Worked examples — Linearization of nonlinear systems

4.6.24 · D3 · Maths › Ordinary Differential Equations › Linearization of nonlinear systems

Shuru karne se pehle, ek reminder vocabulary ka, taaki koi bhi symbol bina explanation ke na aaye:

Recall The four numbers that decide everything

Jacobian $J=\left(\begin{array}{cc}{f}_x& f_y\\ g_x& g_y\end{array}\right)$ ko ek equilibrium par evaluate karke, hum banate hain:

• Trace $\tau =f_x+g_y$ — diagonal ka sum. Socho "badhne ya ghaTne ki total tendency."
• Determinant $\Delta =f_x{g}_y-f_y{g}_x$ — "twisting/scaling" ki strength.
• Eigenvalues: $\lambda =\frac{\tau \pm \sqrt{{\tau }^2-4\Delta }}{2}$.
• Discriminant $D={\tau }^2-4\Delta$ decide karta hai real vs complex: $D>0$ real, $D<0$ complex, $D=0$ repeated. Do shortcuts jinpar hum baar baar depend karte hain: $\tau ={\lambda }_1+{\lambda }_2$ (sum) aur $\Delta ={\lambda }_1{\lambda }_2$ (product).

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The scenario matrix

Har equilibrium jo tum kabhi classify karoge wo in mein se kisi ek cell mein padega. Right column us page par us example ka naam deta hai jo use hit karta hai.

#	Cell (kya special hai)	Eigenvalue signature	Example
C1	Real, dono negative	${\lambda }_1,{\lambda }_2<0$	Ex 1 stable node
C2	Real, dono positive	${\lambda }_1,{\lambda }_2>0$	Ex 2 unstable node
C3	Real, opposite signs	${\lambda }_1<0<{\lambda }_2$	Ex 3 saddle
C4	Complex, negative real part	$\alpha \pm i\beta ,\alpha <0$	Ex 4 stable spiral
C5	Complex, positive real part	$\alpha \pm i\beta ,\alpha >0$	Ex 5 unstable spiral
C6	Degenerate: pure imaginary ($\tau =0$)	$\pm i\beta$	Ex 6 center — inconclusive case
C7	Degenerate: ek zero eigenvalue ($\Delta =0$)	$\lambda =0$ present	Ex 7 non-hyperbolic line
C8	Word problem do equilibria ke saath alag-alag type ke	mixed	Ex 8 Lotka–Volterra
C9	Exam twist: sirf $\tau ,\Delta$ se classify karo + repeated-root ($D=0$) boundary	boundary $D=0$	Ex 9 parameter sweep

Wo rule jo har example guide karta hai wo hai neeche ka boundary map — iska shape yaad kar lo aur tum ek bhi eigenvalue touch kiye bina classify kar sakte ho.

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Ex 1 — Cell C1: stable node (dono eigenvalues negative)

Forecast: Dono variables mein bada negative self-term hai ($-2x$, $-2y$) aur weak coupling. Guess: sab kuch decay karta hai → stable node. Do real negative eigenvalues predict karo.

1. Equilibrium dhundo. Dono ko zero set karo: $-2x+y=0$ aur $x-2y=0$. Solve karne par $x=y=0$ milta hai. Ye step kyun? Equilibria hi wo points hain jahan linearization define bhi hoti hai — yahi wo jagah hai jahan Taylor expansion ka constant term vanish hota hai.
2. Ye system already linear hai, isliye $J$ coefficient matrix hai, jo poori jagah constant hai. Yahan $f=-2x+y$ hai isliye $f_x=-2,f_y=1$; $g=x-2y$ hai isliye $g_x=1,g_y=-2$: $J=\left(\begin{array}{cc}-2& 1\\ 1& -2\end{array}\right).$ Kyun? Jab $f,g$ already linear hon, to partials constant coefficients hote hain — koi evaluation needed nahi.
3. Trace aur determinant. $\tau =-2+-2=-4$, $\Delta =(-2)(-2)-(1)(1)=4-1=3$. Kyun? Ye do numbers hamen figure mein $\tau$–$\Delta$ map par locate karte hain.
4. Discriminant aur eigenvalues. $D={\tau }^2-4\Delta =16-12=4>0$ → real. $\lambda =\frac{-4\pm 2}{2}=-1,-3$. Pehle $D$ kyun? $D$ ka sign hamen "node vs spiral" batata hai roots compute karne se pehle hi.
5. Classify karo. Dono real aur negative → stable node. Ye stable node kyun padha jaata hai? $\Delta >0$ ⇒ eigenvalues ka sign same hai (product positive); $\tau <0$ ⇒ wo shared sign negative hai (sum negative); $D>0$ ⇒ real, spiralling nahi. Negative + real = do directions mein pure decay = stable node.

