Classification — elliptic, parabolic, hyperbolic (discriminant test): Exercises

4.7.1 · D4 · Maths › Partial Differential Equations › Classification — elliptic, parabolic, hyperbolic (discrimina

Yeh page parent classification note ke liye ek self-test ladder hai. Pehle har problem pen se khud solve karo, phir solution kholo. Yahan har symbol parent mein build kiya gaya hai; agar koi step achanak lage, wahan ki derivation dobara padho.

Neeche diya figure poori page ka master reference hai: horizontal axis $Δ$ ki value hai, aur teen coloured bands teen verdicts hain. Jab koi problem finish karo, apna computed $Δ$ is line par locate karo aur check karo ki band tumhare named type se match karta hai.

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

Level 1 — Recognition

Coefficients read off karo, $Δ$ compute karo, type batao.

L1.1 Classify karo $4 u_{xx} + 5 u_{y y} + u_{x} - 3 u = 0$ .

L1.2 Classify karo $u_{xx} - 6 u_{x y} + 9 u_{y y} = 0$ .

L1.3 Classify karo $2 u_{xx} + 3 u_{x y} - u_{y y} + 7 u_{y} = 0$ .

Recall Solutions L1

L1.1 Second-order terms: $4 u_{xx}$ , $0 \cdot u_{x y}$ , $5 u_{y y}$ . Toh $A = 4, B = 0, C = 5$ . $u_{x}$ aur $u$ lower order terms hain — yeh kabhi $Δ$ mein enter nahi karte. $Δ = 0^{2} - 4 (4) (5) = - 80 < 0 \Rightarrow elliptic.$

L1.2 $A = 1, B = - 6, C = 9$ . $Δ = (- 6)^{2} - 4 (1) (9) = 36 - 36 = 0 \Rightarrow parabolic.$ Note karo ki $u_{xx} - 6 u_{x y} + 9 u_{y y} = (u_{x} - 3 u_{y})$ -type structure hai — ek perfect square, isliye ek repeated characteristic.

L1.3 $A = 2, B = 3, C = - 1$ . ( $7 u_{y}$ first order hai → yeh $E$ hai, ignore karo.) $Δ = 3^{2} - 4 (2) (- 1) = 9 + 8 = 17 > 0 \Rightarrow hyperbolic.$

Level 2 — Application

Ab coefficient positions disguised hain, ya equation oddly likhi gayi hai.

L2.1 Heat equation $u_{t} = k u_{xx}$ (jahan $k > 0$ ) ko classify karo, $t$ ko second variable maankar.

L2.2 Classify karo $u_{xx} + 4 u_{x y} + 4 u_{y y} + 3 u_{x} = 0$ .

L2.3 Classify karo $3 u_{xx} + 2 u_{x y} + 3 u_{y y} = 0$ .

Recall Solutions L2

L2.1 Standard form mein rewrite karo: $- k u_{xx} + 0 \cdot u_{x t} + 0 \cdot u_{tt} + u_{t} = 0$ . $u_{tt}$ term nahi hai, toh uska coefficient genuinely zero hai. $A u_{xx} + B u_{x t} + C u_{tt}$ se match karo: $A = - k, B = 0, C = 0$ . $Δ = 0^{2} - 4 (- k) (0) = 0 \Rightarrow parabolic.$ $k$ ka sign yahan matter nahi karta — product $A C$ kyasi bhi zero hai.

L2.2 $A = 1, B = 4, C = 4$ . $Δ = 4^{2} - 4 (1) (4) = 16 - 16 = 0 \Rightarrow parabolic.$ (Phir ek perfect square: $u_{xx} + 4 u_{x y} + 4 u_{y y}$ $(a + 2 b)^{2}$ ko mirror karta hai.)

L2.3 $A = 3, B = 2, C = 3$ . $Δ = 2^{2} - 4 (3) (3) = 4 - 36 = - 32 < 0 \Rightarrow elliptic.$

Level 3 — Analysis

Coefficients ab position par depend karte hain — type poore plane mein vary karti hai.

L3.1 Tricomi-type equation $u_{xx} + y u_{y y} = 0$ ke liye, $x y$ -plane mein woh region(s) dhundho jahan yeh elliptic, parabolic, aur hyperbolic hai.

L3.2 $x u_{xx} + u_{y y} = 0$ ko region ke hisaab se classify karo, aur woh curve(s) do jahan type change hoti hai.

