Question bank — Linearization of nonlinear systems

False. Ye sirf equilibrium ke ek chhote se neighbourhood mein trustworthy hai, jahan displacements $u=x-x^{\ast }$, $v=y-y^{\ast }$ itne chhote hon ki quadratic terms negligible hon. Door jaane par discard kiye gaye terms dominate karte hain aur picture bilkul alag ho sakti hai.

False. Ek symbolic $J(x,y)$ mein abhi bhi $x,y$ hain aur ye ek constant-coefficient linear system nahi hai. Eigenvalues ka koi matlab nikle iske liye pehle $J$ ko ek specific equilibrium par evaluate karke numeric entries prapt karni hongi.

True. Dono eigenvalues ka real part negative hai, isliye point hyperbolic hai; Hartman–Grobman tab guarantee karta hai ki nonlinear flow nearby (stable node) linear flow jaisi dikhti hai.

False. Pure imaginary eigenvalues ka matlab hai $\text{Re}(\lambda )=0$, isliye point non-hyperbolic hai aur dropped quadratic terms "center" ko ek slow inward ya outward spiral mein badal sakte hain. Decide karne ke liye tumhe Lyapunov ya energy arguments chahiye.

True. Saddle ke real eigenvalues opposite sign ke hote hain, isliye kam se kam ek mode $e^{\lambda t}$ jaisi grow karta hai jahan $\lambda >0$ — wo growing direction ise chahe kuch bhi ho unstable banata hai.

False. $f=g=0$ set karna equations ka ek aisa system hai jiske zero, ek, ya kaafi solutions ho sakte hain (pendulum ke infinitely many hain, $x=n\pi$ par). Har equilibrium ko alag se analyse kiya jata hai.

True. $\Delta >0$ ke saath $\tau <0$ dono eigenvalue real parts ko negative force karta hai (unka sum $\tau <0$ hai, product $\Delta >0$ hai), isliye har mode decay karta hai.

True. Dono eigenvalue real parts ab positive hain (sum $\tau >0$, product $\Delta >0$). Ye ek unstable node hai agar ${\tau }^2-4\Delta \ge 0$ (real eigenvalues) aur ek unstable spiral hai agar ${\tau }^2-4\Delta <0$ (complex eigenvalues).

True. Discriminant ${\tau }^2-4\Delta$ ka sign unhe distinguish karta hai: negative complex eigenvalues deta hai (spiral), non-negative real eigenvalues deta hai (node). Dono stable hain agar $\text{Re}(\lambda )<0$ hai.

True — par sirf tab jab ye hyperbolic ho. Ek positive-real-part eigenvalue (hyperbolic) guarantee karta hai ki ek growing direction Hartman–Grobman ke through nonlinear flow mein survive karta hai. Agar "instability" zero-real-part case se aayi ho, to aisi koi guarantee nahi hai.

Error: incomplete root-finding. Factored doosri equation $y(-1+x)=0$ ye bhi allow karti hai $y=0$, jo pehli equation ke saath doosre solutions jaise $(0,0)$ kholti hai. Tumhe $f=g=0$ ki har branch exhaust karni chahiye.

Error: ise 'stable' kehna. Ek center ka $\text{Re}(\lambda )=0$ hota hai, isliye ye na decay karta hai na grow — ye asymptotically stable nahi hai, aur iske alawa linearization yahan sach mein nonlinear behaviour ke liye inconclusive hai.

Error: zero eigenvalue ko ignore karna. Ek zero eigenvalue ka matlab hai $\text{Re}(\lambda )=0$, jo point ko non-hyperbolic banata hai. Hartman–Grobman fail karta hai aur stability linearization se bilkul nahi padhi ja sakti.

Error: kabhi equilibrium substitute nahi kiya. Variables se bhari matrix ke eigenvalues wo constant growth rates nahi hain jo tumhe chahiye; pehle $(x^{\ast },y^{\ast })$ substitute karo, phir characteristic polynomial solve karo.

Error: ye IS zero hai. Equilibrium ki bilkul definition se, $f(x^{\ast },y^{\ast })=0$ hota hai. Agar ye zero nahi hota to tum equilibrium par nahi the aur poora linearization setup apply hi nahi hota.

Error: hyperbolicity ko round off karna. Koi bhi nonzero real part, chahe kitni bhi chhoti ho, point ko hyperbolic banati hai — yahan ek unstable spiral hai. "Basically zero" zero nahi hai; flow genuinely bahar ki taraf spiral karta hai.