Verify: figure par ye left interior region hai (parabola ke neeche, centre ke left). Eigenvalues ka sum $-1+-3=-4=\tau$ ✔, product $(-1)(-3)=3=\Delta$ ✔. Forecast confirm hua.

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Ex 2 — Cell C2: unstable node (dono eigenvalues positive)

Forecast: Positive self-terms $+3x,+3y$; $3x^2$ ek higher-order nonlinearity hai jo origin ke paas mar jaati hai. Guess: unstable node, do real positive eigenvalues.

1. Origin ek equilibrium hai? $f(0,0)=0$, $g(0,0)=0$ ✔ (yaad karo $f=3x+y$, $g=x+3x^2+3y$). Ye step kyun? $(0,0)$ actually ek fixed point hai ya nahi, ye confirm karna zaroori hai uske around expand karne se pehle.
2. Jacobian, general. $f_x=3,f_y=1,g_x=1+6x,g_y=3$. $g_x=1+6x$ kyun compute kiya? $3x^2$ term $6x$ contribute karta hai derivative mein — lekin origin par ye vanish kar jaayega, exactly yahi dikhata hai ki kyun quadratic terms linear stability ko affect nahi karte.
3. $(0,0)$ par evaluate karo. $g_x=1+0=1$, isliye $J=\left(\begin{array}{cc}3& 1\\ 1& 3\end{array}\right).$ Kyun substitute karo? $x$ contain karta Jacobian ek linear system nahi hai — ye parent note ka "steel-man" trap hai.
4. Numbers. $\tau =6,\Delta =9-1=8,D=36-32=4>0$. $\lambda =\frac{6\pm 2}{2}=4,2$.
5. Classify karo. Dono real aur positive → unstable node. Ye unstable node kyun padha jaata hai? $\Delta >0$ ⇒ same sign; $\tau >0$ ⇒ wo sign positive hai; $D>0$ ⇒ real. Positive + real = do directions mein growth = unstable node. Ye Ex 1 ka $\tau =0$ ke paas mirror hai.

Verify: sum $4+2=6=\tau$ ✔, product $4\cdot 2=8=\Delta$ ✔. Figure par: right interior region. Notice karo ki $3x^2$ kabhi nahi aaya — proof ki quadratics drop karna ek hyperbolic point par legitimate tha.

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Ex 3 — Cell C3: saddle (opposite signs)

Forecast: Strong cross-coupling ($2y$, $2x$) usually diagonal ko beat karta hai → ek growing aur ek decaying direction ki expect karta hoon → saddle.

1. Origin ek equilibrium hai: dono RHS $(0,0)$ par vanish karte hain ✔ ($f=x+2y$, $g=2x+y-y^3$).
2. Jacobian. $f_x=1,f_y=2,g_x=2,g_y=1-3y^2$.
3. $(0,0)$ par: $g_y=1-0=1$, isliye $J=\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$. Kyun? $-y^3$ term $-3y^2$ contribute karta hai, jo origin par zero hai.
4. Numbers. $\tau =2,\Delta =1-4=-3$. Yahan key insight hai: $\Delta <0$. $\Delta <0$ saddle guarantee kyun karta hai? $\Delta ={\lambda }_1{\lambda }_2$. Ek negative product forces karta hai do eigenvalues ko opposite signs rakhne par — ek positive, ek negative — chahe $\tau$ kuch bhi ho. Ye definition se saddle hai.
5. Explicitly eigenvalues: $D=4+12=16$, $\lambda =\frac{2\pm 4}{2}=3,-1$.
6. Classify karo: saddle (unstable).

