Visual walkthrough — Phase plane analysis — trajectories, critical points

4.6.22 · D2 · Maths › Ordinary Differential Equations › Phase plane analysis — trajectories, critical points

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Step 1 — Tiny arrows ka ek field (ek "system" ASAL mein kya hai)

WHAT. Hamare paas do rules hain jo ek chalte hue dot ko bataate hain ki wo kitna sideways aur kitna upar jaaye: $\dot{x}=f(x,y),\dot{y}=g(x,y).$ Dots ko "rate of change" padhho: $\dot{x}$ ka matlab hai "abhi $x$ kitni tezi se badal raha hai". Plane ke har point $(x,y)$ pe yeh do numbers milke ek arrow $(\dot{x},\dot{y})$ banaate hain — ek choti si velocity jo bataati hai yahan rakha hua dot aage kahaan jaayega.

WHY. Kyunki right side pe $t$ nahin hai (yeh ek autonomous system hai), arrow sirf iss par depend karta hai ki tum kahaan ho, kabhi iss par nahin ki tum kab pahunche. Toh plane ek fixed arrows ka field ban jaati hai — ek wind map jo kabhi nahin badlta.

PICTURE. Har point ko exactly ek arrow milta hai. Ek dot drop karo, arrows follow karo, aur wo ek curve banaata hai — ek trajectory.

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Step 2 — Wo ek point jahan hawa ruk jaati hai (critical point)

WHAT. Hum ek special location $(x_0,y_0)$ dhundhte hain jahaan dono rules zero dete hain: $\underset{\text{koi sideways push nahin}}{\underbrace{f(x_0,y_0)=0}}\text{aur}\underset{\text{koi up-down push nahin}}{\underbrace{g(x_0,y_0)=0}}.$

WHY. Agar arrow ke dono parts zero hain, toh arrow ki zero length hai — exactly wahaan rakha dot kabhi nahin hilda. Yeh ek critical point (equilibrium) hai. Yeh akela aisa jagah hai jahaan trajectories mil sakti hain, kyunki baaki har jagah ek arrow ek direction force karta hai.

PICTURE. Dead spot ke around surrounding arrows khud ko organize karte hain — andar point karte hue, baahir, ya swirl karte hue. Woh neighbouring arrows is pure poori kahani hai.

Recall Trajectories baaki jagah cross kyun nahin kar sakti?

Do curves ek ordinary point pe cross karna ::: ek point pe do alag arrows maangega — lekin field exactly ek deta hai. Toh sirf zero-arrow (critical) points pe hi crossings ho sakti hain.

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Step 3 — Zoom in karo jab tak curve seedhi na lage (WHY hum linearize karte hain)

WHAT. Hum origin ko critical point par slide karte hain. Equilibrium se displacement define karo: $u=x-x_0,v=y-y_0.$ Toh $(u,v)=(0,0)$ hi critical point hai, aur $u,v$ measure karte hain "kitna off" hum hain.

WHY. Dead spot ke bilkul paas, $u$ aur $v$ tiny hain. Ek smooth vector field, kaafi strong microscope se dekha jaaye, flat lagta hai — jaise koi bhi curve close up apni tangent line jaisi lagti hai. Agar hum sirf woh part rakhein jo $u$ aur $v$ ke proportion mein badhta hai ($u^2$, $uv$, aur chhote terms ko hataa ke), toh hum ek naakaam nonlinear tangle ko ek saaf linear se replace karte hain.

PICTURE. Point ke paas curved field, aur straight-arrow field jo ek chhote disc mein usse chipakti hai — centre ke paas alag nahin dikh ti, edge pe alag hoti jaati hai.

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Step 4 — Woh chaar slopes jo local field banate hain (Jacobian)

WHAT. Pehle ek shorthand. Ek partial derivative measure karta hai ki ek output kitna badalta hai jab tum ek input ko nudge karo aur doosre ko freeze rakhho. Hum likhte hain: $f_x\equiv \frac{\partial f}{\partial x},f_y\equiv \frac{\partial f}{\partial y},g_x\equiv \frac{\partial g}{\partial x},g_y\equiv \frac{\partial g}{\partial y}.$ $f_x$ padhho as "sideways push $f$ kitna badalta hai per unit step $x$ mein, $y$ ko still rakh ke" — ek slope, ek output vs ek input.

Ab: sideways push $\dot{u}$, $u$ mein ek small step par kaise respond karta hai? $v$ mein? Wahi $\dot{v}$ ke liye poochho. Woh chaar slopes, critical point par evaluate kiye gaye, ek matrix mein stack ho jaate hain:

$$\left(\begin{array}{c}\dot{u}\\ \dot{v}\end{array}\right)=\underset{J=\text{[[Linearization and the Jacobian matrix|Jacobian]]}}{\underbrace{\left(\begin{array}{cc}{f}_x& f_y\\ g_x& g_y\end{array}\right)}}\left(\begin{array}{c}u\\ v\end{array}\right).$$

Matrix ko term-by-term padhho:

• $f_x$ = "horizontal push kitna badhta hai jab main right step karta hoon" (top-left),
• $f_y$ = "horizontal push kitna badhta hai jab main upar step karta hoon" (top-right),
• $g_x$ = "vertical push kitna badhta hai jab main right step karta hoon" (bottom-left),
• $g_y$ = "vertical push kitna badhta hai jab main upar step karta hoon" (bottom-right).

