Linearizing an explicit differentiation scheme

[Pietair]

Homework Statement

Consider the following implicit scheme:
$$y_{n+1}=y_{n}+\frac{\Delta t}{2}\left [f(y_{n+1})+f(y_{n})]$$

By linearization one can obtain an explicit scheme which is an approximation to this - with approximation error $$O(\Delta t^{3})$$

Homework Equations

The solution is:
$$y_{n+1}=y_{n}+\Delta t\left [1-\frac{1}{2}\Delta t f'(y_{n}) \right ]^{-1}f(y_{n})$$

And the notation: $$y'(x)=f(x,y(x))$$

The Attempt at a Solution

I think I have to take the Taylor expansion of $$f(y_{n+1})$$

I get:
$$f(y_{n+1})=f(y_{n})+\Delta tf'(y_{n})+O(\Delta t^{2})$$

Substituting in the main scheme:

$$y_{n+1}=y_{n}+\frac{\Delta t}{2}\left [f(y_{n})+\Delta tf'(y_{n})+O(\Delta t^{2})+f(y_{n})]f(y_{n+1})=y_{n}+\frac{\Delta t}{2}\left [f(y_{n})+\Delta ty'(y_{n})+O(\Delta t^{2})+f(y_{n})]f(y_{n+1})$$

This expression is, as far as I can see, not equal to the expression of the solution.

Any help would be appreciated!

 

[SammyS]

Homework Statement

Consider the following implicit scheme:
$$y_{n+1}=y_{n}+\frac{\Delta t}{2}\left [f(y_{n+1})+f(y_{n})\right]$$

By linearization one can obtain an explicit scheme which is an approximation to this - with approximation error $$O(\Delta t^{3})$$

Homework Equations

The solution is:
$$y_{n+1}=y_{n}+\Delta t\left [1-\frac{1}{2}\Delta t f'(y_{n}) \right ]^{-1}f(y_{n})$$

And the notation: $$y'(x)=f(x,y(x))$$

The Attempt at a Solution

I think I have to take the Taylor expansion of $$f(y_{n+1})$$

I get:
$$f(y_{n+1})=f(y_{n})+\Delta tf'(y_{n})+O(\Delta t^{2})$$

Substituting in the main scheme:

$$y_{n+1}=y_{n}+\frac{\Delta t}{2}\left [f(y_{n})+\Delta tf'(y_{n})+O(\Delta t^{2})+f(y_{n})\right]f(y_{n+1})=y_{n}+\frac{\Delta t}{2}\left [f(y_{n})+\Delta ty'(y_{n})+O(\Delta t^{2})+f(y_{n})\right]f(y_{n+1})$$

This expression is, as far as I can see, not equal to the expression of the solution.

Any help would be appreciated!

You are missing the "\right" code with several of your "]" symbols. (Corrected in Quote.)

 

[hunt_mat]

This is what I did, it may help you it may not.
$$\begin{array}{rcl} f(y_{n+1}) & = & f\left( y_{n}+\frac{\delta t}{2}X\right) \\ & = & f(y_{n}+\frac{X\delta t}{2}f'(y_{n}) \\ & = & f(y_{n})+ \frac{\delta t}{2}(f(y_{n+1})+f(y_{n}))f'(y_{n}) \end{array}$$
where $X=f(y_{n+1})+f(y_{n})$ and we can solve for $f(y_{n+1})$, to obtain:
$$f(y_{n+1})=\left( 1-\frac{\delta t}{2}f'(y_{n})\right)^{-1}\left( 2f(y_{n})+\frac{\delta t}{2}f'(y_{n})\right)$$
Substitute this into your equation.