right] h^{2}+O\left(h^{3}\right)

where x_{n-1}<\xi<x_{n+1}.

The order of the difference scheme in (b) is one.