WHY. Ek tiny displacement $(u,v)$ arrow $J\left(\begin{array}{c}u\\ v\end{array}\right)$ produce karta hai. Woh akela matrix $J$ hi local wind machine hai — isko batao tum kahaan ho, woh return karega tum kahaan jaoge.

PICTURE. Har entry ek direction ke against ek push ka slope hai; chaaroon milke grid ko bend karte hain.

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Step 5 — $e^{\lambda t}$ guess karna (WHY exponentials, aur $\lambda$ ka matlab)

WHAT. Hum $\dot{\mathbf{u}}=J\mathbf{u}$ ke solutions dhundhte hain jo waqt ke saath sirf stretch ya shrink hon, same shape rakhte hue. Wo ek function jiska rate of change khud ke proportional ho, exponential hai, toh hum try karte hain: $\mathbf{u}(t)=\mathbf{v}{e}^{\lambda t}.$ Yahaan $\mathbf{v}$ ek fixed direction (eigenvector) hai aur $\lambda$ ek number (eigenvalue) hai jo growth control karta hai.

WHY yeh tool aur koi nahin. Humein ek function $q(t)$ chahiye jisme $\dot{q}=\lambda q$ ho — "apne size ke proportional rate par badalta hai". Yeh exactly $e^{\lambda t}$ ki defining property hai: $\frac{d}{dt}{e}^{\lambda t}=\lambda e^{\lambda t}$. Koi akela polynomial ya trig yeh nahin karta. Substitute karo:

$$\lambda \mathbf{v}{e}^{\lambda t}=J\mathbf{v}{e}^{\lambda t}⟹J\mathbf{v}=\lambda \mathbf{v}.$$

$e^{\lambda t}$ cancel ho jaata hai, eigenvalue equation bachti hai. Toh allowed growth rates $\lambda$ precisely $J$ ke eigenvalues hain.

PICTURE. Special direction $\mathbf{v}$ ke along, motion pure stretch hai: arrow aur position parallel rehte hain, $e^{\lambda t}$ se scale hote hue.

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Step 6 — $J$ ke andar chhupe do numbers (trace aur determinant)

WHAT. $\lambda$ find karne ke liye hum $\det (J-\lambda I)=0$ solve karte hain, jo ek $2\times 2$ matrix ke liye hamesha iss par collapse ho jaata hai:

$${\lambda }^2-\tau \lambda +\Delta =0,\tau =\mathrm{tr}J=f_x+g_y,\Delta =\det J=f_x{g}_y-f_y{g}_x.$$

Term-by-term matlab:

• $\tau$ (trace) $={\lambda }_1+{\lambda }_2$ — do growth rates ka sum,
• $\Delta$ (determinant) $={\lambda }_1{\lambda }_2$ — do growth rates ka product.

Quadratic solve karo:

$${\lambda }_{1,2}=\frac{1}{2}(\underset{\text{sum}}{\underbrace{\tau }}\pm \underset{\text{discriminant }D}{\underbrace{\sqrt{{\tau }^2-4\Delta }}}).$$

WHY. Explicitly eigenvalues compute karne ki bajaye, hum type seedha $\tau$ aur $\Delta$ se padh sakte hain. Root ke neeche ki quantity, $D={\tau }^2-4\Delta$, real-vs-complex decide karti hai; $\tau$ aur $\Delta$ ke signs baaki decide karte hain. Do numbers sab kuch classify karte hain — Trace–determinant plane.

PICTURE. Ek dial: $\Delta$ ek axis par, $\tau$ doosre par, parabola $D=0$ regions kaat rahi hai.

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Step 7 — Do numbers se type padhna (generic cases)

WHAT. "Generic" ka matlab hai ek region ka interior: $\tau$ ya $\Delta$ ko thoda nudge karo toh type nahin badlta. Un robust possibilities se guzro $({\lambda }_1,{\lambda }_2)$ ke liye, yaani $D={\tau }^2-4\Delta$ strictly positive ya strictly negative aur $\tau \ne 0$:

$\Delta$ ka sign	$D={\tau }^2-4\Delta$ ka sign	$\tau$ ka sign	Eigenvalues	Type aur stability
$\Delta <0$	$D>0$ (forced)	—	real, opposite signs	saddle, unstable
$\Delta >0$	$D>0$	$\tau <0$	real, dono $<0$	node sink, stable
$\Delta >0$	$D>0$	$\tau >0$	real, dono $>0$	node source, unstable
$\Delta >0$	$D<0$	$\tau <0$	complex, $\mathrm{Re}<0$	spiral in, stable
$\Delta >0$	$D<0$	$\tau >0$	complex, $\mathrm{Re}>0$	spiral out, unstable

WHY har row.

