Worked examples — Phase plane analysis — trajectories, critical points

4.6.22 · D3 · Maths › Ordinary Differential Equations › Phase plane analysis — trajectories, critical points

Do numbers se sab classify hota hai. Maano $J$ ek critical point par Jacobian hai. Likho

• $\tau =\mathrm{tr}J={\lambda }_1+{\lambda }_2$ (yeh hai trace — diagonal ka sum),
• $\Delta =\det J={\lambda }_1{\lambda }_2$ (yeh hai determinant — cross-multiplied difference),

aur eigenvalues ${\lambda }^2-\tau \lambda +\Delta =0$ ke do roots hain, yaani ${\lambda }_{1,2}=\frac{1}{2}\left(\tau \pm \sqrt{{\tau }^2-4\Delta }\right).$ Root ke andar ki quantity, $D={\tau }^2-4\Delta$, hai discriminant: agar $D\ge 0$ toh roots real hain (straight-line flow), agar $D<0$ toh complex hain (rotating flow). $(\tau ,\Delta ,D)$ apne dimag mein rakho — neeche sab kuch sirf unke signs padhna hai. Poora map dekhne ke liye Trace–determinant plane dekho.

---

Scenario matrix

Har 2-D linear (ya linearized) critical point neeche exactly ek cell mein aata hai. Har column ek alag sign/degeneracy case hai; aakhri do rows woh "story" aur "exam" cells hain jo syllabus pasand karta hai.

Cell	$(\tau ,\Delta ,D)$ ke signs	Type	Worked in
C1	$\Delta <0$	Saddle (unstable)	Ex 1
C2	$\Delta >0,\tau <0,D>0$	Stable node (sink)	Ex 2
C3	$\Delta >0,\tau >0,D>0$	Unstable node (source)	Ex 3
C4	$\Delta >0,\tau <0,D<0$	Stable spiral	Ex 4
C5	$\Delta >0,\tau >0,D<0$	Unstable spiral	Ex 4 (contrast)
C6	$\Delta >0,\tau =0$	Center (pure imaginary) — borderline	Ex 5
C7	$D=0$ (repeated root)	Degenerate / star node	Ex 6
C8	$\Delta =0$	Degenerate: critical points ki poori line	Ex 7
C9	Multiple critical points, nonlinear	Har ek ko alag classify karo	Ex 8 (word problem)
C10	Exam twist: $\Delta >0,\tau <0$ par nonlinear center suspected	Linearization inconclusive	Ex 9

Upar ki picture poori matrix ko ek map ke roop mein dikhati hai: horizontal axis $\tau$ hai, vertical axis $\Delta$ hai, red parabola $D=0$ hai. Map yaad karo, das cells nahi. Apni nazar trace karo: black horizontal line ke neeche jaao ($\Delta <0$) aur tum hamesha saddle zone mein ho; uske upar, left jaana ($\tau <0$) stable hai aur right jaana ($\tau >0$) unstable hai, jabki red parabola ke upar cross karna tumhe node (neeche, real roots) se spiral (upar, rotating roots) mein switch kar deta hai.

---

Ex 1 — Cell C1: saddle ($\Delta <0$)

Forecast: off-diagonal mein do "3" bade hain — guess hai ki cross-coupling jeet jaayegi aur yeh unstable hoga. Saddle ya source?

1. Critical point. Set karo $x+3y=0$ aur $3x+y=0$. Yeh step kyun? equilibrium ke liye dono rates zero chahiye. Add/subtract karne se sirf $(0,0)$ milta hai.
2. Jacobian. Pehle se hi linear hai, toh $J=\left(\begin{array}{cc}1& 3\\ 3& 1\end{array}\right)$. Kyun? $f_x=1,f_y=3,g_x=3,g_y=1$.
3. Do numbers. $\tau =1+1=2$, $\Delta =(1)(1)-(3)(3)=-8$. Kyun? trace = diagonal sum, det = cross-product difference.
4. Sign padho. $\Delta <0$ $\Rightarrow$ saddle, hamesha unstable — $D$ ki zaroorat nahi. Kyun? $\Delta ={\lambda }_1{\lambda }_2<0$ opposite signs force karta hai, toh ek direction badhta hai, ek shrinks hota hai.