Verify: product $3\cdot (-1)=-3=\Delta$ ✔ (negative ⇒ opposite signs, jaise predict kiya tha). Sum $3+(-1)=2=\tau$ ✔. Figure par ye $\Delta =0$ line ke neeche baithta hai — poora lower band saddle territory hai.

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Ex 4 — Cell C4: stable spiral (complex, negative real part)

Forecast: Diagonal negative hai (decay), lekin antisymmetric off-diagonal ($+4y$ vs $-x$) rotation create karta hai. Guess: stable spiral — andar ki taraf spiral karta hai.

1. Already linear hai, isliye $f=-x+4y$, $g=-x-y$ ke saath: $J=\left(\begin{array}{cc}-1& 4\\ -1& -1\end{array}\right)$.
2. Numbers. $\tau =-2,\Delta =(-1)(-1)-(4)(-1)=1+4=5$.
3. Discriminant. $D={\tau }^2-4\Delta =4-20=-16<0$. $D<0$ spiral kyun matlab hai? $\sqrt{D}$ imaginary ho jaata hai, isliye $\lambda =\alpha \pm i\beta$. Imaginary part $\beta$ rotation hai; real part $\alpha =\tau /2$ growth/decay hai.
4. Eigenvalues. $\lambda =\frac{-2\pm \sqrt{-16}}{2}=-1\pm 2i$.
5. Classify karo. Real part $\alpha =-1<0$ → stable spiral. Trajectories rotate karti hain ($\beta =2$ se) jabki shrink karti hain ($\alpha =-1$ se).

Verify: sum $(-1+2i)+(-1-2i)=-2=\tau$ ✔; product $(-1{)}^2-(2i{)}^2=1+4=5=\Delta$ ✔. Map par ($\tau <0,D<0$) ye parabola ke upar left side mein hai — stable-spiral region. Figure s02 mein, blue trajectory dekho: ye equilibrium ke around wind karta hai ($\beta$ rotation) jabki uski radius white dot ki taraf shrink karti hai ($\alpha =-1$ decay). Wo geometric spiral hai $\alpha$ ka sign visible bana diya gaya.

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Ex 5 — Cell C5: unstable spiral (complex, positive real part)

Forecast: Ex 4 jaisi hi rotation lekin positive diagonal → baahir ki taraf spiral, unstable spiral.

1. $f=x+4y$, $g=-x+y$ ke saath: $J=\left(\begin{array}{cc}1& 4\\ -1& 1\end{array}\right)$.
2. $\tau =2,\Delta =1-(4)(-1)=1+4=5$.
3. $D=4-20=-16<0$ → complex.
4. $\lambda =\frac{2\pm 4i}{2}=1\pm 2i$.
5. Classify karo. $\alpha =+1>0$ → unstable spiral. Sirf real part ka sign Ex 4 se flip hua; poora character badal gaya. Ye unstable spiral kyun padha jaata hai? $D<0$ hamen parabola ke upar rakhta hai (spiral); $\tau >0$ hamen uske right half mein rakhta hai, isliye $\alpha =\tau /2>0$ — radius har turn mein badhta hai.

Verify: sum $=2=\tau$ ✔, product $1+4=5=\Delta$ ✔. Figure s01 par: Ex 4 ki same height par lekin $\tau =0$ ke right ki taraf mirror hua. Same $\Delta$, opposite $\tau$ — map ka vertical axis hi stable aur unstable spirals ke beech poora fark hai.

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Ex 6 — Cell C6 (degenerate): center — jahan linearization JHOOTH bol sakti hai

Forecast: Pure rotation, koi bhi decay term nahi → center (closed loops). Lekin: parent ne warn kiya tha ki centers nonlinear systems ke liye inconclusive hote hain.

