4.6.23
Intuition Bada picture (WHY ye matter karta hai)Ek differential equation har point par velocity batata hai. Ek equilibrium woh point hai jahan velocity zero hoti hai — system wahan hamesha reh sakta hai. Lekin asli sawaal yeh hai: agar thoda sa dhakka do, toh kya woh wapas aata hai, door bhagta hai, ya chakkar lagata hai? Yahi stability hai. Hum iska jawab equilibrium ke paas linearise karke aur local matrix ki eigenvalues padhkar dete hain. Eigenvalues system ki secret "growth/rotation rates" hain — unka sign aur complexity poori kahani bata deta hai.
Definition Equilibrium (fixed point)Ek 2D autonomous system ke liye
x ˙ = f ( x , y ) , y ˙ = g ( x , y ) , ( x ∗ , y ∗ ) equilibrium (ya fixed point ) hai agar f ( x ∗ , y ∗ ) = 0 g ( x ∗ , y ∗ ) = 0
Definition Stability ke flavours
Stable (attracting / sink): nearby trajectories t → ∞ Unstable (repelling / source): nearby trajectories door bhaag jaati hain.Saddle : ek direction mein attract karta hai, doosri mein repel karta hai (hamesha unstable).Spiral (focus): trajectories paas aate hue rotate karti hain (stable spiral) ya bahar jaate hue rotate karti hain (unstable spiral).Centre : trajectories iske aas-paas closed loops banati hain — na paas aati hain, na jaati hain.
Intuition WHAT hum karte hain( x ∗ , y ∗ ) f , g tangent plane se approximate karte hain — pehle order tak Taylor expansion.
Maano u = x − x ∗ v = y − y ∗
f ( x , y ) = = 0 f ( x ∗ , y ∗ ) + f x u + f y v + (higher order) , g ( x , y ) = = 0 g ( x ∗ , y ∗ ) + g x u + g y v + … Kyunki u ˙ = x ˙ v ˙ = y ˙ linearised system milta hai:
( u ˙ v ˙ ) = J ( u v ) , J = ( f x g x f y g y ) ( x ∗ , y ∗ ) .
J Jacobian matrix hai jo equilibrium par evaluate ki jaati hai. Yeh flow ki local "linear rulebook" hai.
HOW hum ise padhte hain: w ˙ = J w e λ t e λ eigenvalue hai aur e
Re ( λ ) < 0 ⇒ decay karta hai (stabilising).Re ( λ ) > 0 ⇒ grow karta hai (destabilising).Im ( λ ) = 0 ⇒ rotation (spiralling / circling).
Ek 2 × 2 J det ( J − λ I ) = 0
λ 2 − ( τ = tr J f x + g y ) λ + Δ = d e t J ( f x g y − f y g x ) = 0. Yeh kyun powerful hai: tum equilibrium ko sirf do numbers τ Δ
Intuition WHY ki ek-line memory
Δ Δ > 0 Δ < 0 τ τ < 0 τ > 0 D
Worked example Example 1 — Ek clean saddleSystem: x ˙ = x , y ˙ = − 2 y J = ( 1 0 0 − 2 ) λ 1 = 1 , λ 2 = − 2 Yeh step kyun? Diagonal J Δ = ( 1 ) ( − 2 ) = − 2 < 0 saddle . x y
Worked example Example 2 — Damped oscillator (stable spiral)x ¨ + 2 β x ˙ + ω 0 2 x = 0 0 < β < ω 0 y = x ˙ x ˙ = y , y ˙ = − ω 0 2 x − 2 β y . J = ( 0 − ω 0 2 1 − 2 β ) Yeh step kyun? Yeh standard "phase-space" rewrite hai: position + velocity.
τ = − 2 β < 0 Δ = ω 0 2 > 0 D = 4 β 2 − 4 ω 0 2 < 0 β < ω 0 Re < 0 stable spiral : oscillate karta hai aur khatam hota jaata hai. ✔ real damped springs se match karta hai.
Worked example Example 3 — Undamped oscillator (centre)β = 0 x ˙ = y , y ˙ = − ω 0 2 x τ = 0 , Δ = ω 0 2 > 0 λ = ± i ω 0 centre . Energy conserved, closed orbits — perpetual oscillation.
Worked example Example 4 — Ek nonlinear equilibriumx ˙ = y , y ˙ = − sin x ( 0 , 0 ) J = ( 0 − cos x 1 0 ) ( 0 , 0 ) ( 0 − 1 1 0 ) τ = 0 , Δ = 1 centre (neeche latak raha rest).
