Phase plane analysis — trajectories, critical points: Question bank

4.6.22 · D5 · Maths › Ordinary Differential Equations › Phase plane analysis — trajectories, critical points

Neeche use ki gayi notation ki reminders (sab parent mein build ki gayi hain):

$\overset{x}{˙} = f (x, y), \overset{y}{˙} = g (x, y)$ — ek autonomous system ( $t$ kabhi right side par nahi aata).
Ek critical point wahan hota hai jahan $f = g = 0$ (velocity arrow zero hai — koi motion nahi).
$J$ = Jacobian, first partial derivatives ki matrix $(f_{x} g_{x} f_{y} g_{y})$ jo critical point par evaluate ki jaati hai.
$τ = tr J = λ_{1} + λ_{2}$ (eigenvalues ka sum), $Δ = det J = λ_{1} λ_{2}$ (eigenvalues ka product).
Discriminant $τ^{2} - 4Δ$ real vs complex eigenvalues decide karta hai: jab $Δ > 0$ ho, ek positive discriminant ( $τ^{2} - 4Δ > 0$ ) real eigenvalues deta hai (ek node), ek negative discriminant ( $τ^{2} - 4Δ < 0$ ) complex eigenvalues deta hai (ek spiral).
$r = x^{2} + y^{2}$ origin se distance (radius) hai — sirf last edge-case item mein use hota hai.

True or false — justify

Kisi bhi solution ki trajectory aakhirkar ek critical point tak pahunch jaati hai.

False. Trajectories infinity tak bhi jaati hain, ya phir ek closed orbit (center ya limit cycle) par hamesha loop karti rehti hain, kabhi rest point tak nahi pahunchti.

Do alag trajectories ek hi ordinary (non-critical) point se guzar sakti hain.

False. Ek autonomous field har point par exactly ek velocity arrow deta hai, toh ek hi direction hoti hai — do curves ka wahan se guzarna matlab do directions chahiye honge. Woh sirf critical points par mil sakti hain jahan arrow zero hota hai.

Ek akeli trajectory khud ko cross kar sakti hai (loop bana sakti hai) ek ordinary point par.

False for a simple crossing, true only as a closed loop. Ek trajectory ek closed curve (periodic orbit) ho sakti hai, lekin woh khud ko transversally cross nahi kar sakti — woh phir se ek point par do directions ki maang karti.

Agar

det J > 0

ho toh critical point stable hai.

False.

det J = λ_{1} λ_{2} > 0

ka matlab sirf yeh hai ki eigenvalues ek hi sign share karte hain (node ya spiral), yeh nahi ki woh sign negative hai. Stability

τ = tr J

se decide hoti hai: negative

τ

stable, positive

τ

unstable.

Agar

det J < 0

ho toh sahi trace ke liye point ek stable node ho sakta hai.

False.

det J < 0

ka matlab

λ_{1} λ_{2} < 0

hai, yani real eigenvalues opposite sign ke — hamesha ek saddle, hamesha unstable, trace chahe kuch bhi ho.

Linearization mein pure imaginary eigenvalues guarantee karte hain ki nonlinear system mein center hai.

False. Center borderline case hai

Re λ = 0

; choti si nonlinear terms ise slow spiral mein push kar sakti hain. Linearization yahan inconclusive hai — energy/Lyapunov methods use karo.

Har autonomous 2D system mein kam se kam ek critical point hota hai.

False. Example ke liye

\overset{x}{˙} = 1, \overset{y}{˙} = 0

mein

f = 1 \neq = 0

har jagah hai — koi rest point hi nahi, toh koi critical points nahi.

Ek stable node aur ek stable spiral nearby trajectories ke liye same long-term fate rakhte hain.

True in fate, different in path. Dono nearby trajectories ko point mein kheenchte hain; node straight-ish eigendirections ke saath approach karta hai, spiral wind karke aata hai kyunki eigenvalues mein imaginary part (rotation) hota hai.

Time reverse karna (

t \to - t

) ek stable node ko unstable node mein badal deta hai.

True. Time reverse karna velocity field ko negate karta hai, har eigenvalue ke real part ka sign flip karta hai; sinks sources ban jaate hain aur vice versa (saddle saddle rehta hai, bas uske do signs roles swap karte hain).

Slope relation

\frac{d y}{d x} = \frac{g}{f}

original system ki saari information rakhta hai.

False. Yeh geometry (curves ki shape) rakhta hai lekin timing/speed hata deta hai — ab tum nahi jaante ki point kitni tezi se move karta hai ya curve par time-direction kya hai.

Spot the error

f (x_{0}, y_{0}) = 0

akela

(x_{0}, y_{0})

ko critical point banata hai."

Error: tumhe dono

f = 0

aur

g = 0

chahiye. Agar sirf

f = 0

hai toh

\overset{y}{˙} = g \neq = 0

, toh point abhi bhi vertically move kar raha hai — rest mein nahi.

"Eigenvalues

λ^{2} + τ λ + Δ = 0

ke roots hain."

Error: trace term ka sign galat hai. Characteristic equation

λ^{2} - τ λ + Δ = 0

hai (trace ka minus), jisse

λ = \frac{1}{2} (τ \pm τ^{2} - 4Δ)

milta hai.

"Discriminant

τ^{2} - 4Δ < 0

ka matlab system unstable hai."

Error: negative discriminant ka sirf matlab hai ki eigenvalues complex hain (spiral/center), stability ke baare mein kuch nahi kehta. Stability abhi bhi

τ

ke sign se aati hai.

"Hum

x (1 - x^{2} - y^{2})

x

drop karke linearize karte hain kyunki

x

small hai."

