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Stability of Crank-Nicholson method

  • Thread starter Nusc
  • Start date
754
2
1. Homework Statement
If p(A^-1 B) < 1 then the Crank-Nicholson method is stable for all eigenvalues.


Where p is the spectral radius.

2. Homework Equations
Stability requires that A*U_j=B*U_{j-1} which gives
U_j = A^-1 B U_{j-1}


3. The Attempt at a Solution

Where do I start?
 

Answers and Replies

212
0
Where do I start?
By defining A,B, and the problem more clearly. Are you saying you need to prove that if p(A^-1 B)<1, then the Crank-Nicholson method applied to some equation involving A and B is stable?
 
754
2
Yes.
So for A*U_j = B*U_{j-1} we have:
The finite difference matrix for A:

1+lamda -lamda/2 ... 0
-lamda/2 1+lamda ... 0
0 ........
... -lamda/2
0 .... -lamda/2 1+lamda

Some tridiagonal matrix.
Similarly for B
B:

1-lamda +lamda/2 ... 0
+lamda/2 1-lamda ... 0
0 ........
... +lamda/2
0 .... +lamda/2 1-lamda

This is in general. I'm not sure if this is correct because I assume we need to do it for another problem.
 
212
0
If your equation is
[tex]Au_j = Bu_{j-1}[/tex],
then
[tex]u_j = A^{-1}Bu_{j-1}[/tex]
and
[tex]u_{j+1} = A^{-1}Bu_{j}= \left(A^{-1}B\right)A^{-1}Bu_{j-1}[/tex],
and
[tex]u_{j+2} = A^{-1}Bu_{j-1}u_{j+1} = \left(A^{-1}B\right)^3u_{j-1}[/tex].
You can continue this on forever to get the term after n steps as
[tex]u_{j+n} = \left(A^{-1}B\right)^{n+1}u_{j-1}[/tex].

Now factor A^{-1}B by eigenvalue decomposition to obtain
[tex] A^{-1}B = TDT^{-1}[/tex]
where D is a diagonal matrix containing the eigenvalues and T contains the corresponding eigenvectors.
Note that
[tex]\left(A^{-1}B\right)^2 = \left(TDT^{-1}\right)^2 = TDT^{-1}TDT^{-1}= TD^2T^{-1}[/tex]
And similarly,
[tex] \left( A^{-1}B \right)^n = TD^nT^{-1}[/tex]

Now if you plug this in to the previous equation, you find that
[tex]u_{j+n} = \left(A^{-1}B\right)^{n+1}u_{j-1} = TD^{n+1}T^{-1}u_{j-1}[/tex]

The system is stable if the solution u_{j+n} is bounded for all n. Since D is a diagonal matrix, D^{n+1} is just the diagonal elements raised to the n+1th power. So what happens if a number bigger than one is raised to a large power? And what about when the number is smaller than one? This is why you have the condition on the size of the eigenvalues.
 
754
2
If a number bigger than one is raised to a large power, then the system will become unstable.
If a number smaller than one is raised to a large power, then the system will become stable.

Hence the method is unconditionally stable.

Is that correct?
 

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