Intuition The big picture (WHY this matters)A differential equation tells you the velocity at every point. An equilibrium is a point where the velocity is zero — the system can sit there forever. But the real question is: if you nudge it a tiny bit, does it come back, fly away, or circle around? That's stability . We answer it by linearising near the equilibrium and reading off the eigenvalues of the local matrix. Eigenvalues are the system's secret "growth/rotation rates" — their sign and complexity tell the whole story.
Definition Equilibrium (fixed point)For a 2D autonomous system
x ˙ = f ( x , y ) , y ˙ = g ( x , y ) , \dot{x}=f(x,y),\qquad \dot{y}=g(x,y), x ˙ = f ( x , y ) , y ˙ = g ( x , y ) , ( x ∗ , y ∗ ) (x^*,y^*) ( x ∗ , y ∗ ) equilibrium (or fixed point ) if f ( x ∗ , y ∗ ) = 0 f(x^*,y^*)=0 f ( x ∗ , y ∗ ) = 0 g ( x ∗ , y ∗ ) = 0 g(x^*,y^*)=0 g ( x ∗ , y ∗ ) = 0
Definition Stability flavours
Stable (attracting / sink): nearby trajectories converge to it as t → ∞ t\to\infty t → ∞ Unstable (repelling / source): nearby trajectories run away.Saddle : attracts along one direction, repels along another (always unstable).Spiral (focus): trajectories rotate while approaching (stable spiral) or rotating outward (unstable spiral).Centre : trajectories form closed loops around it — neither approach nor leave.
Near ( x ∗ , y ∗ ) (x^*,y^*) ( x ∗ , y ∗ ) f , g f,g f , g tangent plane — a Taylor expansion to first order.
Let u = x − x ∗ u=x-x^* u = x − x ∗ v = y − y ∗ v=y-y^* v = y − y ∗
f ( x , y ) = f ( x ∗ , y ∗ ) ⏟ = 0 + f x u + f y v + (higher order) , f(x,y)=\underbrace{f(x^*,y^*)}_{=0}+f_x\,u+f_y\,v+\text{(higher order)}, f ( x , y ) = = 0 f ( x ∗ , y ∗ ) + f x u + f y v + (higher order) , g ( x , y ) = g ( x ∗ , y ∗ ) ⏟ = 0 + g x u + g y v + … g(x,y)=\underbrace{g(x^*,y^*)}_{=0}+g_x\,u+g_y\,v+\dots g ( x , y ) = = 0 g ( x ∗ , y ∗ ) + g x u + g y v + … Since u ˙ = x ˙ \dot u=\dot x u ˙ = x ˙ v ˙ = y ˙ \dot v=\dot y v ˙ = y ˙ linearised system :
( u ˙ v ˙ ) = J ( u v ) , J = ( f x f y g x g y ) ( x ∗ , y ∗ ) . \begin{pmatrix}\dot u\\ \dot v\end{pmatrix}
= J\begin{pmatrix}u\\ v\end{pmatrix},
\qquad
J=\begin{pmatrix} f_x & f_y\\ g_x & g_y\end{pmatrix}_{(x^*,y^*)}. ( u ˙ v ˙ ) = J ( u v ) , J = ( f x g x f y g y ) ( x ∗ , y ∗ ) .
J J J Jacobian matrix of partial derivatives evaluated at the equilibrium . It is the local "linear rulebook" of the flow.
HOW we read it: solutions of w ˙ = J w \dot{\mathbf{w}}=J\mathbf{w} w ˙ = J w e λ t e e^{\lambda t}\mathbf{e} e λ t e λ \lambda λ eigenvalue and e \mathbf{e} e
R e ( λ ) < 0 ⇒ \mathrm{Re}(\lambda)<0 \Rightarrow Re ( λ ) < 0 ⇒ decays (stabilising).R e ( λ ) > 0 ⇒ \mathrm{Re}(\lambda)>0 \Rightarrow Re ( λ ) > 0 ⇒ grows (destabilising).I m ( λ ) ≠ 0 ⇒ \mathrm{Im}(\lambda)\neq 0 \Rightarrow Im ( λ ) = 0 ⇒ rotation (spiralling / circling).
