Tc9, Tc10 - Runge-Kutta methods | OCR A-Level Further Mathematics B (MEI)

Use four-stage Runge–Kutta calculations and implement them in a spreadsheet. You will explain the stage points and weights, adapt the differential equation and step length, and read a supplied variation carefully rather than force every method into the standard pattern.

Four trial gradients make one corrected step

Could four gradient readings land exactly on a curved solution after one large step?

Sometimes they can. For $y'=3x^2$ , $y(0)=1$ , the standard four-stage method with $h=1$ gives exactly $y(1)=2$ , matching $y=x^3+1$ . That special success is useful to understand, but it is not a promise of exactness for every differential equation.

You already know

Modified Euler samples a starting gradient and a predicted endpoint gradient, then combines their increments. Each increment already contains the step length. A predicted stage point is an aid to the calculation, not a new confirmed point on the solution curve.

The standard fourth-order Runge–Kutta method uses the start, two different midpoint predictions and one endpoint prediction. Each new stage uses information from the preceding stage.

Standard fourth-order Runge–Kutta method

For $y'=f(x,y)$ , starting from $(x_n,y_n)$ ,

\begin{aligned} k_1&=h f(x_n,y_n)\\ k_2&=h f\left(x_n+\frac h2,y_n+\frac{k_1}{2}\right)\\ k_3&=h f\left(x_n+\frac h2,y_n+\frac{k_2}{2}\right)\\ k_4&=h f(x_n+h,y_n+k_3)\\ x_{n+1}&=x_n+h\\ y_{n+1}&=y_n+\frac{k_1+2k_2+2k_3+k_4}{6} \end{aligned}

All four evaluations must be defined. The $k_i$ are increments; there is no extra multiplication by $h$ in the final weighted average.

In the formula booklet.

The two midpoint stages have the same $x$ -coordinate but generally different predicted $y$ -coordinates. Their doubled weights do not mean that the method takes two actual half steps.

Worked example 1: one four-stage step

Question. For $y'=x+y$ , $y(0)=1$ , calculate one standard Runge–Kutta step with $h=0.2$ .

Plan. Calculate the increments in order, then combine them with weights $1,2,2,1$ .

\begin{aligned} k_1&=0.2(0+1)=0.2 && \text{starting point}\\ k_2&=0.2(0.1+1.1)=0.24 && \text{first midpoint prediction}\\ k_3&=0.2(0.1+1.12)=0.244 && \text{revised midpoint prediction}\\ k_4&=0.2(0.2+1.244)=0.2888 && \text{endpoint prediction}\\ y_1&=1+\frac{0.2+2(0.24)+2(0.244)+0.2888}{6} =1.2428 && \text{weighted correction} \end{aligned}

Why this step? The second midpoint height uses $k_2/2=0.12$ , rather than reusing $k_1/2=0.1$ .

Answer. $y(0.2)\approx1.2428$ from this single step.

Check. The analytical solution $y=2e^x-x-1$ gives $y(0.2)=1.24280552\ldots$ , close to the numerical estimate but not identical.

Change only $h$ to $0.1$ , leaving the equation and starting point unchanged.

Worked example 2: halve the one-step interval

Question. Calculate one standard Runge–Kutta step with $h=0.1$ .

Plan. Recalculate every stage with the new step, rather than halve the earlier increments.

\begin{aligned} k_1&=0.1(0+1)=0.1 && \text{new starting increment}\\ k_2&=0.1(0.05+1.05)=0.11 && \text{new midpoint}\\ k_3&=0.1(0.05+1.055)=0.1105 && \text{revised midpoint}\\ k_4&=0.1(0.1+1.1105)=0.12105 && \text{new endpoint prediction}\\ y_1&=1+\frac{0.1+2(0.11)+2(0.1105)+0.12105}{6} =1.110341666\ldots && \text{weighted correction} \end{aligned}

Why this step? A changed step changes both coordinates of the stage points, so the increments are not simply scaled copies.

Answer. $y(0.1)\approx1.11034167$ . One step now reaches $0.1$ , not $0.2$ .

Check. The exact value is $1.11034184\ldots$ . To compare methods at $x=0.2$ , take a second $0.1$ step instead of comparing these different endpoints directly.

For the opening equation $y'=3x^2$ , the four increments are $0,0.75,0.75,3$ . Their weighted combination is

\frac{0+2(0.75)+2(0.75)+3}{6}=1

so $y_1=2$ , exactly as claimed. The right side depends only on $x$ , so the trial heights do not affect these particular readings. The exact result is a feature of this equation, not a general error rule.

