Partial Differential Equations

An ordinary differential equation tracks a function of one variable; a partial differential equation (PDE) tracks a function of two or more, so its rates of change are partial derivatives. That single step up covers most of the classical physics the board can lean on: heat flowing along a bar, a string vibrating, and a steady potential field. This lesson names the order and linearity of a PDE, introduces the heat, wave, and Laplace equations that the CELE keeps returning to, sorts any second-order PDE into elliptic, parabolic, or hyperbolic with one discriminant, and then solves a rod-cooling problem and a vibrating-string problem all the way to a number with units by separation of variables.

What makes an equation partial

Write the unknown as u and let subscripts denote partial derivatives: u_t is du/dt, u_x is du/dx, u_xx is the second derivative in x, and u_xy is the mixed second derivative. The one-dimensional heat equation u_t = alpha u_xx, for instance, ties the rate of temperature change in time to the curvature of the temperature profile in space. Because u depends on both x and t, no single ordinary derivative can describe it; you need the partials.

Two features you must read off any PDE at a glance are its order and whether it is linear.

Order and linearity

The order of a PDE is the order of the highest partial derivative that appears. All three headline equations below are second order. A PDE is linear when u and every one of its derivatives appear only to the first power, are never multiplied together, and carry coefficients that depend at most on the independent variables. A term like u u_x (the unknown times its own derivative) or (u_x)^2 breaks linearity.

PDE	Order	Linear?
u_t = alpha u_xx (heat)	2	linear
u_tt = c^2 u_xx (wave)	2	linear
u_xx + u_yy = 0 (Laplace)	2	linear
u_t + u u_x = 0 (inviscid Burgers)	1	nonlinear
u_x + (u_y)^2 = 0	1	nonlinear

Linearity is worth guarding jealously, because the whole toolkit that follows (superposition and separation of variables) rests on it. If the board hands you a linear PDE, any sum of solutions is again a solution, and that is exactly what lets a Fourier-style series of simple product pieces build the answer.

The three equations the board keeps returning to

Almost every PDE item on the MSTE paper is one of three second-order linear equations, each attached to a familiar piece of physics.

Equation	Standard form	Models	Conditions needed
Heat / diffusion	u_t = alpha u_xx	temperature or concentration spreading in time	1 initial condition plus boundary conditions
Wave	u_tt = c^2 u_xx	a vibrating string or a traveling disturbance	2 initial conditions plus boundary conditions
Laplace	u_xx + u_yy = 0	a steady state, no time dependence	boundary conditions only

Here alpha is the thermal diffusivity (units of length^2 per time, for example cm^2/s) and c is the wave speed (length per time). Notice the pattern in the last column: the heat equation is first order in time, so it needs one starting snapshot u(x, 0); the wave equation is second order in time, so it needs both an initial shape u(x, 0) and an initial velocity u_t(x, 0); and Laplace's equation has no time in it at all, so it is fixed entirely by what happens on the boundary.

Classification: elliptic, parabolic, hyperbolic

Any second-order linear PDE in two variables can be written A u_xx + B u_xy + C u_yy + (lower-order terms) = 0. The single quantity that sets its character is the discriminant B^2 - 4AC, read exactly like the discriminant of a quadratic.

Discriminant B^2 - 4AC	Type	Model equation
< 0	elliptic	Laplace u_xx + u_yy = 0
= 0	parabolic	heat u_t = alpha u_xx
> 0	hyperbolic	wave u_tt = c^2 u_xx

Check the three headliners. Laplace has A = 1, B = 0, C = 1, so B^2 - 4AC = 0 - 4 = -4, which is negative and elliptic. The wave equation u_tt - c^2 u_xx = 0, read in the variables t and x, has A = 1 (on u_tt), B = 0, C = -c^2, so B^2 - 4AC = 0 - 4(1)(-c^2) = 4 c^2, positive and hyperbolic. The heat equation carries only one second derivative, u_xx, with no u_tt and no mixed term, so A = 0 for the u_tt slot while the u_xx coefficient sits alone; its discriminant works out to 0, the parabolic borderline. The three names are not decoration: elliptic problems are steady and smooth everywhere, parabolic problems smear sharp features out as they diffuse forward in time, and hyperbolic problems carry signals at finite speed along characteristics.

Worked example (classification carried to a number): Classify u_xx - 3 u_xy + 2 u_yy = 0. Match coefficients: A = 1 sits on u_xx, B = -3 sits on u_xy, and C = 2 sits on u_yy. Then B^2 - 4AC = (-3)^2 - 4(1)(2) = 9 - 8 = 1. Since 1 > 0, the equation is hyperbolic. The lower-order terms, had there been any, would not change the answer; only the three leading coefficients A, B, and C enter the discriminant.

Boundary and initial conditions

A PDE alone has a whole family of solutions; the side conditions single out the one you want. Conditions imposed in space, on the edges of the region, are boundary conditions; conditions imposed at the starting time are initial conditions. Two boundary-condition flavors recur: a Dirichlet condition fixes the value of u itself on the boundary (for instance u = 0 at both ends of a rod held in an ice bath), while a Neumann condition fixes the derivative, that is the flux, on the boundary (for instance u_x = 0 at an insulated end that lets no heat through).