1. $f=y$, $g=-x$ ke saath: $J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$.
2. $\tau =0,\Delta =0\cdot 0-(1)(-1)=1$. $\tau =0$ alarm bell kyun hai? $\tau =0$ with $\Delta >0$ deta hai $\lambda =\pm i\sqrt{\Delta }$ — pure imaginary, zero real part. Zero real part = non-hyperbolic = Hartman–Grobman apply nahi hota.
3. $D=0-4=-4<0$; $\lambda =\pm i$. Pure imaginary → linearization ke liye center.
4. Honest conclusion. Is exact linear system ke liye origin sach mein ek center hai. Lekin koi bhi nonlinear term add karo, jaise $\dot{x}=y,\dot{y}=-x-x^{2}y$: linearization abhi bhi $\pm i$ hai, phir bhi extra $-x^{2}y$ energy drain karta hai aur real trajectory andar ki taraf spiral karti hai. Linear picture ne jhooth bola.

Verify: sum $i+(-i)=0=\tau$ ✔, product $i\cdot (-i)=1=\Delta$ ✔. Map par ye exactly vertical $\tau =0$ axis par baithta hai — razor-thin center line. Kyunki $\tau =0$ hai, tumhe ek Lyapunov function ya energy argument ki zaroorat hai — linearization se ek center par kabhi trust mat karo akele.

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Ex 7 — Cell C7 (degenerate): ek zero eigenvalue

Forecast: $\dot{y}=-y$ $y$ ko zero ki taraf kheenchta hai (stable direction), lekin $\dot{x}=x^2$ na linearly stable hai na unstable — iska derivative $0$ par vanish karta hai. Ek zero eigenvalue ki expect karo → non-hyperbolic.

1. Origin ek fixed point hai: $0^2=0$, $-0=0$ ✔ ($f=x^2$, $g=-y$).
2. Jacobian. $f_x=2x,f_y=0,g_x=0,g_y=-1$.
3. $(0,0)$ par: $f_x=0$, isliye $J=\left(\begin{array}{cc}0& 0\\ 0& -1\end{array}\right)$.
4. Numbers. $\tau =-1,\Delta =0$. $\Delta =0$ kyun matter karta hai? $\Delta ={\lambda }_1{\lambda }_2=0$ force karta hai ki kam se kam ek eigenvalue zero ho. Zero real part ⇒ non-hyperbolic ⇒ linearization inconclusive.
5. $\lambda =0,-1$.
6. Nonlinear term se sach. $\dot{x}=x^2\ge 0$ hamesha: $x$ badhta hai $x>0$ ke liye aur $x<0$ ke liye zero ki taraf creep karta hai. Isliye $x$-axis left se stable hai, right se unstable — ek "saddle-node"-like semi-stable behaviour jo linear zero kabhi reveal nahi kar sakta tha.

Verify: sum $0+(-1)=-1=\tau$ ✔, product $0\cdot (-1)=0=\Delta$ ✔. Map par, $\Delta =0$ exactly horizontal axis hai — boundary line jahan classification breakdown ho jaati hai.

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Ex 8 — Cell C8 (word problem): predator–prey, do alag equilibria

Forecast: Coexistence matlab $x>0$ aur $y>0$. Agar ye stable hai, dono populations settle karti hain; agar ye spiral hai, oscillate karti hain. Ecology damped oscillation suggest karti hai → guess stable spiral ya stable node.

