Classification — elliptic, parabolic, hyperbolic (discriminant test): Worked examples

4.7.1 · D3 · Maths › Partial Differential Equations › Classification — elliptic, parabolic, hyperbolic (discrimina

Yeh ek drill page hai the parent classification topic ke liye. Hum discriminant test $Δ = B^{2} - 4 A C$ ke har possible case ko cover karenge — clean signs, sneaky zeros, position-dependent types, aur exam twists. Shuru karne se pehle, ek reminder jo zero se build kiya gaya hai.

Definition "Characteristic" aur coordinate

$ξ$ ka matlab — zero se build kiya Ek characteristic $(x, y)$ -plane mein ek special curve hai jiske along PDE data se second derivatives pin down karne mein fail ho jaata hai — isliye ek solution wahan ek "kink" carry kar sakta hai (iske second derivatives discontinuous ho sakte hain, yani curve ke ek taraf se doosri taraf jump kar sakte hain) jabki $u$ khud continuous rehta hai. Physically yeh woh lines hain jinka wave-front use karta hai.

Aisi curves dhundhne ke liye hum coordinates rename karte hain. Ek smooth function $ϕ (x, y)$ chuno aur iske value ko ek naya coordinate $ξ$ (Greek "ksi") ki tarah use karo: $ξ = ϕ (x, y)$ set karo. Tab ek curve " $ϕ = const$ " sirf ek curve " $ξ = const$ " hai — naye coordinate system ki ek grid-line. Symbol $u_{ξ ξ}$ ka matlab hai $u$ ka is naye coordinate ke respect mein second derivative, bilkul waise jaise $u_{xx}$ $x$ mein second derivative tha.

Intuition Slope formula

$m = \frac{- B \pm Δ}{2 A}$ kahan se aata hai $u$ ko naye coordinate $ξ = ϕ (x, y)$ mein rewrite karo. Chain rule principal part $A u_{xx} + B u_{x y} + C u_{y y}$ ko ek coefficient mein convert kar deta hai $Q (ϕ_{x}, ϕ_{y}) = A ϕ_{x}^{2} + B ϕ_{x} ϕ_{y} + C ϕ_{y}^{2}$ jo $u_{ξ ξ}$ ko multiply karta hai. Ek characteristic exactly woh curve $ϕ = const$ hai jahan yeh coefficient vanish karta hai ( $Q = 0$ ): agar $u_{ξ ξ}$ coefficient zero hai, toh woh second derivative equation se determined nahi hota, jo precisely "kink allowed" condition hai. $ϕ = const$ ke along slope $\frac{d y}{d x} = - \frac{ϕ _{x}}{ϕ _{y}}$ hai; $Q = 0$ ko $ϕ_{y}^{2}$ se divide karne par aur $m = ϕ_{x} / ϕ_{y}$ likhne par quadratic milta hai $A m^{2} + B m + C = 0.$ Isliye quadratic formula aata hai: $m = \frac{- B \pm B ^{2} - 4 A C}{2 A}$ . Square root ke andar wali cheez $Δ$ hai — isliye iska sign directly count karta hai ki kitne real characteristic slopes exist karte hain (do, ek, ya zero). Yahi " $Δ$ ke sign" aur "physics ki shape" ke beech ka bridge hai.

Scenario matrix

Neeche har problem mein exactly woh cell tagged hai jo woh fill karta hai. Agar tum yeh sab kar sako, toh exam ka koi bhi version tumhe surprise nahi kar sakta.

Cell	Kyun tricky hai	Example
C1 Clean hyperbolic	$Δ > 0$ , constant coeffs	Ex. 1
C2 Clean elliptic	$Δ < 0$ , constant coeffs	Ex. 2
C3 Parabolic via missing term	ek 2nd-order term absent → $C = 0$	Ex. 3
C4 Cross-term $B \neq = 0$	nonzero $B$ ko square karna padega	Ex. 4
C5 Convention trap	book $2 B u_{x y}$ likhti hai	Ex. 5
C6 Non-trivial parabolic	$B^{2} = 4 A C$ with all of $A, B, C \neq = 0$	Ex. 6
C7 Position-dependent type	$A, B, C$ functions of $x, y$ ; teenon regions + degenerate line	Ex. 7
C8 Quasilinear / solution-dependent	coeff $u$ par depend karta hai	Ex. 8
C9 Real-world word problem	physics → coefficients + data needed translate karo	Ex. 9
C10 Exam twist / limiting value	parameter $\to$ 0 ya sign flip	Ex. 10

Worked examples

Worked example Ex. 3 — Parabolic via ek missing second-order term (Cell C3)

Heat-type equation $u_{t} = 5 u_{xx}$ , yani $5 u_{xx} - u_{t} = 0$ ko classify karo ( $y \to t$ ; yaad raho $t$ sirf second variable hai, lekin $u_{t}$ sirf first order hai).

