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

This is a drill page for the parent classification topic. We will hit every kind of case the discriminant test $Δ = B^{2} - 4 A C$ can throw at you — clean signs, sneaky zeros, position-dependent types, and exam twists. Before we start, one reminder built from zero.

Definition What a "characteristic" and the coordinate

$ξ$ mean — built from zero A characteristic is a special curve in the $(x, y)$ -plane along which the PDE fails to pin down the second derivatives from the data — so a solution can carry a "kink" there (its second derivatives can be discontinuous, i.e. jump from one side of the curve to the other) even while $u$ itself stays continuous. Physically these are the lines a wave-front rides along.

To find such curves we rename coordinates. Pick a smooth function $ϕ (x, y)$ and use its value as a new coordinate called $ξ$ (Greek "ksi"): set $ξ = ϕ (x, y)$ . Then a curve " $ϕ = const$ " is just a curve " $ξ = const$ " — a grid-line of the new coordinate system. The symbol $u_{ξ ξ}$ means the second derivative of $u$ with respect to this new coordinate, exactly as $u_{xx}$ was the second derivative in $x$ .

Intuition Where the slope formula

$m = \frac{- B \pm Δ}{2 A}$ comes from Rewrite $u$ in the new coordinate $ξ = ϕ (x, y)$ . The chain rule turns the principal part $A u_{xx} + B u_{x y} + C u_{y y}$ into a coefficient $Q (ϕ_{x}, ϕ_{y}) = A ϕ_{x}^{2} + B ϕ_{x} ϕ_{y} + C ϕ_{y}^{2}$ multiplying $u_{ξ ξ}$ . A characteristic is exactly a curve $ϕ = const$ where this coefficient vanishes ( $Q = 0$ ): if the $u_{ξ ξ}$ coefficient is zero, that second derivative is not determined by the equation, which is precisely the "kink allowed" condition. Along $ϕ = const$ the slope is $\frac{d y}{d x} = - \frac{ϕ _{x}}{ϕ _{y}}$ ; dividing $Q = 0$ by $ϕ_{y}^{2}$ and writing $m = ϕ_{x} / ϕ_{y}$ gives the quadratic $A m^{2} + B m + C = 0.$ That is why the quadratic formula appears: $m = \frac{- B \pm B ^{2} - 4 A C}{2 A}$ . The thing under the square root is $Δ$ — so its sign directly counts how many real characteristic slopes exist (two, one, or none). This is the bridge between "sign of $Δ$ " and "shape of the physics."

The scenario matrix

Every problem below is tagged with exactly which cell it fills. If you can do all of these, no exam version can surprise you.

Cell	What makes it tricky	Example
C1 Clean hyperbolic	$Δ > 0$ , constant coeffs	Ex. 1
C2 Clean elliptic	$Δ < 0$ , constant coeffs	Ex. 2
C3 Parabolic via missing term	one 2nd-order term absent → $C = 0$	Ex. 3
C4 Cross-term $B \neq = 0$	must square a nonzero $B$	Ex. 4
C5 Convention trap	book writes $2 B u_{x y}$	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 $x, y$ ; all three regions + degenerate line	Ex. 7
C8 Quasilinear / solution-dependent	coeff depends on $u$	Ex. 8
C9 Real-world word problem	translate physics → coefficients + data needed	Ex. 9
C10 Exam twist / limiting value	parameter $\to$ 0 or sign flip	Ex. 10

Worked examples

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

Classify the heat-type equation $u_{t} = 5 u_{xx}$ , i.e. $5 u_{xx} - u_{t} = 0$ (take $y \to t$ ; recall $t$ is just the second variable, but $u_{t}$ is only first order).

Forecast: many students blurt "elliptic because $C = 1$ ". Is there even a $u_{tt}$ term?

Check the order of each term. $u_{xx}$ is second order; $u_{t}$ is first order. There is no $u_{tt}$ . Why this step? $C$ is the coefficient of $u_{tt}$ — if that term does not exist, $C = 0$ , not $1$ .
Coefficients. $A = 5$ , $B = 0$ , $C = 0$ — and $C = 0$ precisely because step 1 found no $u_{tt}$ term to supply it. Why this step? Reading coefficients term-by-term is the only way to avoid the " $C = 1$ because $t$ is the other variable" trap; the coefficient must come from an actual term.
Compute. $Δ = 0 - 4 (5) (0) = 0$ . Why this step? A single zero factor in $- 4 A C$ kills the whole product, forcing $Δ = 0$ — the parabolic signature.
Read. $Δ = 0 \Rightarrow$ parabolic. Why this step? $Δ = 0$ means the characteristic quadratic has one repeated real root — exactly one smearing direction, which is what diffusion does.

