Modified Euler (Heun's method): Exercises — Maths deep dive

Ten graded problems, from "just spot the pieces" up to "invent a variation." Every one has a full worked solution hidden inside a collapsible box — try first, then reveal.

This page assumes the machinery from the parent note: Heun's method. Recall the two moves we lean on constantly:

Throughout, $k_{1}$ = slope at the start, $k_{2}$ = slope at the predicted end. Every problem is just: find $k_{1}$ , take a peek-step, find $k_{2}$ , average. Keep that picture in front of you.

The figure below is the mental image to hold for every problem on this page — it draws all four objects we juggle (start slope, predicted endpoint, end slope, averaged Heun step) for the running example $y^{'} = x + y, y (0) = 1$ (drawn with a big $h = 1$ so the geometry is visible):

Read it like this: the coral dashed line is the start slope $k_{1}$ — follow it alone and you land at the coral predicted point $y^{(p)}$ (that is exactly the plain-Euler guess in every predictor step below, e.g. L1.1's $3.5$ or L2.1's $0.8$ ). The butter dotted line is the end slope $k_{2}$ read at that predicted point. The lavender line is the Heun step: its slope is the average of coral and butter, and it lands at the lavender $y_{1}$ , which sits far closer to the mint true point than the coral prediction did. Whenever a solution below says "average the two slopes," picture the lavender line splitting the angle between the coral and butter lines.

Level 1 — Recognition

(Can you point at the pieces and plug numbers into the machine?)

L1.1

Given $y^{'} = f (x, y) = 2 x + 1$ with $y (0) = 3$ and $h = 0.5$ . This $f$ does not depend on $y$ . Compute $k_{1}$ , the predictor $y_{1}^{(p)}$ , then $k_{2}$ , and finally $y_{1}$ .

Recall Solution

What we do first: read off the start point. $x_{0} = 0$ , $y_{0} = 3$ .

$k_{1}$ — slope at the start (the coral dashed line in the figure). Plug $(0, 3)$ into $f$ : $k_{1} = 2 (0) + 1 = 1$ .

Predictor — the peek step (walk along coral to the predicted point). $y_{1}^{(p)} = y_{0} + h k_{1} = 3 + 0.5 (1) = 3.5$ . Why? We need a value of $y$ at the endpoint so we can evaluate the end-slope.

$k_{2}$ — slope at the predicted end (the butter dotted line). $x_{1} = 0.5$ . Since $f$ ignores $y$ here, $k_{2} = f (0.5, 3.5) = 2 (0.5) + 1 = 2$ .

Corrector — average and step (the lavender line). $y_{1} = 3 + \frac{0.5}{2} (1 + 2) = 3 + 0.25 (3) = 3.75 .$ Sanity note: because $f$ has no $y$ , Heun here is exactly the trapezoidal rule on $\int_{0}^{0.5} (2 x + 1) d x = [x^{2} + x]_{0}^{0.5} = 0.75$ , so $y_{1} = 3 + 0.75 = 3.75$ is exact. No error at all.

L1.2

For $y^{'} = - y$ , $y (0) = 2$ , $h = 0.1$ , identify $k_{1}$ and $k_{2}$ (numbers), then give $y (0.1)$ .

Recall Solution

$k_{1}$ : $f (0, 2) = - 2$ . Predictor: $y^{(p)} = 2 + 0.1 (- 2) = 1.8$ . $k_{2}$ : $f (0.1, 1.8) = - 1.8$ . Corrector: $y_{1} = 2 + \frac{0.1}{2} (- 2 + (- 1.8)) = 2 + 0.05 (- 3.8) = 2 - 0.19 = 1.81 .$ Exact: $y = 2 e^{- 0.1} = 1.80967$ . Error $\approx 0.0003$ — tiny.

Level 2 — Application

(Run the full method on a real ODE, one and then two steps.)

L2.1

Solve $y^{'} = x - y$ , $y (0) = 1$ , one Heun step to reach $y (0.2)$ with $h = 0.2$ .

Recall Solution

$f (x, y) = x - y$ , $x_{0} = 0$ , $y_{0} = 1$ , $x_{1} = 0.2$ .

$k_{1}$ : $f (0, 1) = 0 - 1 = - 1$ . Predictor: $y^{(p)} = 1 + 0.2 (- 1) = 0.8$ . $k_{2}$ : $f (0.2, 0.8) = 0.2 - 0.8 = - 0.6$ . Corrector: $y_{1} = 1 + \frac{0.2}{2} (- 1 + (- 0.6)) = 1 + 0.1 (- 1.6) = 0.84 .$ Exact solution $y = x - 1 + 2 e^{- x}$ gives $y (0.2) = 0.2 - 1 + 2 e^{- 0.2} = 0.83746$ . Error $\approx 0.0025$ .

