Exercises — Direction fields and Euler's method — visual - numerical intuition first

"Slope at a point" means: plug that point into $f$. Here $f(x,y)=x-2y$. $f(3,1)=3-2(1)=3-2=1.$ The dash at $(3,1)$ tilts upward at slope $1$ (a 45° line). ✔

Stand at $(x_0,y_0)=(0,1)$. Slope $=f(0,1)=1$. Step: $y_1=y_0+hf=1+0.1(1)=1.1$, at $x_1=0.1$. Notice this is the Euler approximation to $e^{0.1}\approx 1.1052$ — a slight undershoot. ✔

Step 0 at $(0,1)$: slope $=0+1=1$. $y_1=1+0.5(1)=1.5$, $x_1=0.5$. Step 1 at $(0.5,1.5)$: slope $=0.5+1.5=2$. $y_2=1.5+0.5(2)=2.5$, $x_2=1$. $y(1)\approx \mathrm{2.5.}$ Exact solution is $y=2e^x-x-1$, so $y(1)=2e-2\approx 3.44$. Euler undershoots. Figure s02 (below) shows these two straight tangent steps (coral, dashed) staying below the true curve (slate solid). ✔

Walk four times, recomputing the slope at each landing point:

• $(0,1)$: $f=1\Rightarrow y_1=1+0.25(1)=1.25$
• $(0.25,1.25)$: $f=1.5\Rightarrow y_2=1.25+0.25(1.5)=1.625$
• $(0.5,1.625)$: $f=2.125\Rightarrow y_3=1.625+0.25(2.125)=2.15625$
• $(0.75,2.15625)$: $f=2.90625\Rightarrow y_4=2.15625+0.25(2.90625)=2.8828125$ $y(1)\approx \mathrm{2.883.}$ Closer to $3.44$ than the coarse $2.5$: halving $h$ shrinks the error, exactly the $O(h)$ behaviour. In figure s02 above, this four-step walk is the green square-marker path — its shorter, more frequent steps hug the true curve much more tightly than the coarse two-step walk. ✔

A negative $h$ just means we walk left. The formula is unchanged. Stand at $(1,2)$: slope $=f(1,2)=1\cdot 2=2$. $y_1=y_0+hf=2+(-0.5)(2)=2-1=1$, at $x_1=1+(-0.5)=0.5$. $y(0.5)\approx 1.$ ✔

Euler walks along the tangent line at each point. For a concave-down curve, the curve bends below its tangents, so the tangent step lands above the true curve: Euler over-shoots. (Mirror of the parent's convex example, where $y^{\prime \prime }>0$ made tangents lie below and Euler under-shot.) Figure s01 (below) shows the true solution (slate) as a frown; the coral tangent step from the start point shoots off above it, and the butter double-arrow marks the over-shoot gap. ✔

Euler's global error is $O(h)$ — proportional to $h$ (recall $O(h)$: error under a straight line, so halving $h$ halves it). $\text{error}(0.05)\approx \frac{1}{2}(0.020)=\mathrm{0.010.}$ ✔ (This is a first-order method; contrast with RK4, where halving $h$ cuts error by $\sim 16$.)

(a) Steps $N=L/h=2/0.01=200$. (b) Local error per step $\approx c{h}^2=0.5(0.01{)}^2=0.5\times 10^{-4}$. Accumulated $\approx N\cdot c{h}^2=200\times 0.5\times 10^{-4}=0.01$. Cross-check with the $O(h)$ formula: global $\approx cLh=0.5\cdot 2\cdot 0.01=0.01$. Same. ✔

(a) Separate: $\frac{dy}{y}=dx\Rightarrow \ln y=x+C$. With $y(0)=1$, $C=0$, so $y=e^x$ and $y(1)=e\approx 2.71828$. (b) $(0,1)$: $f=1\Rightarrow y_1=1+0.5(1)=1.5$. $(0.5,1.5)$: $f=1.5\Rightarrow y_2=1.5+0.5(1.5)=2.25$. $y(1)\approx \mathrm{2.25.}$ (c) Error $=|2.71828-2.25|\approx 0.468$. Euler under-shoots the convex exponential ($y^{\prime \prime }>0$). ✔

