Stability of equilibria — stable, unstable, saddle, spiral, centre: Worked examples

4.6.23 · D3 · Maths › Ordinary Differential Equations › Stability of equilibria — stable, unstable, saddle, spiral,

Do numbers saara kaam karte hain (parent mein build kiye gaye, yahan restate kiye hain taaki kuch assume na karna pade):

Recall Teen numbers jo hum read off karte hain

$τ = tr J = f_{x} + g_{y}$ — eigenvalues ka sum ("average growth rate"). $τ < 0$ andar kheenchta hai, $τ > 0$ bahar dhakelta hai.
$Δ = det J = f_{x} g_{y} - f_{y} g_{x}$ — eigenvalues ka product. $Δ > 0$ matlab same sign, $Δ < 0$ matlab opposite sign.
$D = τ^{2} - 4Δ$ — $λ = \frac{τ \pm D}{2}$ mein square root ke andar discriminant. $D < 0$ ka matlab complex eigenvalues hain, yani rotation.

Yeh sab regions ek saath dekhne ke liye Trace–determinant plane dekho, $J$ kahan se aata hai yeh jaanne ke liye Jacobian matrix dekho, aur $e^{λ t}$ kyun aata hai yeh jaanne ke liye Eigenvalues and eigenvectors dekho.

Scenario matrix

Neeche har cell ek alag behaviour hai jo ek fixed point ka ho sakta hai. Last column is page ka example batata hai jo us cell ko hit karta hai.

#	$τ$	$Δ$	$D = τ^{2} - 4Δ$	Type	Stability	Example
A	koi bhi	$Δ < 0$	$D > 0$ hamesha	Saddle	unstable	Ex 1
B	$τ < 0$	$Δ > 0$	$D > 0$	Stable node	stable	Ex 2
C	$τ > 0$	$Δ > 0$	$D > 0$	Unstable node	unstable	Ex 3
D	$τ < 0$	$Δ > 0$	$D < 0$	Stable spiral	stable	Ex 4
E	$τ > 0$	$Δ > 0$	$D < 0$	Unstable spiral	unstable	Ex 4 (sign flip)
F	$τ = 0$	$Δ > 0$	$D < 0$	Centre	neutral	Ex 5
G	$τ \neq = 0$	$Δ > 0$	$D = 0$	Degenerate / star node	$τ$ ke sign se match karta hai	Ex 6
H	$τ < 0$	$Δ = 0$	$D = τ^{2} > 0$	Line of fixed points	non-isolated	Ex 7
I	nonlinear	—	—	Har equilibrium par alag type	mixed	Ex 8 (word problem)
J	exam twist	parameter $a$	—	Type badalta hai jab parameter threshold cross karta hai	bifurcation	Ex 9

Ab hum inhe order mein walk karte hain. Har example mein pehle tumhe forecast karna hai.

Cell A — the saddle

Worked example Example 1 · Cell A (

$Δ < 0$ ) System: $\overset{x}{˙} = x + 3 y, \overset{y}{˙} = 3 x + y$ . Equilibrium dhundo aur classify karo.

Forecast: do variables ek doosre mein strongly feed kar rahe hain (cross-coupling $3$ self-terms $1$ se bada hai). Aage padhne se pehle guess karo: sink, source, ya saddle?

Step 1 — equilibrium locate karo. Dono velocities zero set karo: $x + 3 y = 0$ aur $3 x + y = 0$ . Yeh step kyun? Equilibrium by definition wahan hota hai jahan $\overset{x}{˙} = \overset{y}{˙} = 0$ ; classification sirf aisi jagah pe sense deta hai. Pehle se, $x = - 3 y$ . Doosre mein substitute karo: $3 (- 3 y) + y = - 8 y = 0 \Rightarrow y = 0$ , isliye $x = 0$ . Toh equilibrium origin $(0, 0)$ hai.

