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AMATH 581 ( c○J. N. Kutz) 15This then gives the maximum error to be|E n |≤ e r + e r∆t+ ∆t22 M = 2e r∆t + ∆t2 M. (1.2.11)2To minimize the error, we require that ∂|E n |/∂(∆t) = 0. Calculating thisderivative gives∂|E n |∂(∆t) = − 2e r+ M∆t =0, (1.2.12)∆t2 so that∆t =( ) 1/3 2er. (1.2.13)MThis gives the step-size resulting in a minimum error. Thus the smallest stepsizeis not necessarily the most accurate. Rather, a balance between round-offerror and truncation error is achieved to obtain the optimal step-size.StabilityThe accuracy of any scheme is certainly important. However, it is meaninglessif the scheme is not stable numerically. The essense of a stable scheme: thenumerical solutions do not blow up to infinity. As an example, consider thesimple differential equationdy= λy (1.2.14)dtwithy(0) = y 0 . (1.2.15)The analytic solution is easily calculated to be y(t) =y 0 exp(λt). However, ifwe solve this problem numerically with a forward Euler method wefindy n+1 = y n +∆t · λy n =(1+λ∆t)y n . (1.2.16)After N steps, we find this iteration scheme yieldsy N =(1+λ∆t) N y 0 . (1.2.17)Given that we have a certain amount of round off error, the numerical solutionwould then be given byy N =(1+λ∆t) N (y 0 + e) . (1.2.18)The error then associated with this scheme is given byE =(1+λ∆t) N e. (1.2.19)