Classification — elliptic, parabolic, hyperbolic (discriminant test): Visual walkthrough

4.7.1 · D2 · Maths › Partial Differential Equations › Classification — elliptic, parabolic, hyperbolic (discrimina

Step 1 — Symbols ka matlab kya hai

KYA. Hamein ek equation di gayi hai jo ek quantity $u$ aur uske rates of change ko mix karti hai. Pehle main har piece explain karta hoon.

$u$ ek number hai jo do cheezein pe depend karta hai: ek position $x$ aur ek doosra variable $y$ (aksar $y$ time hota hai). $u$ ko socho temperature at place $x$ aur time $y$ ke roop mein. To $u = u (x, y)$ .
$u_{x}$ ka matlab hai "kitni tezi se $u$ change hota hai jab tum $x$ ko thoda hatate ho" ( $y$ fixed rakhke). Yeh ek slope hai.
$u_{xx}$ ka matlab hai "woh slope khud kitni tezi se change hota hai" — $x$ -direction mein $u$ ka bending.
$u_{y y}$ = $y$ -direction mein bending. $u_{x y}$ = $x$ -slope kitna change hota hai jab tum $y$ ko nudge karte ho (twist).

YEH KYUN MATTER KARTE HAIN. Parent note claim karta hai sirf bending terms ( $u_{xx}, u_{x y}, u_{y y}$ ) equation ka character decide karte hain. To inhi pe hum dhyan denge. Poori equation hai:

$how much x -bend A u_{xx} + twist B u_{x y} + how much y -bend C u_{y y} + slopes, u itself, constant — we IGNORE these D u_{x} + E u_{y} + F u + G = 0$

Term by term: $A$ hai $x$ -bending pe weight, $C$ hai $y$ -bending pe weight, $B$ hai twist pe weight. Trailing group ( $D, E, F, G$ ) sirf shift aur damp karta hai — yeh character nahi badal sakta, isliye hum ise drop karte hain aur principal part $A u_{xx} + B u_{x y} + C u_{y y}$ rakhte hain.

PICTURE. Figure mein $u$ ko $x$ – $y$ floor ke upar ek bumpy surface ki tarah draw kiya gaya hai, jo dikhata hai ki " $x$ mein bend", " $y$ mein bend", aur "twist" teen alag-alag chhoti surfaces ki tarah kaise dikhte hain.

Figure — Classification — elliptic, parabolic, hyperbolic (discriminant test)

Step 2 — Woh ek sawaal jo sab kuch classify karta hai

KYA. Hum ek single geometric sawaal poochte hain: Kya $x$ – $y$ floor mein koi special curves hain jinke saath solution mein ek achanak "kink" ho sakta hai — uske second derivatives mein ek break? Aise curves ko characteristics kaha jaata hai.

YEH SAWAAL KYUN. Ek PDE kuch certain directions ke saath ek ordinary equation (ODE) ki tarah behave karta hai. Woh directions exactly wahan hain jahan second derivative surrounding data dwara pin down nahi hoti. Aise kitne real directions exist karte hain — yahi count karna poora game hai — aur jawab hoga $0$ , $1$ , ya $2$ . Woh count hi classification hai.

PICTURE. Do smooth solution patches ek curve ke saath milte hain; us curve ke across surface continuous hai lekin uska bending jump karta hai. Woh meeting curve ek candidate characteristic hai.

Step 3 — Special curve ko ek coordinate ki tarah rename karo

KYA. Maan lo special curve kisi function ka level set hai: $ϕ (x, y) = const$ . Hum $ϕ$ ko naya coordinate $ξ = ϕ$ ki tarah use karte hain (Greek "xi", sirf ek naye axis ka naam). Phir hum PDE ko $ξ$ ke terms mein rewrite karte hain.

KYUN. Agar koi curve sach mein kink-line hai, to us coordinate mein jo uske across jaata hai, pure second derivative $u_{ξ ξ}$ ko multiply karne wala coefficient vanish ho jaana chahiye — yehi precisely allow karta hai ki second derivative free (un-pinned) ho. To hum woh coefficient compute karte hain aur use zero set karte hain.

Chain rule (variables change karne ka rule) bending terms ko turn karta hai: $u_{xx} \to ϕ_{x}^{2} u_{ξ ξ} + \dots, u_{x y} \to ϕ_{x} ϕ_{y} u_{ξ ξ} + \dots, u_{y y} \to ϕ_{y}^{2} u_{ξ ξ} + \dots$

Yahan $ϕ_{x}$ = $x$ mein $ϕ$ ka slope, $ϕ_{y}$ = $y$ mein slope. $u_{ξ ξ}$ ko multiply karne wali sab cheezein collect karo:

$Q (ϕ_{x}, ϕ_{y}) = from u_{xx} A ϕ_{x}^{2} + from u_{x y} B ϕ_{x} ϕ_{y} + from u_{y y} C ϕ_{y}^{2}$

PICTURE. Purana grid $(x, y)$ warp kiya gaya hai taaki grid-lines ki ek family curve $ϕ = const$ ke saath lie kare; label $ξ$ un lines ke across jaata hai.

