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

This page is a self-test ladder for the parent classification note. Work each problem with pen down first, then open the solution. Every symbol here was built in the parent; if a step feels sudden, re-read the derivation there.

The figure below is your master reference for the whole page: the horizontal axis is the value of $Δ$ , and the three coloured bands are the three verdicts. When you finish a problem, locate your computed $Δ$ on this line and check the band matches your named type.

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

Level 1 — Recognition

Read off the coefficients, compute $Δ$ , name the type.

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

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

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

Recall Solutions L1

L1.1 Second-order terms: $4 u_{xx}$ , $0 \cdot u_{x y}$ , $5 u_{y y}$ . So $A = 4, B = 0, C = 5$ . The $u_{x}$ and $u$ terms are lower order — they never enter $Δ$ . $Δ = 0^{2} - 4 (4) (5) = - 80 < 0 \Rightarrow elliptic.$

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

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

Level 2 — Application

Now the coefficient positions are disguised, or the equation is written oddly.

L2.1 Classify the heat equation $u_{t} = k u_{xx}$ (with $k > 0$ ), treating $t$ as the second variable.

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

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

Recall Solutions L2

L2.1 Rewrite in standard form: $- k u_{xx} + 0 \cdot u_{x t} + 0 \cdot u_{tt} + u_{t} = 0$ . There is no $u_{tt}$ term, so its coefficient is genuinely zero. Match $A u_{xx} + B u_{x t} + C u_{tt}$ : $A = - k, B = 0, C = 0$ . $Δ = 0^{2} - 4 (- k) (0) = 0 \Rightarrow parabolic.$ Sign of $k$ doesn't matter here — the product $A C$ is zero either way.

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

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

Level 3 — Analysis

Coefficients now depend on position — the type varies across the plane.

L3.1 For the Tricomi-type equation $u_{xx} + y u_{y y} = 0$ , find the region(s) of the $x y$ -plane where it is elliptic, parabolic, and hyperbolic.

L3.2 Classify $x u_{xx} + u_{y y} = 0$ by region, and give the curve(s) where it changes type.

L3.3 For $u_{xx} + 2 x u_{x y} + (1 - y) u_{y y} = 0$ , determine the type at the point $(x, y) = (1, 0)$ and at $(0, 2)$ .

Recall Solutions L3

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

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

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

$x > 0$ : $Δ < 0 \Rightarrow$ elliptic (right half-plane).
$x = 0$ : $Δ = 0 \Rightarrow$ parabolic (the $y$ -axis).
$x < 0$ : $Δ > 0 \Rightarrow$ hyperbolic (left half-plane). The type changes across the curve $x = 0$ .

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

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

The map below shows L3.3 concretely: the violet region is where $Δ < 0$ (elliptic), the orange region where $Δ > 0$ (hyperbolic), and the magenta parabola $x^{2} - 1 + y = 0$ is the exact seam where the type flips. Locate the two test points on it — $(1, 0)$ sits on the parabola (parabolic), $(0, 2)$ sits well inside orange (hyperbolic). This is the picture that makes "type depends on position" concrete.

Level 4 — Synthesis

Combine classification with characteristics and with the quasilinear subtlety.

L4.1 For $u_{xx} - 5 u_{x y} + 6 u_{y y} = 0$ , confirm it is hyperbolic and find both families of real characteristic curves $y = y (x)$ .

L4.2 The quasilinear equation $u_{y y} - u u_{xx} = 0$ has $A = - u, B = 0, C = 1$ . On the solution region where $u > 0$ , is it elliptic, parabolic, or hyperbolic? What if $u < 0$ ? Explain why the type depends on the solution.

L4.3 Find all constants $a$ for which $u_{xx} + a u_{x y} + 4 u_{y y} = 0$ is parabolic. For those $a$ , state the (repeated) characteristic slope.

Recall Solutions L4

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

Now build the characteristic slopes from scratch. A characteristic is a level curve $ϕ (x, y) = const$ (here $ϕ$ is just a label function whose level curves are the characteristics we hunt for). Its slope $m = d y / d x$ is fixed by the fact that $ϕ$ does not change as we move along the curve: $d ϕ = ϕ_{x} d x + ϕ_{y} d y = 0 \Rightarrow ϕ_{x} + ϕ_{y} \frac{d y}{d x} = 0 \Rightarrow m = \frac{d y}{d x} = - \frac{ϕ _{x}}{ϕ _{y}}, so \frac{ϕ _{x}}{ϕ _{y}} = - m .$ The characteristic condition (parent) is $A ϕ_{x}^{2} + B ϕ_{x} ϕ_{y} + C ϕ_{y}^{2} = 0$ . Divide every term by $ϕ_{y}^{2}$ and write $r = ϕ_{x} / ϕ_{y} = - m$ : $A r^{2} + B r + C = 0 \Rightarrow A (- m)^{2} + B (- m) + C = A m^{2} - B m + C = 0.$ Why the minus on $B$ ? The $A$ term is quadratic in $r = - m$ , so $(- m)^{2} = m^{2}$ keeps its sign; the $B$ term is linear in $r = - m$ , so it flips to $- B m$ . That is the whole reason the characteristic quadratic is $A m^{2} - B m + C = 0$ , not $A m^{2} + B m + C = 0$ . Substituting $A = 1, B = - 5, C = 6$ : $m^{2} - (- 5) m + 6 = m^{2} + 5 m + 6 = 0 \Rightarrow (m + 3) (m + 2) = 0 \Rightarrow m = - 3, - 2.$ Equivalently, the boxed formula (top of page) gives $\frac{d y}{d x} = \frac{B \mp Δ}{2 A} = \frac{- 5 \mp 1}{2} = - 3, - 2$ — same roots. Integrating: $y = - 3 x + c_{1}$ and $y = - 2 x + c_{2}$ , i.e. the two families $y + 3 x = const, y + 2 x = const .$ Back-substitution check: $m = - 3$ gives $9 - (- 5) (- 3) + 6 = 9 - 15 + 6 = 0$ ✔.

