Classification and canonical forms

In one sentence

For $Au_{xx} + Bu_{xy} + Cu_{yy} + \cdots = 0$ the sign of $B^2 - 4AC$ decides the type, the type decides the behaviour, and the type can change from point to point.

Why the exam asks it

Because it is a one-line computation that determines everything else about an equation, and because the coefficients may depend on $x$ and $y$ — so "what type is this" is often really "where is it each type", which options rarely make explicit.

The idea

For the second-order equation with principal part $Au_{xx} + Bu_{xy} + Cu_{yy}$:

$$B^2 - 4AC \begin{cases} < 0 & \textbf{elliptic}\\ = 0 & \textbf{parabolic}\\ > 0 & \textbf{hyperbolic}\end{cases}$$

The names come from conic sections, and the analogy is exact: the characteristic equation $A\,dy^2 - B\,dx\,dy + C\,dx^2 = 0$ has no real roots, one, or two, giving zero, one or two families of real characteristic curves.

What each type does

Elliptic — Laplace, $u_{xx}+u_{yy}=0$. No real characteristics. Solutions are as smooth as possible: harmonic functions are analytic, satisfy the mean value property and the maximum principle, and are determined by boundary values on a closed curve. Boundary value problems, not initial value problems.

Parabolic — heat, $u_t = u_{xx}$. One characteristic family, $t =$ constant. Smooths immediately and irreversibly: discontinuous initial data becomes analytic for any $t>0$, and information propagates at infinite speed. Well posed forwards in time only — the backward heat equation is the standard ill-posed problem.

Hyperbolic — wave, $u_{tt} = u_{xx}$. Two families, $x \pm t$. Preserves kinks: a corner in the initial data travels along a characteristic and stays a corner. Finite propagation speed, and the domain of dependence is a triangle. Reversible in time.

The type can change

$$y\,u_{xx} + u_{yy} = 0$$

has $B^2-4AC = -4y$. Elliptic for $y>0$, parabolic on the axis, hyperbolic for $y<0$ — the Tricomi equation, and the standard example that "type" is a local property. When a question gives variable coefficients, the answer is a region.

Canonical forms

Change variables to $\xi, \eta$ along the characteristics:

• Hyperbolic: $u_{\xi\eta} = \cdots$, or $u_{\xi\xi}-u_{\eta\eta} = \cdots$.
• Parabolic: $u_{\eta\eta} = \cdots$.
• Elliptic: $u_{\xi\xi}+u_{\eta\eta} = \cdots$.

The recipe: solve $A\,dy^2 - B\,dx\,dy + C\,dx^2 = 0$ for $dy/dx$, integrate to get the characteristic curves, and use them as the new coordinates. In the parabolic case there is only one family, so the second coordinate is chosen freely — any function independent of the first.

d'Alembert is the hyperbolic canonical form paying off: $u_{tt}=c^2u_{xx}$ becomes $u_{\xi\eta}=0$ in $\xi = x-ct$, $\eta = x+ct$, so

$$u = F(x-ct)+G(x+ct),$$

and with initial data $u(x,0)=f$, $u_t(x,0)=g$,

$$u(x,t) = \frac{f(x-ct)+f(x+ct)}{2} + \frac{1}{2c}\int_{x-ct}^{x+ct} g .$$

Where intuition breaks

"The type is a property of the equation." It is a property of the equation at a point. $yu_{xx}+u_{yy}=0$ is all three types in different regions.

"The heat equation can be solved backwards." It is ill-posed backwards: arbitrarily small changes in the final data produce arbitrarily large changes earlier. The wave equation is reversible; the heat equation is not.

"Discontinuities propagate in all three." Only hyperbolic. Parabolic and elliptic equations smooth data immediately, which is why the heat kernel is analytic for every $t>0$.

"The wave equation has infinite propagation speed." Finite, $c$. The heat equation has infinite speed — a bump anywhere makes $u>0$ everywhere at once, which is physically odd and mathematically true.

"An initial value problem is well posed for Laplace." It is not — Hadamard's example shows small Cauchy data producing unbounded solutions. Elliptic equations take boundary conditions.

"The classification needs constant coefficients." Only the coefficients of the second-order terms matter, and they may vary. Lower-order terms never affect the type.

The exam's angle

1. Compute $B^2-4AC$ first. Note the convention: with $u_{xy}$ having coefficient $B$, not $2B$. Check which the question uses.
2. If the coefficients vary, describe the regions rather than giving a single word.
3. For canonical form, solve the characteristic ODE and use the resulting curves as coordinates.
4. Match the boundary data to the type. Boundary values for elliptic, initial data for parabolic and hyperbolic — a mismatch is often what makes the option wrong.
5. Reach for d'Alembert on the wave equation — the formula answers most one-dimensional questions directly.

The night before

• $B^2-4AC$: negative elliptic, zero parabolic, positive hyperbolic. Only the second-order terms matter.
• Characteristic families: none, one, two respectively.
• Elliptic Laplace: smooth, maximum principle, boundary value problems.
• Parabolic heat: instant smoothing, infinite speed, ill-posed backwards.
• Hyperbolic wave: kinks travel, finite speed $c$, reversible, domain of dependence a triangle.
• Type is local: $yu_{xx}+u_{yy}=0$ is elliptic for $y>0$ and hyperbolic for $y<0$.
• d'Alembert: $u = \frac{f(x-ct)+f(x+ct)}{2} + \frac1{2c}\int_{x-ct}^{x+ct}g$.