The count follows the order in time. The heat equation, first order in t, needs the single initial snapshot u(x, 0) together with a boundary condition at each end. The wave equation, second order in t, needs two initial conditions, the initial shape u(x, 0) and the initial velocity u_t(x, 0), plus a boundary condition at each end. Laplace's equation has no time derivative, so it takes boundary conditions only and describes a steady state that has already settled.

Separation of variables and product solutions

The workhorse method for these linear PDEs is separation of variables. Guess that the solution factors into a product of single-variable functions, u(x, t) = X(x) T(t), substitute, and force the variables apart. For the heat equation u_t = alpha u_xx this gives X(x) T'(t) = alpha X''(x) T(t). Divide both sides by alpha X(x) T(t):

T'(t) / (alpha T(t)) = X''(x) / X(x).

The left side depends only on t and the right side only on x, yet they are equal for all x and t, so each must equal the same constant. Call it -lambda (the minus sign is chosen so the eigenvalues come out positive). This splits the one PDE into two ordinary differential equations:

X''(x) + lambda X(x) = 0 and T'(t) + alpha lambda T(t) = 0.

With Dirichlet conditions X(0) = 0 and X(L) = 0, only the discrete eigenvalues lambda_n = (n pi / L)^2 with X_n(x) = sin(n pi x / L) survive, for n = 1, 2, 3, and so on. The matching time factor is T_n(t) = e^(-alpha lambda_n t). Superposing the product solutions gives the full answer:

u(x, t) = sum over n of B_n sin(n pi x / L) e^(-alpha (n pi / L)^2 t).

Because the exponent grows like n^2, the higher modes vanish first and the profile quickly smooths to its lowest surviving mode. That decaying product structure is exactly what the first figure sketches.

Worked example (rod cooling carried to a number): A copper rod of length L = 10 cm has both ends held at 0 degrees C and a thermal diffusivity alpha = 1.0 cm^2/s. Its initial temperature is u(x, 0) = 50 sin(pi x / 10) degrees C, which is already a single sine mode, so only the n = 1 term is present with B_1 = 50 degrees C and all other B_n = 0. First the eigenvalue: lambda_1 = (pi / L)^2 = (pi / 10)^2 = (0.31416)^2 = 0.0987 per cm^2. The solution is therefore u(x, t) = 50 sin(pi x / 10) e^(-(1.0)(0.0987) t). Ask for the midpoint x = 5 cm at t = 5 s. The space factor is sin(pi (5) / 10) = sin(pi / 2) = 1. The exponent is -(1.0)(0.0987)(5) = -0.4935, and e^(-0.4935) = 0.6105. So u(5, 5) = 50 (1)(0.6105) = 30.5 degrees C. The midpoint has cooled from 50 degrees C to about 30.5 degrees C in five seconds.

The wave equation by separation

The same product guess u(x, t) = X(x) T(t) applied to u_tt = c^2 u_xx splits it into X'' + lambda X = 0 and T'' + c^2 lambda T = 0. The only change from the heat case is that the time equation is now second order, so its solution is an oscillation rather than a decay. With fixed ends the eigenfunctions are again X_n = sin(n pi x / L), and each mode oscillates as a standing wave:

u_n(x, t) = sin(n pi x / L) (a_n cos(n pi c t / L) + b_n sin(n pi c t / L)).

The nth mode vibrates at angular frequency omega_n = n pi c / L, which is the ordinary frequency f_n = n c / (2 L). The n = 1 mode is the fundamental, a single half sine that is pinned at the ends (the nodes) and swings largest at the middle (the antinode).

Worked example (vibrating string carried to a number): A string of linear density mu = 0.005 kg/m is stretched to a tension T = 200 N over a fixed span L = 0.8 m. First the wave speed: c = sqrt(T / mu) = sqrt(200 / 0.005) = sqrt(40000) = 200 m/s. The fundamental frequency is f_1 = c / (2 L) = 200 / (2 (0.8)) = 200 / 1.6 = 125 Hz, and its period is T_1 = 1 / f_1 = 1 / 125 = 0.008 s = 8 ms. The next mode up, the first overtone, sits at f_2 = 2 c / (2 L) = 2 (125) = 250 Hz. So the standing wave on this string vibrates at a fundamental of 125 Hz with an 8 ms period.

Exam-day strategy

• Read the order and the linearity first: the highest partial derivative sets the order, and any product of u with itself or its derivatives, or any squared derivative, makes it nonlinear and kills separation of variables.
• Memorize the three canonical forms cold: heat u_t = alpha u_xx (parabolic), wave u_tt = c^2 u_xx (hyperbolic), Laplace u_xx + u_yy = 0 (elliptic).
• Classify with the quadratic-style discriminant B^2 - 4AC on the coefficients of u_xx, u_xy, and u_yy: negative is elliptic, zero is parabolic, positive is hyperbolic, and lower-order terms never enter it.
• Count your side conditions by order in time: heat wants one initial condition, wave wants two (shape and velocity), and Laplace wants none, only boundary conditions. Remember Dirichlet fixes the value and Neumann fixes the flux.
• For separation of variables, substitute u = X(x) T(t), divide through so each side holds one variable, set both equal to the same constant -lambda, and solve the two ordinary differential equations; with fixed ends the eigenvalues are lambda_n = (n pi / L)^2 and the eigenfunctions are sin(n pi x / L).
• On a heat problem the time factor is a decaying e^(-alpha (n pi / L)^2 t) while on a wave problem it is an oscillating cos and sin at frequency f_n = n c / (2 L); do not swap the two.