L3.3 $u_{xx} + 2 x u_{x y} + (1 - y) u_{y y} = 0$ ke liye, point $(x, y) = (1, 0)$ par aur $(0, 2)$ par type determine karo.

Recall Solutions L3

L3.1 $A = 1, B = 0, C = y$ . Toh $Δ = 0 - 4 (1) (y) = - 4 y .$

$y > 0$ : $Δ = - 4 y < 0 \Rightarrow$ elliptic (upper half-plane).
$y = 0$ : $Δ = 0 \Rightarrow$ parabolic ( $x$ -axis).
$y < 0$ : $Δ = - 4 y > 0 \Rightarrow$ hyperbolic (lower half-plane).

L3.2 $A = x, B = 0, C = 1$ , toh $Δ = 0 - 4 (x) (1) = - 4 x$ .

$x > 0$ : $Δ < 0 \Rightarrow$ elliptic (right half-plane).
$x = 0$ : $Δ = 0 \Rightarrow$ parabolic ( $y$ -axis).
$x < 0$ : $Δ > 0 \Rightarrow$ hyperbolic (left half-plane). Type curve $x = 0$ ke across change hoti hai.

L3.3 $A = 1, B = 2 x, C = 1 - y$ , toh $Δ = (2 x)^{2} - 4 (1) (1 - y) = 4 x^{2} - 4 (1 - y) = 4 (x^{2} - 1 + y) .$

$(1, 0)$ par: $Δ = 4 (1 - 1 + 0) = 0 \Rightarrow$ parabolic.
$(0, 2)$ par: $Δ = 4 (0 - 1 + 2) = 4 > 0 \Rightarrow$ hyperbolic.

Neeche ka map L3.3 ko concretely dikhata hai: violet region woh hai jahan $Δ < 0$ (elliptic), orange region jahan $Δ > 0$ (hyperbolic), aur magenta parabola $x^{2} - 1 + y = 0$ exactly woh seam hai jahan type flip hoti hai. Do test points is par locate karo — $(1, 0)$ parabola par hi baitha hai (parabolic), $(0, 2)$ orange ke andar deep hai (hyperbolic). Yahi picture hai jo "type depends on position" ko concrete banati hai.

Level 4 — Synthesis

Classification ko characteristics aur quasilinear subtlety ke saath combine karo.

L4.1 $u_{xx} - 5 u_{x y} + 6 u_{y y} = 0$ ke liye, confirm karo ki yeh hyperbolic hai aur real characteristic curves $y = y (x)$ ki dono families dhundho.

L4.2 Quasilinear equation $u_{y y} - u u_{xx} = 0$ mein $A = - u, B = 0, C = 1$ hain. Solution region jahan $u > 0$ hai, wahan yeh elliptic hai, parabolic hai, ya hyperbolic? Agar $u < 0$ ho toh? Explain karo ki type solution par kyun depend karti hai.

L4.3 Woh saare constants $a$ dhundho jiske liye $u_{xx} + a u_{x y} + 4 u_{y y} = 0$ parabolic ho. Un $a$ ke liye, repeated characteristic slope batao.

Recall Solutions L4

L4.1 $A = 1, B = - 5, C = 6$ . $Δ = (- 5)^{2} - 4 (1) (6) = 25 - 24 = 1 > 0$ → hyperbolic ✔.

