Boundary Value Problems

A boundary value problem is simply a differential equation where you know the answer's value at two ends, instead of its starting value and slope at one point. (The "boundary" is the edge of the interval, so the conditions sit at its two ends.) It's the natural setup for anything held at both ends, like a jump rope held by two people, a rod with each end at a set temperature, or a beam on two supports. The setup $y'' + p(x)y' + q(x)y = g(x)$ with $y(a) = y_a$ and $y(b) = y_b$ looks formal, but it's just two simple parts: the equation (the rule the curve has to follow everywhere) and the two boundary conditions (the heights $y_a$ and $y_b$ where it's pinned at $x = a$ and $x = b$). The surprise is that sometimes no curve fits, and sometimes infinitely many do. It's less mysterious than it sounds; bear with it.

Every problem so far gave you $y$ and $y'$ at one starting point, and initial value problems have a reliable existence and uniqueness theorem: exactly one solution, every time. Now change the question. Instead of a starting value and slope, fix the value of $y$ at two different places, like a string tied down at both ends.

Boundary value problems don't have that guarantee, because the general solution's constants must satisfy conditions at two different places. Think of it this way: an initial condition sets the constants all at one spot, and the solution just runs from there. A boundary condition asks whether some solution that starts right also happens to land right at the far end, and sometimes the far end is out of reach.

For $y'' + 4y = 0$ (general solution $c_1\cos 2x + c_2\sin 2x$):

• $y(0) = -2$, $y\!\left(\frac\pi4\right) = 10$: $c_1 = -2$, $c_2 = 10$. One solution.
• $y(0) = -2$, $y(2\pi) = -2$: $c_1 = -2$, and $c_2$ is anything. Infinitely many.
• $y(0) = -2$, $y(2\pi) = 3$: $c_1$ would have to be both $-2$ and $3$. None.

Notice what goes wrong at $2\pi$. Both $\cos 2x$ and $\sin 2x$ have period $\pi$, so at $x = 2\pi$ every solution is back where it started: $y(2\pi) = c_1$, and $c_2$ has no say. The second condition either repeats the first (infinitely many solutions) or contradicts it (none).

A common mistake. It's tempting to think two conditions for two constants always give exactly one solution, because that's how it worked for initial value problems. It isn't so: the second and third cases above have two conditions and two constants, and still give infinitely many and none. What matters is whether the two conditions actually pin down different constants.

These are the problems that arise when a physical object (a string, a rod, a beam) is held at both ends. There's an animation further down that slides the right end and shows the solution bending harder and harder to reach it.

The sag of a cable between two poles, the temperature in a wall between two rooms and the deflection of a beam on two supports are boundary value problems.