For a 2 × 2 2\times2 2 × 2 J J J det ( J − λ I ) = 0 \det(J-\lambda I)=0 det ( J − λ I ) = 0
λ 2 − ( f x + g y ⏟ τ = t r J ) λ + ( f x g y − f y g x ) ⏟ Δ = det J = 0. \lambda^2-(\underbrace{f_x+g_y}_{\tau=\mathrm{tr}\,J})\,\lambda+\underbrace{(f_xg_y-f_yg_x)}_{\Delta=\det J}=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. Why this is powerful: you classify the equilibrium using just two numbers τ \tau τ Δ \Delta Δ
Intuition One-line memory of WHY
Sign of Δ \Delta Δ Δ > 0 \Delta>0 Δ > 0 Δ < 0 \Delta<0 Δ < 0
Sign of τ \tau τ τ < 0 \tau<0 τ < 0 τ > 0 \tau>0 τ > 0
Sign of D D D
Worked example Example 1 — A clean saddleSystem: x ˙ = x , y ˙ = − 2 y \dot x = x,\ \dot y=-2y x ˙ = x , y ˙ = − 2 y J = ( 1 0 0 − 2 ) J=\begin{pmatrix}1&0\\0&-2\end{pmatrix} J = ( 1 0 0 − 2 ) λ 1 = 1 , λ 2 = − 2 \lambda_1=1,\ \lambda_2=-2 λ 1 = 1 , λ 2 = − 2 Why this step? Diagonal J J J Δ = ( 1 ) ( − 2 ) = − 2 < 0 \Delta=(1)(-2)=-2<0 Δ = ( 1 ) ( − 2 ) = − 2 < 0 saddle . Grows along x x x y y y
Worked example Example 2 — Damped oscillator (stable spiral)x ¨ + 2 β x ˙ + ω 0 2 x = 0 \ddot x + 2\beta\dot x+\omega_0^2 x=0 x ¨ + 2 β x ˙ + ω 0 2 x = 0 0 < β < ω 0 0<\beta<\omega_0 0 < β < ω 0 y = x ˙ y=\dot x y = x ˙ x ˙ = y , y ˙ = − ω 0 2 x − 2 β y . \dot x=y,\ \dot y=-\omega_0^2 x-2\beta y. x ˙ = y , y ˙ = − ω 0 2 x − 2 β y . J = ( 0 1 − ω 0 2 − 2 β ) J=\begin{pmatrix}0&1\\-\omega_0^2&-2\beta\end{pmatrix} J = ( 0 − ω 0 2 1 − 2 β ) Why this step? This is the standard "phase-space" rewrite: position + velocity.
τ = − 2 β < 0 \tau=-2\beta<0 τ = − 2 β < 0 Δ = ω 0 2 > 0 \Delta=\omega_0^2>0 Δ = ω 0 2 > 0 D = 4 β 2 − 4 ω 0 2 < 0 D=4\beta^2-4\omega_0^2<0 D = 4 β 2 − 4 ω 0 2 < 0 β < ω 0 \beta<\omega_0 β < ω 0 R e < 0 \mathrm{Re}<0 Re < 0 stable spiral : it oscillates while dying out. ✔ matches real damped springs.
Worked example Example 3 — Undamped oscillator (centre)Set β = 0 \beta=0 β = 0 x ˙ = y , y ˙ = − ω 0 2 x \dot x=y,\ \dot y=-\omega_0^2 x x ˙ = y , y ˙ = − ω 0 2 x τ = 0 , Δ = ω 0 2 > 0 \tau=0,\ \Delta=\omega_0^2>0 τ = 0 , Δ = ω 0 2 > 0 λ = ± i ω 0 \lambda=\pm i\omega_0 λ = ± i ω 0 centre . Energy conserved, closed orbits — perpetual oscillation.