Forecast: bahut se students blurt karte hain "elliptic kyunki $C = 1$ ". Kya wahan koi $u_{tt}$ term bhi hai?

Har term ka order check karo. $u_{xx}$ second order hai; $u_{t}$ first order hai. Koi $u_{tt}$ nahi hai. Yeh step kyun? $C$ $u_{tt}$ ka coefficient hai — agar woh term exist hi nahi karta, toh $C = 0$ hai, $1$ nahi.
Coefficients. $A = 5$ , $B = 0$ , $C = 0$ — aur $C = 0$ precisely isliye kyunki step 1 mein koi $u_{tt}$ term supply karne ke liye nahi mila. Yeh step kyun? Term-by-term coefficients read karna hi ek maatra tarika hai " $C = 1$ kyunki $t$ doosra variable hai" trap se bachne ka; coefficient kisi actual term se aana chahiye.
Compute karo. $Δ = 0 - 4 (5) (0) = 0$ . Yeh step kyun? $- 4 A C$ mein ek zero factor poora product zero kar deta hai, $Δ = 0$ force karta hai — yeh parabolic signature hai.
Dekho. $Δ = 0 \Rightarrow$ parabolic. Yeh step kyun? $Δ = 0$ ka matlab hai characteristic quadratic ka ek repeated real root hai — exactly ek smearing direction, jo diffusion karta hai.

Verify: Ek repeated characteristic ( $t = const$ ) — diffusion ki single smearing direction. Heat equation ki known parabolic nature se match karta hai. ✔

Worked example Ex. 5 —

$2 B$ convention trap (Cell C5) Ek textbook apna PDE $A u_{xx} + 2 B u_{x y} + C u_{y y} = 0$ likhti hai aur $Δ = B^{2} - A C$ define karti hai. $3 u_{xx} + 6 u_{x y} + 2 u_{y y} = 0$ ke liye, dono conventions mein type dhundho aur dikhao ki dono agree karte hain.

Forecast: kya dono formulas same number denge? Same sign?

Hamara convention ( $B u_{x y}$ , $Δ = B^{2} - 4 A C$ ). Cross term $6 u_{x y}$ hai, isliye $B = 6$ , $A = 3$ , $C = 2$ . $Δ = 6^{2} - 4 (3) (2) = 36 - 24 = 12 > 0 \Rightarrow$ hyperbolic.
Book convention ( $2 B u_{x y}$ , $Δ = B^{2} - A C$ ). Ab $2 B = 6 \Rightarrow B = 3$ , $A = 3$ , $C = 2$ . $Δ = 3^{2} - (3) (2) = 9 - 6 = 3 > 0 \Rightarrow$ hyperbolic. Yeh step kyun? Is convention mein cross coefficient jo dikhta hai uska half hota hai, isliye tumhe $2 B = 6$ solve karna hoga; $B^{2} - A C$ formula phir missing factor of $4$ ke liye compensate karta hai.
Compare karo. Numbers alag hain ( $12$ vs $3$ ) lekin dono positive hain. Yeh step kyun? Classification sirf sign par depend karti hai, magnitude par kabhi nahi — isliye dono conventions ko, aur karte hain, type par agree karna chahiye.

Verify: $12 = 4 \times 3$ , isliye ek $Δ$ exactly doosre ka $4$ guna hai — same sign guaranteed. Dono hyperbolic kehte hain. ✔

Worked example Ex. 6 — Non-trivial parabolic:

$B^{2} = 4 A C$ jab har coefficient nonzero ho (Cell C6) $u_{xx} + 4 u_{x y} + 4 u_{y y} + u_{x} = 0$ ko classify karo.