Step 2 — Jacobian banao. $f = x + 3 y, g = 3 x + y$ , toh $J = (f_{x} g_{x} f_{y} g_{y}) = (1331) .$ Yeh step kyun? Fixed point ke paas nonlinear flow linear dikhta hai; $J$ woh local linear rulebook hai (yahan system already linear hai, toh $J$ har jagah constant hai).

Step 3 — $τ, Δ$ read karo. $τ = 1 + 1 = 2$ . Determinant ke liye, off-diagonal product subtract karo: $Δ = f_{x} g_{y} - f_{y} g_{x} = (1) (1) - (3) (3) = 1 - 9 = - 8 < 0$ . Yeh step kyun? Cross terms hi $Δ$ ko negative banate hain — exactly isliye "vibes" wale forecasts fail karte hain; tumhe actually $f_{y} g_{x}$ compute karna padega. Aur $Δ < 0$ eigenvalues ko opposite sign forced karta hai (unka product negative hai), jo saddle ki definition hai: ek direction grow karta hai, ek decay karta hai.

Verify: eigenvalues $λ = \frac{τ \pm τ ^{2} - 4Δ}{2} = \frac{2 \pm 4 + 32}{2} = \frac{2 \pm 6}{2} = {4, - 2}$ . Opposite signs ✔ → saddle, unstable. Mode $e^{4 t}$ blow up karta hai, toh almost har nudge trajectory ko fly off kar deta hai.

Figure yeh saddle dikhata hai: orange line $y = x$ (eigenvalue $λ = 4$ ) outward-fleeing direction hai, plum line $y = - x$ (eigenvalue $λ = - 2$ ) incoming direction hai, aur teal streamlines ek se andar sweep karte hain aur doosre se bahar — signature "riding-saddle" shape.

Figure — Stability of equilibria — stable, unstable, saddle, spiral, centre

Cells B & C — real nodes

Cells D & E — spirals (figure ke saath)

Worked example Example 4 · Cells D and E (spirals,

$D < 0$ ) Damped-oscillator-style system lo: $\overset{x}{˙} = - x + ω y, \overset{y}{˙} = - ω x - y$ with $ω = 2$ .

Forecast: antisymmetric $\pm ω$ terms flow ko rotate karte hain. Diagonal $- 1$ hai. Spiral andar ya bahar?

Step 1 — Jacobian. $J = (- 1 - 2 2 - 1)$ . Yeh step kyun? Skew part ( $+ 2, - 2$ ) hi rotation force karta hai; diagonal $- 1$ decay/growth hai. Inhe saath padhne se ek decaying rotation dikhta hai.

Step 2 — numbers. $τ = - 2$ , $Δ = (- 1) (- 1) - (2) (- 2) = 1 + 4 = 5$ , $D = 4 - 20 = - 16 < 0$ . Yeh step kyun? $D < 0$ ⇒ eigenvalues complex conjugates $α \pm i β$ hain; imaginary part circling speed hai, real part $α = τ /2$ spiral ki growth hai.

Step 3 — eigenvalues. $λ = \frac{- 2 \pm - 16}{2} = - 1 \pm 2 i$ . Real part $- 1 < 0$ . Yeh step kyun? $Re λ < 0$ ⇒ amplitude $e^{- t}$ shrink karta hai jabki phase $e^{2 i t}$ rotate karta hai → stable spiral (Cell D).

Sign flip → Cell E. Diagonals $+ 1$ mein change karo: $\overset{x}{˙} = x + 2 y, \overset{y}{˙} = - 2 x + y$ . Tab $τ = 2$ , $Δ = 5$ , $D = - 16$ , $λ = 1 \pm 2 i$ , $Re = + 1 > 0$ → unstable spiral. Same rotation, outward growth.

Verify: stable case product $(- 1)^{2} + 2^{2} = 5 = Δ$ ✔ ( $α \pm i β$ ke liye, product $= α^{2} + β^{2}$ ), sum $= 2 α = - 2 = τ$ ✔.

