Consistency of Crank Nicolson method

[mbsl5]

Homework Statement

Analyze the consistency of the Crank Nicolson method:

u_n+1 = u_n + h/2[f_n + f_n+1]

where f_n = f(t_n, u_n)

Homework Equations

How does this method work if you have to have u_n+1 to calculate f_n+1, and vice versa? Which comes first?

The Attempt at a Solution

I have worked other consistency analysis problems with simpler approximations, such as Forward Euler Method, but I'm still struggling with this question in general (order, stability, consistency). I know (I think) that we want to see that the difference (error) between the approximation of the method and the actual solution goes to zero as h(the fixed interval size) goes to zero, which says the method is consistent. I have been doing that by deriving the value of f from a Taylor series expansion of u around x_n. So,

y_n+1 = y_n + hf_n + (h^2/2)*f'_n + (h^3/6)*f''_n + . . .

and by assuming the method is true for actual y:

(y_n+1 - y_n)/h = (1/2)*[f_n+1 + f_n]

Not sure where to go from there. Any help appreciated.

 

[mighty2000]

Thank you for your question about the consistency of the Crank Nicolson method. This is a very important aspect to consider when evaluating the accuracy and reliability of any numerical method.

As you correctly stated, the goal of consistency analysis is to show that the error between the approximation and the actual solution decreases as the step size h goes to zero. In order to do this, we need to look at the Taylor series expansions for both f(t_n, u_n) and u_n+1.

Starting with the Taylor series expansion for f(t_n, u_n), we have:

f(t_n, u_n) = f_n + h*f'_n + (h^2/2)*f''_n + (h^3/6)*f'''_n + . . .

Now, for the Taylor series expansion for u_n+1, we have:

u_n+1 = u_n + h*u'_n + (h^2/2)*u''_n + (h^3/6)*u'''_n + . . .

Since we are using the Crank Nicolson method, we know that u_n+1 = u_n + h/2[f_n + f_n+1]. Substituting this into the Taylor series expansion for u_n+1, we get:

u_n+1 = u_n + h*u'_n + (h^2/2)*u''_n + (h^3/6)*u'''_n + (h/2)*[f_n + f_n+1]

= u_n + h/2[f_n + f_n+1] + (h^2/2)*u''_n + (h^3/6)*u'''_n + (h/2)*[f_n + f_n+1]

= u_n + h*f_n + (h^2/2)*[f'_n + u''_n] + (h^3/6)*[f''_n + u'''_n] + . . .

Now, we can compare this to the Taylor series expansion for u_n+1 that we derived earlier, and we can see that they are consistent up to terms of order h^2. This means that as h goes to zero, the error between the approximation and the actual solution will also go to zero, and the method is consistent.

To answer your question about which comes first,