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

Step 1 — What the symbols even mean

WHAT. We are handed an equation that mixes a quantity $u$ and its rates of change. Let me first say what each piece is.

$u$ is a number that depends on two things: a position $x$ and a second variable $y$ (often $y$ is time). Think of $u$ as temperature at place $x$ and time $y$ . So $u = u (x, y)$ .
$u_{x}$ means "how fast $u$ changes as you nudge $x$ " (holding $y$ fixed). It is a slope.
$u_{xx}$ means "how fast that slope itself changes" — the bending of $u$ in the $x$ -direction.
$u_{y y}$ = bending in the $y$ -direction. $u_{x y}$ = how the $x$ -slope changes when you nudge $y$ (twist).

WHY these matter. The parent note claims only the bending terms ( $u_{xx}, u_{x y}, u_{y y}$ ) decide the character of the equation. So those are the ones we watch. The full equation is:

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

Term by term: $A$ is the weight on $x$ -bending, $C$ the weight on $y$ -bending, $B$ the weight on the twist. The trailing group ( $D, E, F, G$ ) only shifts and damps — it cannot change the character, so we drop it and keep the principal part $A u_{xx} + B u_{x y} + C u_{y y}$ .

PICTURE. The figure draws $u$ as a bumpy surface over the $x$ – $y$ floor, showing what "bend in $x$ ", "bend in $y$ ", and "twist" look like as three separate little surfaces.

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

Step 2 — The one question that classifies everything

WHAT. We ask a single geometric question: Are there special curves in the $x$ – $y$ floor along which the solution is allowed to have a sudden "kink" — a break in its second derivatives? Such curves are called characteristics.

WHY this question. A PDE turns out to behave like an ordinary equation (an ODE) along certain directions. Those directions are exactly where the second derivative is not pinned down by the surrounding data. Counting how many such real directions exist is the entire game — and the answer will be $0$ , $1$ , or $2$ . That count is the classification.

PICTURE. Two smooth solution patches meeting along a curve; across that curve the surface is continuous but its bending jumps. That meeting curve is a candidate characteristic.

Step 3 — Rename the special curve as a coordinate

WHAT. Suppose the special curve is a level set of some function: $ϕ (x, y) = const$ . We use $ϕ$ as a new coordinate $ξ = ϕ$ (Greek "xi", just a name for a new axis). We then rewrite the PDE in terms of $ξ$ .

WHY. If a curve really is a kink-line, then in the coordinate that runs across it, the coefficient multiplying the pure second derivative $u_{ξ ξ}$ must vanish — that's precisely what lets the second derivative be free (un-pinned). So we compute that coefficient and set it to zero.

The chain rule (the rule for changing variables) turns the bending terms into: $u_{xx} \to ϕ_{x}^{2} u_{ξ ξ} + \dots, u_{x y} \to ϕ_{x} ϕ_{y} u_{ξ ξ} + \dots, u_{y y} \to ϕ_{y}^{2} u_{ξ ξ} + \dots$

Here $ϕ_{x}$ = slope of $ϕ$ in $x$ , $ϕ_{y}$ = slope in $y$ . Collect everything multiplying $u_{ξ ξ}$ :

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

PICTURE. The old grid $(x, y)$ warped so one family of grid-lines lies along the curve $ϕ = const$ ; the label $ξ$ rides across those lines.

Step 4 — Set the coefficient to zero, get a slope

WHAT. A characteristic is exactly where $Q = 0$ : $A ϕ_{x}^{2} + B ϕ_{x} ϕ_{y} + C ϕ_{y}^{2} = 0.$

Now convert this into the slope of the curve. Along $ϕ = const$ , moving a tiny step $(d x, d y)$ keeps $ϕ$ unchanged, so: $d ϕ = ϕ_{x} d x + ϕ_{y} d y = 0 ⟹ \frac{d y}{d x} = - \frac{ϕ _{x}}{ϕ _{y}} .$

WHY divide by $ϕ_{y}^{2}$ . We want the equation to speak in terms of one thing — a slope. Divide $Q = 0$ by $ϕ_{y}^{2}$ and name the ratio $m = \frac{ϕ _{x}}{ϕ _{y}}$ :

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

Term by term: this is an ordinary quadratic in the single unknown $m$ . That's the payoff — a fearsome PDE question ("kink curves?") has collapsed into "solve a quadratic."

PICTURE. A characteristic curve with its little step $(d x, d y)$ drawn as a right triangle, the ratio $d y / d x$ marked as the slope $m$ .

Step 5 — Solve the quadratic; the discriminant appears

WHAT. Apply the quadratic formula to $A m^{2} + B m + C = 0$ :

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

The thing living under the square root is the discriminant: $Δ = B^{2} - 4 A C$

Because $m = ϕ_{x} / ϕ_{y} = - d y / d x$ (up to sign), each solution $m$ is a direction of a characteristic. The characteristic slopes themselves are $\frac{d y}{d x} = \frac{B \mp B ^{2} - 4 A C}{2 A} .$

WHY the square root controls everything. A square root of a positive number is real; of zero is a single value; of a negative number is imaginary. So the sign of $Δ$ decides how many real slopes exist — and that is literally the count of real kink-lines from Step 2.