Verify: One repeated characteristic ( $t = const$ ) — the single smearing direction of diffusion. Matches the heat equation's known parabolic nature. ✔

Worked example Ex. 5 — The

$2 B$ convention trap (Cell C5) A textbook writes its PDE as $A u_{xx} + 2 B u_{x y} + C u_{y y} = 0$ and defines $Δ = B^{2} - A C$ . For $3 u_{xx} + 6 u_{x y} + 2 u_{y y} = 0$ , find the type in both conventions and show they agree.

Forecast: will the two formulas give the same number? The same sign?

Our convention ( $B u_{x y}$ , $Δ = B^{2} - 4 A C$ ). The cross term is $6 u_{x y}$ , so $B = 6$ , $A = 3$ , $C = 2$ . $Δ = 6^{2} - 4 (3) (2) = 36 - 24 = 12 > 0 \Rightarrow$ hyperbolic.
Book convention ( $2 B u_{x y}$ , $Δ = B^{2} - A C$ ). Now $2 B = 6 \Rightarrow B = 3$ , $A = 3$ , $C = 2$ . $Δ = 3^{2} - (3) (2) = 9 - 6 = 3 > 0 \Rightarrow$ hyperbolic. Why this step? In this convention the cross coefficient is half of what you see, so you must solve $2 B = 6$ ; the $B^{2} - A C$ formula then compensates for the missing factor of $4$ .
Compare. Numbers differ ( $12$ vs $3$ ) but both are positive. Why this step? Classification depends on the sign only, never the magnitude — so both conventions must, and do, agree on the type.

Verify: $12 = 4 \times 3$ , so one $Δ$ is exactly $4$ times the other — same sign guaranteed. Both say hyperbolic. ✔

Worked example Ex. 6 — Non-trivial parabolic:

$B^{2} = 4 A C$ with every coefficient nonzero (Cell C6) Classify $u_{xx} + 4 u_{x y} + 4 u_{y y} + u_{x} = 0$ .

Forecast: every second-order term is present and nonzero, so it cannot be parabolic-by-missing-term. Guess the type before computing — is a full $A, B, C$ ever parabolic?

Coefficients. $A = 1$ , $B = 4$ , $C = 4$ ; all three are nonzero. Why this step? This is the whole point of the example — parabolic here is not because a term is absent, so we must test the actual arithmetic $B^{2} - 4 A C$ .
Compute. $Δ = 4^{2} - 4 (1) (4) = 16 - 16 = 0$ . Why this step? Parabolic means the characteristic quadratic $A m^{2} + B m + C = 0$ has a repeated real root, which happens exactly when its discriminant $B^{2} - 4 A C$ vanishes — a perfect-square principal part $u_{xx} + 4 u_{x y} + 4 u_{y y} = (\partial_{x} + 2 \partial_{y})^{2} u$ .
Read. $Δ = 0 \Rightarrow$ parabolic. Why this step? One repeated slope $m = - B / (2 A) = - 2$ means a single characteristic direction, the hallmark of parabolic even with a full set of coefficients.

Verify: Repeated slope $m = \frac{- B}{2 A} = \frac{- 4}{2} = - 2$ , and indeed $A m^{2} + B m + C = 1 (4) + 4 (- 2) + 4 = 0$ . The principal part factors as $(\partial_{x} + 2 \partial_{y})^{2} u$ — a genuine perfect square, confirming parabolic. ✔

Worked example Ex. 7 — Position-dependent type: three regions AND the degenerate line (Cell C7)

Classify the Tricomi-type equation $u_{xx} + x u_{y y} = 0$ in every part of the plane.

Forecast: the type is not one word here. Where is it elliptic, parabolic, hyperbolic?

