Phase plane analysis — trajectories, critical points

1. What is a trajectory? (WHAT)

WHY autonomous matters: because $t$ is absent, the velocity vector $(\dot x,\dot y)$ depends only on where you are, not when. So through each point there is a unique direction → trajectories never cross (except at critical points).

2. Critical points (WHAT & WHY)

HOW to find them: solve the simultaneous algebraic equations $f=0,\ g=0$. These are the only points where the direction $(\dot x,\dot y)$ is undefined as a "where to go" arrow, hence the only places trajectories may meet.

3. Classifying critical points — linearization (the 80/20 core)

Derivation from scratch. Shift origin: let $u=x-x_0,\ v=y-y_0$. Taylor expand: $\dot u = f_x u + f_y v + \text{(higher order)},\qquad \dot v = g_x u + g_y v + \cdots$ all partials evaluated at the critical point. Drop higher-order terms: $\begin{pmatrix}\dot u\\ \dot v\end{pmatrix}=J\begin{pmatrix}u\\ v\end{pmatrix},\qquad J=\begin{pmatrix} f_x & f_y\\ g_x & g_y\end{pmatrix}.$ $J$ is the Jacobian at the critical point.

WHY $e^{\lambda t}$: try $\mathbf u = \mathbf v\,e^{\lambda t}$. Then $\lambda \mathbf v e^{\lambda t} = J\mathbf v e^{\lambda t}\Rightarrow J\mathbf v = \lambda\mathbf v$. So $\lambda$ must be an eigenvalue. The sign of $\operatorname{Re}\lambda$ says grow (unstable) or decay (stable); imaginary part says rotate.

Classification table

4. Worked Example A — a linear system (full derivation)

System: $\dot x = x + y,\quad \dot y = -2x + 4y$? Let's instead use clean one: $\dot x = -x + 3y,\qquad \dot y = -3x - y?$

Let me do a transparent saddle: $\dot x = x + 2y,\qquad \dot y = 2x + y.$

Step 1 — critical point. Set $x+2y=0,\ 2x+y=0$. Why? equilibrium needs both rates zero. Subtract: $3x+3y=0\Rightarrow$... solve: only solution $(0,0)$.

Step 2 — Jacobian. Here $f,g$ are already linear so $J=\begin{pmatrix}1&2\\2&1\end{pmatrix}$. Why? $f_x=1,f_y=2,g_x=2,g_y=1$.

Step 3 — eigenvalues. $\tau=2,\ \Delta=1\cdot1-2\cdot2=-3$. $\lambda^2 -2\lambda -3=0\Rightarrow(\lambda-3)(\lambda+1)=0\Rightarrow \lambda=3,-1.$ Step 4 — classify. Real, opposite signs $\Rightarrow$ saddle, unstable. Why this step? opposite signs mean one direction grows, one shrinks — the hallmark of a saddle.

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

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

Step 1 — critical point. At origin the cubic terms vanish: $\dot x=-y,\ \dot y=x$, both zero only at $(0,0)$. Why drop cubics? near origin $x^2+y^2$ is tiny.

Step 2 — Jacobian at origin. Linear part is $\dot x=-y+x,\ \dot y=x+y$ (since $1-x^2-y^2\to1$). So $J=\begin{pmatrix}1&-1\\1&1\end{pmatrix}.$ Why this step? differentiate including the $x\cdot1$ term from $x(1-\dots)$.

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 $\operatorname{Re}\lambda=1>0\Rightarrow$ unstable spiral outward. (In fact trajectories spiral out toward a circular limit cycle $r=1$ — that's the full nonlinear truth.)

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

Concept Map

Hinglish (regional understanding)