# Euler (explicit or implicit)

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 May 7, 2006, 05:11 Euler (explicit or implicit) #1 anybody Guest   Posts: n/a Hi, I have written a little program which solves an ordinary differential equation y'=B*y-A*C*exp(B*x)*sin(C*x). y(x=0) = A. For the integration I have used an explicit and implicit Euler-scheme (first order). Since both schemes are first order accurate the integration fault should be identical. Do you think so too? How can I prove the first order for the implicit scheme? (explicit is easy using a taylor-expansion around x). To get the derivative dy/dx|_(x+dx) (IMPLICIT), I have to do an explicit forward step to get y_(x+dx). Right? Thanks.

 May 7, 2006, 07:32 Re: Euler (explicit or implicit) #2 ganesh Guest   Posts: n/a Dear Anybody, The fact that the time integration is first order accurate, does not essentially mean that the integration fault is the same. All it means is that the accuracy is first order in time, which means a smaller delt would give a smaller fault, and the rate of decrease would be at unity, the schemes being to first orde accuracy. However, the magnitude of the fault would be determined by the coefficients of the error terms, and therefore could be different. In general, an implicit scheme is expected to give a lower fault than the explicit scheme. Hope this helps Regards, Ganesh

 May 8, 2006, 02:12 Re: Euler (explicit or implicit) #3 anybody Guest   Posts: n/a Hi, thanks for your comments - it seems logical to me. But could you briefly deschribe how I have to determine the derivative at (n+1). I thought have to integrate forward which leads me to f_i+1. Since f'=A*f+..... this leads to df/dx_(n+1).

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