Coefficients (now functions of position). $A = 1$ , $B = 0$ , $C = x$ . Why this step? $C$ is literally the variable $x$ , so $Δ$ becomes a function of position instead of a single number — you cannot answer with one word.
Discriminant as a function. $Δ (x) = 0 - 4 (1) (x) = - 4 x$ . Why this step? Writing $Δ$ as an expression in $x$ lets us see the sign flip: the type is decided point-by-point, so we track $Δ$ 's sign across the plane.
Case $x > 0$ : $Δ = - 4 x < 0 \Rightarrow$ elliptic (right half-plane). Why this step? For positive $x$ , $- 4 x$ is negative, and negative $Δ$ is the elliptic (no real characteristics) case.
Case $x = 0$ : $Δ = 0 \Rightarrow$ parabolic — the degenerate line where the type switches. Why this step? $x = 0$ is exactly where $Δ$ crosses zero; a continuous $Δ$ must pass through the parabolic value $0$ to change sign between the two regions.
Case $x < 0$ : $Δ = - 4 x > 0 \Rightarrow$ hyperbolic (left half-plane). Why this step? For negative $x$ , $- 4 x$ is positive, and positive $Δ$ is the hyperbolic (two real characteristics) case.

Read the figure: the vertical white line

x = 0

splits the blueprint into a cyan elliptic zone on the right (

Δ < 0

) and an amber hyperbolic zone on the left (

Δ > 0

); the line itself is the lone parabolic boundary (

Δ = 0

). The three white sample dots show

Δ = - 4, 0, + 4

x = 1, 0, - 1

— watch the sign flip as you cross the line. This is the mechanism of transonic flow, where a flow changes character crossing the sonic surface.

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

Worked example Ex. 8 — Quasilinear: type depends on the solution (Cell C8)

A quasilinear equation is $u u_{xx} + u_{y y} = 0$ . Classify it where $u = 2$ and where $u = - 3$ .

Forecast: the coefficient of $u_{xx}$ is $u$ itself. Can the type flip depending on the solution's value?

Coefficients. $A = u$ , $B = 0$ , $C = 1$ . Why this step? In a quasilinear PDE the top-derivative coefficient may depend on $u$ ; it still enters linearly, so the discriminant test still applies pointwise — but now the "point" includes the value of $u$ there.
Discriminant. $Δ = 0 - 4 (u) (1) = - 4 u$ . Why this step? Because $A = u$ , the sign of $Δ$ is controlled by the sign of the solution $u$ , so the type cannot be named until we know $u$ .
Where $u = 2$ : $Δ = - 8 < 0 \Rightarrow$ elliptic. Why this step? Substituting the actual solution value $u = 2$ turns the symbolic $Δ$ into a number whose sign we can read.
Where $u = - 3$ : $Δ = 12 > 0 \Rightarrow$ hyperbolic. Why this step? A different solution value flips the sign of $- 4 u$ , and therefore flips the type — the defining feature of quasilinear classification.

Verify: $u = 2$ : $- 4 (2) = - 8 < 0$ elliptic ✔. $u = - 3$ : $- 4 (- 3) = 12 > 0$ hyperbolic ✔. So you truly cannot name the type until you know the solution's sign. ✔

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

A thin metal rod is heated. Physicists model its temperature $T (x, t)$ by "the rate of temperature change in time equals a constant $κ > 0$ times the curvature in space": $T_{t} = κ T_{xx}$ . Classify it, and state what data you must supply to solve it.

Forecast: "in time" and "curvature in space" — which single type describes spreading heat, and does it need data all around, or just a starting condition?

