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Several slope samples inside one step, weighted so the error terms cancel, why the classical four-stage method is Simpson's rule in disguise, and what an order costs.
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By the end of this lesson you will be able to carry out a classical Runge-Kutta step stage by stage, say where its weights come from, state what fourth order does and does not promise, and compare solvers at equal evaluation cost.
You have seen a second slope sample raise Euler's method by an order, and you know that Simpson's rule integrates a cubic exactly using three samples. Runge-Kutta methods are those two ideas together: more samples inside one step, weighted so that more error terms cancel.
A stage is one slope evaluation inside a step; a method is $s$-stage when it uses $s$ of them. The weights say how the stages are combined and the nodes say where in the step each is sampled. The classical Runge-Kutta method is the four-stage, fourth-order one. A Butcher barrier is a theorem saying how many stages an order needs.
Take several slope samples inside a single step and weight them so that the leading error terms cancel. Euler is the one-stage member and Heun a two-stage one; the classical method takes four: $k_1 = f(t_n, y_n)$; $k_2 = f(t_n + \tfrac{h}{2}, y_n + \tfrac{h}{2}k_1)$; $k_3 = f(t_n + \tfrac{h}{2}, y_n + \tfrac{h}{2}k_2)$; $k_4 = f(t_n + h, y_n + hk_3)$; and $y_{n+1} = y_n + \tfrac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4)$. It is fourth order: halving the step divides the global error by sixteen. Where those weights come from is worth seeing once — for an equation whose slope depends on $t$ alone, $k_2$ and $k_3$ coincide and the combination becomes $\tfrac{h}{6}(f(t_n) + 4f(t_n + \tfrac h2) + f(t_n + h))$, which is Simpson's rule. Two cautions. Up to order four an extra order costs one extra stage, and after that the deal collapses: order five needs six stages and order eight needs eleven, which is why the four-stage method is the one in every library. And order is not stability — the classical method's stability region is bounded and barely larger than Heun's, so a step chosen for accuracy alone can still produce a solution that grows without limit.
Another way: picture
One step, four soundings: one at the near edge, two in the middle taken from slightly different guesses, and one at the far edge reached using the third. The step is taken along a weighted average of the four directions, with the middle pair counting double because the middle of the step is where the average slope most nearly lives.
Another way: steps
'Fourth order' is often heard as 'accurate'. It is a statement about a rate, and it carries two conditions: the step must be small enough for the asymptotic regime, and the solution must actually have the fifth derivative the error term names. On a problem with a kink, a fourth-order method delivers whatever order the kink allows and the extra three stages are simply wasted — which is why a serious solver measures the order it is getting rather than assuming the one it was sold.
For $y' = f(t)$ the stages do not depend on $y$, so $k_2 = k_3$.
Both sample the midpoint.
The combination becomes $\tfrac{h}{6}(f_0 + 4f_{1/2} + f_1)$.
Simpson's rule.
So the fourth order is quadrature's fourth order, arriving by a different road.
One idea, two subjects.
A solution with a corner has no fifth derivative there.
The error term does not exist.
Measured order across the corner falls to one, whatever the method claims.
Four stages, first-order results.
The cure is to stop the step at the corner and restart, not to raise the order further.
Match the method to the solution.
$k_1 = 0$, and the midpoint is at $t = 1$, so $k_2 = k_3 = 1$.
$k_4 = 2$, at the far end.
The combination is $\tfrac{0 + 2 + 2 + 2}{6} = 1$, so the step adds $2 \times 1 = 2$ — and the exact answer is $\tfrac{2^{2}}{2} = 2$.
Take one classical Runge-Kutta step on $y' = t$ from $t = 6$, $y = 3$, with step $1$. Fill in the four stage slopes and the value the step produces.
| Value | |
|---|---|
| First stage | |
| Second stage | |
| Third stage | |
| Fourth stage | |
| Value after the step |
A solver advertises the classical Runge-Kutta method: four stages a step, fourth order. A run of $15$ steps is planned. What does the claim actually establish?
Consider the classical Runge-Kutta method, which is order $4$ and uses $4$ slope evaluation(s) a step, and three others. Match each method to its order and cost.
| Order 1, for one slope evaluation a step | Order 2, for two slope evaluations a step | Order 4, for four slope evaluations a step | Order 8, but eleven evaluations: order and stage count have parted company | |
|---|---|---|---|---|
| Euler's method | ||||
| Heun's method | ||||
| The classical Runge-Kutta method | ||||
| An eighth-order Runge-Kutta method |
Put one step of the classical Runge-Kutta method, inside a loop of $44$ steps, into order.
Number the steps in order (write the number in the box):
The classical Runge-Kutta method with step $0.25$ costs the same number of slope evaluations as Euler's method with step $\tfrac{0.25}{4}$. Runge-Kutta's error is proportional to the fourth power of its step and Euler's to the first power of its step, with the same constant. What is the ratio of the Runge-Kutta error to Euler's? Give a fraction.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Take one classical Runge-Kutta step on $y' = t$ from $t = 7$, $y = 0$, with step $2$. What is the value after the step?
Answer:
You can run a four-stage step and state the order and cost of each solver you have met. Say in your own words why order and stage count stop matching above order four.
8. Your turn: the four stages for $y' = t$ from $t = 0$ with $h = 2$, step 3