Phase plane analysis — trajectories, critical points

4.6.22 · Maths › Ordinary Differential Equations

1. Trajectory kya hoti hai? (WHAT)

WHY autonomous matters: kyunki $t$ absent hai, velocity vector $(\dot{x},\dot{y})$ sirf is baat pe depend karta hai ki aap kahan hain, kab nahi. Toh har point se ek unique direction hoti hai → trajectories kabhi cross nahi hoti (siwaaye critical points ke).

2. Critical points (WHAT & WHY)

HOW inhe find karte hain: simultaneous algebraic equations $f=0,g=0$ solve karo. Yahi woh akele points hain jahan direction $(\dot{x},\dot{y})$ ek "kahan jaana hai" arrow ke roop mein undefined hoti hai, isliye yahi akele jagahein hain jahan trajectories mil sakti hain.

3. Critical points classify karna — linearization (80/20 core)

Derivation scratch se. Origin shift karo: maano $u=x-x_0,v=y-y_0$. Taylor expand karo: $\dot{u}=f_{x}u+f_{y}v+\text{(higher order)},\dot{v}=g_{x}u+g_{y}v+\cdots$ saare partials critical point pe evaluate kiye gaye hain. Higher-order terms drop karo: $\left(\begin{array}{c}\dot{u}\\ \dot{v}\end{array}\right)=J\left(\begin{array}{c}u\\ v\end{array}\right),J=\left(\begin{array}{cc}{f}_x& f_y\\ g_x& g_y\end{array}\right).$ $J$ critical point pe Jacobian hai.

WHY $e^{\lambda t}$: try karo $\mathbf{u}=\mathbf{v}{e}^{\lambda t}$. Tab $\lambda \mathbf{v}{e}^{\lambda t}=J\mathbf{v}{e}^{\lambda t}\Rightarrow J\mathbf{v}=\lambda \mathbf{v}$. Toh $\lambda$ eigenvalue hona chahiye. $\mathrm{Re}\lambda$ ka sign bolta hai grow (unstable) ya decay (stable); imaginary part rotation batata hai.

Classification table

4. Worked Example A — ek linear system (poori derivation)

System: $\dot{x}=x+y,\dot{y}=-2x+4y$? Ek clean wala use karte hain: $\dot{x}=-x+3y,\dot{y}=-3x-y?$

Ek transparent saddle karte hain: $\dot{x}=x+2y,\dot{y}=2x+y.$

Step 1 — critical point. Set karo $x+2y=0,2x+y=0$. Kyun? equilibrium ke liye dono rates zero chahiye. Subtract karo: $3x+3y=0\Rightarrow$... solve karo: sirf solution $(0,0)$ hai.

Step 2 — Jacobian. Yahan $f,g$ already linear hain toh $J=\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$. Kyun? $f_x=1,f_y=2,g_x=2,g_y=1$.

Step 3 — eigenvalues. $\tau =2,\Delta =1\cdot 1-2\cdot 2=-3$. ${\lambda }^2-2\lambda -3=0\Rightarrow (\lambda -3)(\lambda +1)=0\Rightarrow \lambda =3,-1.$ Step 4 — classify. Real, opposite signs $\Rightarrow$ saddle, unstable. Yeh step kyun? opposite signs matlab ek direction grow karta hai, ek shrink — saddle ki pehchaan yehi hai.

5. Worked Example B — ek spiral (nonlinear, linearized)

$\dot{x}=-y+x(1-x^2-y^2),\dot{y}=x+y(1-x^2-y^2).$

Step 1 — critical point. Origin pe cubic terms vanish ho jaate hain: $\dot{x}=-y,\dot{y}=x$, dono zero sirf $(0,0)$ pe. Cubics kyun drop karte hain? origin ke paas $x^2+y^2$ bahut chota hota hai.

Step 2 — Jacobian at origin. Linear part hai $\dot{x}=-y+x,\dot{y}=x+y$ (kyunki $1-x^2-y^2\to 1$). Toh $J=\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right).$ Yeh step kyun? $x\cdot 1$ term ko $x(1-\dots )$ se differentiate karo.

Step 3 — eigenvalues. $\tau =2,\Delta =1+1=2$, discriminant ${\tau }^2-4\Delta =4-8=-4<0$. $\lambda =1\pm i.$ Step 4 — classify. Complex with $\mathrm{Re}\lambda =1>0\Rightarrow$ unstable spiral baahir ki taraf. (Actually trajectories ek circular limit cycle $r=1$ ki taraf baahir spiral karti hain — yeh poori nonlinear sachai hai.)

6. Common mistakes (Steel-man + fix)

Flashcards

Connections

• Autonomous systems of ODEs
• Linearization and the Jacobian matrix
• Eigenvalues and eigenvectors
• Stability and Lyapunov functions
• Limit cycles and Poincaré–Bendixson theorem
• Trace–determinant plane

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