[DIAGRAM: asset_name: Four Runge-Kutta stage points; asset_slug: rk4_stage_points; caption: Two midpoint predictions share x but need not share y; description: A schematic mathematical stage plot for y'=x+y, h=0.2, starting at (0,1), labels stage 1 at (0,1), stage 2 at (0.1,1.1), stage 3 at (0.1,1.12), stage 4 at (0.2,1.244). The final corrected point (0.2,1.2428) is shown separately using a magnified inset or a side table so it is not confused with the predicted endpoint. A note gives weights 1,2,2,1.]
Two midpoint predictions share x but need not share y

Watch out

Wrong: set $k_3=k_2$ because both use the midpoint, or use $y_n+k_4$ as the next height. Right: use the distinct predicted heights and combine all four increments with the stated weights.

Apply the standard rule to one new step.

Make each stage a visible spreadsheet column

A spreadsheet makes a long chain of four-stage steps manageable. It must still show enough formulae to reveal the design: the starting conditions, parameter references, each stage and the next-row update.

Consider the family

y'=x-y+ay^2,\qquad y(0)=1

Store $h$ in $\text{J1}$ and $a$ in $\text{J2}$ . Set $\text{A2}=0$ , $\text{B2}=1$ , and use columns $\text{C:F}$ for the increments. Three auxiliary cells in $\text{G:I}$ hold the trial heights, keeping the formulae short. These literal entries use repeated multiplication for the square.

Cell	Formula
$\text{C2}$	$=\text{\$ J$1}(\text{A2}-\text{B2}+\text{$J$2}\text{B2}*\text{B2})$
$\text{G2}$	$=\text{B2}+\text{C2}/2$
$\text{D2}$	$=\text{\$ J$1}((\text{A2}+\text{$J$1}/2)-\text{G2}+\text{$J$2}\text{G2}*\text{G2})$
$\text{H2}$	$=\text{B2}+\text{D2}/2$
$\text{E2}$	$=\text{\$ J$1}((\text{A2}+\text{$J$1}/2)-\text{H2}+\text{$J$2}\text{H2}*\text{H2})$
$\text{I2}$	$=\text{B2}+\text{E2}$
$\text{F2}$	$=\text{\$ J$1}((\text{A2}+\text{$J$1})-\text{I2}+\text{$J$2}\text{I2}*\text{I2})$
$\text{A3}$	$=\text{A2}+\text{\$ J$1}$
$\text{B3}$	$=\text{B2}+(\text{C2}+2\text{D2}+2\text{E2}+\text{F2})/6$

Copy $\text{C2:I2}$ down, and copy $\text{A3:B3}$ down. Keep full stored precision. The stage formulae are applications of the supplied standard rule, not extra formulae to memorise; this worksheet layout is not printed in the booklet.

Worked example 3: check the spreadsheet against an analytical case

Question. Set $a=0$ , $h=0.1$ . Use the spreadsheet to estimate $y(0.4)$ .

Plan. Recalculate the four stages at each row, take four complete steps and compare with the analytical solution of $y'=x-y$ .

\begin{aligned} k_1&=0.1(0-1)=-0.1 && \text{first row}\\ k_2&=0.1(0.05-0.95)=-0.09 && \text{first midpoint}\\ k_3&=0.1(0.05-0.955)=-0.0905 && \text{revised midpoint}\\ k_4&=0.1(0.1-0.9095)=-0.08095 && \text{endpoint prediction}\\ y_1&=1+\frac{-0.1+2(-0.09)+2(-0.0905)-0.08095}{6} =0.909675 && \text{corrected value} \end{aligned}

$x_n$	Displayed $y_n$ , $8$ d.p.
$0.1$	$0.90967500$
$0.2$	$0.83746180$
$0.3$	$0.78163684$
$0.4$	$0.74064058$

Why this step? A known analytical case checks the worksheet before an equation without a convenient exact answer is investigated.

Answer. The estimate is $0.74064058$ , or $0.7406$ to $4$ decimal places.

Check. $y=2e^{-x}+x-1$ gives $y(0.4)=0.74064009\ldots$ , confirming this rounded result.

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Tc9, Tc10 - Runge-Kutta methods

Four trial gradients make one corrected step

Make each stage a visible spreadsheet column

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