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Runge Kutta Method, Lecture Notes - Mathematics, Study notes of Calculus

ranga kutta method, second order, computation solution, fourth order, formulation, multi steps, taylor expand the local truncation error

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2010/2011

Uploaded on 10/09/2011

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Download Runge Kutta Method, Lecture Notes - Mathematics and more Study notes Calculus in PDF only on Docsity! Runge-Kutta Method Adrian Down April 25, 2006 1 Second-order Runge-Kutta method 1.1 Review Last time, we began to develop methods to obtain approximate solutions to the differential equation ẋ = f(t, x(t)) with error of O(h2), where h is the spacing of the mesh used to calculate the approximation. To construct these methods, our goal was to minimize the local truncation error. We saw that, in general, the global truncation error should be one order less in h. Last time, we introduced an approximation that generalized the Modified Euler method. The general form of the expression was, y(t + h)− y(t) = ω1hF (t) + ω2hf (t + αh, y(t) + βhF (t)) where F (t) ≡ f(t, y(t)). Our proposal was to choose ω1, ω2, α and β such that the local truncation error of the approximation is O(h3), from which we expect the desired global truncation error to be O(h2). 1.2 Computation We began to evaluate this condition last time by Taylor expanding the func- tion f up to second order in h. Since f is a function of two variables, we used the Taylor formula for multiple dimensions. We obtained, x(t + h)− x(t) = h(ω1 + ω2)F (t) + h2ω2 (αf1?βFfx) + O(h3) where subscripts indicate partial differentiation. The partial derivatives are to be evaluated at (t, x(t)). 1 The Taylor series of the left side is easily computed, x(t + h)− x(t) = hẋ(t) + h 2 2 ẍ(t) + O(h3) Since these two expression are equal, match terms in h and cancel coef- ficients, such that only terms of O(h3) remain. Matching terms in h yields two equations, (ω1 + ω2)F (t) = ẋ(t) ω2 (αft + βFfx) = 1 2 ẍ(t) The first equation can be simplified using the definition of f , F (t) = f(t, x(t)) = ẋ ⇒ (ω1 + ω2) = 1 The second equation can be solved by taking the time derivative of the differential equation. This yields partial derivatives, which can be compared with those on the left of the equation, ẋ = f(t, x(t)) ⇒ ẍ = ft + fxẋ = ft + Ffx ⇒ ω2αft + ω2βFfx = 1 2 ft + 1 2 Ffx This equation must hold for all t and x, so it must be that the coefficients of the partial derivatives are separately equal, ω2α = ω2β = 1 2 1.3 Solutions We now have three equations and four unknowns. This system is under- determined, meaning that we should expect a family of solutions in one parameter. One possible choice of parameters is, ω1 = 1 2 ω2 = 1 2 α = 1 β = 1 2
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