Exercises — Stability of equilibria — stable, unstable, saddle, spiral, centre

KYA HAI: Matrix diagonal hai, isliye dono seedhe coordinate axes eigen-directions hain aur eigenvalues simply diagonal entries hain: ${\lambda }_1=-3$, ${\lambda }_2=-1$. Woh shortcut kyun kaam karta hai: diagonal matrix ke liye $\det (J-\lambda I)=(-3-\lambda )(-1-\lambda )=0$ turant factor ho jaata hai. Classify: $\tau =-3+(-1)=-4<0$, $\Delta =(-3)(-1)=3>0$, $D=(-4{)}^2-4(3)=4>0$. Real, distinct, dono negative $\Rightarrow$ stable node (sink). Aas-paas sab kuch andar slide karta hai.

$\tau =0+0=0$, $\Delta =(0)(0)-(2)(-2)=4>0$. Kyunki $\tau =0$ aur $\Delta >0$ hai, eigenvalues pure imaginary hain: $\lambda =\pm \sqrt{-4}=\pm 2i$. Centre — closed loops, neutrally stable (asymptotically stable nahi!).

Opposite signs $\Rightarrow$ product $\Delta ={\lambda }_1{\lambda }_2=(4)(-1)=-4<0$. Jab bhi $\Delta <0$ hota hai toh saddle hota hai — ${\lambda }_2$ eigen-direction ke saath attract karta hai, ${\lambda }_1$ wale ke saath repel karta hai. Hamesha unstable.

KYA HAI: Partial derivatives se milta hai $J=\left(\begin{array}{cc}{\partial }_x(2x+y)& {\partial }_y(2x+y)\\ {\partial }_x(x+2y)& {\partial }_y(x+2y)\end{array}\right)=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right).$ $\tau =4$, $\Delta =(2)(2)-(1)(1)=3$, $D=16-12=4>0$. Real distinct, dono positive (kyunki $\tau >0,\Delta >0,D>0$) $\Rightarrow$ unstable node (source). Eigenvalues $\lambda =\frac{4\pm 2}{2}=3,1$.

$J=\left(\begin{array}{cc}-1& 4\\ -2& -5\end{array}\right)$. $\tau =-1+(-5)=-6<0$, $\Delta =(-1)(-5)-(4)(-2)=5+8=13>0$, $D=36-52=-16<0$. Complex eigenvalues with $\mathrm{Re}=\tau /2=-3<0$ $\Rightarrow$ stable spiral. Yeh andar ki taraf rotate karta hai.

$J=\left(\begin{array}{cc}0& 1\\ -4& -4\end{array}\right)$. $\tau =-4$, $\Delta =(0)(-4)-(1)(-4)=4$, $D=16-16=0$. $D=0$ boundary case hai: repeated eigenvalue $\lambda =\tau /2=-2$ (dono negative) $\Rightarrow$ degenerate / critically-damped stable node. Physically yeh critical damping hai — bina overshoot ke sabse tez return.

Fixed points dhundo: $\dot{y}=0\Rightarrow y=0$; $\dot{x}=x(1-x)=0\Rightarrow x=0$ ya $x=1$. Toh do points: $(0,0)$ aur $(1,0)$. General mein Jacobian: $J=\left(\begin{array}{cc}1-2x& 0\\ 0& -2\end{array}\right)$. $(0,0)$ par: $J=\left(\begin{array}{cc}1& 0\\ 0& -2\end{array}\right)$, $\Delta =-2<0\Rightarrow$ saddle. $(1,0)$ par: $J=\left(\begin{array}{cc}-1& 0\\ 0& -2\end{array}\right)$, $\tau =-3<0,\Delta =2>0,D=9-8=1>0\Rightarrow$ stable node. Alag-alag evaluate kyun karte hain: ek hi nonlinear system alag-alag types host kar sakta hai — Jacobian is par depend karta hai ki tum kahan ho.

Fixed points: $y=0$ aur $x-x^3=x(1-x^2)=0\Rightarrow x=0,\pm 1$. Teen points: $(0,0),(1,0),(-1,0)$. Jacobian: $J=\left(\begin{array}{cc}0& 1\\ 1-3x^2& 0\end{array}\right)$. $(0,0)$ par: $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, $\tau =0,\Delta =-1<0\Rightarrow$ saddle (dono wells ke beech ki pahadi ki choti). $(\pm 1,0)$ par: $1-3(1)=-2$, toh $\left(\begin{array}{cc}0& 1\\ -2& 0\end{array}\right)$, $\tau =0,\Delta =2>0\Rightarrow$ centre (har well ka bottom). Linear kehta hai centre; kyunki $\tau =0$ yeh borderline hai — lekin yahan system conservative hai (energy $E=\frac{1}{2}{y}^2-\frac{1}{2}{x}^2+\frac{1}{4}{x}^4$ constant hai), isliye centres genuine hain.

Fixed points: $y=0$, $\sin x=0\Rightarrow x=0$ ya $x=\pi$. $J=\left(\begin{array}{cc}0& 1\\ -\cos x& -\frac{1}{2}\end{array}\right)$. $(0,0)$ par: $\left(\begin{array}{cc}0& 1\\ -1& -\frac{1}{2}\end{array}\right)$, $\tau =-\frac{1}{2}<0$, $\Delta =1>0$, $D=\frac{1}{4}-4=-\frac{15}{4}<0\Rightarrow$ stable spiral (neeche latkaa rest position, ab friction ki wajah se decay karta hai). $(\pi ,0)$ par: $-\cos \pi =1$, $\left(\begin{array}{cc}0& 1\\ 1& -\frac{1}{2}\end{array}\right)$, $\Delta =(0)(-\frac{1}{2})-(1)(1)=-1<0\Rightarrow$ saddle (seedha khada, abhi bhi unstable — damping ek balancing broomstick ko nahi bacha sakta).

