Partial Differential Equations

PDE classification (parabolic, hyperbolic, elliptic)

For a second-order linear PDE in two independent variables, write it as A u_{xx} + 2B u_{xy} + C u_{yy} + (\text{lower-order terms}) = 0. The type is decided only by the principal part, through the discriminant \Delta = B^2 - AC. The equation is hyperbolic where \Delta > 0, parabolic where \Delta = 0, and elliptic where \Delta < 0.

Lower-order terms such as D u_x + E u_y + F u have no effect on the type. Be careful with the factor 2 in front of B. If the mixed term is written as b\,u_{xy}, then B = b/2 and the discriminant is (b/2)^2 - AC.

When the coefficients depend on x and y, the type changes from point to point. The sets where \Delta = 0 are curves that separate hyperbolic and elliptic regions. To find them, set \Delta = 0 and complete the square to recognise the curve. For example, in u_{xx} + 2x\,u_{xy} + u_{yy} = 0 we get \Delta = x^2 - 1. This is parabolic on the lines x = \pm 1, hyperbolic for |x| > 1 and elliptic for |x| < 1.

When \Delta factors, check the sign of each factor carefully. For instance, \Delta = (y-1)(x-3)^2 has a square factor that is never negative. Its sign is therefore set by (y-1) alone, and every point on the line x = 3 is parabolic.

A useful reference table: the wave equation u_{tt} = c^2 u_{xx} is hyperbolic (\Delta = c^2 > 0), the heat equation u_t = k\,u_{xx} is parabolic (no u_{tt} term, so \Delta = 0), and the Laplace equation u_{xx} + u_{yy} = 0 is elliptic (\Delta = -1 < 0). For example, u_{xx} + 2u_{xy} - 3u_{yy} = 0 has \Delta = 1 + 3 = 4 > 0, so it is hyperbolic.

Heat equation solution by separation of variables

The heat equation u_t = k\,u_{xx} on 0 < x < L with homogeneous Dirichlet conditions u(0,t) = u(L,t) = 0 is solved by separation of variables. Put u = X(x)T(t). The equation gives \dfrac{T'}{kT} = \dfrac{X''}{X} = -\lambda. The boundary conditions force X_n(x) = \sin\left(\dfrac{n\pi x}{L}\right) with \lambda_n = \left(\dfrac{n\pi}{L}\right)^2 for n = 1, 2, \dots. The time factor is T_n(t) = e^{-k(n\pi/L)^2 t}, so every mode decays.

The general solution is u(x,t) = \sum_{n\ge 1} b_n \sin\left(\dfrac{n\pi x}{L}\right) e^{-k(n\pi/L)^2 t}, where b_n = \dfrac{2}{L}\displaystyle\int_0^L f(x)\sin\left(\dfrac{n\pi x}{L}\right)dx for initial data u(x,0) = f(x). Two shortcuts follow. First, if the initial data is already a single sine mode, only that term survives and no integral is needed. Second, higher modes decay faster, so for large t only the lowest surviving mode remains.