Linearization of nonlinear systems

4.6.24 · Maths › Ordinary Differential Equations

HOW hum linearize karte hain — scratch se derive

Hum jaanna chahte hain ki ek choti disturbance kaise behave karti hai. Let $u = x - x^{*}, v = y - y^{*}$ equilibrium se tiny displacements hain. Tab $\overset{u}{˙} = \overset{x}{˙} = f (x, y)$ .

Taylor expand karo $f$ ko $(x^{*}, y^{*})$ ke baare mein (yahi poora trick hai): $f (x, y) = = 0 f (x^{*}, y^{*}) + f_{x} u + f_{y} v + higher order \frac{1}{2} f_{xx} u^{2} + \dots$

Pehla term 0 kyun hai? Kyunki $(x^{*}, y^{*})$ ek equilibrium hai — definition se $f (x^{*}, y^{*}) = 0$ .
Quadratic terms kyun drop karte hain? Equilibrium ke paas $u, v$ chote hote hain, toh $u^{2}, uv, v^{2}$ tiny-squared hain — linear terms ke comparison mein negligible. Yeh 80/20 move hai: woh part rakho jo dominate karta hai.

$g$ ke liye bhi yahi karo:

Figure — Linearization of nonlinear systems

Equilibrium ko classify karna

$det (J - λ I) = 0$ solve karo eigenvalues $λ_{1, 2}$ ke liye. Trace $τ = f_{x} + g_{y}$ aur determinant $Δ = f_{x} g_{y} - f_{y} g_{x}$ ke saath: $λ = \frac{τ \pm τ ^{2} - 4Δ}{2}$

Eigenvalues	Type	Stability
real, dono $< 0$	stable node	stable
real, dono $> 0$	unstable node	unstable
real, opposite signs	saddle	unstable
complex, $Re < 0$	stable spiral	stable
complex, $Re > 0$	unstable spiral	unstable
pure imaginary	center	linearization inconclusive

Worked Example 1 — ek simple nonlinear pair

System: $\overset{x}{˙} = x - x^{2} - x y, \overset{y}{˙} = - y + x y$ (ek Lotka–Volterra-style model).

Step 1 — equilibria dhundo. Dono ko 0 set karo.

Kyun? Equilibria woh jagah hain jahan analysis hoti hai. $x (1 - x - y) = 0$ aur $y (- 1 + x) = 0$ . Doosri equation se: $y = 0$ ya $x = 1$ .
$y = 0$ : pehli equation se $x (1 - x) = 0 \Rightarrow (0, 0), (1, 0)$ .
$x = 1$ : pehli equation se $1 - 1 - y = 0 \Rightarrow y = 0$ (yeh pehle se hai). Toh equilibria: $(0, 0)$ aur $(1, 0)$ .

Step 2 — Jacobian (general). $J = (1 - 2 x - y y - x - 1 + x)$ Yeh entries kyun? $f_{x} = \partial_{x} (x - x^{2} - x y) = 1 - 2 x - y$ , etc.

Step 3 — $(0, 0)$ par evaluate karo. $J = (10 0 - 1) \Rightarrow λ = 1, - 1.$ Opposite signs → saddle (unstable). Kyun? Ek mode grow karta hai ( $e^{t}$ ), ek decay karta hai.

Step 4 — $(1, 0)$ par evaluate karo. $J = (- 1 0 - 1 0) \Rightarrow λ = - 1, 0.$ Ek zero eigenvalue → non-hyperbolic, linearization inconclusive (neeche steel-man flag).

Worked Example 2 — pendulum

$\ddot{θ} + sin θ = 0$ . $x = θ, y = \dot{θ}$ ke saath system likho: $\overset{x}{˙} = y, \overset{y}{˙} = - sin x .$

Step 1 — equilibria. $y = 0$ aur $sin x = 0 \Rightarrow x = nπ$ . Do physical cases: $(0, 0)$ = neeche latakna, $(π, 0)$ = seedha balanced.

Step 2 — Jacobian. $J = (0 - cos x 10)$ $- cos x$ kyun? $\partial_{x} (- sin x) = - cos x$ .

Step 3 — bottom $(0, 0)$ : $J = (0 - 1 10)$ , $λ = \pm i$ . Pure imaginary → center (predicted oscillation). Physically sahi! Lekin formally linearization inconclusive hai — energy conservation confirm karta hai ki yeh sach mein center hai.

Step 4 — top $(π, 0)$ : $cos π = - 1$ , toh $J = (0110)$ , $λ = \pm 1$ . Saddle → unstable. Bilkul theek hai: pendulum ko seedha balance karna unstable hota hai. ✔

Common mistakes (Steel-man + fix)

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho ek marble ek ulti-seedhi landscape mein hai. Poori landscape complicated hai, lekin ek katora ke bottom par (ya pahad ki choti par), tumhare aas-paas ki zameen ek simple flat slope jaisi dikhti hai. Linearization yeh hai ki resting spot ke itna paas zoom karo ki ulti-seedhi zameen ek simple ramp jaisi lage. Agar ramp itni jhuki ho ki marble wapas aata hai → yeh stable jagah hai. Agar door jaata hai → unstable. Jacobian bas ek number hai jo measure karta hai "yahaan zameen kis taraf jhuki hai?"

Active-recall flashcards

\overset{x}{˙} = f, \overset{y}{˙} = g

ka equilibrium point kya define karta hai?

Woh point jahan

f = g = 0

(kuch nahi hilta).

Linearize karte waqt quadratic terms kyun drop kar sakte hain?

Equilibrium ke paas displacements chote hote hain, toh

u^{2}, uv, v^{2}

linear terms ke comparison mein negligible hain.

Konsa matrix nonlinear system ko linearize karta hai?

Jacobian

(f_{x} g_{x} f_{y} g_{y})

equilibrium par evaluate kiya gaya.

Eigenvalues real aur opposite signs ke saath → konsa type?

Saddle (unstable).

Eigenvalues complex aur negative real part ke saath → ?

Stable spiral.

Hartman–Grobman kya require aur guarantee karta hai?

Require karta hai ki koi eigenvalue zero real part wala na ho (hyperbolic); guarantee karta hai ki nonlinear flow ≈ linear flow topologically point ke paas.

Linearization inconclusive kab hota hai?

Jab kisi eigenvalue ka zero real part ho (centers, pure imaginary, zero eigenvalue).

Pendulum mein upright equilibrium

(π, 0)

kaisa type hai?

Saddle (unstable), eigenvalues

\pm 1

J

ka Trace aur determinant eigenvalues se kaise relate karta hai?

λ = \frac{τ \pm τ ^{2} - 4Δ}{2}

jahan

τ =

trace,

Δ =

det.

Connections

Phase Plane Analysis — jahan trajectories rehti hain.
Eigenvalues and Eigenvectors — classification ka engine.
Stability of Equilibria — node/saddle/spiral/center catalogue.
Taylor Series — woh approximation jo hum truncate karte hain.
Lyapunov Functions — non-hyperbolic cases ke liye rescue tool.
Hartman-Grobman Theorem — legal justification.
Lotka-Volterra Predator Prey — classic application.

Hum kya study kar rahe hain