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Nice link Krab. What I fail to see there is any formal error analysis of the methods. The claim that the "failure to conserve energy" property of Euler is characteristic to periodic functions is not proven. They only show 1 step size, what would happen if they choose a smaller step? What would happen if they chose a DIFFERENT function. The local discretization error of Euler is ~step size, this holds for ALL applications of Euler,not just periodic.
Here is an expression derived in Conte and de Boor which relates step size and local error.
|f_y(x,y)|\leq L
|y^{\prime\prime}(x)|<Y
|{error}_n|= \frac{hY}{2L}(e^{(x_n-x_0)L}-1)
He points out that this is a "Upper bound rather then a realistic bound" so you may be able to achieve the same error with a smaller step. But a step size computed with this formula will yield the desired error at the specified end point.
Notice that this error is indeed monotonic. So for the pendulum it shows up as a steadily increasing amplitude. Some one with no deeper vision could indeed say that it looks like energy is not conserved. But that is failing to look at the mechanism of the tool being applied. (Failure: Car won't run, solution: Replace engine, root cause: out of gas) Further, if you were apply this method to a different system (perhaps financial) would it still fail to conserve energy? Of course not, because energy is not involved, but it would still show a steady increase in the final solution. So to say Euler fails to conserve energy is to broad a statement. It can only be make for specific models. While when I say Euler has local discretization of ~step this applies to ALL applications of Euler, and is the root cause of the failure to conserve energy phenomena in the pendulum problem.
Here is an expression derived in Conte and de Boor which relates step size and local error.
|f_y(x,y)|\leq L
|y^{\prime\prime}(x)|<Y
|{error}_n|= \frac{hY}{2L}(e^{(x_n-x_0)L}-1)
He points out that this is a "Upper bound rather then a realistic bound" so you may be able to achieve the same error with a smaller step. But a step size computed with this formula will yield the desired error at the specified end point.
Notice that this error is indeed monotonic. So for the pendulum it shows up as a steadily increasing amplitude. Some one with no deeper vision could indeed say that it looks like energy is not conserved. But that is failing to look at the mechanism of the tool being applied. (Failure: Car won't run, solution: Replace engine, root cause: out of gas) Further, if you were apply this method to a different system (perhaps financial) would it still fail to conserve energy? Of course not, because energy is not involved, but it would still show a steady increase in the final solution. So to say Euler fails to conserve energy is to broad a statement. It can only be make for specific models. While when I say Euler has local discretization of ~step this applies to ALL applications of Euler, and is the root cause of the failure to conserve energy phenomena in the pendulum problem.
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