Worked example Example 4 — A nonlinear equilibriumx ˙ = y , y ˙ = − sin x \dot x=y,\ \dot y=-\sin x x ˙ = y , y ˙ = − sin x ( 0 , 0 ) (0,0) ( 0 , 0 ) J = ( 0 1 − cos x 0 ) J=\begin{pmatrix}0&1\\-\cos x&0\end{pmatrix} J = ( 0 − cos x 1 0 ) ( 0 , 0 ) (0,0) ( 0 , 0 ) ( 0 1 − 1 0 ) \begin{pmatrix}0&1\\-1&0\end{pmatrix} ( 0 − 1 1 0 ) τ = 0 , Δ = 1 \tau=0,\Delta=1 τ = 0 , Δ = 1 centre (hanging-down rest).
At ( π , 0 ) (\pi,0) ( π , 0 ) − cos π = 1 -\cos\pi=1 − cos π = 1 J = ( 0 1 1 0 ) J=\begin{pmatrix}0&1\\1&0\end{pmatrix} J = ( 0 1 1 0 ) Δ = − 1 < 0 \Delta=-1<0 Δ = − 1 < 0 saddle (balanced upright — unstable, as expected!).
Why this matters: the same system has different equilibria of different types — always evaluate J J J at each fixed point separately .
Common mistake "Negative determinant means stable because negative is calming."Why it feels right: we associate "negative" with decay. The fix: Δ < 0 \Delta<0 Δ < 0 opposite signs → one grows → saddle = unstable . Stability is governed by τ < 0 \tau<0 τ < 0 and Δ > 0 \Delta>0 Δ > 0
Common mistake "A centre is stable — orbits don't blow up."Why it feels right: trajectories stay bounded forever. The fix: A centre is only neutrally (marginally) stable — perturbations neither grow nor decay, they just stay nearby on a new orbit. It does NOT return to the equilibrium, so it is not asymptotically stable .
Common mistake "Linearisation always tells the truth."Why it feels right: Taylor approximation is excellent for small nudges. The fix: when τ = 0 \tau=0 τ = 0 centre ) the nonlinear terms decide — the true behaviour might be a weak spiral. Linearisation is reliable only for hyperbolic points (R e ( λ ) ≠ 0 \mathrm{Re}(\lambda)\neq0 Re ( λ ) = 0
Recall Predict before checkingFor x ˙ = y , y ˙ = − x − 3 y \dot x=y,\ \dot y=-x-3y x ˙ = y , y ˙ = − x − 3 y Forecast: damping present ⇒ probably stable node/spiral.
Verify: J = ( 0 1 − 1 − 3 ) J=\begin{pmatrix}0&1\\-1&-3\end{pmatrix} J = ( 0 − 1 1 − 3 ) τ = − 3 < 0 \tau=-3<0 τ = − 3 < 0 Δ = 1 > 0 \Delta=1>0 Δ = 1 > 0 D = 9 − 4 = 5 > 0 D=9-4=5>0 D = 9 − 4 = 5 > 0 stable node (overdamped — too much friction to oscillate). ✔
What makes a point an equilibrium of x ˙ = f , y ˙ = g \dot x=f,\dot y=g x ˙ = f , y ˙ = g f = g = 0 f=g=0 f = g = 0 there (zero velocity).
Which matrix governs local stability? The Jacobian
J J J of partials evaluated at the equilibrium.
Eigenvalue formula for a 2×2 system? λ = τ ± τ 2 − 4 Δ 2 \lambda=\frac{\tau\pm\sqrt{\tau^2-4\Delta}}{2} λ = 2 τ ± τ 2 − 4Δ with
τ = t r J , Δ = det J \tau=\mathrm{tr}J,\ \Delta=\det J τ = tr J , Δ = det J .
Δ < 0 \Delta<0 Δ < 0 Saddle (eigenvalues of opposite sign) — always unstable.
Condition for a stable node/spiral? τ < 0 \tau<0 τ < 0 AND
Δ > 0 \Delta>0 Δ > 0 (negative trace, positive determinant).
What distinguishes spiral from node? Discriminant
D = τ 2 − 4 Δ < 0 D=\tau^2-4\Delta<0 D = τ 2 − 4Δ < 0 → complex eigenvalues → rotation → spiral.
