5.4 Runge-Kutta 4 Method

Martin Wilhelm Kutta (1867-1944) Carl Runge (1856-1927) Martin Wilhelm Kutta (1867-1944)

Motivation With Euler’s method, error is described by a straight line: i.e., it is proportionate to (linear in) Dt. We say that error is O(Dt) : “Order Dt”, or “Big-O Dt” Can we do better?

First Estimate ∂1 Recall update rule from Euler’s Method: Pn ← Pn-1 + f(tn-1, Pn-1 ) ∆t where f is the derivative function We call f(tn-1, Pn-1 ) = 80 the first estimate, or ∂1 ∂1 = 80 ∆t = 8

Second Estimate ∂2 Second estimate ∂2 uses the halfway point along the line segment to ∂1

Second Estimate ∂2 Combined with slope 0.10 (from dP/dt = 0.10P), this gives us a new endpoint = (0.1)(140)(8) = 112, which is the second estimate ∂2. ∂2 = f(tn-1+0.5t, Pn-1+0.5∂1) t ∂2= 112 ∆t = 8

Third Estimate ∂3 Third estimate ∂3 uses the halfway point along the line segment from ∂2 ∂3 =(0.1)(156)(8) = 124.8

Third Estimate ∂3 (0.1)(156)(8) = 124.8, which is the third estimate ∂3. ∂3 = f(tn-1+0.5t, Pn-1+0.5∂2) t ∂3= 124.8

Fourth Estimate ∂4 ∂4 =(0.1)(224.8)(8) = 179.84 Fourth estimate ∂4 is taken at end of interval ∂4 =(0.1)(224.8)(8) = 179.84

Fourth Estimate ∂4 ∂4 =(0.1)(224.8)(8) = 179.84, which is the fourth estimate ∂4. ∂4 = f(tn-1+ t, Pn-1+∂3) t ∂4= 179.8

Runge-Kutta 4 Estimate Bring it all together: a weighted average that privileges the middle values: Pn = Pn-1 + (∂1 + 2∂2 + 2∂3 + ∂4) / 6 = 100 + (80 + 2*112 + 2*124.8 + 179.84) / 6 = 222.24 Relative error = |222.4 - 222.55| / |222.55| = 0.14% (Compare 19% for Euler’s Method) RK4 error is O(Dt4) : a very small number, because error is < 1.

Runge-Kutta 4 Algorithm P(t0)← P0 Initialize NumberOfSteps for n going from 1 to NumberOfSteps do the following: tn ← t0 + n∆t ∂1 = f(tn-1, Pn-1 ) ∂2 = f(tn-1+0.5t, Pn-1+0.5∂1) t ∂3 = f(tn-1+0.5t, Pn-1+0.5∂2) t ∂4 = f(tn-1+ t, Pn-1+∂3) t

Runge-Kutta 4 in Excel

Runge-Kutta 4 in Excel

Runge-Kutta 4 in Excel

Runge-Kutta 4 in Excel

Runge-Kutta 4 in Excel

Runge-Kutta 4 in Excel