Step 4 — Coefficient ko zero set karo, slope lo

KYA. Characteristic exactly wahan hoti hai jahan $Q = 0$ : $A ϕ_{x}^{2} + B ϕ_{x} ϕ_{y} + C ϕ_{y}^{2} = 0.$

Ab ise curve ke slope mein convert karo. $ϕ = const$ ke saath ek tiny step $(d x, d y)$ lete hue $ϕ$ unchanged rehta hai, isliye: $d ϕ = ϕ_{x} d x + ϕ_{y} d y = 0 ⟹ \frac{d y}{d x} = - \frac{ϕ _{x}}{ϕ _{y}} .$

$ϕ_{y}^{2}$ se divide kyun karte hain. Hum chahte hain ki equation ek cheez — ek slope — ke terms mein bole. $Q = 0$ ko $ϕ_{y}^{2}$ se divide karo aur ratio ko naam do $m = \frac{ϕ _{x}}{ϕ _{y}}$ :

$A m^{2} + B m + C = 0.$

Term by term: yeh single unknown $m$ mein ek ordinary quadratic hai. Yahi payoff hai — ek daunting PDE sawaal ("kink curves?") ek simple cheez mein collapse ho gaya: "ek quadratic solve karo."

PICTURE. Ek characteristic curve jisme uska chhota sa step $(d x, d y)$ ek right triangle ke roop mein draw kiya gaya hai, ratio $d y / d x$ slope $m$ ke roop mein marked hai.

Step 5 — Quadratic solve karo; discriminant appear hota hai

KYA. $A m^{2} + B m + C = 0$ pe quadratic formula apply karo:

$m = \frac{- B \pm B ^{2} - 4 A C Δ}{2 A} .$

Square root ke neeche wali cheez discriminant hai: $Δ = B^{2} - 4 A C$

Kyunki $m = ϕ_{x} / ϕ_{y} = - d y / d x$ (sign tak), har solution $m$ ek characteristic ki direction hai. Characteristic slopes khud hain: $\frac{d y}{d x} = \frac{B \mp B ^{2} - 4 A C}{2 A} .$

Square root sab kuch kyun control karta hai. Ek positive number ka square root real hota hai; zero ka ek single value hoti hai; ek negative number ka imaginary. To $Δ$ ka sign decide karta hai kitne real slopes exist karte hain — aur yahi literally Step 2 se real kink-lines ka count hai.

PICTURE. $Δ$ ke liye ek number-line: right side par ( $Δ > 0$ ) do real roots; $0$ pe double root; left side par ( $Δ < 0$ ) roots real line se complex plane mein chale jaate hain.

Step 6 — Teen worlds, drawn

KYA. Teen cases padho aur unhe real characteristics se match karo.

$Δ > 0$ : do distinct real slopes → straight kink-lines ki do families. Yeh hyperbolic hai. Signals lines $x \pm c t = const$ pe finite speed se ride karte hain.
$Δ = 0$ : $\pm Δ$ collapse ho jaata hai, ek repeated real slope deta hai → lines ki ek akeli family. Yeh parabolic hai. Data $t = const$ ke saath smear ho jaata hai; infinite signal speed lekin sab kuch smooth ho jaata hai.
$Δ < 0$ : $Δ$ imaginary hai → koi real slopes nahi. Yeh elliptic hai. Koi preferred lines nahi; har interior point poori boundary ko feel karta hai; solution perfectly smooth hota hai.

NAMES KYUN. Wahi $B^{2} - 4 A C$ conic sections $A x^{2} + B x y + C y^{2} = const$ ko hyperbola / parabola / ellipse mein classify karta hai. Do real asymptote directions ↔ hyperbola ↔ do characteristics; aur aisa hi. Same algebra, isliye same names.

PICTURE. Teen side-by-side floors: do crossing lines (hyperbolic), ek repeated line (parabolic), ek swirl-with-no-lines (elliptic), har ek apni equation ke saath labelled.

Real characteristic lines exactly woh paths hain jo Method of Characteristics mein study ki jaati hain, aur har world ko jo data chahiye woh Well-posedness and Boundary Conditions mein set hota hai.

Step 7 — Degenerate aur position-dependent cases (skip mat karo)

KYA. Hum un corners ko cover karna chahte hain jo tidy formula chhupa deta hai.