( π , 0 ) − cos π = 1 J = ( 0 1 1 0 ) Δ = − 1 < 0 saddle (seedha balance — unstable, jaise expect tha!).
Yeh kyun matter karta hai: ek hi system ke alag equilibria alag types ke hote hain — hamesha J har fixed point par alag evaluate karo.
Common mistake "Negative determinant matlab stable hai kyunki negative calming hota hai."Kyun sahi lagta hai: hum "negative" ko decay se associate karte hain. Fix: Δ < 0 opposite signs hain → ek grow karta hai → saddle = unstable . Stability τ < 0 aur Δ > 0
Common mistake "Centre stable hai — orbits blow up nahi hoti."Kyun sahi lagta hai: trajectories hamesha bounded rehti hain. Fix: Centre sirf neutrally (marginally) stable hota hai — perturbations na grow hote hain na decay, woh bas kareeb ek nayi orbit par rehte hain. Yeh equilibrium par wapas nahi aata, isliye yeh asymptotically stable nahi hai.
Common mistake "Linearisation hamesha sach batata hai."Kyun sahi lagta hai: chhote nudges ke liye Taylor approximation excellent hota hai. Fix: jab τ = 0 centre ) toh nonlinear terms decide karte hain — asli behaviour weak spiral ho sakta hai. Linearisation sirf hyperbolic points ke liye reliable hai (Re ( λ ) = 0
Recall Check karne se pehle predict karox ˙ = y , y ˙ = − x − 3 y Forecast: damping present hai ⇒ shayad stable node/spiral.
Verify: J = ( 0 − 1 1 − 3 ) τ = − 3 < 0 Δ = 1 > 0 D = 9 − 4 = 5 > 0 stable node (overdamped — oscillate karne ke liye zyada friction). ✔
x ˙ = f , y ˙ = g Wahan f = g = 0
Local stability kaun sa matrix govern karta hai? Jacobian J
2×2 system ke liye eigenvalue formula kya hai? λ = 2 τ ± τ 2 − 4Δ jahan
τ = tr J , Δ = det J .
Δ < 0 Saddle (opposite sign eigenvalues) — hamesha unstable.
Stable node/spiral ke liye condition kya hai? τ < 0 Δ > 0
Spiral aur node mein kya difference hai? Discriminant D = τ 2 − 4Δ < 0
Centre kab milta hai? τ = 0 , Δ > 0 ± iω
Kya centre asymptotically stable hai? Nahi, sirf neutrally/marginally stable hai; perturbations bane rehte hain.
λ 1 + λ 2 λ 1 λ 2 Krama se τ Δ
Linearisation kab fail ho sakta hai? Non-hyperbolic points par (kuch Re λ = 0
Recall Feynman: 12-saal ke bachche ko samjhaoEk marble ek landscape mein socho. Equilibrium ek flat spot hai jahan woh rest kar sakta hai. Agar woh bowl ke bottom par hai, dhakka do aur woh wapas roll karta hai — stable . Pahad ki choti par, dhakka do aur woh door roll karta hai — unstable . Mountain pass (saddle) par, ridge ke saath dhakka do toh wapas aata hai lekin valley ke neeche dhakka do toh door — saddle . Agar bowl slippery aur spinning hai, marble spiral karta hua neeche aata hai — spiral . Ek bilkul frictionless circular track par woh bas hamesha chakkar lagata rehta hai — centre . "Jacobian" bas resting spot par har direction mein ground kitni steep hai yeh measure karne ka tarika hai.
Mnemonic Decision tree yaad karo"Delta decides the sign-game, Tau decides the fate, D decides the dance."
Δ < 0 saddle (sign-game hari gayi).warna τ < 0 > 0 fate .
D < 0 spiral/dance karta hai; D > 0 τ = 0 centre .
Jacobian matrix — linearisation ka engine.Eigenvalues and eigenvectors — sign aur complexity = stability.Phase portraits — fixed points ke paas trajectories visualise karna.Linearisation and Hartman–Grobman theorem — kab linear ≈ nonlinear.Lyapunov stability — eigenvalues ke bina stability prove karna.Damped harmonic oscillator — physical spiral/centre example.Trace–determinant plane — master classification chart.
Linearise near equilibrium
lambda squared minus tau lambda plus Delta
