Boundary Value Problems

IVP vs BVP — why they're different

An initial value problem (IVP) pins both the value and (usually) the derivative at the same point: $y(a) = \alpha$, $y'(a) = \beta$. Picard–Lindelöf guarantees a unique solution nearby.

A boundary value problem (BVP) pins values at different points: $y(a) = \alpha$, $y(b) = \beta$. This is a totally different beast. A BVP may have

• Zero solutions,
• A unique solution, or
• Infinitely many solutions — depending on the ODE and the boundary values. There's no Picard–Lindelöf analogue.

When a BVP admits only the trivial solution for generic data but infinitely many for special data, those special data points are eigenvalues and the corresponding solutions are eigenfunctions. This is the structure that turns up in quantum mechanics, vibrating-string / drum-head problems, and PDEs solved by separation of variables.

The given BVP

$y'' + y = 0, \qquad y(0) = 0, \; y(\pi) = 0.$

Step-by-step analysis

Step 1 — General solution of the ODE.

Characteristic equation $r^{2} + 1 = 0 \implies r = \pm i$. See #184. $y(x) = C_1 \cos x + C_2 \sin x.$

Step 2 — Apply boundary condition at $x = 0$.

$y(0) = C_1 \cdot 1 + C_2 \cdot 0 = C_1 = 0 \implies C_1 = 0.$

So $y(x) = C_2 \sin x$.

Step 3 — Apply boundary condition at $x = \pi$.

$y(\pi) = C_2 \sin \pi = C_2 \cdot 0 = 0.$

This is satisfied for every value of $C_2$. So the BVP has infinitely many solutions: $y(x) = C_2 \sin x$ for any constant $C_2$.

$\boxed{\, y(x) = C_2 \sin x, \quad C_2 \in \mathbb{R} \,}$

Why this happened — the eigenvalue structure

The BVP is part of a family: $y'' + \lambda \, y = 0, \qquad y(0) = 0, \; y(\pi) = 0,$ parameterised by $\lambda$. For generic $\lambda$, the unique solution satisfying both boundary conditions is $y \equiv 0$ (the trivial solution). For special values of $\lambda$, non-trivial solutions exist.

Step A. General solution for $\lambda > 0$: $y = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda} x)$.

Step B. $y(0) = 0 \implies A = 0$, so $y = B \sin(\sqrt{\lambda} x)$.

Step C. $y(\pi) = B \sin(\sqrt{\lambda} \pi) = 0$. Either $B = 0$ (trivial solution) or $\sin(\sqrt{\lambda} \pi) = 0$. The latter requires $\sqrt{\lambda} \pi = n \pi$ for integer $n$: $\sqrt{\lambda_n} = n \implies \lambda_n = n^{2}, \quad n = 1, 2, 3, \ldots$

These are the eigenvalues, and the corresponding eigenfunctions are $\phi_n(x) = \sin(n x), \quad n = 1, 2, 3, \ldots$

Our original problem has $\lambda = 1 = 1^{2}$, so it's the first eigenvalue ($n = 1$) and the first eigenfunction $\phi_1 = \sin x$.

Why there are infinitely many eigenvalues

Every integer $n$ works because $\sin(n \pi) = 0$ for every integer $n$. So we get an infinite sequence of eigenvalue–eigenfunction pairs:

• $\lambda_1 = 1$, $\phi_1 = \sin x$ (one half-wave on $[0, \pi]$).
• $\lambda_2 = 4$, $\phi_2 = \sin 2x$ (two half-waves).
• $\lambda_3 = 9$, $\phi_3 = \sin 3x$ (three half-waves).
• $\ldots$
• $\lambda_n = n^{2}$, $\phi_n = \sin n x$ ($n$ half-waves).

These are the standing-wave modes of a guitar string of length $\pi$ pinned at both ends.

Sturm–Liouville perspective

Any linear BVP of the form $\frac{d}{dx}\!\left[p(x) y'\right] + q(x) y + \lambda \, r(x) y = 0,$ with appropriate boundary conditions, is a Sturm–Liouville problem. Under mild hypotheses, its eigenvalues form a discrete, increasing, unbounded sequence $\lambda_1 < \lambda_2 < \ldots \to \infty$, and its eigenfunctions are orthogonal under the weight $r(x)$: $\int_{a}^{b} \phi_n(x) \phi_m(x) r(x) \, dx = 0 \quad \text{for } n \ne m.$

This orthogonality is the foundation of Fourier series (the $[-\pi, \pi]$ case of $y'' + \lambda y = 0$) and generalises to Legendre polynomials, Bessel functions, spherical harmonics, etc.

When a BVP has no solution

Consider $y'' = 0$, $y(0) = 0$, $y(1) = 1$. General solution $y = a + b x$, boundary conditions give $a = 0$ and $a + b = 1$ → $b = 1$. Unique solution $y(x) = x$.

Now change to $y'' = 0$, $y(0) = 0$, with $y(0) = 1$ as the second boundary condition (both at the same point, a contradiction). No solution — the boundary data is inconsistent.

Non-homogeneous BVPs at resonance with an eigenvalue can also fail to have a solution. The Fredholm alternative makes this precise: a BVP at an eigenvalue has solutions iff the forcing is orthogonal to the homogeneous eigenfunctions.

Physical meaning — standing waves on a string

The BVP $y'' + \lambda y = 0$, $y(0) = y(L) = 0$ describes a taut string fixed at both ends of length $L$. Allowed modes are $\phi_n(x) = \sin(n \pi x / L)$ with frequencies $\omega_n = n \pi c / L$ (for wave speed $c$). The fundamental ($n = 1$) is the lowest pitch; $n = 2, 3, \ldots$ are overtones. This is literally the math of musical instruments.

Initial conditions vs boundary conditions

IVP: "Given where you are and how fast you're moving right now, where will you be later?" — a forward-in-time propagation.

BVP: "You start here, and you must end there — how should you get there?" — a constraint across space/time; no unique answer unless the ODE plus endpoints cooperate.

Common mistakes

• Using Picard–Lindelöf for BVPs. The existence–uniqueness theorem is specifically for IVPs. BVPs can have many or no solutions even for smooth ODEs.
• Missing negative-$\lambda$ and $\lambda = 0$ cases. In the systematic eigenvalue search for $y'' + \lambda y = 0$, one has to rule out $\lambda \le 0$ (by showing only the trivial solution satisfies the BCs for those values) before pinning down the positive eigenvalue ladder.
• Forgetting orthogonality. Sturm–Liouville eigenfunctions are orthogonal — a miraculous fact that makes Fourier-style expansions possible. Knowing this shapes how you solve associated PDEs.
• Confusing eigenvalue of a BVP with eigenvalue of a matrix. They're related (one is the spectrum of a differential operator, the other of a matrix) but involve very different machinery.

Try it in the visualization

Sweep $\lambda$ continuously and watch how the solution satisfying $y(0) = 0$ evolves. As $\lambda$ hits $1, 4, 9, \ldots$, the solution's value at $x = \pi$ is forced to zero — the eigenvalue condition — and you unlock a whole line of non-trivial solutions. Between eigenvalues, only the flat $y \equiv 0$ can satisfy both BCs.