Verify karo: eigenvalues ${\lambda }^2-2\lambda -8=0\Rightarrow (\lambda -4)(\lambda +2)=0\Rightarrow \lambda =4,-2$ se. Opposite signs saddle confirm karte hain. ✓

---

Ex 2 — Cell C2: stable node ($\Delta >0,\tau <0,D>0$)

Forecast: dono diagonal terms $-3$ hain (andar ki taraf kheench rahe hain). Guess: sab kuch origin mein drain ho jaata hai — stable node.

1. Critical point. $-3x+y=0,x-3y=0\Rightarrow$ sirf $(0,0)$. Kyun? dono rates vanish hone chahiye.
2. Jacobian. $J=\left(\begin{array}{cc}-3& 1\\ 1& -3\end{array}\right)$. Kyun? linear right-hand side ke partials.
3. Teen numbers. $\tau =-6$, $\Delta =9-1=8$, $D={\tau }^2-4\Delta =36-32=4>0$. Ab $D$ kyun? $\Delta >0$ matlab same-sign eigenvalues — $D$ chahiye node (real, $D\ge 0$) aur spiral ($D<0$) mein distinguish karne ke liye.
4. Classify karo. $D>0$ real roots, $\tau <0$ dono negative $\Rightarrow$ stable node (sink). Kyun? dono eigenvalues negative matlab har direction decay karta hai.

Verify karo: ${\lambda }^2+6\lambda +8=0\Rightarrow (\lambda +2)(\lambda +4)=0\Rightarrow \lambda =-2,-4$. Dono real aur negative. ✓

---

Ex 3 — Cell C3: unstable node ($\Delta >0,\tau >0,D>0$)

Forecast: Ex 2 ke saare signs flip karo — same shape, opposite time direction. Guess: source.

1. Jacobian. $J=\left(\begin{array}{cc}3& 1\\ 1& 3\end{array}\right)$ (linear, partials padho). Pehle kyun? linear system → constant Jacobian.
2. Teen numbers. $\tau =6$, $\Delta =9-1=8$, $D=36-32=4>0$. Kyun? Ex 2 jaisi hi recipe.
3. Classify karo. $D>0$ real, $\tau >0$ dono positive $\Rightarrow$ unstable node (source). Kyun? dono eigenvalues positive → har direction grow karta hai.

Verify karo: ${\lambda }^2-6\lambda +8=0\Rightarrow (\lambda -2)(\lambda -4)=0\Rightarrow \lambda =2,4$. Positive reals. ✓ (Ex 2 ka exact mirror — isliye time-reversal sinks ko sources mein badal deta hai.)

---

Ex 4 — Cells C4 & C5: stable vs unstable spiral ($D<0$)

Forecast: off-diagonal terms $\pm 2$ equal-magnitude opposite-sign hain — yeh rotation hai. Diagonal ka sign decide karta hai andar ya bahar.

1. Jacobians. (a) $J_a=\left(\begin{array}{cc}-1& -2\\ 2& -1\end{array}\right)$, (b) $J_b=\left(\begin{array}{cc}1& -2\\ 2& 1\end{array}\right)$. Kyun? partials padho.
2. Numbers (a). $\tau =-2$, $\Delta =1+4=5$, $D=4-20=-16<0$. $D$ kyun? rotation detect karne ke liye ($D<0$ ⇒ complex roots).
3. Numbers (b). $\tau =+2$, $\Delta =1+4=5$, $D=4-20=-16<0$.
4. Classify karo. Dono mein $D<0$ ⇒ spirals. (a) $\tau <0$ ⇒ stable spiral (inward). (b) $\tau >0$ ⇒ unstable spiral (outward). Kyun? eigenvalue ka real part $=\tau /2$ grow/decay set karta hai; imaginary part swirl set karta hai.

Neeche ke do panels exactly yahi dikhate hain. Panel (a) mein red trajectory bahut door se shuru hoti hai aur origin par kale dot ki taraf andar ki taraf coil karti hai — yeh $\tau <0$ hai jo har loop mein energy drain karta hai. Panel (b) mein yeh origin ke paas se shuru hoti hai aur bahar fling hoti hai, wohi swirl ulta chalta hai kyunki $\tau >0$ hai. Dhyan do dono ek hi taraf spin karte hain (rotation $\pm 2$ off-diagonals se aata hai, jo unchanged hai); sirf in/out sense $\tau$ ke sign ke saath flip karta hai.