L2.2

Same ODE $y^{'} = x - y$ , $y (0) = 1$ , but take two steps of $h = 0.1$ to reach $y (0.2)$ . Compare with the single-step answer above.

Recall Solution

Step 1 (from $x = 0$ to $0.1$ ): $k_{1} = f (0, 1) = - 1$ ; predictor $= 1 + 0.1 (- 1) = 0.9$ ; $k_{2} = f (0.1, 0.9) = 0.1 - 0.9 = - 0.8$ ; $y (0.1) = 1 + \frac{0.1}{2} (- 1 - 0.8) = 1 + 0.05 (- 1.8) = 0.91.$

Step 2 (from $x = 0.1$ , $y = 0.91$ to $x = 0.2$ ): $k_{1} = f (0.1, 0.91) = 0.1 - 0.91 = - 0.81$ ; predictor $= 0.91 + 0.1 (- 0.81) = 0.829$ ; $k_{2} = f (0.2, 0.829) = 0.2 - 0.829 = - 0.629$ ; $y (0.2) = 0.91 + \frac{0.1}{2} (- 0.81 - 0.629) = 0.91 + 0.05 (- 1.439) = 0.83805.$

Compare: two-step $\approx 0.83805$ vs single-step $0.84$ ; exact $0.83746$ . Errors: single-step $0.00254$ , two-step $0.00059$ . Halving $h$ cut the error by about $4 \times$ — exactly the $O (h^{2})$ promise.

Level 3 — Analysis

(Reason about the method itself, not just crank numbers.)

L3.1

For the linear ODE $y^{'} = λ y$ , show that one Heun step is $y_{1} = (1 + hλ + \frac{1}{2} (hλ)^{2}) y_{0},$ i.e. Heun reproduces the exponential $e^{hλ}$ up to and including the $(hλ)^{2}$ term.

Recall Solution

$f (x, y) = λ y$ , so the value of $x$ is irrelevant.

$k_{1}$ : $λ y_{0}$ . Predictor: $y^{(p)} = y_{0} + hλ y_{0} = (1 + hλ) y_{0}$ . $k_{2}$ : $λ y^{(p)} = λ (1 + hλ) y_{0}$ . Corrector:

= y_0 + \frac{h\lambda}{2}\big[1 + 1 + h\lambda\big]y_0.$$ $$y_1 = y_0\Big[1 + \frac{h\lambda}{2}(2 + h\lambda)\Big] = y_0\Big[1 + h\lambda + \tfrac{1}{2}(h\lambda)^2\Big].$$ **What this means:** the true step is $y_1 = e^{h\lambda}y_0 = y_0(1 + h\lambda + \tfrac{1}{2}(h\lambda)^2 + \tfrac{1}{6}(h\lambda)^3 + \dots)$. Heun matches the first **three** terms and first disagrees at $(h\lambda)^3$ — exactly the fingerprint of a second-order method. This is the linear-case proof behind [[Order of Accuracy and Step Size]].

L3.2

Take $y^{'} = λ y$ with $λ = - 20$ (a fast-decaying, "stiff" problem), $y_{0} = 1$ . Using the amplification factor $G (hλ) = 1 + hλ + \frac{1}{2} (hλ)^{2}$ from L3.1, decide whether Heun stays bounded (does not blow up) for $h = 0.05$ and for $h = 0.15$ .

Recall Solution

Each step multiplies $y$ by $G = 1 + z + \frac{1}{2} z^{2}$ where $z = hλ$ . The numerical solution stays bounded iff $∣ G ∣ \leq 1$ .

$h = 0.05$ : $z = 0.05 (- 20) = - 1$ . $G = 1 + (- 1) + \frac{1}{2} (1) = 0.5$ . Since $∣0.5∣ \leq 1$ , stable — values shrink, matching the true decaying solution.

$h = 0.15$ : $z = 0.15 (- 20) = - 3$ . $G = 1 + (- 3) + \frac{1}{2} (9) = 1 - 3 + 4.5 = 2.5$ . Since $∣2.5∣ > 1$ , unstable — each step grows the value by $2.5 \times$ , even though the true solution is racing to zero.