PICTURE. A number-line for $Δ$ : to the right ( $Δ > 0$ ) two real roots; at $0$ a double root; to the left ( $Δ < 0$ ) the roots leave the real line into the complex plane.

Step 6 — The three worlds, drawn

WHAT. Read off the three cases and match them to real characteristics.

$Δ > 0$ : two distinct real slopes → two families of straight kink-lines. This is hyperbolic. Signals ride the lines $x \pm c t = const$ at finite speed.
$Δ = 0$ : the $\pm Δ$ collapses, giving one repeated real slope → a single family of lines. This is parabolic. Data smears along $t = const$ ; infinite signal speed but everything smooths.
$Δ < 0$ : $Δ$ is imaginary → no real slopes. This is elliptic. No preferred lines; every interior point feels the whole boundary; the solution is perfectly smooth.

WHY the names. The very same $B^{2} - 4 A C$ classifies conic sections $A x^{2} + B x y + C y^{2} = const$ into hyperbola / parabola / ellipse. Two real asymptote directions ↔ hyperbola ↔ two characteristics; and so on. Same algebra, so the same names.

PICTURE. Three side-by-side floors: two crossing lines (hyperbolic), one repeated line (parabolic), a swirl-with-no-lines (elliptic), each labelled with its equation.

The real characteristic lines are exactly the paths studied in the Method of Characteristics, and the data each world needs is set by Well-posedness and Boundary Conditions.

Step 7 — The degenerate & position-dependent cases (do not skip)

WHAT. We must cover the corners the tidy formula hides.

Case $A = 0$ . Then $A m^{2} + B m + C = 0$ is no longer a genuine quadratic — dividing by $2 A$ is illegal. Handle it directly: if $A = 0, B \neq = 0$ the equation is linear in $m$ , $B m + C = 0$ , giving one finite slope $m = - C / B$ and one "vertical" characteristic (the second root ran off to infinity). The sign test still works: $Δ = B^{2} - 4 (0) C = B^{2} \geq 0$ , so $A = 0$ is always hyperbolic (if $B \neq = 0$ ) or parabolic (if $B = 0$ too).

Case $A = C = 0, B \neq = 0$ (e.g. $u_{x y} = 0$ ). Then $Δ = B^{2} > 0$ : hyperbolic, with the two coordinate axes as characteristics.

Missing $u_{y y}$ (the heat trap). If there is no $u_{y y}$ term then $C = 0$ , not $C = 1$ . For the heat equation $u_{t} - k u_{xx} = 0$ : with $y \to t$ , $A = - k, B = 0, C = 0$ , so $Δ = 0 - 4 (- k) (0) = 0$ → parabolic. Assigning $C = 1$ out of habit is the classic error.

Position-dependent type (Tricomi). When $A, B, C$ depend on $(x, y)$ , $Δ$ is a function of location. For $y u_{xx} + u_{y y} = 0$ : $A = y, B = 0, C = 1$ , so $Δ = - 4 y$ .

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

WHY these belong here. The parent warns that type is not always fixed. For quasilinear equations the coefficients may even depend on $u$ , so the type can shift with the solution. Always evaluate $Δ$ at the point (and state).

PICTURE. The Tricomi plane split into an upper elliptic band, the parabolic line $y = 0$ , and a lower hyperbolic band with two characteristics sketched only below the line.

The one-picture summary

One figure compresses the whole chain: bending terms → ask for kink-lines → new coordinate makes coefficient $Q$ → set $Q = 0$ , divide, get a quadratic in the slope $m$ → its discriminant $B^{2} - 4 A C$ → sign gives 0/1/2 real lines → elliptic / parabolic / hyperbolic.

Recall Feynman retelling — explain the whole walkthrough in plain words

I have a quantity $u$ that bends across a floor with two directions. Only its bending ( $u_{xx}, u_{x y}, u_{y y}$ ) sets its personality, so I keep just $A, B, C$ . I ask one question: where can the surface have a hidden crease? Along such a crease line, if I re-label it as a fresh coordinate, the number in front of the pure second derivative must be zero. That number is $A ϕ_{x}^{2} + B ϕ_{x} ϕ_{y} + C ϕ_{y}^{2}$ . Setting it to zero and turning it into a slope gives a plain quadratic $A m^{2} + B m + C = 0$ . Squares mean at most two answers. The quadratic formula puts $B^{2} - 4 A C$ under a square root, and a square root only cares about a sign: — positive → two real creases → wave world (hyperbolic), — zero → one crease → heat world (parabolic), — negative → no real creases, forced smooth → equilibrium world (elliptic). If $A, B, C$ depend on where I stand, I just check the sign at that spot — that's how one equation (Tricomi) can be all three in different regions.

Recall Quick self-test

Why is only the principal part used? ::: Lower-order terms shift/damp but cannot create or destroy characteristics. What equation does setting $Q = 0$ reduce to? ::: A quadratic $A m^{2} + B m + C = 0$ in the slope $m$ . Why does the sign of $Δ$ matter, not its value? ::: A square root is real / single / imaginary according only to whether its argument is $+$ / $0$ / $-$ . Heat equation coefficients? ::: $A = - k, B = 0, C = 0$ , so $Δ = 0$ (parabolic) — there is no $u_{tt}$ . Tricomi type at $y = - 3$ ? ::: $Δ = - 4 (- 3) = 12 > 0$ → hyperbolic.