Case $A = 0$ . Tab $A m^{2} + B m + C = 0$ genuine quadratic nahi raha — $2 A$ se divide karna illegal hai. Directly handle karo: agar $A = 0, B \neq = 0$ to equation $m$ mein linear hai, $B m + C = 0$ , ek finite slope $m = - C / B$ deta hai aur ek "vertical" characteristic (doosra root infinity tak chala gaya). Sign test phir bhi kaam karta hai: $Δ = B^{2} - 4 (0) C = B^{2} \geq 0$ , isliye $A = 0$ always hyperbolic hota hai (agar $B \neq = 0$ ) ya parabolic (agar $B = 0$ bhi ho).

Case $A = C = 0, B \neq = 0$ (jaise $u_{x y} = 0$ ). Tab $Δ = B^{2} > 0$ : hyperbolic, dono coordinate axes characteristics ke roop mein.

$u_{y y}$ missing (the heat trap). Agar koi $u_{y y}$ term nahi hai to $C = 0$ hai, $C = 1$ nahi. Heat equation $u_{t} - k u_{xx} = 0$ ke liye: $y \to t$ ke saath, $A = - k, B = 0, C = 0$ , isliye $Δ = 0 - 4 (- k) (0) = 0$ → parabolic. Aadat se $C = 1$ assign karna classic galti hai.

Position-dependent type (Tricomi). Jab $A, B, C$ $(x, y)$ pe depend karte hain, $Δ$ location ka ek function hai. $y u_{xx} + u_{y y} = 0$ ke liye: $A = y, B = 0, C = 1$ , isliye $Δ = - 4 y$ .

$y > 0 \Rightarrow Δ < 0$ → elliptic.
$y = 0 \Rightarrow Δ = 0$ → parabolic (single line $y = 0$ ).
$y < 0 \Rightarrow Δ > 0$ → hyperbolic.

YEH YAHAN KYUN BELONG KARTE HAIN. Parent note warn karta hai ki type hamesha fixed nahi hoti. Quasilinear equations ke liye coefficients $u$ pe bhi depend kar sakte hain, isliye type solution ke saath shift ho sakta hai. Hamesha $Δ$ ko us point (aur state) par evaluate karo.

PICTURE. Tricomi plane upar elliptic band, parabolic line $y = 0$ , aur neeche hyperbolic band mein split, sirf line ke neeche do characteristics sketch ki gayi hain.

Ek picture mein poora summary

Ek figure poori chain compress karta hai: bending terms → kink-lines ke liye poochho → naya coordinate coefficient $Q$ banata hai → $Q = 0$ set karo, divide karo, slope $m$ mein quadratic milti hai → uska discriminant $B^{2} - 4 A C$ → sign 0/1/2 real lines deta hai → elliptic / parabolic / hyperbolic.

Recall Feynman retelling — poora walkthrough plain words mein explain karo

Mere paas ek quantity $u$ hai jo do directions wale floor ke across bend karti hai. Sirf uska bending ( $u_{xx}, u_{x y}, u_{y y}$ ) uski personality set karta hai, isliye main sirf $A, B, C$ rakhta hoon. Main ek sawaal poochta hoon: surface mein hidden crease kahan ho sakta hai? Aise crease line ke saath, agar main use fresh coordinate ki tarah re-label karoon, to pure second derivative ke aage number zero hona chahiye. Woh number hai $A ϕ_{x}^{2} + B ϕ_{x} ϕ_{y} + C ϕ_{y}^{2}$ . Use zero set karo aur slope mein convert karo to ek plain quadratic milti hai $A m^{2} + B m + C = 0$ . Squares ka matlab hai at most do answers. Quadratic formula $B^{2} - 4 A C$ ko square root ke neeche rakhta hai, aur square root sirf sign ki parwah karta hai: — positive → do real creases → wave world (hyperbolic), — zero → ek crease → heat world (parabolic), — negative → koi real creases nahi, forced smooth → equilibrium world (elliptic). Agar $A, B, C$ mere khade hone ki jagah pe depend karte hain, to main wahan sign check karta hoon — aise ek equation (Tricomi) alag-alag regions mein teeno ho sakta hai.

Recall Quick self-test

Sirf principal part kyun use hota hai? ::: Lower-order terms shift/damp karte hain lekin characteristics create ya destroy nahi kar sakte. $Q = 0$ set karne se kaun si equation milti hai? ::: Slope $m$ mein ek quadratic $A m^{2} + B m + C = 0$ . $Δ$ ki value nahi, sign kyun matter karta hai? ::: Square root real / single / imaginary hota hai sirf is hisaab se ki uska argument $+$ / $0$ / $-$ hai. Heat equation coefficients kya hain? ::: $A = - k, B = 0, C = 0$ , isliye $Δ = 0$ (parabolic) — koi $u_{tt}$ nahi hai. Tricomi type at $y = - 3$ ? ::: $Δ = - 4 (- 3) = 12 > 0$ → hyperbolic.