Forecast: har second-order term present aur nonzero hai, isliye yeh missing-term-se-parabolic nahi ho sakta. Type guess karo computing se pehle — kya full $A, B, C$ kabhi parabolic hota hai?

Coefficients. $A = 1$ , $B = 4$ , $C = 4$ ; teenon nonzero hain. Yeh step kyun? Yahi example ka poora point hai — parabolic yahan isliye nahi hai kyunki koi term absent hai, isliye hum actual arithmetic $B^{2} - 4 A C$ test karte hain.
Compute karo. $Δ = 4^{2} - 4 (1) (4) = 16 - 16 = 0$ . Yeh step kyun? Parabolic ka matlab hai characteristic quadratic $A m^{2} + B m + C = 0$ ka ek repeated real root, jo exactly tab hota hai jab iska discriminant $B^{2} - 4 A C$ vanish kare — ek perfect-square principal part $u_{xx} + 4 u_{x y} + 4 u_{y y} = (\partial_{x} + 2 \partial_{y})^{2} u$ .
Dekho. $Δ = 0 \Rightarrow$ parabolic. Yeh step kyun? Ek repeated slope $m = - B / (2 A) = - 2$ ka matlab hai ek single characteristic direction, jo full set of coefficients ke saath bhi parabolic ki hallmark hai.

Verify: Repeated slope $m = \frac{- B}{2 A} = \frac{- 4}{2} = - 2$ , aur sach mein $A m^{2} + B m + C = 1 (4) + 4 (- 2) + 4 = 0$ . Principal part $(\partial_{x} + 2 \partial_{y})^{2} u$ factor karta hai — ek genuine perfect square, parabolic confirm karta hai. ✔

Worked example Ex. 7 — Position-dependent type: teen regions AUR degenerate line (Cell C7)

Tricomi-type equation $u_{xx} + x u_{y y} = 0$ ko plane ke har hisse mein classify karo.

Forecast: type yahan ek word nahi hai. Kahan elliptic, parabolic, hyperbolic hai?

Coefficients (ab position ke functions hain). $A = 1$ , $B = 0$ , $C = x$ . Yeh step kyun? $C$ literally variable $x$ hai, isliye $Δ$ ek single number ki jagah position ka function ban jaata hai — tum ek word mein jawab nahi de sakte.
Discriminant as a function. $Δ (x) = 0 - 4 (1) (x) = - 4 x$ . Yeh step kyun? $Δ$ ko $x$ mein expression likhne se sign flip dikhta hai: type point-by-point decide hota hai, isliye hum $Δ$ ka sign plane ke across track karte hain.
Case $x > 0$ : $Δ = - 4 x < 0 \Rightarrow$ elliptic (right half-plane). Yeh step kyun? Positive $x$ ke liye, $- 4 x$ negative hai, aur negative $Δ$ elliptic (koi real characteristics nahi) case hai.
Case $x = 0$ : $Δ = 0 \Rightarrow$ parabolic — degenerate line jahan type switch karta hai. Yeh step kyun? $x = 0$ exactly woh jagah hai jahan $Δ$ zero cross karta hai; ek continuous $Δ$ ko parabolic value $0$ se guzarna hi hoga jab woh dono regions ke beech sign change kare.
Case $x < 0$ : $Δ = - 4 x > 0 \Rightarrow$ hyperbolic (left half-plane). Yeh step kyun? Negative $x$ ke liye, $- 4 x$ positive hai, aur positive $Δ$ hyperbolic (do real characteristics) case hai.

Figure padho: vertical white line

x = 0

blueprint ko right par cyan elliptic zone (

Δ < 0

) aur left par amber hyperbolic zone (

Δ > 0

) mein split karti hai; woh line khud akeli parabolic boundary hai (

Δ = 0

). Teen white sample dots

x = 1, 0, - 1

par

Δ = - 4, 0, + 4

dikhate hain — line cross karte waqt sign flip dekho. Yeh transonic flow ka mechanism hai, jahan ek flow sonic surface cross karte waqt character change karta hai.

Verify: Sample points. $x = 1$ par: $Δ = - 4 < 0$ elliptic ✔. $x = 0$ par: $Δ = 0$ parabolic ✔. $x = - 1$ par: $Δ = 4 > 0$ hyperbolic ✔. ✔

Worked example Ex. 8 — Quasilinear: type solution par depend karti hai (Cell C8)

Ek quasilinear equation hai $u u_{xx} + u_{y y} = 0$ . Ise classify karo jahan $u = 2$ hai aur jahan $u = - 3$ hai.

