a boundary value problem
/ BVP /
Suppose you want to know the shape of a cable strung between two poles. You do not know its slope at either end, and you do not start at one pole and march along; instead you know just two facts — the height of the cable where it is fastened at each pole. The shape in between is whatever the physics of sagging dictates, pinned down by those two end conditions. That is the spirit of a boundary value problem: you control the solution at the two ends of an interval rather than at a single starting point.
Precisely, you have a differential equation on an interval, say a y'' + b y' + c y = g(x) for a <= x <= L, together with conditions imposed at the two endpoints a and L — for example y(a) = A and y(L) = B (fixed values), or y'(a) = 0 and y'(L) = 0 (fixed slopes), or a mixture. These are the boundary conditions. The method is to write the general solution of the equation with its two arbitrary constants, then force those constants to satisfy the two end conditions. You end up not integrating forward in time but solving a small system of equations for the constants.
Boundary value problems are how steady-state and equilibrium phenomena are posed: the deflection of a loaded beam clamped at both ends, the temperature along a rod held at fixed temperatures at each end, the standing waves on a guitar string anchored at both bridges. They are also the gateway to eigenvalue theory and to solving partial differential equations by separation of variables, where the spatial part of the problem is always a boundary value problem.
Solve y'' + y = 0 on 0 <= x <= 1 with y(0) = 0 and y(1) = 2. The general solution is y = c1 cos(x) + c2 sin(x). The condition y(0) = 0 gives c1 = 0, and y(1) = 2 gives c2 sin(1) = 2, so c2 = 2/sin(1). Hence y = (2/sin(1)) sin(x) — a single, unique solution.
Two end conditions pin down the two constants in the general solution — but, unlike an initial value problem, this is not always solvable.
A boundary value problem can have exactly one solution, no solution at all, or infinitely many — there is no general existence-and-uniqueness guarantee like the one for initial value problems. Which case occurs depends delicately on the equation and the endpoints.