The full picture (complex $z$ ). In general $λ$ can be complex (oscillatory or spiral solutions), so $z = hλ$ is a point in the complex plane. The region of absolute stability of RK2 is the set of complex $z$ with $1 + z + \frac{1}{2} z^{2} \leq 1$ — a bounded egg-shaped region hugging the origin in the left half-plane. Our real cases are just the slice along the negative real axis: it turns out $∣ G ∣ \leq 1$ there exactly for $- 2 \leq z \leq 0$ , so $z = - 1$ sits inside (stable) and $z = - 3$ falls outside (unstable). For a stiff problem you must keep $z = hλ$ inside that region.

What this teaches: accuracy order is not the whole story. For stiff problems the step $h$ must be small enough to keep $z$ in the stability region, or the answer oscillates and explodes.

Level 4 — Synthesis

(Combine ideas — iterated correctors, weighted variants, error estimation.)

L4.1 (Iterated corrector)

For $y^{'} = x + y$ , $y (0) = 1$ , $h = 0.1$ : do the predictor + corrector, then iterate the corrector twice more (re-feeding the latest $y_{1}$ into $k_{2}$ each time). Report the sequence and where it settles.

Recall Solution

$k_{1} = f (0, 1) = 1$ (fixed — it only uses the start point).

The implicit equation we are solving. The true trapezoidal corrector is implicit because the unknown $y_{1}$ appears on both sides:

= 1 + 0.05\big[1 + (0.1 + y_1)\big] = 1.055 + 0.05\,y_1.$$ We solve it by **fixed-point iteration**: write $y_1^{(n+1)} = 1.055 + 0.05\,y_1^{(n)}$ and feed each output back in. **Iteration 0 (predictor):** $y^{(0)} = 1 + 0.1(1) = 1.1$. **Iteration 1:** $k_2 = f(0.1, 1.1) = 1.2$; $\;y^{(1)} = 1 + 0.05(1 + 1.2) = 1.11$. **Iteration 2:** $k_2 = f(0.1, 1.11) = 1.21$; $\;y^{(2)} = 1 + 0.05(1 + 1.21) = 1.1105$. **Iteration 3:** $k_2 = f(0.1, 1.1105) = 1.2105$; $\;y^{(3)} = 1 + 0.05(1 + 1.2105) = 1.110525$. **Where it settles.** Because the map has slope $0.05$ (well below $1$), the iteration is a contraction and converges to the exact fixed point $y_1^{*} = \dfrac{1.055}{1-0.05} = \dfrac{1.055}{0.95} = 1.1105263\ldots$ — the true solution of the implicit trapezoidal equation. The iterates $1.1,\,1.11,\,1.1105,\,1.110525,\dots$ march straight to it. **What we learn:** iterating solves the implicit corrector more exactly, but it converges to the trapezoid's own answer, which still carries the method's $O(h^2)$ truncation error against the ODE's exact $y = 2e^{0.1}-1.1 = 1.110342$ — iterating cannot beat the underlying rule. This is the heart of [[Predictor-Corrector Methods]].

L4.2 (Error estimate from predictor vs corrector)

A cheap error indicator per step is the gap $∣ y_{1}^{(corr)} - y_{1}^{(p)} ∣$ . Compute it for $y^{'} = x + y$ , $y (0) = 1$ , $h = 0.1$ , and comment.

Recall Solution

From L4.1: predictor $y^{(p)} = 1.1$ , first corrector $y_{1} = 1.11$ . $∣ y_{1} - y^{(p)} ∣ = ∣1.11 - 1.1∣ = 0.01.$ The actual global error at this point (against exact $1.110342$ ) is $\approx 0.00034$ . The predictor–corrector gap ( $0.01$ ) is larger than the true error but tracks the same order in $h$ — it is proportional to $h^{2} f_{y}$ and is used in adaptive step-size control to decide whether to shrink or grow $h$ . It's a conservative alarm, not the exact error.

Level 5 — Mastery

(Prove/derive a general result, or build a new variant.)

