微分方程数值解作业 6

微分方程

Problem 1

Question (a)

The equation to solve is

Use a rectangular grid

and the central difference scheme for first and second order derivatives

where is the truncation error. Dropping the error term and rearranging the equation, we have

where and . Consider the boundary conditions described as

The boundary conditions can be written as

FDEs at the top row (where ) transform to

Let

then the equation can be written as

where . The above equation can be written in matrix form

where and is a matrix. has the following structure

where is a tridiagonal matrix

and and are diagonal matrices

is a vector

It is easy to see that and can be reduced to those in (6.20) when .

Question (b)

With ,

and

To use the conjugate gradient method, must be symmetric and positive definite. The symmetric property is obvious in this case. Then we use Test 1 in 6.2.2 to derive a necessary condition for to be positive definite. The condition is

And derive a sufficient condition for using Test 3.

.
, covered by the first condition.
The first row of satifies when the first condition is satisfied.
is irreducible, since the directed graph of is equivalent to the grid graph, which is connected.

The test suggests that is a sufficient condition for to be positive definite.

Using the same test, we can discover that

is a sufficient condition for to be positive definite, which also allows the use of the conjugate gradient method.

Question (c)

The symmetric condition requires , and then with , the matrix falls back to that in (6.20), which is proved to be positive definite. Therefore the necessary and sufficient condition to use the conjugate gradient method is .

Question (d)

Combining the results from (b) and (c), we can find a sufficient condition for the utilization of the conjugate gradient method is

Problem 2

Question (a)

Equation (6.17)

for and . Use centered difference at to handle the bottom boundary condition

The FDEs for are

With the boundary condition , the above equation can be simplified to

Let

we can express the above FDE system as

where is the vectorized form of . has the following structure

is a tridiagonal matrix

is a vector

It is obvious that is not symmetric.

Question (b)

Use second ordered forward difference at to handle the bottom boundary condition

and (6.17) at becomes

That is

With

we express the above FDE system as

where is the vectorized form of . has the following structure

where is a tridiagonal matrix

It is clear that is not symmetric.

Problem 3

Question (a)

It is easy to see that the directed graph of forms a grid graph. The grid graph is connected, so the matrix is irreducible.

Question (b)

The Gershgorin-Taussky theorem states that the eigenvalues of a real symmetric matrix lies in the union of intervals

For the given matrix ,

Therefore

Problem 4

Question (a)

From equation (6.15) we can derive with

Substracting the above equation with Equation (6.17)

and use the definition of error , we have

The above equation has the same coefficient matrix as Equation (6.17). Same simplification can be applied to the boundary conditions, where the left side of the equaiton has the same coefficient matrix, and the constants on the right side are discarded, leaving only . Therefore the error FDE system can be expressed with the same matrix as in (6.21), and the system is