Visual walkthrough — Stability of equilibria — stable, unstable, saddle, spiral, centre

4.6.23 · D2 · Maths › Ordinary Differential Equations › Stability of equilibria — stable, unstable, saddle, spiral,

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Step 1 — Ek rule jo har point pe ek arrow assign karta hai

KYA: hum plane ke har point pe ek arrow paint karte hain. KYUN: ek differential equation hota hi arrows ka ek field hai — arrows ko follow karo aur tum ek trajectory trace kar lete ho. PICTURE: neeche, har blue arrow dikhata hai ki us spot pe maujood ek particle aage kidhar drift karega. Akela red dot special hai — uske paas koi arrow nahi hai.

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Step 2 — Itna zoom in karo ki curved field seedha dikhe

Resting point se tiny deviation ko naam dein: $u=x-x^{\ast },v=y-y^{\ast }.$

• $u,v$ ::: hum kitna nudge hue hain, horizontally aur vertically. Dono chhote hain.

KYA: hum $f$ aur $g$ ko unki linear approximation (first-order Taylor) se likhte hain:

$$f(x,y)\approx \underset{=0}{\underbrace{f(x^{\ast },y^{\ast })}}+\underset{\text{slope in }x}{\underbrace{f_x}}u+\underset{\text{slope in }y}{\underbrace{f_y}}v.$$

• $f_x$ ::: partial derivative — "$y$ ko fixed rakh ke $x$ mein thoda move karo to $f$ kitna change hota hai." Yeh $x$-direction mein $f$ ki steepness hai.
• $f_y$ ::: same, lekin $y$ mein ek step ke liye.
• Pehla term $0$ hai kyunki yeh ek equilibrium hai — yahi woh cheez hai jo sab kuch linear bana deti hai.

DERIVATIVE AUR KOI CHEEZ KYUN NAHI: hum dot par arrow-field ka slope chahte hain — derivative precisely woh tool hai jo batata hai "output kitni tezi se change hota hai per unit nudge?" PICTURE: curved graph aur uski tangent line point ke paas agree karti hain.

$g$ ke liye same karke aur dono rules ko stack karke linear system milta hai.

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Step 3 — Jacobian: ek box mein local rulebook

• $2\times 2$ box $J$ ::: Jacobian matrix, equilibrium par evaluate kiya gaya. Ek row ko across padhो: yeh kehta hai "$\dot{u}$, $f_x$ times $u$ plus $f_y$ times $v$ se bana hai."
• Har entry arrow-field ka ek slope hai, red dot par frozen.

KYA: humne chaar numbers ek matrix mein pack kiye. KYUN: ek matrix exactly woh object hai jo ek vector ko stretch aur rotate karta hai — aur stretch/rotate karna hi woh hai jo flow ek tiny nudge ke saath karta hai. PICTURE: $J$ nudge-arrow $(u,v)$ leta hai aur velocity-arrow $(\dot{u},\dot{v})$ output karta hai — position ko motion mein badalne wali ek machine.

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Step 4 — Special directions: eigenvectors seedhi lines mein travel karte hain

• $e^{\lambda t}$ ::: exponential — yeh isliye aata hai kyunki "rate of change current size ke proportional" ($\dot{}=\lambda \cdot$) exactly woh equation hai jise exponential solve karta hai. Isliye yeh tool aur koi nahi.
• Agar $\lambda >0$: $e^{\lambda t}$ grow karta hai → nudge bahar ud jaata hai. Agar $\lambda <0$: yeh decay karta hai → nudge wapas aata hai.

PICTURE: do special axes; green wale par arrow dot ki taraf shrink hota hai, coral wale par door stretch hota hai.

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Step 5 — Sirf do numbers se $\lambda$ dhundhna: trace aur determinant

KYA: $J\mathbf{e}=\lambda \mathbf{e}$ same hai jaise $(J-\lambda I)\mathbf{e}=0$. Kisi non-zero direction $\mathbf{e}$ ko zero mein squash karne ke liye, matrix $J-\lambda I$ singular honi chahiye, yaani $\det (J-\lambda I)=0$. $2\times 2$ determinant multiply karke:

$${\lambda }^2-\underset{\tau }{\underbrace{(f_x+g_y)}}\lambda +\underset{\Delta }{\underbrace{(f_x{g}_y-f_y{g}_x)}}=0.$$

YEH DO NUMBERS KYUN: yeh exactly quadratic ke coefficients hain, isliye unme ${\lambda }_{1,2}$ ki sari information hai — classify karne ke liye eigenvectors ki kabhi zaroorat nahi. PICTURE: Trace–determinant plane par plot kiya gaya ek single point, $\tau$ across aur $\Delta$ upar.

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Step 6 — $\Delta$ ka sign: same team ya opposite team?

KYA: $\Delta ={\lambda }_1{\lambda }_2$. YEH KYUN MATTER KARTA HAI: ek product batata hai ki do eigenvalues ek hi sign share karte hain ya nahi.

• $\Delta <0$ ::: product negative → ek $\lambda >0$, ek $\lambda <0$ → ek taraf grow, doosri taraf decay → saddle (hamesha unstable).
• $\Delta >0$ ::: product positive → dono same sign → phir trace $\tau$ decide karta hai kaunsa sign.

PICTURE: saddle — arrows ek axis se andar aate hain aur perpendicular wale se bahar blast karte hain, jaise kisi mountain pass ki surface.

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Step 7 — $D$ ka sign: real slide ya complex swirl?

KYA: root $\sqrt{D}$ real hota hai jab $D>0$ aur imaginary jab $D<0$. Yahan $\omega$ (Greek letter "omega") rotation frequency ko naam deta hai — nudge dot ke around ek unit time mein kitne radians sweep karta hai; yeh $\omega =\frac{1}{2}\sqrt{4\Delta -{\tau }^2}$ hai, jo $\lambda$ ke imaginary part ka size hai.

