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Crank-Nicolson vs Heun's method

by Smed
Tags: cranknicolson, heun, method
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Smed
#1
Apr6-10, 09:49 AM
P: 30
Hi, can someone tell me the difference between the Crank-Nicolson and Heun numerical methods? For Heun's method I'm looking here http://en.wikipedia.org/wiki/Heun%27s_method and for the Crank-Nicolson method I'm looking here http://en.wikipedia.org/wiki/Crank%E...icolson_method . When I actually carry out a calculation with equal timesteps for both methods and f(t,u)=-.5*u, I get the exact same solution.

The equation I have for both is:

u[tex]^{n+1}[/tex] = u[tex]^{n}[/tex] - [tex]\frac{1}{2}[/tex]u[tex]^{n}[/tex]dt - [tex]\frac{1}{8}[/tex]u[tex]^{n}[/tex]dt[tex]^{2}[/tex]
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Lord Crc
#2
Apr9-10, 10:02 AM
P: 88
I'm no expert, but from what I can gather Heun's method is for ODE's while Crank-Nicolson is for PDE's?
matematikawan
#3
Apr9-10, 10:39 AM
P: 333
Heun's method is an improvement of the forward Euler's method which is an explicit method.
While Crank-Nicolson method is an implicit method. Probably the improvement for the backward Euler method. This is the Crank-Nicolson method for ODE.

But of course the Crank-Nicolson method is verypopular in PDE.

Lord Crc
#4
Apr10-10, 09:46 AM
P: 88
Crank-Nicolson vs Heun's method

Ah, in all cases I've come across Crank-Nicolson, it has been to solve PDEs.

So, if I read my notes correctly, while both methods take an average of the current state and the state at the next timestep, the main difference between Heun's method and Crank-Nicolson is that for Heun's method you use a predictor for the next timestep, keeping it explicit, while for Crank-Nicolson it is used implicitly instead. At least that's my understanding.

Using this I get some different results from yours, both with Crank-Nicolson and Heun's method, are you sure you do Heun's method correctly?
Redbelly98
#5
Apr10-10, 05:57 PM
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Quote Quote by Smed View Post
The equation I have for both is:

u[tex]^{n+1}[/tex] = u[tex]^{n}[/tex] - [tex]\frac{1}{2}[/tex]u[tex]^{n}[/tex]dt - [tex]\frac{1}{8}[/tex]u[tex]^{n}[/tex]dt[tex]^{2}[/tex]
If we make that a +⅛ instead, I agree that Heun's method gives that equation. But I get something different for Crank-Nicolson.

For Crank-Nicolson, ignore the x-dependence of u and we have
(un+1 - un) / Δt = ( - un+1 - un)
Solve that for un+1 and we get something different than the Heun's method equation. (Though they do agree up to order Δt2.)


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