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Linearizing an explicit differentiation scheme

  • Thread starter Pietair
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  • #1
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Homework Statement


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

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

Homework Equations


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

And the notation: [tex]y'(x)=f(x,y(x))[/tex]

The Attempt at a Solution


I think I have to take the Taylor expansion of [tex]f(y_{n+1})[/tex]

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

Substituting in the main scheme:

[tex]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})[/tex]

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

Any help would be appreciated!
 

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
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11,315
1,005

Homework Statement


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

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

Homework Equations


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

And the notation: [tex]y'(x)=f(x,y(x))[/tex]

The Attempt at a Solution


I think I have to take the Taylor expansion of [tex]f(y_{n+1})[/tex]

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

Substituting in the main scheme:

[tex]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})[/tex]

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.)
 
  • #3
hunt_mat
Homework Helper
1,741
25
This is what I did, it may help you it may not.
[tex]
\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}
[/tex]
where [itex]X=f(y_{n+1})+f(y_{n})[/itex] and we can solve for [itex]f(y_{n+1})[/itex], to obtain:
[tex]
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)
[/tex]
Substitute this into your equation.
 
Last edited:

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