Phase plane analysis — trajectories, critical points: Exercises

4.6.22 · D4 · Maths › Ordinary Differential Equations › Phase plane analysis — trajectories, critical points

Do tools jinhe hum constantly use karenge. Main unhe simple words mein dobara state karta hoon taaki tumhe kabhi guess na karna pade.

Figure — Phase plane analysis — trajectories, critical points

Upar wali picture tumhara decision map hai (Trace–determinant plane). Neeche ke har exercise mein kuch isi par land karta hai. Ise khula rakho.

Level 1 — Recognition

Exercise 1.1

Diya gaya Jacobian $J = (- 2 0 0 - 5)$ ek critical point par, point ko classify karo (type + stability) bina kuch bhi haath se solve kiye, sirf diagonal padh ke.

Recall Solution 1.1

KYA karte hain: $τ$ aur $Δ$ seedha padho. $τ = - 2 + (- 5) = - 7, Δ = (- 2) (- 5) - 0 = 10.$ Discriminant $τ^{2} - 4Δ = 49 - 40 = 9 > 0$ → real eigenvalues. Kyunki matrix diagonal hai, eigenvalues hain hi diagonal entries: $λ_{1} = - 2, λ_{2} = - 5$ . YEH KAISA DIKHTA HAI: dono eigenvalues real hain, same sign, dono negative → har direction decay karta hai. Yeh ek stable node (sink) hai. Paths seedha andar kheench liye jaate hain.

Exercise 1.2

$J = (30 0 - 1)$ ko classify karo.

Recall Solution 1.2

$λ_{1} = 3, λ_{2} = - 1$ (diagonal entries). Ek badhta hai, ek ghatta hai. $Δ = (3) (- 1) = - 3 < 0$ . KYO: real eigenvalues opposite sign ke → saddle, hamesha unstable. Negative determinant akela hi "saddle" chillata hai (mnemonic dekho: negative det = saddle).

Level 2 — Application

Exercise 2.1

In sabke critical points dhundo: $\overset{x}{˙} = x + 2 y, \overset{y}{˙} = 2 x + y .$

Recall Solution 2.1

KYA: dono rates ko zero set karo (kuch nahi hilta). $x + 2 y = 0, 2 x + y = 0.$ KYO: critical point wahan hota hai jahan $f = g = 0$ . Pehle se, $x = - 2 y$ . Doosre mein substitute karo: $2 (- 2 y) + y = - 3 y = 0 \Rightarrow y = 0$ , isliye $x = 0$ . Answer: ek hi critical point hai $(0, 0)$ .

Exercise 2.2

Usi system ke liye, Jacobian banao, $τ, Δ$ , eigenvalues nikalo, aur classify karo.

Recall Solution 2.2

KYA — Jacobian. Right-hand sides pehle se hi linear hain, isliye partial derivatives sirf coefficients hain: $J = (f_{x} g_{x} f_{y} g_{y}) = (1221) .$ KYA — trace/det. $τ = 1 + 1 = 2$ , $Δ = (1) (1) - (2) (2) = 1 - 4 = - 3$ . KYA — eigenvalues. $λ^{2} - τ λ + Δ = λ^{2} - 2 λ - 3 = 0$ solve karo. $(λ - 3) (λ + 1) = 0 \Rightarrow λ = 3, - 1.$ KYO classify karein: opposite signs → saddle, unstable. Yeh parent note ke Worked Example A se exactly match karta hai.

Exercise 2.3

Decoupled system ke liye critical points aur classification: $\overset{x}{˙} = - x, \overset{y}{˙} = - 4 y .$

Recall Solution 2.3

Critical point: $- x = 0, - 4 y = 0 \Rightarrow (0, 0)$ . $J = (- 1 0 0 - 4)$ , isliye $λ = - 1, - 4$ . Dono real, negative, unequal → stable node (sink), jise improper node kehte hain kyunki dono decay rates alag hain.

Level 3 — Analysis

Exercise 3.1

Nonlinear system $\overset{x}{˙} = - y + x (1 - x^{2} - y^{2}), \overset{y}{˙} = x + y (1 - x^{2} - y^{2})$ ka origin par ek critical point hai. Wahan linearize karo aur classify karo.

Recall Solution 3.1

KYA — origin ke paas cubics hatao. Chhote $x, y$ ke liye, factor $1 - x^{2} - y^{2} \to 1$ . Isliye linear part hai $\overset{x}{˙} \approx - y + x, \overset{y}{˙} \approx x + y .$ KYO: origin ke paas, $x^{2} + y^{2}$ second-order-small hai — linear terms ke against negligible. KYA — Jacobian. $f = x - y \Rightarrow f_{x} = 1, f_{y} = - 1$ ; $g = x + y \Rightarrow g_{x} = 1, g_{y} = 1$ . $J = (11 - 1 1) .$ KYA — trace/det/discriminant. $τ = 2$ , $Δ = (1) (1) - (- 1) (1) = 2$ , discriminant $= 4 - 8 = - 4 < 0$ → complex eigenvalues (rotation). $λ = \frac{1}{2} (2 \pm - 4) = 1 \pm i .$ KYO classify karein: complex with real part $+ 1 > 0$ → unstable spiral (outward). Poora nonlinear sach: trajectories radius $r = 1$ par ek limit cycle ki taraf baahir spiral karte hain.

Exercise 3.2

$J = (0 - 4 10)$ ke liye, linearized behaviour classify karo aur woh caution batao jo apply hoti hai.