Verify karo: (a) ${\lambda }^2+2\lambda +5=0\Rightarrow \lambda =-1\pm 2i$, $\mathrm{Re}<0$. (b) ${\lambda }^2-2\lambda +5=0\Rightarrow \lambda =1\pm 2i$, $\mathrm{Re}>0$. ✓

---

Ex 5 — Cell C6: center ($\tau =0,\Delta >0$)

Forecast: diagonal terms bilkul nahi — kuch andar nahi kheench raha, kuch bahar nahi dhakkel raha. Guess: closed loops, ek center.

1. Jacobian. $J=\left(\begin{array}{cc}0& -4\\ 1& 0\end{array}\right)$. Kyun? $f_x=0,f_y=-4,g_x=1,g_y=0$.
2. Numbers. $\tau =0$, $\Delta =0-(-4)(1)=4>0$, $D=0-16=-16<0$. $\tau =0$ note kyun karo? yeh stable aur unstable spirals ke beech exact borderline par baitha hai.
3. Classify karo. $\tau =0,\Delta >0$ ⇒ center: eigenvalues pure imaginary $\Rightarrow$ neutral closed orbits, na grow na decay. "Neutral" kyun? $\mathrm{Re}\lambda =0$ matlab koi exponential factor hi nahi.

Verify karo: ${\lambda }^2+4=0\Rightarrow \lambda =\pm 2i$, purely imaginary. Trajectories ellipses hain $x^2+4y^2=\text{const}$ ($\frac{dy}{dx}=\frac{x}{-4y}\Rightarrow 4ydy=-xdx$ se, integrate karo). ✓

---

Ex 6 — Cell C7: repeated eigenvalue ($D=0$, star node)

Forecast: dono variables same rate par independently decay karte hain. Guess: sab kuch seedha origin ki taraf shrink karta hai.

1. Jacobian. $J=\left(\begin{array}{cc}-2& 0\\ 0& -2\end{array}\right)$. Kyun? dono equations uncoupled hain.
2. Numbers. $\tau =-4$, $\Delta =4$, $D=16-16=0$. $D=0$ flag kyun karo? yeh exact boundary hai — ek repeated root.
3. Classify karo. $\lambda =-2$ (double), aur $J$ pehle se hi diagonal hai ⇒ stable star node: trajectories seedhi rays hain origin mein jaati hui. Star kyun, spiral nahi? $D=0$ with diagonalizable $J$ radial lines deta hai; agar $J$ not diagonalizable hota toh improper node milta.

Verify karo: $(\lambda +2{)}^2=0\Rightarrow \lambda =-2,-2$. Trajectory shape: $\frac{dy}{dx}=\frac{-2y}{-2x}=\frac{y}{x}\Rightarrow y=cx$, straight lines. ✓

---

Ex 7 — Cell C8: $\Delta =0$, critical points ki poori line

Forecast: doosri equation pehli ki sirf double hai — degenerate. Guess: koi isolated point nahi balki poori line rest par hogi.

1. Critical points. $x+y=0$ aur $2x+2y=0$ same equation hain. Kyun? row 2 = 2 × row 1. Toh line $y=-x$ par har point ek critical point hai.
2. Jacobian. $J=\left(\begin{array}{cc}1& 1\\ 2& 2\end{array}\right)$, toh $\tau =3$, $\Delta =(1)(2)-(1)(2)=0$. Yeh do kyun compute karo? $\tau$ aur $\Delta$ hi classifier ke inputs hain, aur yahan $\Delta =0$ degeneracy ki tell-tale sign hai — ek eigenvalue zero honi hi chahiye.
3. Classify karo. $\Delta =0$ ⇒ ek eigenvalue $0$ hai. Kyun? $\det J={\lambda }_1{\lambda }_2=0$ ek zero eigenvalue force karta hai. Solve karo ${\lambda }^2-3\lambda =0\Rightarrow \lambda =0,3$.
4. Matlab. $\lambda =0$ direction equilibria ki line ke along hai (koi motion nahi). $\lambda =3>0$ direction points ko line se door push karta hai ⇒ line unstable hai. Yeh reading kyun? motion sirf non-zero-eigenvalue direction mein hota hai.

Verify karo: ${\lambda }^2-3\lambda =0\Rightarrow \lambda =0,3$. Ek zero (the line), ek positive (repulsion). ✓

---

Ex 8 — Cell C9: multiple critical points wala word problem

Forecast: guess hai ki kai rest states honge — extinction, rabbits-only, aur ek coexistence point jahan dono survive karte hain; coexistence aisa models mein aksar spiral ya center hoti hai.

