Approximating y ′ ( t n ) at the relation y ′ ( t n...

Adrianna Macias

Adrianna Macias

Answered

2022-07-20

Approximating y ( t n ) at the relation y ( t n ) = f ( t n , y ( t n ) ) with the difference quotient [ y ( t n + 1 ) y ( t n ) h ] we get to the Euler method.
Approximating the same derivative with the quotient [ y ( t n ) y ( t n 1 ) h ] we get to the backward Euler method
y n + 1 = y n + h f ( t n + 1 , y n + 1 ) , n = 0 , , N 1
where y 0 := y 0 .
In order to find the formula for the forward Euler method, we use the limit lim h 0 y ( x 0 + h ) y ( x 0 ) h for x 0 = t n , h = t n + 1 t n .
In order to find the formula for the backward Euler method, could we pick h = t n 1 t n although it is negative?
Or how do we get otherwise to the approximation:
y ( t n ) y ( t n ) y ( t n 1 ) h

Answer & Explanation

iljovskint

iljovskint

Expert

2022-07-21Added 18 answers

obviously, you still choose t n + 1 = t n + h n , thus h = h n = t n + 1 t n for the step from y n to y n + 1 .

The mean value theorem only tells us that
y ( t n + 1 ) y ( t n ) t n + 1 t n = y ( t n + θ ( t n + 1 t n ) )
where for most of the usual functions θ ( 0 , 1 ) can be found close to 1/2. Setting, for approximation purposes, θ = 0 or θ = 1 has thus about the same degree of inaccuracy.

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