1. Equilibria. Yahan $f=x(3-x-2y)$ aur $g=y(x-1)$ hain. $g=0$ se: $y=0$ ya $x=1$. Coexistence ke liye $y\ne 0$ chahiye, isliye $x=1$. $f=0$ mein plug karo: $1(3-1-2y)=0\Rightarrow 2-2y=0\Rightarrow y=1$. Coexistence point $(1,1)$. DONO factors kyun solve karo? Parent ki teesri mistake — ek equilibrium bhool jaana — tab hoti hai jab tum ek root dhundhke ruk jaate ho; $f=g=0$ ke kai solutions ho sakte hain (yahan $(0,0)$ aur $(3,0)$ bhi hain), har ek ki apni stability hai. Hum jaanboojh kar coexistence root pursue karte hain $y\ne 0$ rakhkar.
2. Jacobian, general. $f=3x-x^2-2xy$, isliye $f_x=3-2x-2y,f_y=-2x$. $g=xy-y$, isliye $g_x=y,g_y=x-1$. Pehle products expand kyun karo? $x(3-x-2y)$ ko directly differentiate karna error-prone hai; multiply out karna har partial ko mechanical bana deta hai.
3. $(1,1)$ par evaluate karo. $J=\left(\begin{array}{cc}3-2-2& -2\\ 1& 1-1\end{array}\right)=\left(\begin{array}{cc}-1& -2\\ 1& 0\end{array}\right).$ Specific point kyun substitute karo? $x,y$ contain karta Jacobian ek linear system nahi hai — humen pure numbers chahiye.
4. Numbers. $\tau =-1+0=-1,\Delta =(-1)(0)-(-2)(1)=0+2=2$. Ye pehle kyun? Ye hamen map par locate karte hain aur directly stability sign dete hain.
5. Discriminant. $D={\tau }^2-4\Delta =1-8=-7<0$ → complex. $\lambda =\frac{-1\pm i\sqrt{7}}{2}$.
6. Classify karo. Real part $-\frac{1}{2}<0$ → stable spiral. Dono populations coexistence mein spiral karte hain: damped oscillation, ecologically sensible. Ye signs stable spiral kyun force karte hain? $D<0$ ⇒ complex eigenvalues ⇒ spiral (node nahi); $\tau <0$ ⇒ real part $\tau /2<0$ ⇒ spiral shrinks ⇒ stable. Figure s01 par ye parabola ke upar, vertical axis ke just left mein land karta hai — Ex 4 jaisa hi neighbourhood.

Verify: eigenvalues ka sum $=-1=\tau$ ✔; product $={\left(\frac{-1}{2}\right)}^2+{\left(\frac{\sqrt{7}}{2}\right)}^2=\frac{1}{4}+\frac{7}{4}=2=\Delta$ ✔. Forecast (damped oscillation) confirm hua. $(1,1)$ par dono populations positive hain ✔ — ek physically valid equilibrium.

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Ex 9 — Cell C9 (exam twist): $D=0$ boundary aur ek parameter sweep

Forecast: Chhote $a$ ke liye damping weak hai — rotation likely (spiral). Bade $a$ ke liye damping dominate karta hai — shayad node. $D=0$ par ek boundary ki expect karo jahan ek repeated real root appear karta hai.

