Visual walkthrough — Linearization of nonlinear systems

4.6.24 · D2 · Maths › Ordinary Differential Equations › Linearization of nonlinear systems

Hum assume karte hain tum sirf yeh jaante ho: curve kya hota hai, "slope" ka matlab kya hai, aur ki arrows yeh dikha sakte hain ki cheezein kis taraf jaati hain. Baaki sab hum yahan kamaenge.

Step 1 — "Flow" hota kya hai

KYA. Hamare paas do rules hain jo ek hilte hue dot ko batate hain ki woh kitni tezi se jaaye, depending on ki woh kahan baitha hai: $\dot{x}=f(x,y),\dot{y}=g(x,y).$

Har symbol ko use karne se pehle decode karte hain.

• $x,y$ — sirf ek flat sheet pe dot ki position, jaise graph paper pe coordinates.
• Upar ka chhota dot, $\dot{x}$, shorthand hai "==$x$ abhi kitni tezi se badal raha hai==" — iska $x$-direction mein velocity. $\dot{x}$ ko "x-dot" padho.
• $f(x,y)$ — ek machine: isko ek position do, yeh woh $x$-velocity wापस deta hai. $g$ wahi karta hai $y$ ke liye.

YEH DO RULES KYUN. Kyunki sheet ke har point pe yeh tumhe ek chhota arrow $(\dot{x},\dot{y})$ dete hain — woh direction aur speed jis par dot driftकरेगा. Poori sheet ko in arrows se dhaanp do aur tumhe milega ek vector field: ek wind map. Dot bas us wind pe sawari karta hai.

PICTURE. Neeche arrows dekho. Har ek $(f(x,y),g(x,y))$ apne point pe planted hai. Ek trajectory (amber) woh path hai jo ek dot trace karta hai agar woh hamesha local arrow follow kare.

Step 2 — Woh ek jagah jahan koi wind nahi

KYA. Hum ek aisi point dhundhte hain jahan dono velocities khatam ho jaayein: $f(x^{\ast },y^{\ast })=0\text{and}g(x^{\ast },y^{\ast })=0.$

• Star ${}^{\ast }$ ek special, fixed location mark karta hai — variable nahi. Socho "calm spot ka address."
• Dono equations ek hi point pe hold karni chahiye. Akele ek kaafi nahi.

KYUN. Agar dono velocities zero hain, toh wahan ka arrow zero length ka hai — koi wind nahi. Ek dot jo exactly wahan rakha jaaye woh kabhi nahi hilega. Yeh ek equilibrium hai (ise fixed ya critical point bhi kehte hain). Stability ek sawaal hai ऐसे point ke paas ke behaviour ke baare mein, isliye yahan hi sab kuch hota hai (compare karo Stability of Equilibria).

PICTURE. Neeche cyan dot wahan hai jahan har aas-paas ka arrow sikar jaata hai. Gaur karo ki kareeb ke arrows ab bhi kisi na kisi taraf point karte hain — shanti exactly centre mein hai.

Step 3 — Rename karo taaki calm spot origin pe aa jaaye

KYA. Position equilibrium ke relative measure karo: $u=x-x^{\ast },v=y-y^{\ast }.$

• $u$ = "main calm spot se kitna daayein hoon?" $v$ = "main uske kitna upar hoon?"
• Equilibrium pe khud $u=v=0$. Humne origin ko calm point pe slide kar diya.

KYUN. Do reasons. Pehla, ek chhota disturbance exactly ek chhota $(u,v)$ hai — toh "near equilibrium" ka matlab bas "$u,v$ tiny hain." Doosra, kyunki $x^{\ast },y^{\ast }$ constants hain, yeh time ke saath nahi badalte, isliye $\dot{u}=\dot{x}=f(x,y),\dot{v}=\dot{y}=g(x,y).$ Displacement ki velocity dot ki velocity ke barabar hai. Kuch lost nahi, bookkeeping cleaner.

PICTURE. Wahi wind map, naya grid calm point pe centred. Chhota amber displacement arrow hi $(u,v)$ hai.

Step 4 — Zoom in karo jab tak curve ek seedha ramp na lage

KYA. $f$ ko ek height surface ki tarah picture karo: har point $(x,y)$ ke upar surface height $f(x,y)$ pe baithti hai. Kyunki $f(x^{\ast },y^{\ast })=0$ hai, yeh surface calm spot ke theek upar zero height ko touch karti hai. Ab zoom in karo.

YEH TOOL KYUN — tangent plane. Ek curvy surface, ek point ke aas paas kaafi zoom karne par, flat dikhti hai — ek ramp ki tarah. Woh flat ramp surface ki best linear approximation hai wahan. Surface ko ramp se replace karna linearization ki poori trick hai: ramp ke saath hum algebra kar sakte hain; curve ke saath nahi. Yeh "zoom until flat" wahi hai jo Taylor Series formally karta hai.

PICTURE. Baayein: $f$ ki sach mein curvy surface. Daayein: zoom in karo, yeh ek tilted flat plane se alag nahi dikhti calm point se hoke.

Step 5 — Do slopes ke naam rakho (partial derivatives)

KYA. Har direction mein plane ka tilt ek partial derivative se measure hota hai: $f_x=\frac{\partial f}{\partial x}{|}_{(x^{\ast },y^{\ast })},f_y=\frac{\partial f}{\partial y}{|}_{(x^{\ast },y^{\ast })}.$

• $f_x$ — "$x$-direction mein sirf ek step chalo ($y$ freeze karo); $f$ kitna utha?" Yeh $x$ ke saath ramp ki steepness hai.
• $f_y$ — wahi, lekin $y$-direction mein step karo aur $x$ frozen rakho.
• Bar ${|}_{(x^{\ast },y^{\ast })}$ kehta hai: yeh slopes calm point pe measure karo, isliye yeh plain numbers hain, functions nahi.