• $D>0$ ::: do distinct real eigenvalues → motion pure stretch/shrink hai do fixed axes ke along → node (ya saddle agar $\Delta <0$).
• $D<0$ ::: $\sqrt{D}=i\sqrt{4\Delta -{\tau }^2}$ $\lambda$ mein ek imaginary part laata hai. Exponent mein imaginary ka matlab $e^{i\omega t}$ — aur yahi rotation hai (Euler: $e^{i\omega t}=\cos \omega t+i\sin \omega t$). To $D<0$ → spirals ya centre.

IMAGINARY PART = ROTATION KYUN: ek real exponent scale karta hai; ek imaginary exponent spin karta hai. Nudge dot ke around circle karta hai seedha chalna chhod ke.

PICTURE: left, ek node (seedha radial approach); right, ek spiral (rotating approach). Switch $D$ ka sign hai.

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Step 8 — Knife-edge $D=0$: nodes jo ek direction khо dete hain

KYA: $D=0$ par tumhe poochhna hoga ki eigenbasis full hai ya nahi. KYUN: ek defective matrix ke paas koi doosra straight-line direction nahi hota, isliye picture radial spokes ki jagah ek curved improper node hai. PICTURE: left, star node (radial); right, improper node (saari curves ek akeli direction ki taraf tangent ho ke funnel karti hain).

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Step 9 — $\tau$ ka sign aur knife-edge $\tau =0$

KYA: jab $\Delta >0$ "dono eigenvalues same sign" fix kar deta hai, tab trace sign pick karta hai: $\tau ={\lambda }_1+{\lambda }_2.$

• $\tau <0$ ::: total growth negative → dono decay → stable (node ya spiral) — arrows spiral/slide andar karti hain.
• $\tau >0$ ::: total growth positive → dono grow → unstable — baahر.
• $\tau =0$ with $\Delta >0$ ::: eigenvalues $\pm i\omega$ hain, pure imaginary → na growth, na decay, sirf rotation → centre: perfect closed loops.

PICTURE: teen panels — inward spiral ($\tau <0$), closed loops ($\tau =0$), outward spiral ($\tau >0$).

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Ek picture mein poora summary

Upar sab kuch ek map par rehta hai: Trace–determinant plane. $(\tau ,\Delta )$ plot karo, aur tum jis region mein land karo wahi jawaab hai — lekin yaad rakho boundaries special hoti hain:

• Parabola $\Delta =\frac{{\tau }^2}{4}$ (yaani $D=0$) nodes (neeche) ko spirals (upar) se alag karti hai, aur uski par star / improper nodes rehte hain (Step 8).
• Horizontal axis $\Delta =0$ saddles (neeche) ko baaki sab se alag karta hai, aur uski par non-hyperbolic zero-eigenvalue case baithta hai (Step 6 ka edge box).
• Vertical axis $\tau =0$ (with $\Delta >0$) centre knife-edge hai, stable (left) ko unstable (right) se alag karti hai.

Recall Feynman retelling — plain words mein bol do

Poori kahani ::: Ek rule har point pe ek arrow paint karta hai; jahan arrow zero hota hai wahan system rest kar sakta hai. Ise nudge karo, zoom in karo jab tak arrows seedhe na lagein, aur chaar local slopes ko Jacobian kehte ek box mein pack karo. Us box mein special directions (eigenvectors) hain jise woh sirf factors $\lambda$ (eigenvalues) se stretch karta hai; har direction $e^{\lambda t}$ ki tarah evolve hoti hai. $\lambda$'s ke sum $\tau$ aur product $\Delta$ — yeh do summary numbers — sab decide karte hain: $\Delta <0$ matlab opposite signs (saddle); $\Delta >0$ matlab same sign, aur phir $\tau <0$ andar kheenchta hai (stable) jabki $\tau >0$ bahar dhakelta hai (unstable); discriminant $D$ decide karta hai seedha slide (node) ya swirl (spiral), aur razor line $\tau =0$ endless loops deta hai (centre). Teen border curves apne khud ke delicate cases hain: $\Delta =0$ (ek zero eigenvalue, near-rest points ki poori ek line), $D=0$ (repeated eigenvalue → star ya defective improper node), aur $\tau =0$ (ek marginal centre).

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Recall drill

Recall Quick self-test

$\Delta <0$ hamesha unstable kyun hota hai? ::: $\Delta ={\lambda }_1{\lambda }_2<0$ opposite signs force karta hai, isliye ek eigenvalue positive hota hai → woh mode grow karta hai → saddle. Exactly $\Delta =0$ par kya hota hai? ::: Ek eigenvalue $0$ hota hai aur doosra $\tau$ → non-hyperbolic, ek one-dimensional centre manifold; linearisation stability decide nahi kar sakta. $D=0$ par, kaun se teen node types ho sakte hain? ::: Full eigenbasis ke saath star (proper) node, ya sirf ek eigen-direction ke saath defective improper node (extra $t{e}^{\lambda t}$ term). Kaunsa single sign ek spiral ko stable se unstable flip karta hai? ::: $\tau$ (trace = total growth rate) ka sign. Ek spiral mein rotation algebraically kahan se aata hai? ::: $\lambda$ ka imaginary part $\omega$ jab $D={\tau }^2-4\Delta <0$, jo $e^{i\omega t}$ = rotation deta hai. Trace–determinant map par, nodes ko spirals se kaunsa curve alag karta hai? ::: Parabola $\Delta ={\tau }^2/4$, yaani $D=0$.