1. $f=-x+y$, $g=-x-ay$ ke saath: $J=\left(\begin{array}{cc}-1& 1\\ -1& -a\end{array}\right)$.
2. Trace aur determinant. $\tau =-1-a,\Delta =(-1)(-a)-(1)(-1)=a+1$. $\tau ,\Delta$ se kyun lead karo? Exam trick: tum sirf signs se stability decide kar sakte ho. Yahan $a>0$ ⇒ $\tau =-1-a<0$ aur $\Delta =a+1>0$ ⇒ hamesha stable, har positive $a$ ke liye.
3. Node vs spiral: discriminant. $D={\tau }^2-4\Delta =(1+a{)}^2-4(a+1)=(a+1)(a+1-4)=(a+1)(a-3).$ Factor kyun karo? Ye sign change ko saaf expose karta hai. Kyunki $a+1>0$ hai, $D$ ka sign $(a-3)$ ka sign hai.
4. Critical value. $D=0$ at $a=3$.
   • $0<a<3$: $D<0$ → stable spiral.
   • $a=3$: $D=0$ → repeated real eigenvalue $\lambda =\tau /2=-2$ (dono roots equal). Ye boundary case hai — step 5 dekho.
   • $a>3$: $D>0$ → stable node.
5. Repeated root ka matlab (star vs improper node). Jab $D=0$ ho to do eigenvalues $\lambda =\tau /2$ par coincide karte hain. Do sub-cases hain:
   • Agar $J$ us point par identity ka scalar multiple ho ($J=\lambda I$), har direction ek eigenvector hai → ek star (proper) node: trajectories seedhi origin mein jaati hain.
   • Warna (sirf ek independent eigenvector) → ek improper (degenerate) node: trajectories bend karti hain aate hue, ek single tangent line share karte hue. Yahan $a=3$ par, $J=\left(\begin{array}{cc}-1& 1\\ -1& -3\end{array}\right)\ne -2I$, isliye iska sirf ek eigenvector hai → ek improper stable node. Ye check kyun karo? "Repeated eigenvalue" akele tumhe shape nahi batata — tumhe test karna padega ki eigenvector space 2D hai (star) ya 1D (improper). Eigenvector count deciding tool hai.
6. Limiting behaviour (honest asymptotics). Consider karo $a\to \infty$. Eigenvalues satisfy karte hain ${\lambda }_1+{\lambda }_2=-1-a$ aur ${\lambda }_1{\lambda }_2=a+1$. Fast root badhta hai like ${\lambda }_1\approx -a$ (ye lagbhag poora trace carry karta hai). Slow root tab hai ${\lambda }_2=\frac{\Delta }{{\lambda }_1}\approx \frac{a+1}{-a}\to -1$ jaise $a\to \infty$ — isliye slow eigenvalue $-1$ ki taraf tend karta hai, $0$ ki taraf nahi. Dono negative rehte hain: ek increasingly anisotropic stable node. Doosri extreme par $a\to 0^{+}$: $\tau \to -1,\Delta \to 1,D\to -3<0$ — ek stable spiral with real part $\tau /2\to -\frac{1}{2}$.

Verify: boundary $a=3$ par: $\tau =-4,\Delta =4$, isliye $D=16-16=0$ ✔, repeated eigenvalue $\lambda =\tau /2=-2$ ✔, aur $J=\left(\begin{array}{cc}-1& 1\\ -1& -3\end{array}\right)$ ka ek single eigenvector hai (geometric multiplicity 1) ✔ → improper node. $a=1$ par: $\tau =-2,\Delta =2,D=4-8=-4<0$ → stable spiral ✔. $a=5$ par: $\tau =-6,\Delta =6,D=36-24=12>0$ → stable node ✔.

Wrap-up (exam ka moral). Poori family har $a>0$ ke liye stable hai — stability kabhi nahi badlti. Jo parameter sweep change karta hai wo sirf approach ki geometry hai: $0<a<3$ ke liye state spiral karke andar aata hai (oscillatory settling), $a=3$ par ye exactly $D=0$ parabola par ek degenerate improper node ki tarah baithta hai, aur $a>3$ ke liye ye seedha bina overshoot ke decay karta hai. Exam skills demonstrate kiye: $\tau ,\Delta$ ke signs se stability instantly padho, shape pin karne ke liye $D$ use karo, aur — $D=0$ boundary par — star ko improper node se alag karne ke liye eigenvectors count karo.

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Recall One-line self-test har cell ke liye

Answers cover karo. $\Delta <0$ ka sign tumhe kya batata hai? ::: Saddle (opposite-sign real eigenvalues) — $\tau$ chahe kuch bhi ho. $\tau <0,\Delta >0,D>0$? ::: Stable node. $\tau >0,\Delta >0,D<0$? ::: Unstable spiral. $\tau =0,\Delta >0$? ::: Center — linearization inconclusive. $\Delta =0$? ::: Ek zero eigenvalue — non-hyperbolic, inconclusive. $D=0$ with $\tau <0$ aur ek eigenvector? ::: Degenerate (repeated-root) improper stable node. $D=0$ with $J=\lambda I$? ::: Star (proper) node — har direction ek eigenvector.