PARTIAL KYUN, ORDINARY DERIVATIVE KYUN NAHI. Ek ordinary slope $\frac{df}{dx}$ assume karta hai ek input. Haare paas do hain. Partial derivative answer deta hai "steepness agar main sirf ek axis ke saath chalun" — exactly woh do tilts jo ek plane ko chahiye.

PICTURE. Calm point se hokar surface ke do cross-sections: $x$ ke saath slice karne se slope $f_x$ dikhta hai; $y$ ke saath slice karne se $f_y$.

Step 6 — Approximation assemble karo (Taylor first order tak)

KYA. Calm point ke paas ramp ki height hai base height + tilt·distance: $f(x,y)\approx \underset{=0}{\underbrace{f(x^{\ast },y^{\ast })}}+\underset{\text{tilt in }x}{\underbrace{f_x}}\underset{\text{step in }x}{\underbrace{u}}+\underset{\text{tilt in }y}{\underbrace{f_y}}\underset{\text{step in }y}{\underbrace{v}}+\underset{\text{curvature: tiny}}{\underbrace{\frac{1}{2}{f}_{xx}{u}^2+\dots }}$

Term by term:

• $f(x^{\ast },y^{\ast })=0$ — base height vanish ho jaata hai kyunki hum calm point pe baithe hain. Gaya.
• $f_{x}u$ — $x$-tilt ek step $u$ pe kitni height add karta hai. Rise = slope × run.
• $f_{y}v$ — wahi $y$-tilt ke liye step $v$ par.
• $\frac{1}{2}{f}_{xx}{u}^2+\dots$ — curved correction.

CURVED PART KYUN CHHODEIN. Agar $u$ ek chhota fraction hai, maano $0.01$, toh $u^2=0.0001$ — sau guna chhota. Calm point ke kaafi paas, squared (curvature) terms straight (linear) terms se dabbe jaate hain. Sirf wahi rakhna jo dominate karta hai woh honest approximation hai.

PICTURE. Sach wala curve versus uski straight-ramp approximation. Vertical amber gap woh discarded curvature hai — dekho yeh centre ke paas jaate jaate zero ho jaata hai.

Step 7 — Dono ramps ko ek matrix mein stack karo, the Jacobian

KYA. Step 6 $g$ ke liye bhi karo, aur dono straight-line rules saath likhо: $\dot{u}\approx f_{x}u+f_{y}v,\dot{v}\approx g_{x}u+g_{y}v.$ "Slope × displacement" ki do rows. Chaaron slopes ko ek grid mein pack karo: $\left(\begin{array}{c}\dot{u}\\ \dot{v}\end{array}\right)=\underset{J=\text{Jacobian}}{\underbrace{\left(\begin{array}{cc}{f}_x& f_y\\ g_x& g_y\end{array}\right){|}_{(x^{\ast },y^{\ast })}}}\left(\begin{array}{c}u\\ v\end{array}\right).$

Matrix product ko row by row padhna literally upar wali do equations hai:

• Top row $(f_x,f_y)$ dotted with $(u,v)$ deta hai $\dot{u}$.

• Bottom row $(g_x,g_y)$ dotted with $(u,v)$ deta hai $\dot{v}$.

• $J$ — Jacobian: chaaron tilts, calm point pe evaluate kiye, isliye saare entries numbers hain.

• Kyunki entries constant hain, yeh ek linear system hai — jaise wala hum Eigenvalues and Eigenvectors se completely crack kar sakte hain.

YEH FINISH LINE KYUN HAI. Tedha-medha curvy wind map, calm point ke paas, ab bas "displacement ko ek fixed matrix se multiply karo" hai. Poora nonlinear problem sikar ke matrix multiplication ban gaya.

PICTURE. Jacobian ek machine ki tarah: displacement arrow $(u,v)$ andar jaata hai, matrix $J$ act karta hai, velocity arrow $(\dot{u},\dot{v})$ bahar aata hai — chhote arrow ko moda aur khicha jaata hai.

Step 8 — Edge case: agar linear part vanish ho jaaye toh?

KYA. Kabhi kabhi do tilts aisa milte hain ki $J$ ke eigenvalues pure imaginary (ek center) ya zero ho jaate hain. Tab $\mathrm{Re}(\lambda )=0$: koi clear growth nahi, koi clear decay nahi.

• $\lambda$ — $J$ ka ek eigenvalue; iska real part $\mathrm{Re}(\lambda )$ batata hai ki ek mode grow karta hai ($>0$), decay karta hai ($<0$), ya kuch nahi ($=0$).

YEH SPECIAL KYUN HAI. Step 6 mein humne curvature uda di thi kyunki linear terms dominant the. Lekin agar linear terms exactly balanced circling dein ($\mathrm{Re}(\lambda )=0$), toh ramp outcome decide nahi kar sakta — woh tiny curvature jo humne uda di woh ab tie-breaker hai. Linearization honestly jhooth bol sakti hai (ek "center" secretly spiral ho sakta hai).

PICTURE. Same starting arrows: linear model ek perfect closed loop draw karta hai, lekin sach wala nonlinear flow dheere dheere andar spiral karta hai — uda di gayi curvature kaam kar rahi hai.

Ek-picture summary

Upar sab kuch, compress kiya: ek curved surface → tangent plane → chaaron slopes → ek matrix jo displacements ko velocities mein badalta hai.

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