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

"Kisi point par slope" ka matlab hai: us point ko $f$ mein plug karo. Yahan $f(x,y)=x-2y$. $f(3,1)=3-2(1)=3-2=1.$ $(3,1)$ par dash slope $1$ par upar ki taraf tilt karta hai (45° line). ✔

$(x_0,y_0)=(0,1)$ par khade raho. Slope $=f(0,1)=1$. Step: $y_1=y_0+hf=1+0.1(1)=1.1$, $x_1=0.1$ par. Dhyaan do yeh $e^{0.1}\approx 1.1052$ ka Euler approximation hai — thoda undershoot. ✔

Step 0 $(0,1)$ par: slope $=0+1=1$. $y_1=1+0.5(1)=1.5$, $x_1=0.5$. Step 1 $(0.5,1.5)$ par: 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 $y=2e^x-x-1$ hai, toh $y(1)=2e-2\approx 3.44$. Euler undershoot karta hai. Figure s02 (neeche) yeh do seedhe tangent steps (coral, dashed) dikhata hai jo sahi curve (slate solid) se neeche rehte hain. ✔

Chaar baar chalo, har landing point par slope recompute karte hue:

• $(0,1)$: $f=1\Rightarrow y_1=1+0.25(1)=1.25$
• $(0.25,1.25)$: $f=1.5\Rightarrow y_2=1.25+0.25(1.5)=1.625$
• $(0.5,1.625)$: $f=2.125\Rightarrow y_3=1.625+0.25(2.125)=2.15625$
• $(0.75,2.15625)$: $f=2.90625\Rightarrow y_4=2.15625+0.25(2.90625)=2.8828125$ $y(1)\approx \mathrm{2.883.}$ Coarse $2.5$ se zyada $3.44$ ke paas: $h$ aadha karne se error shrink hota hai, bilkul $O(h)$ behaviour ki tarah. Figure s02 upar mein, yeh chaar-step walk green square-marker path hai — iske chhote, zyada frequent steps sahi curve ko coarse do-step walk se kahin zyada tightly hug karte hain. ✔

Negative $h$ ka matlab sirf yeh hai ki hum left chalte hain. Formula unchanged rehta hai. $(1,2)$ par khade raho: slope $=f(1,2)=1\cdot 2=2$. $y_1=y_0+hf=2+(-0.5)(2)=2-1=1$, $x_1=1+(-0.5)=0.5$ par. $y(0.5)\approx 1.$ ✔

Euler har point par tangent line ke saath chalta hai. Concave-down curve ke liye, curve apni tangents se neeche bend karti hai, toh tangent step sahi curve se upar land karta hai: Euler over-shoot karta hai. (Parent ke convex example ka mirror, jahan $y^{\prime \prime }>0$ ne tangents ko neeche rakha aur Euler under-shoot kiya.) Figure s01 (neeche) sahi solution (slate) ko frown ki tarah dikhata hai; start point se coral tangent step iske upar shoot karta hai, aur butter double-arrow over-shoot gap ko mark karta hai. ✔

Euler ka global error $O(h)$ hai — $h$ ke proportional ($O(h)$ yaad karo: error seedhi line ke neeche, toh $h$ aadha karne se woh aadhi ho jaati hai). $\text{error}(0.05)\approx \frac{1}{2}(0.020)=\mathrm{0.010.}$ ✔ (Yeh first-order method hai; RK4 se compare karo, jahan $h$ aadha karne se error $\sim 16$ se cut hoti hai.)

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

(a) Separate karo: $\frac{dy}{y}=dx\Rightarrow \ln y=x+C$. $y(0)=1$ ke saath, $C=0$, toh $y=e^x$ aur $y(1)=e\approx 2.71828$. (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 convex exponential ko under-shoot karta hai ($y^{\prime \prime }>0$). ✔

$(0,2)$ par slope: $f=0^2+2=2$. Euler kehta hai $y_1=y_0+hf$: $2.6=2+h(2)\Rightarrow 2h=0.6\Rightarrow h=\mathrm{0.3.}$ Phir $x_1=x_0+h=0+0.3=0.3$. ✔