When do you get a centre? τ = 0 , Δ > 0 \tau=0,\ \Delta>0 τ = 0 , Δ > 0 → pure imaginary eigenvalues
± i ω \pm i\omega ± iω → closed orbits.
Is a centre asymptotically stable? No, only neutrally/marginally stable; perturbations persist.
λ 1 + λ 2 \lambda_1+\lambda_2 λ 1 + λ 2 λ 1 λ 2 \lambda_1\lambda_2 λ 1 λ 2 τ \tau τ and
Δ \Delta Δ respectively (Vieta).
When can linearisation fail? At non-hyperbolic points (some
R e λ = 0 \mathrm{Re}\lambda=0 Re λ = 0 , e.g. a centre); nonlinear terms then decide.
Recall Feynman: explain to a 12-year-oldImagine a marble in a landscape. An equilibrium is a flat spot where it can rest. If it's at the bottom of a bowl , push it and it rolls back — stable . On top of a hill , push it and it rolls away — unstable . On a mountain pass (saddle), it rolls back if you push along the ridge but away if you push down the valley — saddle . If the bowl is slippery and spinning , the marble spirals down — spiral . On a perfectly frictionless circular track it just goes round and round forever — centre . The "Jacobian" is just a way of measuring how steep the ground is in each direction right at the resting spot.
Mnemonic Remember the decision tree"Delta decides the sign-game, Tau decides the fate, D decides the dance."
Δ < 0 \Delta<0 Δ < 0 saddle (sign-game lost).else τ \tau τ < 0 <0 < 0 > 0 >0 > 0 fate .
D < 0 D<0 D < 0 spirals/dances ; D > 0 D>0 D > 0 τ = 0 \tau=0 τ = 0 centre .
Jacobian matrix — the engine of linearisation.Eigenvalues and eigenvectors — sign & complexity = stability.Phase portraits — visualising trajectories near fixed points.Linearisation and Hartman–Grobman theorem — when linear ≈ nonlinear.Lyapunov stability — proving stability without eigenvalues.Damped harmonic oscillator — physical spiral/centre example.Trace–determinant plane — the master classification chart.
Linearise near equilibrium
lambda squared minus tau lambda plus Delta
Intuition Hinglish mein samjhoSocho ek system jiska behaviour ek differential equation se control hota hai. Jahan velocity zero ho jaati hai, wahan ek equilibrium (fixed point) hota hai — system wahan ruk sakta hai. Lekin asli sawaal yeh hai: agar hum thoda sa nudge karein, to system wapas aata hai (stable), bhaag jaata hai (unstable), ya gol gol ghoomta hai (spiral/centre)? Yeh decide karne ke liye hum equation ko fixed point ke aas-paas linearise karte hain, yaani Taylor expansion se sirf first-order term rakhte hain. Isse milta hai Jacobian matrix J J J
Ab J J J trace τ \tau τ determinant Δ \Delta Δ λ = τ ± τ 2 − 4 Δ 2 \lambda=\frac{\tau\pm\sqrt{\tau^2-4\Delta}}{2} λ = 2 τ ± τ 2 − 4Δ
Classification yaad rakhne ka simple rule: agar Δ < 0 \Delta<0 Δ < 0 saddle (unstable, ek direction mein khinchta, doosri mein bhagaata). Agar Δ > 0 \Delta>0 Δ > 0 τ < 0 \tau<0 τ < 0 τ > 0 \tau>0 τ > 0 D = τ 2 − 4 Δ D=\tau^2-4\Delta D = τ 2 − 4Δ spiral (ghoomta hua), positive ho to seedha node . Aur jab τ = 0 \tau=0 τ = 0 Δ > 0 \Delta>0 Δ > 0 centre — bilkul friction-less jhoole jaisa, gol gol orbit, na paas aata na door jaata.
Yeh topic isliye important hai kyunki physics, biology, engineering — sab jagah systems ka long-term behaviour isi se samajh aata hai. Damped spring stable spiral hai, ulta pendulum saddle hai, perfect oscillator centre hai. Ek important warning: centre sirf neutrally stable hota hai, aur jab τ = 0 \tau=0 τ = 0