Figure dono ko side by side rakhta hai: left, stable spiral ( $Re λ = - 1$ ) origin ki taraf andar wind karta hai; right, unstable spiral ( $Re λ = + 1$ ) bahar wind karta hai. Rotation direction dono mein identical hai — sirf $τ$ ka sign (aur isliye $Re λ$ ka) decide karta hai andar-vs-bahar.

Cell F — the centre (borderline)

Worked example Example 5 · Cell F (

$τ = 0, Δ > 0$ ) $\overset{x}{˙} = 3 y, \overset{y}{˙} = - 3 x$ .

Forecast: pure rotation, koi decay term nahi kahin bhi. Kya yeh spiral karta hai ya sirf loop?

Step 1 — Jacobian. $J = (0 - 3 30)$ (origin par equilibrium, kyunki $\overset{x}{˙} = \overset{y}{˙} = 0$ sirf tab jab $x = y = 0$ ). Yeh step kyun? Purely skew-symmetric $J$ ka zero trace hota hai — koi net inward ya outward pull nahi, sirf rotation.

Step 2 — numbers. $τ = 0$ , $Δ = 0 - (3) (- 3) = 9 > 0$ , $D = 0 - 36 = - 36 < 0$ . Yeh step kyun? $τ = 0$ with $Δ > 0$ stable aur unstable spirals ke beech ka knife-edge hai.

Step 3 — eigenvalues. Formula mein plug karo: $λ = \frac{τ \pm D}{2} = \frac{0 \pm - 36}{2} = \frac{\pm 6 i}{2} = \pm 3 i$ . Purely imaginary. Yeh step kyun? Koi real part nahi ⇒ amplitude na grow karta hai na shrink ⇒ closed orbits = centre, neutrally stable.

Verify: $λ_{1} λ_{2} = (3 i) (- 3 i) = 9 = Δ$ ✔, sum $= 0 = τ$ ✔. Centre. Caution: linearisation sirf linear system ke liye centre guarantee karta hai; nonlinear systems ke liye $τ = 0$ point weak spiral ho sakta hai (dekho Linearisation and Hartman–Grobman theorem).

Cell G — degenerate / star node ( $D = 0$ )

Worked example Example 6 · Cell G (repeated eigenvalue)

$\overset{x}{˙} = - 2 x, \overset{y}{˙} = - 2 y$ .

Forecast: dono equations same rate par decay karte hain — trajectories ka shape kya hoga?

Step 1 — Jacobian. $J = (- 2 0 0 - 2) = - 2 I$ . Yeh step kyun? Jab $J$ identity ka scalar multiple ho, toh har direction ek eigenvector hota hai.

Step 2 — numbers. $τ = - 4$ , $Δ = 4$ , $D = 16 - 16 = 0$ . Yeh step kyun? $D = 0$ nodes aur spirals ke beech ki boundary hai — ek repeated eigenvalue.

Step 3 — eigenvalue. $λ = \frac{- 4 \pm 0}{2} = - 2$ (double). Yeh step kyun? Dono modes $e^{- 2 t}$ rate par decay karte hain; kyunki har direction ek eigenvector hai, trajectories origin mein straight rays hain → star node (ek special stable node).

Verify: $λ^{2} = 4 = Δ$ ✔, $2 λ = - 4 = τ$ ✔. Stable star node. (Agar off-diagonal nonzero hota, jaise $J = (- 2 0 1 - 2)$ , phir bhi $D = 0$ lekin sirf ek eigenvector → ek improper/degenerate node, phir bhi stable kyunki $λ = - 2 < 0$ .)

Cell H — fixed points ki poori line ( $Δ = 0$ )

Worked example Example 7 · Cell H (non-isolated equilibria)

$\overset{x}{˙} = - x + y, \overset{y}{˙} = x - y$ .

Forecast: dono right-hand sides ek doosre ke exact negatives hain. Determinant ka kya hoga jab $J$ ki rows proportional hon?