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

• Plain Euler ek step: $y_1=1+1(1)=2$ (error $1.44$).
• Heun: $y_1=3$ (error $0.44$). Leaving-slope aur arriving-slope ko average karne se leading $O(h^2)$ term cancel ho jaata hai — yahi seed idea hai RK4 ke peeche. ✔

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

• $y_1=-3(1)=-3$
• $y_2=-3(-3)=9$
• $y_3=-3(9)=-27$ Magnitude explode karta hai aur sign har step flip karta hai, jabki sahi answer quietly $0$ ki taraf decay karta hai. Kyun: explicit Euler ki $y^{\prime }=\lambda y$ par stability require karti hai $|1+h\lambda |<1$. Yahan $1+h\lambda =1-4=-3$, jiska size $3>1$ hai, toh errors grow karti hain. Isko fix karne ke liye chhota $h$ ya implicit method chahiye — Stability and Stiff Equations dekho. ✔

(a) $y=\frac{1}{1-x}\to \infty$ jab $x\to 1^{-}$: solution finite time mein blow up kar jaata hai. Existence sirf $x<1$ par guaranteed hai — yahi finite-time-escape hai jo Existence and Uniqueness (Picard–Lindelöf) warn karta hai ki smooth $f$ ke liye bhi ho sakta hai. (b) Euler $x=1$ ke dono taraf khushi se finite numbers return karta hai aur koi warning nahi deta — woh bas step karta rehta hai. Numerically:

• $(0,1)$: $f=1\Rightarrow y_1=1+0.25(1)=1.25$
• $(0.25,1.25)$: $f=1.5625\Rightarrow y_2=1.25+0.25(1.5625)=1.640625$ Values tezi se badhti hain lekin finite rehti hain; method vertical asymptote "dekh" nahi sakta. Lesson: existence hamesha analytically check karo — numerics nahi karengi. ✔

Convergence proof ke liye $f$ ko $y$ mein Lipschitz hona chahiye: ek constant $M$ exist karta hai ki $|f(x,y_1)-f(x,y_2)|\le M|y_1-y_2|\text{for all }{y}_1,y_2.$ Yeh kya deta hai: agar do nearby curves close start karti hain, toh ODE unhe $e^{M(x-x_0)}$ factor se zyada tezi se door nahi fling kar sakta. Toh tiny per-step local truncation errors $\frac{h^2}{2}{y}^{\prime \prime }(\xi )$ (upar define ki gayi, seedhi Taylor's Theorem se) controlled tarike se accumulate hoti hain explode karne ke bajaye, aur total (constant)$\times h=O(h)$ se bounded hota hai. Yahi Lipschitz condition hai jo Existence and Uniqueness (Picard–Lindelöf) pehle se ek single solution curve exist karne ke liye use karta hai. Kya todta hai: $f(x,y)=y^2$ (L5.3) globally Lipschitz nahi hai — iska $y$-slope $2y$ without bound grow karta hai — toh guarantee void hai aur solution infinity tak escape kar sakta hai. Koi Lipschitz bound nahi, koi convergence promise nahi. ✔

One-step Euler formula ::: $y_{n+1}=y_n+hf(x_n,y_n)$ Error ke liye $O(h)$ ka kya matlab hai? ::: Error $\le$ constant $\times h$; $h$ aadha karo toh error aadhi hoti hai (first order) Local truncation error kya hai aur kahan se aati hai? ::: One-step slip $\frac{h^2}{2}{y}^{\prime \prime }(\xi )$, wo term jo Taylor's theorem rakhta hai lekin Euler chhod deta hai; size $O(h^2)$ $y^{\prime \prime }<0$ curve ke baare mein kya batata hai? ::: Yeh neeche ki taraf bend karta hai (concave, ek frown); Euler over-shoot karta hai Test equation $y^{\prime }=\lambda y$ mein $\lambda$ kya hai? ::: Ek growth/decay rate; $\lambda <0$ decay karta hai, $\lambda >0$ grow karta hai Explicit Euler ki $y^{\prime }=\lambda y$ par stability condition ::: $|1+h\lambda |<1$ $f$ par wo condition jo convergence guarantee kare ::: Lipschitz in $y$: $|f(x,y_1)-f(x,y_2)|\le M|y_1-y_2|$ $y^{\prime }=x+y,y(0)=1,h=1$ ke liye Heun ka one-step value ::: $3$