# Linearizing an explicit differentiation scheme

• Pietair
In summary, the solution to the implicit scheme is to take the Taylor expansion of f(y_{n+1}), which results in:y'(x)=f(x,y(x))+\frac{\delta t}{2}f'(y_{n}).
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!

Pietair said:

## 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.)

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:
$$f(y_{n+1})=\left( 1-\frac{\delta t}{2}f'(y_{n})\right)^{-1}\left( 2f(y_{n})+\frac{\delta t}{2}f'(y_{n})\right)$$

Last edited:

## 1. What is linearization in the context of explicit differentiation schemes?

Linearization is a technique used in numerical analysis to approximate a nonlinear function with a linear one. In the context of explicit differentiation schemes, linearization is used to simplify the computation of derivatives by approximating the derivative of a function at a particular point with the slope of a tangent line.

## 2. Why is linearization important in explicit differentiation schemes?

Linearization allows for the use of simpler and more efficient algorithms for computing derivatives. This is especially useful in numerical methods such as finite difference methods, where the computation of derivatives is a crucial step.

## 3. How is linearization achieved in explicit differentiation schemes?

Linearization is achieved by using the first-order Taylor series approximation of a function. This involves finding the tangent line to the function at a particular point and using its slope to approximate the derivative at that point.

## 4. What are the limitations of linearization in explicit differentiation schemes?

Linearization can only approximate the derivative at a particular point and may not accurately represent the behavior of the function in the rest of the domain. This can lead to errors in the overall solution, especially for highly nonlinear functions.

## 5. Are there any alternative methods to linearization in explicit differentiation schemes?

Yes, there are other methods such as automatic differentiation and symbolic differentiation that can be used to compute derivatives without the need for linearization. However, these methods may be more computationally expensive and may not be suitable for all types of functions.

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