Step 1 — equilibria locate karo. $- x + y = 0$ aur $x - y = 0$ set karo. Dono same baat kehte hain, $y = x$ . Toh har point line $y = x$ par ek equilibrium hai — sirf ek isolated point nahi. Yeh step kyun? Jab do zero-velocity conditions coincide karein, fixed points ek poori line banate hain; classification mein yeh acknowledge karna zaruri hai ki yeh ek non-isolated case hai $J$ ko touch karne se pehle.

Step 2 — Jacobian aur numbers. $J = (- 1 1 1 - 1)$ , toh $τ = - 1 - 1 = - 2$ aur $Δ = (- 1) (- 1) - (1) (1) = 1 - 1 = 0$ . Yeh step kyun? Yeh exactly Cell H mein land karta hai: $τ < 0$ (ek decaying direction hai) lekin $Δ = 0$ (ek eigenvalue exactly zero hai, rest points ki line ko reflect karta hai).

Step 3 — eigenvalues. $λ = \frac{- 2 \pm 4 - 0}{2} = \frac{- 2 \pm 2}{2} = {0, - 2}$ . Yeh step kyun? $0$ eigenvalue ka matlab line $y = x$ ke along koi motion nahi; $- 2$ eigenvalue har off-line point ko us line par pull karta hai. Equilibria non-isolated hain aur sirf marginally stable.

Verify: product $0 \cdot (- 2) = 0 = Δ$ ✔, sum $0 + (- 2) = - 2 = τ$ ✔.

Cell I — real-world word problem (nonlinear, mixed types)

Worked example Example 8 · Cell I — competing species

Do species $x, y \geq 0$ follow karte hain $\overset{x}{˙} = x (3 - x - 2 y), \overset{y}{˙} = y (2 - x - y)$ . Coexistence equilibrium classify karo.

Forecast: competition aksar matlab hota hai ek species jeet jaati hai (unke beech saddle). Guess karo: stable coexistence ya saddle?

Step 1 — interior equilibrium dhundo. Chahiye $3 - x - 2 y = 0$ aur $2 - x - y = 0$ . Subtract karo: $(3 - x - 2 y) - (2 - x - y) = 1 - y = 0 \Rightarrow y = 1$ , phir $x = 2 - y = 1$ . Toh $(x^{*}, y^{*}) = (1, 1)$ . Yeh step kyun? Hum woh fixed point chahte hain jahan dono species survive karein; ek word problem mein kai equilibria ho sakte hain — jo manga gaya hai woh chuno.

Step 2 — Jacobian. $f = x (3 - x - 2 y) = 3 x - x^{2} - 2 x y$ , $g = y (2 - x - y) = 2 y - x y - y^{2}$ ke saath: $J = (3 - 2 x - 2 y - y - 2 x 2 - x - 2 y) .$ $(1, 1)$ par: $J = (3 - 2 - 2 - 1 - 2 2 - 1 - 2) = (- 1 - 1 - 2 - 1)$ . Yeh step kyun? $J$ ko specific fixed point par evaluate karo — parent warn karta hai ki har equilibrium alag ho sakta hai.

Step 3 — numbers. $τ = - 1 - 1 = - 2$ , $Δ = (- 1) (- 1) - (- 2) (- 1) = 1 - 2 = - 1 < 0$ . Yeh step kyun? $Δ < 0$ ⇒ opposite-sign eigenvalues ⇒ saddle: coexistence unstable hai — ek tiny imbalance ek species ko dominate kar deta hai.

Verify: $λ = \frac{- 2 \pm 4 + 4}{2} = - 1 \pm 2 \approx {0.414, - 2.414}$ ; product $(- 1)^{2} - 2 = - 1 = Δ$ ✔. Saddle — ecological intuition se match karta hai ki strong mutual competition rarely stable coexistence allow karta hai.

Cell J — exam twist: a bifurcation

Worked example Example 9 · Cell J — parameter crossing a threshold

$\overset{x}{˙} = y, \overset{y}{˙} = - x + a y$ , ek tunable damping/forcing parameter $a$ ke saath. Origin ko sab $a$ ke liye classify karo.

