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 &lt; 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?

Question

An upper bound for the global error in Euler&#039;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  &amp;lt;  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?

broeifl · Accepted Answer

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.

Alisa Taylor · Accepted Answer

didn&#039;t think it would be like that, thanks

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