Exercises — ODE solvers — Euler's method (derivation, global error)

• $f(x,y)=3x-y$ — this is the right-hand side of the ODE, the machine that returns a slope for any point.
• $(x_0,y_0)=(1,2)$ — the known starting value.
• Slope at start $=f(1,2)=3(1)-2=1$.

What we did: we did not solve anything — we only read off the parts. Recognising $f$, the seed point, and evaluating the slope once is the whole skill at L1.

Slope at start: $f(0,1)=0+1=1$. $y_1=y_0+hf(x_0,y_0)=1+0.1(1)=1.1,x_1=0+0.1=\mathrm{0.1.}$ Why: "New = Old + step × slope-at-old." One application of the boxed formula, nothing more.

So $y_3=1.362$ at $x_3=0.3$. What we did: each row reuses the previous $y$ — this is the "self-bootstrapping" the parent note described. What it looks like: three short straight segments, each starting where the last ended (see figure below).

Here $f(x,y)=-y$, so each step is $y_{n+1}=y_n+0.25(-y_n)=0.75y_n$. $y_1=0.75,y_2=0.5625,y_3=0.421875,y_4=\mathrm{0.31640625.}$ True value: $e^{-1}=0.367879\dots$ Error $=0.367879-0.316406=0.051473$. Why the multiplication trick? When $f=-y$, the update becomes a constant factor $0.75$ per step — Euler turns the continuous decay into geometric decay $0.75^n$. It shrinks, but too fast, because each straight step undershoots the curving-up-toward-zero real solution.

True value: $y(0.2)=2e^{0.2}-0.2-1=2(1.221402\dots )-1.2=1.242806$. Global error at $x=0.2$ with $h=0.1$: $|1.242806-1.22|=\mathrm{0.022806.}$ Prediction for $h=0.05$: Euler is first order, GE $=O(h)$. Halving $h$ should roughly halve the error, so we expect $\approx 0.0114$. Why: the endpoint $x=0.2$ is fixed. Shrinking $h$ from $0.1$ to $0.05$ doubles the number of steps ($2\to 4$), but each local error $\frac{h^2}{2}{y}^{\prime \prime }$ shrinks by a factor of $4$. Net effect: $2\times \frac{1}{4}=\frac{1}{2}$. One power of $h$ survives → linear, halving behaviour.

$y^{\prime \prime }=2e^x$ is increasing, so its max on $[0,0.2]$ is at the right end: $M=2e^{0.2}=2.442806$. Bound at $x_n=0.2$: $\frac{(0.1)(2.442806)}{2(1)}(e^{1(0.2)}-1)=0.1221403\times (1.221403-1)=0.1221403\times 0.221403=\mathrm{0.027043.}$ Actual error was $0.022806$. Since $0.022806\le 0.027043$, the actual error is within the bound. ✓ Why the bound is bigger: it is a worst-case guarantee — it uses the largest $|y^{\prime \prime }|$ and assumes errors amplify at the full Lipschitz rate $e^{L(x-x_0)}$. Real error is usually a bit smaller.

Update: $y_{n+1}=y_n+0.5(-2y_n)=y_n(1+0.5(-2))=y_n(1-1)=0$. $y_1=1(0)=0,y_2=0(0)=0.$ True: $e^{-2}=0.135335$. Error $=0.135335$ — enormous relative to the answer. Stability reasoning: for a decay problem $y^{\prime }=\lambda y$, one Euler step multiplies by the factor $1+h\lambda$. Here $1+h\lambda =1+0.5(-2)=0$. A factor of exactly $0$ collapses the solution to zero in a single step — the tangent at the start is so steep it overshoots straight through the axis. This is a stability failure, not just accuracy: see Numerical Stability. For real decay we need $|1+h\lambda |<1$, i.e. $h<1$ here; $h=0.5$ sits right on the edge case where the factor vanishes.

Factor per step: $1+0.25(-2)=0.5$. Since $|0.5|<1$, the numerical solution decays sensibly. $y_4=1\times 0.5^4=\mathrm{0.0625.}$ True: $e^{-2}=0.135335$. Error $=0.072835$ — still large, but now the solution decays toward zero from above instead of collapsing to exactly zero. Smaller $h$ restored qualitatively correct behaviour. Why: the factor $|1+h\lambda |=0.5<1$ guarantees each step shrinks $y$ without overshoot, so the numerical solution now looks like the true exponential decay even if the magnitude is not yet accurate.

First order ⇒ global error scales linearly: $E(h)\approx Ch$ for some constant $C$. From $E_1=C(0.1)=0.040$ we get $C=0.4$. Require $Ch\le 0.005\Rightarrow 0.4h\le 0.005\Rightarrow h\le 0.0125$. Largest convenient $h$ dividing the interval: take $h=0.0125$, giving $N=\frac{1-0}{0.0125}=80\text{ steps}.$ What we did: we used only the fact GE $\propto h$ to extrapolate from one measured error to the step size that hits a target. Why this works: the constant $C$ (which hides $M$, $L$, the $e^{L(x-x_0)}$ factor) cancels in the ratio — the order alone gives the scaling law. This is exactly why knowing "$O(h)$" is practically valuable.

Second order: $E(h)=C_2{h}^2$. From $C_2(0.1{)}^2=0.040\Rightarrow C_2=4.0$. Require $4.0h^2\le 0.005\Rightarrow h^2\le 0.00125\Rightarrow h\le 0.035355$. Steps: $N=\frac{1}{0.035355}=28.28\dots \Rightarrow$ round up to $N=29$ steps ($h\approx 0.03448$). Lesson: to gain a factor of $8$ in accuracy ($0.040\to 0.005$), Euler needs $8\times$ more steps ($80$), but the second-order method needs only about $\sqrt{8}\approx 2.83\times$ more ($29$). Higher order pays off dramatically as tolerances tighten — this is precisely the motivation for Runge-Kutta Methods.

L1 read off $f,x_0,y_0$ → L2 crank the table → L3 measure global error and see it halve → L4 stability gates accuracy → L5 the order $p$ gives a scaling law that drives step-count decisions.