Ab characteristic slopes scratch se build karo. Ek characteristic level curve $ϕ (x, y) = const$ hoti hai (yahan $ϕ$ sirf ek label function hai jiske level curves woh characteristics hain jo hum dhundh rahe hain). Iski slope $m = d y / d x$ is baat se fix hoti hai ki $ϕ$ curve ke along move karte waqt change nahi hota: $d ϕ = ϕ_{x} d x + ϕ_{y} d y = 0 \Rightarrow ϕ_{x} + ϕ_{y} \frac{d y}{d x} = 0 \Rightarrow m = \frac{d y}{d x} = - \frac{ϕ _{x}}{ϕ _{y}}, so \frac{ϕ _{x}}{ϕ _{y}} = - m .$ Characteristic condition (parent) hai $A ϕ_{x}^{2} + B ϕ_{x} ϕ_{y} + C ϕ_{y}^{2} = 0$ . Har term ko $ϕ_{y}^{2}$ se divide karo aur $r = ϕ_{x} / ϕ_{y} = - m$ likho: $A r^{2} + B r + C = 0 \Rightarrow A (- m)^{2} + B (- m) + C = A m^{2} - B m + C = 0.$ $B$ par minus kyun? $A$ term $r = - m$ mein quadratic hai, toh $(- m)^{2} = m^{2}$ apna sign rakhta hai; $B$ term $r = - m$ mein linear hai, toh yeh $- B m$ ho jaata hai. Yahi poori wajah hai ki characteristic quadratic $A m^{2} - B m + C = 0$ hai, $A m^{2} + B m + C = 0$ nahi. $A = 1, B = - 5, C = 6$ substitute karo: $m^{2} - (- 5) m + 6 = m^{2} + 5 m + 6 = 0 \Rightarrow (m + 3) (m + 2) = 0 \Rightarrow m = - 3, - 2.$ Equivalently, boxed formula (page ke top par) deta hai $\frac{d y}{d x} = \frac{B \mp Δ}{2 A} = \frac{- 5 \mp 1}{2} = - 3, - 2$ — same roots. Integrate karne par: $y = - 3 x + c_{1}$ aur $y = - 2 x + c_{2}$ , yaani do families $y + 3 x = const, y + 2 x = const .$ Back-substitution check: $m = - 3$ deta hai $9 - (- 5) (- 3) + 6 = 9 - 15 + 6 = 0$ ✔.

L4.2 $Δ = B^{2} - 4 A C = 0 - 4 (- u) (1) = 4 u$ .

Jahan $u > 0$ : $Δ = 4 u > 0 \Rightarrow$ hyperbolic.
Jahan $u < 0$ : $Δ = 4 u < 0 \Rightarrow$ elliptic.
Jahan $u = 0$ : $Δ = 0 \Rightarrow$ parabolic.

Coefficient $A = - u$ khud unknown $u$ par depend karta hai (yahi quasilinear ka matlab hai). Toh equation ko label karna possible nahi jab tak solution pata na ho — wahi PDE solution ke ek hisse mein hyperbolic aur doosre mein elliptic hoti hai. Yahi Linear vs Quasilinear PDEs distinction ka crux hai.

L4.3 $A = 1, B = a, C = 4$ . Parabolic ke liye $Δ = 0$ chahiye: $a^{2} - 4 (1) (4) = 0 \Rightarrow a^{2} = 16 \Rightarrow a = \pm 4.$ Boxed formula se repeated slope ( $Δ = 0$ ke saath): $\frac{d y}{d x} = \frac{B}{2 A} = \frac{a}{2}$ .

$a = 4$ : slope $= 2$ , characteristic $y - 2 x = const$ .
$a = - 4$ : slope $= - 2$ , characteristic $y + 2 x = const$ .

Level 5 — Mastery

Full pipeline: classify karo, characterise karo, aur boundary data aur physics se connect karo.

L5.1 Ek PDE hai $u_{xx} + 2 u_{x y} + u_{y y} + u_{x} = 0$ . (a) Ise classify karo. (b) Yeh ek degenerate perfect-square principal part hai — single repeated characteristic family dhundho. (c) Yeh kis classic equation se sabse zyada milta-julta hai, aur tum kya data supply karoge?

L5.2 Equation $(1 + x^{2}) u_{xx} + 2 u_{x y} + (1 + y^{2}) u_{y y} = 0$ ke liye, prove karo ki yeh origin ko chhodkar har jagah elliptic hai (jahan yeh parabolic mein degenerate ho jaata hai). (Hint: dikhaao ki $Δ \leq 0$ sabhi $x, y$ ke liye, equality sirf $(0, 0)$ par.)

L5.3 $u_{xx} + (x^{2} - 4) u_{y y} = 0$ consider karo. Parabolic curve(s) dhundho, phir batao ki region $∣ x ∣ < 2$ mein kaunsa physical archetype (Wave / Heat / Laplace) govern karta hai aur $∣ x ∣ > 2$ mein kaunsa, aur har region mein sahi tarah ka well-posed data do.

Recall Solutions L5

L5.1 (a) $A = 1, B = 2, C = 1$ . $Δ = 2^{2} - 4 (1) (1) = 4 - 4 = 0 \Rightarrow$ parabolic. ( $u_{x}$ lower order hai, ignore karo.) (b) Repeated slope $\frac{d y}{d x} = \frac{B}{2 A} = \frac{2}{2} = 1$ , toh single family hai $y - x = const$ . Indeed $u_{xx} + 2 u_{x y} + u_{y y} = (\partial_{x} + \partial_{y})^{2} u$ , jiska ek characteristic direction $y = x + c$ hai. (c) Yeh heat/diffusion archetype (Heat Equation — separation of variables) se milta-julta hai: parabolic, ek real characteristic. Tum ek characteristic-transverse surface par data supply karte ho aur aage march karte ho — ek initial condition plus side conditions, full closed-boundary specification nahi.

