# Runge Kutta Method, Lecture Notes - Mathematics, Study notes for Calculus I. University of California (CA) - UCLA

## Calculus I

Description: Ranga kutta method, second order, computation solution, fourth order, formulation, multi steps, taylor expand the local truncation error
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Runge-Kutta Method
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 ˙x=f(t, x(t)) with error of O(h2), where his 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 βsuch
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 fup to second order in h. Since fis 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?βF fx) + 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) = h˙x(t) + h2
2¨x(t) + O(h3)
Since these two expression are equal, match terms in hand cancel coef-
ﬁcients, such that only terms of O(h3) remain. Matching terms in hyields
two equations,
(ω1+ω2)F(t) = ˙x(t)ω2(αft+βF fx) = 1
2¨x(t)
The ﬁrst equation can be simpliﬁed using the deﬁnition of f,
F(t) = f(t, x(t)) = ˙x
(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,
˙x=f(t, x(t)) ¨x=ft+fx˙x
=ft+F fx
ω2αft+ω2βF fx=1
2ft+1
2F fx
This equation must hold for all tand x, so it must be that the coeﬃcients of
the partial derivatives are separately equal,
ω2α=ω2β=1
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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
2ω2=1
2α= 1 β= 1
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