$J=\left(\begin{array}{cc}\mu & -1\\ 1& \mu \end{array}\right)$. $\tau =2\mu$, $\Delta ={\mu }^2-(-1)(1)={\mu }^2+1>0$ hamesha. $D=4{\mu }^2-4({\mu }^2+1)=-4<0$ hamesha $\Rightarrow$ hamesha complex eigenvalues (hamesha spiral ya centre).

• (a) Stable spiral: chahiye $\tau <0\Rightarrow \mu <0$.
• (b) Centre: chahiye $\tau =0\Rightarrow \mu =0$.
• (c) Unstable spiral: $\tau >0\Rightarrow \mu >0$. Jab $\mu$ $0$ se upar jaata hai, eigenvalues ka real part $\mathrm{Re}\lambda =\mu$ zero cross karta hai — fixed point apni stability switch karta hai. Woh crossing ek Hopf bifurcation ka seed hai.

Vieta reverse karo: chahiye $\tau ={\lambda }_1+{\lambda }_2=8$ aur $\Delta ={\lambda }_1{\lambda }_2=15$. Koi bhi matrix jo woh trace/determinant rakhta ho kaam karega. Sabse simple diagonal choice: $J=\left(\begin{array}{cc}3& 0\\ 0& 5\end{array}\right).$ Check: $\tau =8>0$, $\Delta =15>0$, $D=64-60=4>0$ — real, dono positive $\Rightarrow$ unstable node. ✔ (Ek non-diagonal answer jaise $\left(\begin{array}{cc}4& 1\\ 1& 4\end{array}\right)$ ka bhi $\tau =8,\Delta =15$ hai.)

$\tau =-2$ fixed hai, $\Delta =k$ move kar raha hai:

• $k<0$: $\Delta <0\Rightarrow$ saddle.
• $k=0$: $\Delta =0$ — non-isolated / fixed points ki line (degenerate). Transition value $k=0$.
• $0<k<1$: $\Delta >0,\tau <0,D=4-4k>0\Rightarrow$ stable node.
• $k=1$: $D=4-4(1)=0$ — degenerate stable node (critical). Transition value $k=1$.
• $k>1$: $D<0$, $\tau <0\Rightarrow$ stable spiral. Toh ordered path hai: saddle $\to$ (line) $\to$ stable node $\to$ (degenerate) $\to$ stable spiral. Parabola $D=0$, yaani $\Delta ={\tau }^2/4=1$, node–spiral divide hai.

$J(0,0)=\left(\begin{array}{cc}0& 1\\ -1& -b\end{array}\right)$, toh $\tau =-b$, $\Delta =1$, $D=b^2-4$.

• $b=0$: $\tau =0,\Delta =1\Rightarrow$ centre (undamped pendulum, closed orbits).
• $0<b<2$: $\tau <0$, $D<0\Rightarrow$ stable spiral (underdamped — decay karte hue swing karta hai).
• $b=2$: $D=0\Rightarrow$ degenerate stable node (critical damping).
• $b>2$: $D>0$, dono eigenvalues real negative $\Rightarrow$ stable node (overdamped — bina swing ke ghar creep karta hai). Exact threshold: spiral→node switch $b=2$ par hota hai (jahan $\Delta =b^2/4$, yaani $b^2=4\Delta =4$). Yeh wahi kahani hai Damped harmonic oscillator ki: underdamped, critical, overdamped.

Fixed points. $\dot{x}=x(-1+y)=0\Rightarrow x=0$ ya $y=1$.

• Agar $x=0$: $\dot{y}=y=0\Rightarrow y=0$. Point $(0,0)$.
• Agar $y=1$: $\dot{y}=1-x^2=0\Rightarrow x=\pm 1$. Points $(1,1)$ aur $(-1,1)$. General Jacobian: $J=\left(\begin{array}{cc}-1+y& x\\ -2x& 1\end{array}\right)$.
• $(0,0)$ par: $\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)$, $\Delta =-1<0\Rightarrow$ saddle.
• $(1,1)$ par: $\left(\begin{array}{cc}0& 1\\ -2& 1\end{array}\right)$, $\tau =1$, $\Delta =(0)(1)-(1)(-2)=2>0$, $D=1-8=-7<0\Rightarrow$ unstable spiral.
• $(-1,1)$ par: $\left(\begin{array}{cc}0& -1\\ 2& 1\end{array}\right)$, $\tau =1$, $\Delta =(0)(1)-(-1)(2)=2>0$, $D=1-8=-7<0\Rightarrow$ unstable spiral. Ek central saddle ke dono taraf do mirror-image unstable spirals.

Linearisation yahan kyun bharosemand nahi hai: Hartman–Grobman guarantee karta hai ki linear picture nonlinear flow se match karti hai sirf hyperbolic fixed points par — woh jahan har eigenvalue ka $\mathrm{Re}(\lambda )\ne 0$ ho. Yahan $\lambda =\pm i$ ka $\mathrm{Re}=0$ hai: point non-hyperbolic hai, exactly woh excluded case. Chhote nonlinear terms neutral circles ko dheere andar (weak stable spiral) ya bahar (weak unstable spiral) bend kar sakte hain. Kya settle karta hai: ek Lyapunov function $V$ banao (ek energy-jaisa $V>0$ jisme $V(0)=0$ ho) aur trajectories ke saath $\dot{V}$ dekho. $\dot{V}<0$ $\Rightarrow$ truly stable; $\dot{V}=0$ $\Rightarrow$ genuine centre / conserved orbits; $\dot{V}>0$ $\Rightarrow$ unstable. Phase portrait phir visually confirm karta hai.