Since , the right inequality is always true. For the left inequality, we have

which holds irrespective of the value of under the condition

Problem 2

Question (a)

The stencil is shown below.

tj+1tjtj¡1xi¡1xixi+1

Limits on and are

Question (b)

FDE for the Dufort-Frankel scheme is

Extract and use Taylor expansion ( subscript is omitted), we have

Use and drop higher order terms, we have

This is of order . Due to the presence of term, the method is not consistent when , where when . Therefore the method is only conditionally consistent.

Question (c)

The method is explicit. can be directly calculated from , and without solving a linear system or inverse function.

Question (d)

Use notation, the FDE can be written as

Extract , we have

Substitute with the generic solution , we have

Assume , and divide both sides by , we have

which yields the following 2nd order equation

The roots of this equation are

For the method to be stable, we need . For real roots where , we have

For complex roots where , we have

and

Note that guarantees that , and the method is unconditionally stable.

Question (e)

From , we have

This indicates that though deacys exponentially, it does so slowly when is large. Therefore, the method is stable but not L-stable.

Problem 3

Question (a)

From equation (3.22)

we can replace terms with Taylor expansion at , and obtain

Cancel duplicated terms, we have

where has an order of . Note that the fifth spatial derivative terms in also cancel out, so the remain in is instead of .