• $\Delta <0$ ⇒ product ${\lambda }_1{\lambda }_2<0$ ⇒ opposite signs ⇒ ek direction repel karta hai, ek attract karta hai ⇒ saddle. (Negative product $D>0$ force karta hai, toh eigenvalues yahaan hamesha real hote hain.)
• $\Delta >0,D>0$ ⇒ do distinct real roots same sign ke; unka common sign $\tau$ ka sign hai ⇒ dono andar (sink) ya dono baahir (source): ek node.
• $D<0$ ⇒ root mein imaginary part hai ⇒ rotation ⇒ spiral; andar wind karega ya baahir yeh $\tau$ ka sign hai (jo $2\mathrm{Re}\lambda$ ke barabar hai).

PICTURE. Teeno canonical generic portraits — node, saddle, spiral — flow dikhate arrows ke saath.

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Step 8 — Borderline cases jo tum skip nahin kar sakte (degenerate inputs)

WHAT. Har cheez jo trace–determinant plane ke kisi boundary curve par baithe. Har ek ko handle karo, apne flow picture ke saath:

(a) $\Delta =0$ — resting points ki poori line. Ek eigenvalue $0$ hai. Linearized system ke liye product ${\lambda }_1{\lambda }_2=0$, toh linear picture mein ek poori eigen-direction hai jahaan kuch nahin hilda — equilibria ki ek line, doosri eigen-direction ke along flow ke saath. Caution: yeh sirf linearization describe karta hai. Original nonlinear system mein shayad phir bhi sirf ek isolated equilibrium ho — extra fixed points tabhi appear hote hain jab higher-order terms bhi us line ke along vanish ho jaayein. Jab $\Delta =0$ ho, linear approximation ne exactly woh terms throw away kar diye hain jo sach decide karte the, toh yahaan woh unreliable hai.

(b) $D=0$ exactly — ek repeated real root ${\lambda }_1={\lambda }_2=\lambda$. Yahaan tumhe do bahut alag sub-cases mein split karna hoga, aur distinction yeh hai ki $J$ diagonalizable hai (do independent eigen-directions hain) ya defective (sirf ek hai):

Dono stable hain agar $\lambda <0$, unstable agar $\lambda >0$. Figure do portraits side by side dikhata hai.

(c) $\tau =0,\Delta >0$ — center. Linear theory perfect closed loops predict karta hai, lekin yeh woh ek case hai jahaan linearization jhooth bol sakti hai: tiny nonlinear terms center ko slow spiral mein badal sakti hain. Lyapunov/energy argument ya Poincaré–Bendixson machinery se confirm karo.

WHY. Yeh trace–determinant plane ke regions ke boundaries par rehte hain, jahaan arbitrarily small nudge type flip kar sakta hai. Ek reader jo sirf interior cases jaanta hai woh ek real system ko galat classify karega jo fence par baitha ho — aur, $D=0$ par, galat assume karega ki sirf ek eigen-direction hai jabki $J=\lambda I$ actually infinitely many deta hai.

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Ek-picture summary

Sab kuch trace–determinant plane mein collapse ho jaata hai: apna critical point lo, $(\tau ,\Delta )$ compute karo, pin lagao, type padho. Parabola ${\tau }^2=4\Delta$ nodes ko spirals se alag karti hai (aur yahi $D=0$ line hai star/improper nodes ki); axis $\Delta =0$ tumhe neeche saddles mein le jaata hai; axis $\tau =0$ (upar $\Delta =0$ ke) neutral centre line hai.

Recall Feynman: poora walkthrough simple words mein

Ek aisi windy floor imagine karo jahan har spot ki hawa time ke saath kabhi nahin badlti. Ek patta chodo aur woh ek path par drift karta hai — ek trajectory. Woh ek spot dhundho jahan bilkul hawa nahin: patta wahaan bas baitha rehta hai — ek critical point. Nearby patto ke saath kya hota hai yeh jaanne ke liye, us dead spot par magnifying glass rakho — swirling hawa close up seedhi lagti hai. Chaar numbers (har push kitna badalta hai jab tum right ya upar step karo) woh seedhi hawa capture karte hain — yeh Jacobian hai. Poochho "kin directions mein patta seedha andar ya seedha baahir push hota hai, aur kitna fast?" Woh "kitna fast" numbers eigenvalues hain, kyunki sirf woh motion jo khud ko scale kare woh exponential hai. Do summary numbers — un speeds ka sum ($\tau$) aur product ($\Delta$) — bina kuch solve kiye type bata dete hain: negative product matlab saddle (ek taraf andar, ek taraf baahir); same-sign product matlab node (dono andar = sink, dono baahir = source); hidden rotation ise spiral banata hai; positive product ke saath perfectly balanced sum ek nazuk center hai jise real, curved hawa quietly slow spiral mein kharaab kar sakti hai. Aur jab do speeds exactly equal hon, aur bhi close dekho: agar hawa har direction mein seedha andar push karti hai toh yeh ek star node hai, lekin agar sirf ek seedhi direction bachti hai, toh saare patte us single line ke tangent ban jaate hain — ek improper node.