Question bank — Classification — elliptic, parabolic, hyperbolic (discriminant test)

False — only the principal (second-order) part $A{u}_{xx}+B{u}_{xy}+C{u}_{yy}$ enters $\Delta$; the lower-order terms $D{u}_x,E{u}_y$ shift and damp solutions but never change whether real characteristics exist.

False — if $A,B,C$ depend on $x,y$ (like Tricomi $y{u}_{xx}+u_{yy}=0$), $\Delta$ changes sign across the plane, so the type is a property of the point, not the equation as a whole.

False — it is parabolic; there is no $u_{tt}$ term so $C=0$, giving $\Delta =0-4(-k)(0)=0$. Its type comes from having exactly one real characteristic.

True — $\Delta <0$ makes the two characteristic slopes $m=\frac{-B\pm \sqrt{\Delta }}{2A}$ complex (square root of a negative number), so there are no real curves along which information travels; every boundary point influences every interior point instead.

True — the top-derivative coefficients may depend on $u$ itself, so once you plug in a solution, $\Delta$ can turn out elliptic in one region and hyperbolic in another depending on that $u$. See Linear vs Quasilinear PDEs.

False — that single sign flip changes $\Delta$ from positive to negative, turning finite-speed wave propagation (hyperbolic) into smooth all-at-once equilibrium (elliptic). A tiny algebraic change is a total physical change.

False — the repeated root gives exactly one real characteristic direction; parabolic problems smear data along that single family of lines, as in heat diffusion.

False — scaling $A,B,C$ by $-1$ gives $(-B{)}^2-4(-A)(-C)=B^2-4AC$, the same $\Delta$; the sign of the discriminant is invariant under scaling the equation.

True — with no characteristics carrying information inward, the only way to pin down the solution is to prescribe values (or normal derivatives) all the way around; see Well-posedness and Boundary Conditions.

The conclusion is right. The deeper reason it doesn't matter which second derivative you call $A$ versus $C$ is that $\Delta =B^2-4AC$ is symmetric in $A$ and $C$: swapping them leaves $B^2-4AC$ unchanged, so either labelling gives $\Delta =4c^2>0$.

Error of mixed conventions. The $\Delta =B^2-AC$ form assumes the cross term is written $2B{u}_{xy}$, so there $B=\frac{3}{2}$. Pick one convention: with $B{u}_{xy}$ use $B^2-4AC$; the sign comes out the same either way.

$u_x^2$ is a squared first derivative — it is neither second-order nor even linear. The term contributes nothing to $A,B,C$ (which are second-order coefficients), and the presence of $u_x^2$ makes the whole equation nonlinear.

There is no $u_{tt}$, so $C=0$, not $1$; inventing a $C$ term is the mistake. Correctly $\Delta =0-4(-k)(0)=0$, parabolic — check the order of each term before assigning a coefficient.

You are actually correct — $\Delta =16>0$ is hyperbolic. The trap is doubting a right computation; Tricomi is genuinely hyperbolic for $y<0$, elliptic for $y>0$, parabolic on $y=0$.

The discriminant test is built for second-order PDEs. A pure first-order transport equation has no principal second-order part; it is classified by Method of Characteristics directly, not by $\Delta$.

The character — diffuse, propagate, or equilibrate — is set by the fastest-varying part of the solution; lower-order terms can shift or damp but cannot create or destroy the sharp lines (characteristics) along which information travels.

The characteristic slopes solve $A{m}^2+Bm+C=0$, a quadratic in the slope $m$; a positive discriminant gives two distinct real roots, hence two real slope families along which second derivatives can jump.

Hyperbolic (wave) has two real characteristics $x\pm ct=\text{const}$ that bound where a disturbance can reach; parabolic (heat) has no such bounding lines, so a change at one point is instantly felt everywhere, however faintly. See Wave Equation — d'Alembert solution and Heat Equation — separation of variables.

Because both are classified by the identical quantity $B^2-4AC$: the algebra of the principal part of the PDE is the same quadratic form as the conic, so ellipse/parabola/hyperbola map onto elliptic/parabolic/hyperbolic. See Conic Sections.

With no real characteristics there is no direction to "march" the solution forward; every interior value depends on the entire boundary, so the boundary data must be complete and closed — a Laplace/harmonic property, see Laplace Equation — harmonic functions.

Marching (Cauchy) data on an open surface is ill-posed for elliptic equations — tiny high-frequency errors in the data blow up exponentially. Elliptic equilibrium requires a closed boundary, not forward evolution.

Here $\varphi (x,y)=\text{const}$ is the candidate characteristic curve, and we adopt the new coordinate $\xi =\varphi$; its partial derivatives ${\varphi }_x,{\varphi }_y$ are the rates at which $\varphi$ changes in $x$ and $y$. The chain rule turns the second derivatives into a coefficient $Q=A{\varphi }_x^2+B{\varphi }_x{\varphi }_y+C{\varphi }_y^2$ multiplying $u_{\xi \xi }$; where $Q=0$, that second derivative is unconstrained and free to jump — exactly what defines a characteristic.

With $A=0$ the characteristic equation $A{m}^2+Bm+C=0$ drops to the linear $Bm+C=0$, giving one finite slope $m=-C/B$; the "missing" second root corresponds to a vertical characteristic (infinite slope $dy/dx$, i.e. $dx/dy=0$). Counting both, there are two real characteristics and $\Delta =B^2-4AC=B^2>0$, so it is hyperbolic — the standard quadratic just has to be read as a degenerate (one root at infinity) case.

The characteristic equation $A{m}^2+Bm+C=0$ collapses to $C=0$, which has no finite root — instead both roots sit at infinite slope, meaning the characteristics are the vertical lines $x=\text{const}$ (a repeated family). Meanwhile $\Delta =B^2-4AC=0-0=0$, so it is parabolic: one real (repeated) characteristic direction, consistent with $u_{yy}=0$ meaning $u$ is linear in $y$ along each vertical line.

On $y=0$, $\Delta =-4y=0$, so it is parabolic exactly on that line — the degenerate frontier separating the elliptic region ($y>0$) from the hyperbolic region ($y<0$).

There is no principal (second-order) part, so the discriminant test doesn't apply; it is really a first-order PDE and must be handled by Method of Characteristics.

The type is defined pointwise: elliptic where $\Delta <0$, hyperbolic where $\Delta >0$, and parabolic exactly on the zero-set curve. Such "mixed-type" equations (transonic flow) genuinely change character across that curve.

No — swapping exchanges $A$ and $C$, and $\Delta =B^2-4AC=B^2-4CA$ is unchanged. Classification is invariant under relabelling coordinates.

Only along that solution-dependent curve at that instant; elsewhere it may be elliptic or hyperbolic. For quasilinear equations the very location of the parabolic frontier moves with the solution, so type is a local, solution-dependent statement.