Error: tum sirf higher-order cubic

- x^{3} - x y^{2}

drop karte ho. Linear term

x \cdot 1 = x

rehna chahiye — woh Jacobian mein belong karta hai. Ise drop karne se answer poori tarah badal jaata hai.

"Kyunki trajectories cross nahi kar sakti, phase portrait mein koi closed curves nahi hain."

Error: ek closed curve kisi aur trajectory ko cross nahi karti — woh khud ek continuous trajectory hai jo apni starting point par wapas aati hai. Non-crossing loops permit karta hai (centers, limit cycles); yeh alag orbits ke beech intersections forbid karta hai.

"Ek saddle ek direction mein stable hai toh woh 'half stable' hai."

Error: single incoming (stable) direction se koi bhi perturbation outgoing direction ke saath grow karta hai, toh saddle unstable hai, full stop — almost every nearby trajectory aakhirkar chali jaati hai.

Why questions

t

eliminate karne par trajectories ke liye first-order ODE kyun milta hai?

Kyunki

\frac{d y}{d x} = \frac{y ˙}{x ˙} = \frac{g ( x , y )}{f ( x , y )}

directly

y

aur

x

ko ek single derivative se relate karta hai — ek equation, first order — do time-ODEs ko ek shape-ODE se trade karta hai.

J

mein partial derivatives critical point par evaluate kyun karni chahiye?

Linearization us specific point par field ko uski tangent plane se approximate karta hai; slopes

f_{x}, f_{y}, g_{x}, g_{y}

jagah-jagah change hote hain, toh sirf

(x_{0}, y_{0})

par unki values us rest point ke paas flow describe karti hain.

\dot{u} = J u

ke solutions

e^{λ t}

form kyun lete hain?

u = v e^{λ t}

try karne par ODE

J v = λ v

mein badal jaata hai — eigenvalue equation. Toh har exponential mode ko ek eigenvector par ride karna hota hai, aur

λ

necessarily

J

ka eigenvalue hona chahiye.

Re λ

ka sign stability kyun decide karta hai jabki imaginary part shape decide karta hai?

e^{λ t} = e^{(Re λ) t} e^{i (Im λ) t}

: real part growth/decay envelope set karta hai (andar vs bahar), aur imaginary part rotation set karta hai (spiral vs straight-line node).

Center ko "borderline" case kyun kehte hain?

Yeh Trace–determinant plane ki

τ = 0

line par exactly

Δ > 0

ke saath baitha hai, stable aur unstable spirals ke beech —

τ

ko zero se koi bhi nudge ise decisively ek taraf tip kar deta hai, toh yeh structurally fragile hai.

Aam taur par critical point ke paas higher-order terms kyun ignore kar sakte hain?

Point ke paas displacement tiny hai, aur ek tiny number ke squares/cubes number se bahut chhote hote hain, toh linear terms dominate karte hain — sivaay borderline cases (center, zero eigenvalue) ke jahan chhoti terms tie decide karti hain.

Edge cases

Agar

J

ka repeated real eigenvalue ho (

τ^{2} - 4Δ = 0

) toh kya?

Yeh node aur spiral ke beech ki boundary hai — ek degenerate/improper node. Uski stability abhi bhi us repeated

λ

ke sign se set hoti hai (equivalently

τ

ke sign se).

Agar ek eigenvalue exactly zero ho (

Δ = 0

) toh kya?

Linearization degenerate hai: near-equilibria ki poori ek line hai aur flow clearly na attract karta hai na repel karta hai us par. Linearization inconclusive hai; higher-order terms rakhne padenge.

Jab tum ek critical point par

\frac{d y}{d x} = \frac{g}{f}

compute karne ki koshish karte ho toh kya hota hai?

Dono

f = 0

aur

g = 0

hain, jisse

\frac{0}{0}

milta hai — slope genuinely undefined hai wahan. Yahi exact reason hai ki multiple trajectories ek critical point par approach karke mil sakti hain.

\overset{x}{˙} = - y + x (1 - x^{2} - y^{2}), \overset{y}{˙} = x + y (1 - x^{2} - y^{2})

ke liye, kya outward spiral poori kahani hai?

Nahi. Linear part origin par ek unstable spiral deta hai, lekin full nonlinear system mein circle

r = x^{2} + y^{2} = 1

par ek stable limit cycle hai jis par trajectories spiral karte hain — linearization sirf local origin behaviour describe karta hai.

Agar field

\overset{x}{˙} = y, \overset{y}{˙} = - x

hai (ek linear center), kya hum closed loops par trust kar sakte hain?

Haan, specifically yahan, kyunki system exactly linear hai — koi higher-order terms nahi hain jo ise spoil karein. Centers ke baare mein warning sirf tab apply hoti hai jab nonlinear terms exist karein jo linearization ne discard ki hon.

Time-reversal ek saddle ke saath kya karta hai?

Yeh dono eigenvalues negate karta hai, toh incoming aur outgoing directions swap ho jaate hain — lekin ek abhi bhi positive aur ek abhi bhi negative rehta hai, toh woh saddle rehta hai (bas uske "in" aur "out" arms exchange ho jaate hain).

Kya ek trajectory finite time mein critical point tak pahunch sakti hai?

Aam taur par nahi hyperbolic sink ke liye: approach exponential hai (

e^{λ t}

with

λ < 0

), toh actually pahunchne mein infinite time lagta hai — woh sirf arbitrarily close hoti jaati hai.

Recall Traps ki ek-line summary

Do zindagi bhar ki galtiyan yahan reduce ho jaati hain: (1) dono chahiye $f = 0$ aur $g = 0$ ; (2) det type deta hai, trace stability deta hai; (3) centers aur zero eigenvalues borderline hain — linearization jhooth bol sakta hai.