L5.1 (Order-2 condition for a general two-slope method)

Consider a general two-stage scheme

y_1 = y_0 + h(a\,k_1 + b\,k_2).$$ Show that Heun ($p=1,\,a=b=\tfrac12$) satisfies the two conditions needed for **second order**, and find *another* choice of $(p,a,b)$ that is also second order. > [!recall]- Solution > **Notation first.** Let $f_x := \dfrac{\partial f}{\partial x}$ and $f_y := \dfrac{\partial f}{\partial y}$, > both **evaluated at the start point** $(x_0,y_0)$, and write $f := f(x_0,y_0)$. These measure how > fast $f$ changes as you nudge $x$ (holding $y$) or nudge $y$ (holding $x$). > > **Why a two-variable Taylor expansion?** The true step's Taylor series involves $y'' $, and by > the chain rule $y'' = \dfrac{d}{dx}f(x,y(x)) = f_x + f_y\,y' = f_x + f_y f$. So to compare the > scheme with the truth at order $h^2$ we must expand $f$ at a *shifted* point $(x_0+ph,\,y_0+ph\,k_1)$. > The multivariable first-order Taylor expansion is > $$f(x_0+\Delta x,\ y_0+\Delta y) = f + f_x\,\Delta x + f_y\,\Delta y + O(\Delta^2),$$ > i.e. "value plus (rate in $x$)$\times$(step in $x$) plus (rate in $y$)$\times$(step in $y$)." > > **Apply it to $k_2$.** Here $\Delta x = ph$ and $\Delta y = ph\,k_1 = ph\,f$, so > $$k_2 = f + f_x(ph) + f_y(ph\,f) + O(h^2) = f + ph\,f_x + ph\,f\,f_y + O(h^2).$$ > > **Build the scheme's step and match Taylor.** True step: > $$y(x_1) = y_0 + h f + \tfrac{h^2}{2}(f_x + f_y f) + O(h^3).$$ > Scheme: > $$y_1 = y_0 + h\big(a f + b[f + ph f_x + ph f f_y]\big) + O(h^3) > = y_0 + (a+b)hf + bph^2(f_x + f f_y) + O(h^3).$$ > > **Match term by term:** > - $h^1$ term: $(a+b) = 1.$ > - $h^2$ term: $bp = \tfrac12.$ > > These are the **two order-2 conditions**. Heun uses $p=1,\ b=\tfrac12$: check $a+b = \tfrac12+\tfrac12 = 1$ ✓ and $bp = \tfrac12\cdot1 = \tfrac12$ ✓. > > **Another valid choice — the midpoint method.** Take $p=\tfrac12$. Then $bp=\tfrac12$ forces > $b = 1$, and $a+b=1$ forces $a = 0$: > $$k_1 = f(x_0,y_0),\quad k_2 = f\big(x_0+\tfrac{h}{2},\,y_0+\tfrac{h}{2}k_1\big),\quad y_1 = y_0 + h\,k_2.$$ > This is the **midpoint (modified Euler-midpoint) method** — also second order, using only the > single mid-step slope. Both live inside the RK2 family (see [[Runge-Kutta Methods]]). ### L5.2 (Apply the midpoint variant) Use the midpoint method from L5.1 on $y' = x+y$, $y(0)=1$, $h=0.1$, and compare its error to Heun's ($1.11$). > [!recall]- Solution > $k_1 = f(0,1) = 1$. > Mid-point value: $x_0 + \tfrac{h}{2} = 0.05$, $\;y_0 + \tfrac{h}{2}k_1 = 1 + 0.05 = 1.05$. > $k_2 = f(0.05, 1.05) = 0.05 + 1.05 = 1.1$. > $$y_1 = 1 + 0.1(1.1) = \mathbf{1.11}.$$ > Exact $= 1.110342$; error $\approx 0.00034$ — essentially the same accuracy as Heun's, as both > are order 2. They differ only in the third-order term, so for this smooth problem they land at > the same rounded value. > [!mistake] L5 trap: thinking there is only ONE second-order method > **Why it feels right:** Heun is the one everybody names, so it feels canonical/unique. > **Why it's wrong:** the order-2 conditions $a+b=1$, $bp=\tfrac12$ have a **one-parameter > family** of solutions (pick any $p\neq0$). Heun ($p=1$), midpoint ($p=\tfrac12$), and Ralston's > method ($p=\tfrac23$) are all second order. > **Fix:** "second order" is a property (matching Taylor to $h^2$), satisfied by infinitely many > two-stage schemes — Heun is just the equal-weight member. --- ## Connections - [[Euler's Method]] — every predictor here is one Euler step. - [[Trapezoidal Rule]] — the corrector; L1.1 is exactly this rule. - [[Runge-Kutta Methods]] — L5.1's family is RK2; midpoint and Ralston live there too. - [[Order of Accuracy and Step Size]] — L2.2 and L3.1 demonstrate the $O(h^2)$ scaling. - [[Predictor-Corrector Methods]] — L4.1's iteration is the prototype. - [[Taylor Series Methods]] — L5.1 matches the Taylor expansion term by term.