L5.2 $A = 1 + x^{2}, B = 2, C = 1 + y^{2}$ . $Δ = B^{2} - 4 A C = 4 - 4 (1 + x^{2}) (1 + y^{2}) .$ Kyunki $x^{2} \geq 0$ aur $y^{2} \geq 0$ , humein $(1 + x^{2}) (1 + y^{2}) \geq 1$ milta hai, equality sirf tab jab $x^{2} = 0$ aur $y^{2} = 0$ , yaani $(0, 0)$ par. Isliye $Δ = 4 [1 - (1 + x^{2}) (1 + y^{2})] \leq 0,$ jahan $Δ = 0$ exactly origin par aur $Δ < 0$ baaki har jagah. Toh equation $R^{2} ∖ {(0, 0)}$ par elliptic hai aur single point $(0, 0)$ par parabolic (degenerate). Origin par check: $Δ = 4 - 4 (1) (1) = 0$ ✔.

L5.3 $A = 1, B = 0, C = x^{2} - 4$ . $Δ = 0 - 4 (1) (x^{2} - 4) = - 4 (x^{2} - 4) = 4 (4 - x^{2})$ .

Parabolic jahan $Δ = 0$ : $x^{2} = 4 \Rightarrow x = \pm 2$ (do vertical lines).
Region $∣ x ∣ < 2$ : $4 - x^{2} > 0 \Rightarrow Δ > 0 \Rightarrow$ hyperbolic → Wave archetype (Wave Equation — d'Alembert solution). Well-posed data: ek non-characteristic surface par initial $u$ aur $u_{t}$ , aage march karo (ek Cauchy problem via Method of Characteristics).
Region $∣ x ∣ > 2$ : $4 - x^{2} < 0 \Rightarrow Δ < 0 \Rightarrow$ elliptic → Laplace archetype (Laplace Equation — harmonic functions). Well-posed data: poori closed boundary par values (Dirichlet/Neumann), per Well-posedness and Boundary Conditions.

Neeche ka strip map L5.3 dikhata hai: central orange band $∣ x ∣ < 2$ hyperbolic hai (wave physics), do violet outer bands $∣ x ∣ > 2$ elliptic hain (Laplace physics), aur magenta lines $x = \pm 2$ parabolic seams hain. Note karo ki wahi equation $x = \pm 2$ ke dono taraf bilkul alag well-posed data maangti hai — yahi poori classification ka practical punchline hai.

Quick self-check

Recall One-line answers (right side cover karo)

L1.1 ::: elliptic ( $Δ = - 80$ ) L1.2 ::: parabolic ( $Δ = 0$ ) L1.3 ::: hyperbolic ( $Δ = 17$ ) L2.1 (heat) ::: parabolic ( $C = 0$ , $Δ = 0$ ) L2.2 ::: parabolic ( $Δ = 0$ ) L2.3 ::: elliptic ( $Δ = - 32$ ) L3.1 ::: elliptic $y > 0$ , parabolic $y = 0$ , hyperbolic $y < 0$ L3.2 ::: elliptic $x > 0$ , parabolic $x = 0$ , hyperbolic $x < 0$ L3.3 $(1, 0)$ / $(0, 2)$ par ::: parabolic ( $Δ = 0$ ) / hyperbolic ( $Δ = 4$ ) L4.1 ::: hyperbolic; characteristics $y + 3 x =$ const aur $y + 2 x =$ const L4.2 ::: hyperbolic jahan $u > 0$ , elliptic jahan $u < 0$ ( $Δ = 4 u$ ) L4.3 ::: parabolic iff $a = \pm 4$ ; repeated slope $a /2 = \pm 2$ L5.1 ::: parabolic; single family $y - x =$ const; heat archetype L5.2 ::: elliptic har jagah except origin, parabolic at $(0, 0)$ L5.3 ::: $∣ x ∣ < 2$ hyperbolic (wave), $∣ x ∣ > 2$ elliptic (Laplace), parabolic at $x = \pm 2$