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微分方程数值解作业 5

微分方程

Problem 1

Question Dispersion Relation Phase Velocity Group Velocity Dispersive Dissipative
(a) Yes No
(b) Yes Yes
(c) Yes No

Question (a)

With the plane wave solution

we have

That simplifies to

which holds for all if and only if

and this is the dispersion relation. The phase velocity

The group velocity

The equation is dispersive since depends on . The equation is nondissipative since requires to be zero.

Question (b)

With the plane wave solution, we have

That simplifies to

Therefore the dispersion relation is

Since are all real, is real. The phase velocity

The phase velocity

The group velocity

It can be shown that

Also,

which is true since is positive.

The equation is dispersive since depends on . It is also disspative because in , is not necessarily real.

Question (c)

With the plane wave solution, we have

which simplifies to

The dispersion relation is

The phase velocity

The group velocity

The equation is dispersive since depends on . It is also nondissipative since is zero.

Problem 2

Question (a)

Question (b)

From (5.31) we have

With the initial condition in we have

That is

All desired can be solved by the tridiagonal matrix equation

And then compute using , and so on.

Question (c)

The method satisfies CFL condition, since the system described by requires depends on all of , so the numerical domain of dependence is the entire domain. As shown in previous chapters, an unnecessarily large domain of dependence can lead to less accurate results. Using other approximations in space can not lift the need to solve the tridiagonal matrix equation.

Question (d)

From (5.31) we have

With the initial condition in we have

Therefore can be determined explicitly by

The stencil for is , which is a 3-point stencil.

t1t0xi¡1xixi+1

The CFL condition is the same as in the textbook.

Problem 3

Question (a)

  1. Use a uniform grid where and
  2. Evaluate the difference equation at
  3. Using centered differences to approximate the derivatives we have This can be rearranged to where is the truncation error.
  4. Drop the error term to get the finite difference approximation for and . The first boundary condition in (5.2) can be approximated by The second boundary condition in (5.3) can be handled by introducing ghost points at , described in . With we can calculate for

The stencil for this method is

tj+1tjtj¡1xi¡1xixi+1

and

t1t0xi¡1xixi+1

at the first time step The numerical domain of dependence consists of the points