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

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

The coefficient $A = - u$ depends on the unknown $u$ itself (this is what quasilinear means). So you cannot label the equation before you know the solution — the same PDE is hyperbolic in one part of the solution and elliptic in another. This is the crux of the Linear vs Quasilinear PDEs distinction.

L4.3 $A = 1, B = a, C = 4$ . Parabolic requires $Δ = 0$ : $a^{2} - 4 (1) (4) = 0 \Rightarrow a^{2} = 16 \Rightarrow a = \pm 4.$ Repeated slope from the boxed formula (with $Δ = 0$ ): $\frac{d y}{d x} = \frac{B}{2 A} = \frac{a}{2}$ .

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

Level 5 — Mastery

Full pipeline: classify, characterise, and connect to boundary data and physics.

L5.1 A PDE reads $u_{xx} + 2 u_{x y} + u_{y y} + u_{x} = 0$ . (a) Classify it. (b) It is a degenerate perfect-square principal part — find the single repeated characteristic family. (c) Which classic equation does it most resemble, and what data would you supply?

L5.2 For the equation $(1 + x^{2}) u_{xx} + 2 u_{x y} + (1 + y^{2}) u_{y y} = 0$ , prove it is elliptic everywhere except at the origin (where it degenerates to parabolic). (Hint: show $Δ \leq 0$ for all $x, y$ , with equality only at $(0, 0)$ .)

L5.3 Consider $u_{xx} + (x^{2} - 4) u_{y y} = 0$ . Find the parabolic curve(s), then state which physical archetype (Wave / Heat / Laplace) governs the region $∣ x ∣ < 2$ and which governs $∣ x ∣ > 2$ , and give the correct type of well-posed data in each region.

Recall Solutions L5

L5.1 (a) $A = 1, B = 2, C = 1$ . $Δ = 2^{2} - 4 (1) (1) = 4 - 4 = 0 \Rightarrow$ parabolic. (The $u_{x}$ is lower order, ignored.) (b) Repeated slope $\frac{d y}{d x} = \frac{B}{2 A} = \frac{2}{2} = 1$ , so the single family is $y - x = const$ . Indeed $u_{xx} + 2 u_{x y} + u_{y y} = (\partial_{x} + \partial_{y})^{2} u$ , whose one characteristic direction is $y = x + c$ . (c) It resembles the heat/diffusion archetype (Heat Equation — separation of variables): parabolic, one real characteristic. You supply data on one characteristic-transverse surface and march forward — an initial condition plus side conditions, not a full closed-boundary specification.

L5.2 $A = 1 + x^{2}, B = 2, C = 1 + y^{2}$ . $Δ = B^{2} - 4 A C = 4 - 4 (1 + x^{2}) (1 + y^{2}) .$ Since $x^{2} \geq 0$ and $y^{2} \geq 0$ , we have $(1 + x^{2}) (1 + y^{2}) \geq 1$ , with equality only when $x^{2} = 0$ and $y^{2} = 0$ , i.e. at $(0, 0)$ . Therefore $Δ = 4 [1 - (1 + x^{2}) (1 + y^{2})] \leq 0,$ with $Δ = 0$ exactly at the origin and $Δ < 0$ everywhere else. So the equation is elliptic on $R^{2} ∖ {(0, 0)}$ and parabolic (degenerate) at the single point $(0, 0)$ . Check at the origin: $Δ = 4 - 4 (1) (1) = 0$ ✔.

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 where $Δ = 0$ : $x^{2} = 4 \Rightarrow x = \pm 2$ (two vertical lines).
Region $∣ x ∣ < 2$ : $4 - x^{2} > 0 \Rightarrow Δ > 0 \Rightarrow$ hyperbolic → the Wave archetype (Wave Equation — d'Alembert solution). Well-posed data: initial $u$ and $u_{t}$ on a non-characteristic surface, marched forward (a Cauchy problem via Method of Characteristics).
Region $∣ x ∣ > 2$ : $4 - x^{2} < 0 \Rightarrow Δ < 0 \Rightarrow$ elliptic → the Laplace archetype (Laplace Equation — harmonic functions). Well-posed data: values on the entire closed boundary (Dirichlet/Neumann), per Well-posedness and Boundary Conditions.

The strip map below shows L5.3: the orange central band $∣ x ∣ < 2$ is hyperbolic (wave physics), the two violet outer bands $∣ x ∣ > 2$ are elliptic (Laplace physics), and the magenta lines $x = \pm 2$ are the parabolic seams. Notice how the same equation demands totally different well-posed data on either side of $x = \pm 2$ — that is the practical punchline of the whole classification.

Quick self-check

Recall One-line answers (cover the right side)

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 at $(1, 0)$ / $(0, 2)$ ::: parabolic ( $Δ = 0$ ) / hyperbolic ( $Δ = 4$ ) L4.1 ::: hyperbolic; characteristics $y + 3 x =$ const and $y + 2 x =$ const L4.2 ::: hyperbolic where $u > 0$ , elliptic where $u < 0$ ( $Δ = 4 u$ ) 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 everywhere except origin, parabolic at $(0, 0)$ L5.3 ::: $∣ x ∣ < 2$ hyperbolic (wave), $∣ x ∣ > 2$ elliptic (Laplace), parabolic at $x = \pm 2$