Mh/2L (e^(L(t_(i+1)−a))−1), Having L small is good for convergence and stability (right?) in the sense that L small implies little change. If L=0 then the function f(t,y) (on the right hand side of the ODE) is constant on y. If L<1 the function is contracting. So why is L in the denominator of the bound? My intuition (which is not working good) says that the error bound should diminish as L diminshes.

Lara Cortez 2022-10-15 Answered
An upper bound for the global error in Euler's method to solve a first order ODE numerically is given by the equation
M h 2 L ( e L ( t i + 1 a ) 1 ) ,
where t i + 1 is the i + 1 cell in the abcsisa and a is the first abcisa. M is a bound for the second derivative of the ODE unknown y and L is the Lipschitz continuity property bound. This is derived in most of numerical analysis books.

My question is the following: Having L small is good for convergence and stability (right?) in the sense that L small implies little change. If L = 0 then the function f ( t , y ) (on the right hand side of the ODE) is constant on y. If L < 1 the function is contracting. So why is L in the denominator of the bound? My intuition (which is not working good) says that the error bound should diminish as L diminshes.

What is wrong with my intuition here?
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Answers (2)

broeifl
Answered 2022-10-16 Author has 11 answers
First this bound is not a good estimation. It is useful to say the error is bounded, but in practice it is completely inaccurate. Then the error does not explode when L 0. You can develop 1 L ( e L ( t i + 1 a ) 1 ) = 1 L ( 1 + L ( t i + 1 a ) + O ( L 2 ) 1 ) = t i + 1 a + O ( L ). So you see that the error diminishes as L tends to 0, which is logical.
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Alisa Taylor
Answered 2022-10-17 Author has 4 answers

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