Kyunki equilibrium ke paas $u,v$ chhote hain, isliye unke squares bahut chhote hain ($0.01^2=0.0001$). Linear terms dominate karte hain, isliye unhe rakhna leading behaviour capture karta hai — ye tangent-plane approximation hai.

Solutions $e^{\lambda t}\mathbf{v}$ ke sums hain; exponent $\lambda$ control karta hai ki har mode grow karega ya decay karega. Eigenvector $\mathbf{v}$ sirf us mode ki direction set karta hai, ye nahi ki wo time ke saath survive karega ya nahi.

Jacobian first partial derivatives $f_x,f_y,g_x,g_y$ collect karta hai — bilkul equilibrium par har surface ke tangent plane ki slopes. Ye by construction vector field ki best linear approximation hai wahan.

Ek zero real part ek knife-edge hai: linear flow na grow karta hai na decay, isliye wo chhote quadratic terms jo humne discard kiye the decisive ho jaate hain. Sirf jab har mode strictly grow ya strictly decay kare tabhi hum unhe safely ignore kar sakte hain.

Sirf linearization ye certify nahi kar sakta, lekin pendulum energy conserve karta hai, aur closed energy contours closed orbits force karte hain — ek independent argument jo inconclusive linear center ko real center mein upgrade karta hai.

Bottom par restoring force ek nudged pendulum ko wapas kheenchti hai (oscillation → center); top par gravity ek nudge ko amplify karta hai, ek growing aur ek shrinking direction deta hai (mixed-sign real eigenvalues → saddle).

Ek equilibrium ko dono velocities ek saath vanish karni hoti hain; ek point jo $f=0$ satisfy karta hai par $g\ne 0$ abhi bhi move karta hai. Sirf dono equations ke common solutions hi sach mein fixed points hain.

Kam se kam ek eigenvalue zero hai (unka product $\Delta =0$ hai), isliye point non-hyperbolic hai aur linearization inconclusive hai — near-equilibria ki ek poori line ya curve exist ho sakti hai, jiske liye deeper analysis chahiye.

Linearization ek center hai (closed elliptical loops), lekin $\text{Re}(\lambda )=0$ ise non-hyperbolic banata hai, isliye ye inconclusive hai: sach mein nonlinear point ek genuine center ya ek slow spiral ho sakta hai. Pendulum ke bottom ki tarah energy ya Lyapunov argument se confirm karo.

Ye ek unstable improper node hai (dono $\text{Re}(\lambda )>0$, isliye hyperbolic aur unstable). Agar $J$ diagonalizable hai to ye ek unstable star node hai; agar $J$ defective hai to ye ek unstable improper node hai jiske trajectories ek single shear direction ke along bahar bhagte hain — koi rotation nahi.

Haan, lekin check karo ki $J$ defective to nahi hai. Ek repeated negative eigenvalue $\text{Re}(\lambda )<0$ rakhta hai, isliye point hyperbolic aur stable hai, aur Hartman–Grobman apply hota hai. Agar $J$ diagonalizable hai to ye ek star node hai; agar $J$ defective hai (non-diagonalizable, jisme Jordan block chahiye) to ye ek improper node hai jiske trajectories real-axis shear se ek single eigendirection ke saath bend karte hain — koi rotation ya spiralling nahi hai, geometry alag hai lekin stability unchanged hai, isliye Jordan form inspect karo.

Jacobian ke liye partial derivatives ka exist karna zaroori hai; agar $f$ ya $g$ wahan non-smooth hai, to koi Jacobian exist nahi karta aur standard linearization simply apply nahi hoti — tumhe doosre methods chahiye.

Worse. Ek double-zero eigenvalue highly degenerate hai: ek rotation bhi guarantee nahi hai, aur ye genuinely imaginary nahi hai (jiske liye nonzero $\omega$ chahiye hota). Is strongly non-hyperbolic case ke liye Lyapunov ya centre-manifold techniques chahiye.

Ye ek saddle hai (opposite signs → unstable), lekin "magnitude" sirf ye set karta hai ki trajectories har direction mein kitni tezi se chalti hain, qualitative type nahi. Instability ke liye ek growing direction kaafi hai.

Sirf linearization se nahi — ye non-hyperbolic hai. Classic Lotka–Volterra model ke liye ek conserved quantity independently prove karta hai ki orbits sach mein closed hain, lekin ek generic system linear center ke saath secretly spiral kar sakta hai.