Slope at $(0,2)$: $f=0^2+2=2$. Euler says $y_1=y_0+hf$: $2.6=2+h(2)\Rightarrow 2h=0.6\Rightarrow h=\mathrm{0.3.}$ Then $x_1=x_0+h=0+0.3=0.3$. ✔

$k_1=f(0,1)=1$. Predictor step: $y_0+h{k}_1=1+1(1)=2$ at $x=1$. $k_2=f(1,2)=1+2=3$. Average the two slopes: $y_1=1+\frac{1}{2}(1+3)=1+2=3.$

• Plain Euler one step: $y_1=1+1(1)=2$ (error $1.44$).
• Heun: $y_1=3$ (error $0.44$). Averaging the leaving-slope and the arriving-slope cancels the leading $O(h^2)$ term — the seed idea behind RK4. ✔

Update: $y_{n+1}=y_n+h(-20y_n)=y_n(1-20\cdot 0.2)=y_n(1-4)=-3y_n$.

• $y_1=-3(1)=-3$
• $y_2=-3(-3)=9$
• $y_3=-3(9)=-27$ The magnitude explodes and the sign flips each step, while the true answer decays quietly to $0$. Why: stability of explicit Euler on $y^{\prime }=\lambda y$ requires $|1+h\lambda |<1$. Here $1+h\lambda =1-4=-3$, whose size $3>1$, so errors grow. Fixing this needs a tiny $h$ or an implicit method — see Stability and Stiff Equations. ✔

(a) $y=\frac{1}{1-x}\to \infty$ as $x\to 1^{-}$: the solution blows up in finite time. Existence is only guaranteed on $x<1$ — this is exactly the finite-time-escape that Existence and Uniqueness (Picard–Lindelöf) warns can happen even for smooth $f$. (b) Euler happily returns finite numbers on both sides of $x=1$ and gives no warning — it just keeps stepping. Numerically:

• $(0,1)$: $f=1\Rightarrow y_1=1+0.25(1)=1.25$
• $(0.25,1.25)$: $f=1.5625\Rightarrow y_2=1.25+0.25(1.5625)=1.640625$ The values grow fast but stay finite; the method cannot "see" the vertical asymptote. Lesson: always check existence analytically — the numerics won't. ✔

The convergence proof needs $f$ to be Lipschitz in $y$: there is a constant $M$ such that $|f(x,y_1)-f(x,y_2)|\le M|y_1-y_2|\text{for all }{y}_1,y_2.$ What this buys you: if two nearby curves start close, the ODE cannot fling them apart faster than the factor $e^{M(x-x_0)}$. So the tiny per-step local truncation errors $\frac{h^2}{2}{y}^{\prime \prime }(\xi )$ (defined at the top, straight from Taylor's Theorem) accumulate in a controlled way rather than exploding, and the total is bounded by (constant)$\times h=O(h)$. This is the same Lipschitz condition that Existence and Uniqueness (Picard–Lindelöf) uses to guarantee a single solution curve exists in the first place. What breaks it: $f(x,y)=y^2$ (L5.3) is not globally Lipschitz — its $y$-slope $2y$ grows without bound — so the guarantee is void and the solution can escape to infinity. No Lipschitz bound, no convergence promise. ✔

One-step Euler formula ::: $y_{n+1}=y_n+hf(x_n,y_n)$ What does $O(h)$ mean for the error? ::: Error $\le$ constant $\times h$; halving $h$ halves it (first order) What is the local truncation error and where does it come from? ::: The one-step slip $\frac{h^2}{2}{y}^{\prime \prime }(\xi )$, the term Taylor's theorem keeps but Euler throws away; size $O(h^2)$ What does $y^{\prime \prime }<0$ tell you about the curve? ::: It bends downward (concave, a frown); Euler over-shoots What is $\lambda$ in the test equation $y^{\prime }=\lambda y$? ::: A growth/decay rate; $\lambda <0$ decays, $\lambda >0$ grows Stability condition of explicit Euler on $y^{\prime }=\lambda y$ ::: $|1+h\lambda |<1$ Condition on $f$ that guarantees convergence ::: Lipschitz in $y$: $|f(x,y_1)-f(x,y_2)|\le M|y_1-y_2|$ Heun's one-step value for $y^{\prime }=x+y,y(0)=1,h=1$ ::: $3$