Recall Solution 3.2

$τ = 0 + 0 = 0$ , $Δ = (0) (0) - (1) (- 4) = 4 > 0$ . Discriminant $= 0 - 16 = - 16 < 0$ → complex, aur kyunki $τ = 0$ hai, real part zero hai: $λ = \frac{1}{2} (0 \pm - 16) = \pm 2 i .$ Classification: pure imaginary → center (closed loops), linear picture mein neutrally stable. Caution: center borderline case hai. Nonlinear terms ise ek slow spiral mein tip kar sakte hain, isliye linearization inconclusive hai. Ensure karne ke liye ki yeh sach mein center hai, tumhe ek energy / Lyapunov argument chahiye.

Level 4 — Synthesis

Exercise 4.1

$\overset{x}{˙} = y, \overset{y}{˙} = - x - x^{2}$ consider karo. Sabhi critical points dhundo, aur har ek ko linearization se classify karo.

Recall Solution 4.1

KYA — critical points. $y = 0$ aur $- x - x^{2} = 0$ set karo. $- x (1 + x) = 0 \Rightarrow x = 0$ ya $x = - 1$ . $y = 0$ dono baar: $critical points: (0, 0) aur (- 1, 0) .$ KYA — general Jacobian. $f = y \Rightarrow f_{x} = 0, f_{y} = 1$ . $g = - x - x^{2} \Rightarrow g_{x} = - 1 - 2 x, g_{y} = 0$ . $J (x, y) = (0 - 1 - 2 x 10) .$

$(0, 0)$ par: $J = (0 - 1 10)$ . $τ = 0, Δ = 1$ , discriminant $= - 4 < 0$ , $λ = \pm i$ → center (linear); borderline, caution se treat karo.

$(- 1, 0)$ par: $g_{x} = - 1 - 2 (- 1) = 1$ , isliye $J = (0110)$ . $τ = 0, Δ = (0) (0) - (1) (1) = - 1 < 0$ . $Δ < 0$ → saddle, unstable. (Eigenvalues $λ = \pm 1$ .)

Bada picture: flow ka origin par ek delicate center hai aur uski left mein ek saddle — ek classic nonlinear phase portrait.

Exercise 4.2

Ek system ka critical point par Jacobian $J = (2112)$ hai. Ise classify karo, aur batao ki trajectories point se door aur pass mein kis eigendirection ke saath align hoti hain.

Recall Solution 4.2

$τ = 4, Δ = 4 - 1 = 3$ , discriminant $= 16 - 12 = 4 > 0$ → real. $λ = \frac{1}{2} (4 \pm 2) = 3, 1$ . Dono positive → unstable node (source). Eigenvectors. $λ = 3$ ke liye: $(J - 3 I) v = (- 1 1 1 - 1) v = 0 \Rightarrow v = (1, 1)$ . $λ = 1$ ke liye: $(1111) v = 0 \Rightarrow v = (1, - 1)$ . Geometry: trajectories origin se nikalti hain. Fast direction $(1, 1)$ (bada $λ = 3$ ) door jakar dominate karta hai; trajectories origin ke paas slow direction $(1, - 1)$ ke tangent ho jaati hain. Yahi ek node ki pehchaan hai.

Level 5 — Mastery

Exercise 5.1

Ek linear system $\overset{x}{˙} = a x + b y, \overset{y}{˙} = c x + d y$ design karo jiska origin ek stable spiral ho. Explicit numbers do aur prove karo ki yeh kaam karta hai.

Recall Solution 5.1

KYA chahiye (target region): stable spiral ke liye:

complex eigenvalues → discriminant $τ^{2} - 4Δ < 0$ ;
decaying rotation → $Re λ < 0$ , yaani $τ < 0$ . KYO: complex ⇒ rotation; negative real part ⇒ shrinking radius. Ek construct karo. $J = (- 1 2 - 2 - 1)$ lo, yaani $a = - 1, b = - 2, c = 2, d = - 1$ . Check karo: $τ = - 1 + (- 1) = - 2 < 0$ ✓. $Δ = (- 1) (- 1) - (- 2) (2) = 1 + 4 = 5 > 0$ . Discriminant $= 4 - 20 = - 16 < 0$ ✓ complex. $λ = \frac{1}{2} (- 2 \pm - 16) = - 1 \pm 2 i$ . Real part $- 1 < 0$ ✓. Answer: $\overset{x}{˙} = - x - 2 y, \overset{y}{˙} = 2 x - y$ ka origin par ek stable spiral hai. Trajectories andar spiral karti hain.

Exercise 5.2

Ek system ka origin par Jacobian $J = (1 - 1 α 1)$ hai jahan $α$ ek real parameter hai. Woh value(s) of $α$ dhundo jahan type change hoti hai, aur har regime ka naam batao.

Recall Solution 5.2

$τ = 1 + 1 = 2$ ( $α$ se independent), $Δ = (1) (1) - (α) (- 1) = 1 + α$ . Discriminant $= τ^{2} - 4Δ = 4 - 4 (1 + α) = - 4 α$ . KYO discriminant track karein: iska sign real ↔ complex flip karta hai — wahi jagah type badlti hai.

$α < 0$ : discriminant $- 4 α > 0$ → real eigenvalues. Saath mein $Δ = 1 + α$ negative bhi ho sakta hai agar $α < - 1$ .
- $- 1 < α < 0$ : $Δ > 0, τ > 0$ → unstable node.
- $α < - 1$ : $Δ < 0$ → saddle.
- $α = - 1$ : $Δ = 0$ → degenerate (ek zero eigenvalue, fixed points ki ek line).
$α = 0$ : discriminant $= 0$ → repeated eigenvalue $λ = 1$ (borderline node).
$α > 0$ : discriminant $< 0$ → complex, aur $τ = 2 > 0$ → unstable spiral. Bifurcation values: $α = 0$ (node ↔ spiral boundary) aur $α = - 1$ (saddle ↔ node boundary $Δ = 0$ ke through).