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An overview of runge-kutta methods for solving differential equations. It covers the euler method, trapezoidal method, and the improved euler method, as well as the general idea of constructing accurate explicit methods. The document also includes the derivation of the local error for a 2nd order runge-kutta method and examples of various runge-kutta methods.
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SUMMARY (so far) To solve dy dx
= f (x, y) starting from (x 0 , y 0 ).
Euler:
yn+1 = yn + hf (xn, yn) Explicit but not very accurate.
Trapezoidal:
yn+1 = yn + h 2
[ f (xn, yn) + f (xn+1, yn+1)
]
More accurate but implicit.
Yes, Runge-Kutta methods.
Consider Euler-trapezoidal predictor-corrector method with just one correction step:
yn+1 = yn + h^1 2
[ f (xn, yn) + f
( xn+1, yn + hf (xn, yn)
)]
x (^) n x +hn
n n
(x +h,y +hf(x ,y ))n n n n
slope = f(x ,y )n n
slope = f(x +h,y +hf(x ,y ))n n n n
slope = 0.5*(f(x ,y ) + f(x +h,y +hf(x ,y )))
(x ,y )
n n n n n n
2nd order Runge-Kutta method
Step 1: Take an Euler step from xn to a second point in the interval xn+αh (α ≤ 1).
Euler gives approximate value of y at this point as
yn + αhf (xn, yn)
Step 2: Take a weighted average of the slopes at the initial point and the point given by the Euler method estimate:
yn+1 = yn + h [w 1 f (xn, yn) + w 2 f (xn + αh, yn + αhf (xn, yn))]
Choose α, w 1 and w 2 to make the local error as small as possible.
[Improved Euler method corresponds to α = 1, w 1 = w 2 = 12 .]
To calculate the local error we must substitute in the analytic expression and Taylor expand:
Error = y(xn+1)−y(xn)−w 1 hf (xn, y(xn))−w 2 hf (xn + αh, y(xn) + αhf (xn, y(xn)))
First note, (Taylor series)
y(xn+1) = y(xn) + hy′(xn) + h
2 2
y′′(xn) + 0(h^3 ) We know y′(x) = f (x, y(x)) (†) so y′(xn) = f (xn, y(xn)) ≡ fn for short. To find y′′(x) differentiate (†):
y′′(x) = d dx
f (x, y(x)) = ∂f ∂x
dy dx
= ∂f ∂x
f
So, in the obvious notation,
y(xn+1) = y(xn) + hfn + h
2 2
[ ∂fn ∂x +^
∂fn ∂y fn
]
Examples Write k = hf (xn, yn)
i) α = 1 ⇒ w 2 = 12 ⇒ w 1 = (^12)
yn+1 = yn + h 2
[f (xn, yn) + f (xn + h, yn + k)] This is our previous improved Euler method.
ii) α = 12 ⇒ w 2 = 1 ⇒ w 1 = 0
yn+1 = yn + hf (xn + 12 h, yn + 12 k) This is sometimes called the modified Euler method.
iii) α = 23 ⇒ w 2 = 34 ⇒ w 1 = (^14)
yn+1 = yn + 4 h
[ f (xn, yn) + 3f (xn + 23 h, yn + 23 k)
]
These methods will, in general, give different answers but they will have comparable errors.
Geometrical interpretation of the modified Euler method
x x +h
slope = f(x ,y )
slope = f(x +0.5h,y +0.5hf(x ,y ))) (x ,y )n n
n n n n
n n
n (^) n
⇒ y 1 = y 0 + hf (0. 1 , y 0 + 12 hf (0, y 0 )) = 1 + 0. 2 × (− 0 .9) = 0. 82 ⇒ y 2 = y 1 + h · f (0. 3 , y 1 + 12 hf (0. 2 , y 1 )) = 0 .82 + 0. 2 × (− 0 .738) = 0. 6724
⇒ Error = 0. 6703 − 0 .6724 = − 0. 0021
Conclusion: Euler → 2% accuracy Second order RK → 0 .3% accuracy
with the same computational effort.
Fourth order Runge-Kutta
Often called the Runge-Kutta method. In calculating yn+1 starting from (xn, yn) the function f is evaluated four times.
i.e. four “slopes” are calculated and combined to give a “mean slope”,
yn+1 = yn + h × “mean slope”, in such a way that the error is O(h^5 ).
There are many ways of doing this (just like the second order case) but one has been adopted as standard.
Geometrical interpretation of the standard 4th order Runge-Kutta method
A
B
C
D
slope=f(A)
slope=f(B)^ slope=f(C)
slope=(f(A)+2f(B)+2f(C)+f(D))/