Forecast: $u_{xx}$ ka coefficient $u$ khud hai. Kya type solution ki value par depend karke flip ho sakti hai?

Coefficients. $A = u$ , $B = 0$ , $C = 1$ . Yeh step kyun? Ek quasilinear PDE mein top-derivative coefficient $u$ par depend kar sakta hai; woh phir bhi linearly enter karta hai, isliye discriminant test pointwise apply hota hai — lekin ab "point" mein wahan $u$ ki value bhi shamil hai.
Discriminant. $Δ = 0 - 4 (u) (1) = - 4 u$ . Yeh step kyun? Kyunki $A = u$ hai, $Δ$ ka sign solution $u$ ke sign se control hota hai, isliye type tab tak naam nahi de sakte jab tak $u$ pata na ho.
Jahan $u = 2$ : $Δ = - 8 < 0 \Rightarrow$ elliptic. Yeh step kyun? Actual solution value $u = 2$ substitute karne par symbolic $Δ$ ek aisa number ban jaata hai jiska sign hum padh sakte hain.
Jahan $u = - 3$ : $Δ = 12 > 0 \Rightarrow$ hyperbolic. Yeh step kyun? Ek alag solution value $- 4 u$ ka sign flip kar deti hai, aur isliye type bhi flip ho jaati hai — yeh quasilinear classification ki defining feature hai.

Verify: $u = 2$ : $- 4 (2) = - 8 < 0$ elliptic ✔. $u = - 3$ : $- 4 (- 3) = 12 > 0$ hyperbolic ✔. Toh sach mein type tab tak naam nahi de sakte jab tak solution ka sign pata na ho. ✔

Worked example Ex. 9 — Real-world word problem + data needed (Cell C9)

Ek patla metal rod garam kiya jaata hai. Physicists iske temperature $T (x, t)$ ko model karte hain "time mein temperature change ki rate space mein curvature ke constant $κ > 0$ gune ke barabar hai": $T_{t} = κ T_{xx}$ . Ise classify karo, aur batao ki solve karne ke liye kaunsa data dena hoga.

Forecast: "time mein" aur "space mein curvature" — kaunsa single type spreading heat describe karta hai, aur kya ise data har jagah chahiye, ya sirf ek starting condition?

Standard form mein translate karo. $κ T_{xx} - T_{t} = 0$ , jahan $(x, t)$ do variables hain ( $t$ $y$ ka role play karta hai). Yeh step kyun? Coefficients padhne se pehle physics ko $A u_{xx} + B u_{x t} + C u_{tt}$ ke saath line up karna zaroori hai.
Coefficients. $A = κ$ , $B = 0$ , $C = 0$ — aur $C = 0$ isliye kyunki model mein $T_{t}$ hai (time mein first order) lekin koi $T_{tt}$ term nahi hai jo coefficient supply kare. Yeh step kyun? "Time mein rate of change" ek first derivative hai; $C = 0$ assign karna tabhi sense banata hai jab hum notice karein ki physics mein koi second time-derivative exist nahi karta.
Discriminant. $Δ = 0 - 4 (κ) (0) = 0 \Rightarrow$ parabolic. Yeh step kyun? Missing $T_{tt}$ se zero factor $Δ = 0$ force karta hai, parabolic (diffusion) signature — is intuition se match karta hai ki heat sharp fronts ki tarah propagate karne ki jagah smooth karta hai.
Required data batao. Parabolic problems ko ek initial condition $T (x, 0) = f (x)$ chahiye (starting temperature profile) plus rod ke dono ends par boundary conditions (e.g. fixed end temperatures $T (0, t)$ aur $T (L, t)$ ), aur phir tum $t$ mein march forward karte ho. Tum "final time" par data prescribe nahi karte, aur karna bhi nahi chahiye — woh ill-posed hoga. Contrast: ek wave equation additionally initial velocity $T_{t} (x, 0)$ bhi maangta, aur ek elliptic (Laplace) problem ko poori closed boundary par data chahiye. (Dekho Well-posedness and Boundary Conditions aur Heat Equation — separation of variables.)