Forecast: jab $a$ negative se positive hota hai, origin ka character zarur badlega. Kahan switch hoga, aur kis cheez mein?

Step 1 — Jacobian. $J = (0 - 1 1 a)$ (origin hamesha ek equilibrium hai kyunki $\overset{x}{˙} = \overset{y}{˙} = 0$ wahan).

Step 2 — $a$ ke functions ke roop mein numbers. $τ = a$ , $Δ = (0) (a) - (1) (- 1) = 1 > 0$ hamesha, $D = a^{2} - 4$ . Yeh step kyun? $Δ > 0$ fixed matlab hum kabhi saddle nahi paenge; sirf $τ = a$ aur $D = a^{2} - 4$ move karte hain — toh type entirely $a$ se steer hoti hai.

Step 3 — har case sweep karo.

$a < - 2$ : $τ < 0, D > 0$ → stable node.
$a = - 2$ : $D = 0$ → degenerate stable node (boundary).
$- 2 < a < 0$ : $τ < 0, D < 0$ → stable spiral.
$a = 0$ : $τ = 0, Δ > 0$ → centre (bifurcation point!).
$0 < a < 2$ : $τ > 0, D < 0$ → unstable spiral.
$a = 2$ : $D = 0$ → degenerate unstable node.
$a > 2$ : $τ > 0, D > 0$ → unstable node. Yeh step kyun? $a = 0$ cross karna $Re λ$ ko negative se positive flip karta hai — stable spiral se unstable spiral mein ek Hopf-type change centre ke through.

Verify ( $a = 1$ par): $λ = \frac{1 \pm 1 - 4}{2} = \frac{1}{2} \pm \frac{3}{2} i$ , toh $Re = \frac{1}{2} > 0$ → unstable spiral ✔; product $= \frac{1}{4} + \frac{3}{4} = 1 = Δ$ ✔. Verify ( $a = - 3$ par): $λ = \frac{- 3 \pm 9 - 4}{2} = \frac{- 3 \pm 5}{2} \approx {- 0.382, - 2.618}$ — do real negative roots → stable node ✔; product $\approx (- 0.382) (- 2.618) \approx 1 = Δ$ ✔. Conclusion: single parameter $a$ origin ko paanch alag types se walk karata hai. Figure dekho: jaise $a$ right slide karta hai, $τ = a$ zero ke through climb karta hai (stability lost) jabki $D = a^{2} - 4$ $a = - 2$ aur $a = 2$ ke beech zero se neeche dip karta hai (spiral region). $a = - 2, 0, 2$ par dots teen boundaries mark karte hain jahan type badalta hai.

Consolidation

Recall Har example kaunse cell mein hai?

Ex 1 saddle (A) ::: $Δ < 0$ , opposite-sign eigenvalues, hamesha unstable. Ex 2 vs Ex 3 ::: stable node (B) aur uska time-reversed unstable node (C). Ex 4 ::: stable spiral (D) aur, sign flip ke baad, unstable spiral (E) — $D < 0$ . Ex 5 ::: centre (F) — $τ = 0, Δ > 0$ , purely imaginary eigenvalues. Ex 6 ::: star node (G) — repeated eigenvalue, $D = 0$ . Ex 7 ::: line of fixed points (H) — $Δ = 0$ , ek zero eigenvalue. Ex 8 ::: nonlinear ecology (I) — coexistence par saddle; har equilibrium ke liye $J$ evaluate karo. Ex 9 ::: bifurcation (J) — type badalta hai jab $a$ , $- 2, 0, 2$ cross karta hai.

In sab cells ki poori geography ke liye Trace–determinant plane dekho; physical archetype (Ex 4/5/9) Damped harmonic oscillator hai; aur Phase portraits dikhate hain ki har cell flow ke roop mein kaisa dikhta hai. Ek stability proof ke liye jo $τ = 0$ survive kare, Lyapunov stability use karo.