Translate to standard form. $κ T_{xx} - T_{t} = 0$ , with $(x, t)$ as the two variables ( $t$ plays the role of $y$ ). Why this step? You must line the physics up with $A u_{xx} + B u_{x t} + C u_{tt}$ before reading coefficients.
Coefficients. $A = κ$ , $B = 0$ , $C = 0$ — and $C = 0$ because the model has $T_{t}$ (first order in time) but no $T_{tt}$ term to supply a coefficient. Why this step? "Rate of change in time" is a first derivative; assigning $C = 0$ only makes sense once we notice no second time-derivative exists in the physics.
Discriminant. $Δ = 0 - 4 (κ) (0) = 0 \Rightarrow$ parabolic. Why this step? The zero factor from the missing $T_{tt}$ forces $Δ = 0$ , the parabolic (diffusion) signature — matching the intuition that heat smooths rather than propagates as sharp fronts.
State the data required. Parabolic problems need one initial condition $T (x, 0) = f (x)$ (the starting temperature profile) plus boundary conditions at the rod's two ends (e.g. fixed end temperatures $T (0, t)$ and $T (L, t)$ ), and you then march forward in $t$ . You do not, and should not, prescribe data at a "final time" — that would be ill-posed. Contrast: a wave equation would additionally need the initial velocity $T_{t} (x, 0)$ , and an elliptic (Laplace) problem needs data on the whole closed boundary. (See Well-posedness and Boundary Conditions and Heat Equation — separation of variables.)

Verify: With $κ = 1$ : $Δ = - 4 (1) (0) = 0$ , parabolic — the correct, well-known character of diffusion (infinite signal speed but smoothing). "One initial condition + two boundary conditions, march forward in $t$ " is exactly the standard well-posed heat problem. ✔

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

A PDE contains a knob $ε$ : $ε u_{tt} + u_{xx} - u_{t} = 0$ (variables $x, t$ ; here $A$ is the $u_{xx}$ coefficient, $C$ is the $u_{tt}$ coefficient). Classify for $ε > 0$ , then examine the limit $ε \to 0^{+}$ .

Forecast: what happens to the type as $ε$ shrinks to zero — does it change identity?

Coefficients. $A = 1$ (from $u_{xx}$ ), $B = 0$ , $C = ε$ (from $u_{tt}$ ). The $- u_{t}$ is first order, discarded. Why this step? We must identify the second-order coefficients before we can track the limit; only $u_{xx}, u_{x t}, u_{tt}$ enter $Δ$ .
Discriminant. $Δ = 0 - 4 (1) (ε) = - 4 ε$ . Why this step? Keeping $ε$ symbolic lets us watch how the sign of $Δ$ behaves as the knob is turned toward zero, rather than losing that dependence too early.
For $ε > 0$ : $Δ = - 4 ε < 0 \Rightarrow$ elliptic.
Limit $ε \to 0^{+}$ : the $u_{tt}$ term vanishes, leaving $u_{xx} - u_{t} = 0$ , i.e. a heat equation. Now $C = 0$ and $Δ = 0 \Rightarrow$ parabolic. Why this step? A singular limit can change the order of the equation (the top time-derivative disappears), and thus its type — a favourite exam trap (a "singular perturbation").

Verify: $ε = 0.5$ : $Δ = - 2 < 0$ elliptic ✔. $ε = 0.01$ : $Δ = - 0.04 < 0$ still elliptic ✔. Exactly at $ε = 0$ : term disappears, $Δ = 0$ parabolic ✔. The type does not approach parabolic smoothly — it snaps at the moment the top term dies. ✔

Active recall

Recall Cover and answer

Which cell had $C = 0$ because a term was missing, giving $Δ = 0$ ? ::: Cell C3 (heat-type), Ex. 3. Which example is parabolic even though ALL of $A, B, C$ are nonzero? ::: Ex. 6, where $B^{2} = 4 A C$ ( $16 = 16$ ) and the principal part is a perfect square $(\partial_{x} + 2 \partial_{y})^{2} u$ . In Ex. 4 the cross term $B = 4$ overrode same-sign $A, C$ . What was $Δ$ ? ::: $Δ = 16 - 4 = 12 > 0$ , hyperbolic. In the Tricomi Ex. 7, along which line is the type parabolic? ::: The line $x = 0$ , where $Δ = - 4 x = 0$ . For quasilinear $u u_{xx} + u_{y y} = 0$ , is it elliptic or hyperbolic where $u = - 3$ ? ::: Hyperbolic, since $Δ = - 4 (- 3) = 12 > 0$ . Why can two $Δ$ conventions ( $B^{2} - 4 A C$ vs $B^{2} - A C$ ) both be correct? ::: They always share the same sign; only the sign classifies. What must you check before ever using $Δ$ ? ::: That at least one of $A, B, C$ is nonzero — if all three are zero the equation is not second order. What data does a parabolic (heat) problem need? ::: One initial condition plus boundary conditions at each end, then march forward in time.