1. Critical points. $\dot{y}=0\Rightarrow y=0$ ya $x=1$. Factor kyun karo? products zero hote hain jab koi factor zero ho.
   • $y=0$: tab $x(3-x)=0\Rightarrow x=0$ ya $x=3$. Points $(0,0),(3,0)$.
   • $x=1$: tab $1\cdot (3-1-2y)=0\Rightarrow 2-2y=0\Rightarrow y=1$. Point $(1,1)$.
2. General Jacobian. $f=x(3-x-2y)=3x-x^2-2xy$ aur $g=xy-y$ ke saath: $J=\left(\begin{array}{cc}3-2x-2y& -2x\\ y& x-1\end{array}\right).$ Pehle symbolic kyun? ek formula har point par evaluate karo.
3. Coexistence $(1,1)$ par. $J=\left(\begin{array}{cc}3-2-2& -2\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}-1& -2\\ 1& 0\end{array}\right)$. Yeh point kyun? yeh biologically interesting wala hai.
4. $(1,1)$ classify karo. $\tau =-1$, $\Delta =(-1)(0)-(-2)(1)=2$, $D=1-8=-7<0$. Complex with $\mathrm{Re}=\tau /2=-\frac{1}{2}<0$ ⇒ stable spiral: populations oscillate karte hain par coexistence par settle ho jaate hain. Spiral kyun? $D<0$ rotation, $\tau <0$ decay.

Verify karo: $(1,1)$ par: ${\lambda }^2+\lambda +2=0\Rightarrow \lambda =\frac{-1\pm \sqrt{-7}}{2}$, $\mathrm{Re}=-1/2$. Aur $(0,0)$: $J=\left(\begin{array}{cc}3& 0\\ 0& -1\end{array}\right)\Rightarrow \lambda =3,-1$ (saddle); $(3,0)$: $J=\left(\begin{array}{cc}-3& -6\\ 0& 2\end{array}\right)\Rightarrow \lambda =-3,2$ (saddle). ✓

---

Ex 9 — Cell C10: exam trap — linearization inconclusive

Forecast: parent note ne warn kiya tha: centers borderline hote hain. Guess: cubic term secretly decide karta hai, toh $a\ne 0$ ke liye "center" galat hai.

1. Linearize karo. Cubic terms drop karo: $\dot{x}=-y,\dot{y}=x$, $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. Cubics drop kyun karo? origin ke paas $x^2+y^2$ negligible hai.
2. Numbers. $\tau =0$, $\Delta =1$, $D=-4<0$ ⇒ pure imaginary $\lambda =\pm i$ ⇒ linear center. Suspicious kyun? $\mathrm{Re}\lambda =0$ exactly — higher-order terms ab negligible nahi rehe.
3. Radius se test karo. Maano $r^2=x^2+y^2$. Tab $\frac{1}{2}\frac{d(r^2)}{dt}=x\dot{x}+y\dot{y}=x(-y+ax{r}^2)+y(x+ay{r}^2)=a{r}^2(x^2+y^2)=a{r}^4$. Toh $\dot{r}=a{r}^3$. $\dot{r}$ kyun compute karo? yeh directly inward/outward drift reveal karta hai jo linear picture ne chhupaya tha. Stability and Lyapunov functions dekho.
4. Verdict. $a>0$: $\dot{r}>0$ ⇒ unstable spiral. $a<0$: $\dot{r}<0$ ⇒ stable spiral. $a=0$: genuine center. Toh linearization inconclusive hai — cubic term decide karta hai. Yeh matter kyun karta hai? yeh classic exam warning hai centers ke liye linearization par rely karne ke baare mein (aur kaise limit cycles appear ho sakte hain).

Verify karo: $\dot{r}=a{r}^3$ toh $\mathrm{sign}(\dot{r})=\mathrm{sign}(a)$. Linear eigenvalues $\pm i$ $a$ se independent. ✓

---

Recall Ek-line decision procedure (mujhe collapse karo aur recite karo)

$\Delta$ compute karo: agar $<0$ ⇒ saddle, khatam. Warna $\tau$ compute karo (stable agar $<0$) aur $D={\tau }^2-4\Delta$ (node agar $\ge 0$, spiral agar $<0$). Agar $\tau =0,\Delta >0$ ⇒ linear center — par maanne se pehle nonlinear terms check karo.