Verify: $κ = 1$ ke saath: $Δ = - 4 (1) (0) = 0$ , parabolic — diffusion ka correct, well-known character (infinite signal speed lekin smoothing). "Ek initial condition + do boundary conditions, $t$ mein march forward" exactly standard well-posed heat problem hai. ✔

Worked example Ex. 10 — Exam twist with a limiting parameter (Cell C10)

Ek PDE mein ek knob $ε$ hai: $ε u_{tt} + u_{xx} - u_{t} = 0$ (variables $x, t$ ; yahan $A$ $u_{xx}$ ka coefficient hai, $C$ $u_{tt}$ ka coefficient hai). $ε > 0$ ke liye classify karo, phir limit $ε \to 0^{+}$ examine karo.

Forecast: jab $ε$ zero tak shrink hota hai type ka kya hota hai — kya woh identity change karta hai?

Coefficients. $A = 1$ ( $u_{xx}$ se), $B = 0$ , $C = ε$ ( $u_{tt}$ se). $- u_{t}$ first order hai, discard karo. Yeh step kyun? Limit track karne se pehle second-order coefficients identify karne zaroori hain; sirf $u_{xx}, u_{x t}, u_{tt}$ hi $Δ$ mein enter karte hain.
Discriminant. $Δ = 0 - 4 (1) (ε) = - 4 ε$ . Yeh step kyun? $ε$ ko symbolic rakhne se hum dekh sakte hain ki knob zero ki taraf turn karne par $Δ$ ka sign kaise behave karta hai, na ki woh dependence bahut jaldi lose karne par.
$ε > 0$ ke liye: $Δ = - 4 ε < 0 \Rightarrow$ elliptic.
Limit $ε \to 0^{+}$ : $u_{tt}$ term vanish ho jaata hai, chhod kar $u_{xx} - u_{t} = 0$ , yani ek heat equation. Ab $C = 0$ aur $Δ = 0 \Rightarrow$ parabolic. Yeh step kyun? Ek singular limit equation ka order change kar sakta hai (top time-derivative disappear ho jaata hai), aur isliye type bhi — ek favourite exam trap (ek "singular perturbation").

Verify: $ε = 0.5$ : $Δ = - 2 < 0$ elliptic ✔. $ε = 0.01$ : $Δ = - 0.04 < 0$ still elliptic ✔. Exactly $ε = 0$ par: term disappear hota hai, $Δ = 0$ parabolic ✔. Type smoothly parabolic ki taraf approach nahi karta — yeh us moment snap karta hai jab top term mar jaata hai. ✔

Active recall

Recall Cover karo aur jawab do

Kaunsa cell tha jahan $C = 0$ isliye tha kyunki ek term missing tha, jisse $Δ = 0$ mila? ::: Cell C3 (heat-type), Ex. 3. Kaunsa example parabolic hai jabki $A, B, C$ TEENO nonzero hain? ::: Ex. 6, jahan $B^{2} = 4 A C$ ( $16 = 16$ ) aur principal part ek perfect square hai $(\partial_{x} + 2 \partial_{y})^{2} u$ . Ex. 4 mein cross term $B = 4$ ne same-sign $A, C$ ko override kiya. $Δ$ kya tha? ::: $Δ = 16 - 4 = 12 > 0$ , hyperbolic. Tricomi Ex. 7 mein, kis line par type parabolic hai? ::: Line $x = 0$ par, jahan $Δ = - 4 x = 0$ . Quasilinear $u u_{xx} + u_{y y} = 0$ ke liye, kya yeh elliptic hai ya hyperbolic jahan $u = - 3$ hai? ::: Hyperbolic, kyunki $Δ = - 4 (- 3) = 12 > 0$ . Do $Δ$ conventions ( $B^{2} - 4 A C$ vs $B^{2} - A C$ ) dono sahi kyun ho sakte hain? ::: Woh hamesha same sign share karte hain; sirf sign classify karta hai. $Δ$ use karne se pehle kya check karna zaroori hai? ::: Ki $A, B, C$ mein se kam se kam ek nonzero ho — agar teeno zero hain toh equation second order nahi hai. Ek parabolic (heat) problem ko kaunsa data chahiye? ::: Ek initial condition plus har end par boundary conditions, phir time mein march forward karo.