Boundary Value Problems

So far we have dealt with initial-value problems, where all conditions are imposed at one point and a solution always exists and is unique. Boundary-value problems, with conditions at two or more points, behave completely differently: the same equation may have exactly one solution, infinitely many, or none at all, depending only on where the conditions are placed and what values they demand.

Quick Reference

Item	Statement
Initial value problem	All conditions at one point: y'' + 3y' - y = \sin x, $y (0) = - 1$ , y'(0) = 0
Boundary value problem	Conditions at two or more points: $y (x_{0}) = α$ , y'(x_1) = \beta
IVP existence	Guaranteed and unique for linear equations with continuous coefficients
BVP existence	May have a unique solution, infinitely many, or none
Dirichlet condition	Prescribes the value, $y (a) = α$
Neumann condition	Prescribes the derivative, y'(a) = \alpha
Robin (mixed) condition	Prescribes a combination, \alpha y(a) + \beta y'(a) = \gamma
Eigenvalue	A parameter value for which a nontrivial solution exists
Eigenfunction	The corresponding nontrivial solution

Recalling the Distinction

An initial value problem is a differential equation together with conditions all specified at a single point. For example,

y'' + 3y' - y = \sin x, \quad y(0) = -1, \quad y'(0) = 0

is an initial value problem, because the value of $y$ and of its derivative are both specified at $x = 0$ .

A boundary value problem (BVP) is a differential equation together with conditions in which the values of $y$ , or its derivatives, or combinations of them, are specified at the boundaries of an interval. For example,

y'' + P(x)y' + Q(x)y = f(x), \quad y(x_0) = \alpha, \quad y'(x_1) = \beta

is a boundary value problem, because the conditions are specified at the two different points $x_{0}$ and $x_{1}$ .

Boundary value problems arise throughout engineering and the sciences, typically wherever a steady state is described across a region rather than an evolution forward in time. The deflection of a beam supported at both ends, the temperature distribution in a rod held at fixed temperatures at each end, and the shape of a hanging cable are all boundary value problems.

Existence and Uniqueness Fail

Unlike initial value problems, for which existence and uniqueness are guaranteed by the theorem of Chapter 9, a boundary value problem may have a unique solution, infinitely many solutions, or no solution at all.

The reason is structural. An initial value problem fixes the solution's value and slope at one point, and the equation then determines everything downstream, so nothing can conflict. A boundary value problem instead demands that the solution, having left the first boundary, arrive at the second with a prescribed value. What happens in between is governed by the equation, not by us, so the arrival value is not free to be chosen. The following example shows all three outcomes for a single equation.

Consider y'' + y = 0 subject to each of the following sets of boundary conditions.

(a) $y (0) = 1$ and $y (π / 2) = 3$

(b) $y (0) = 1$ and $y (π) = - 1$

(c) $y (0) = 1$ and $y (π) = 2$

Solution

The general solution of y'' + y = 0 is

y = A \sin x + B \cos x .

To determine $A$ and $B$ we apply the boundary conditions in each case. It helps to note in advance that $\sin 0 = 0$ , $\cos 0 = 1$ , $\sin (π / 2) = 1$ , $\cos (π / 2) = 0$ , $\sin π = 0$ , and $\cos π = - 1$ .

(a) With $y (0) = 1$ and $y (π / 2) = 3$ ,

\left\{ \begin{aligned} y(0) = 1 &\ \Longrightarrow\ 1 = 0 + B \\ y(\pi/2) = 3 &\ \Longrightarrow\ 3 = A + 0 \end{aligned} \right. \quad \Longrightarrow \quad A = 3, \ B = 1 .

Each condition determined one constant independently, so the problem has a unique solution

y = 3 \sin x + \cos x .

(b) With $y (0) = 1$ and $y (π) = - 1$ ,

\left\{ \begin{aligned} y(0) = 1 &\ \Longrightarrow\ 1 = 0 + B \\ y(\pi) = -1 &\ \Longrightarrow\ -1 = 0 - B \end{aligned} \right. \quad \Longrightarrow \quad B = 1 .

Both equations say the same thing, $B = 1$ , and $A$ never appears. It is therefore arbitrary, so the problem has infinitely many solutions

y = A \sin x + \cos x

one for each real $A$ . Note that $\sin x$ vanishes at both endpoints, so adding any multiple of it disturbs neither condition.

(c) With $y (0) = 1$ and $y (π) = 2$ ,

\left\{ \begin{aligned} y(0) = 1 &\ \Longrightarrow\ 1 = 0 + B \\ y(\pi) = 2 &\ \Longrightarrow\ 2 = 0 - B \end{aligned} \right.

The two equations demand $B = 1$ and $B = - 2$ simultaneously, which is impossible. This problem has no solution.

The three cases are not accidents of arithmetic. Cases (b) and (c) share the same endpoints $0$ and $π$ , which are exactly the points where $\sin x$ vanishes. That coincidence is what makes the coefficient $A$ invisible to both conditions, and it forces the all-or-nothing behavior: either the two demands on $B$ agree, giving a whole family, or they conflict, giving nothing.

Eigenvalues and Eigenfunctions

The phenomenon in case (b) is important enough to have a name.

Consider a boundary value problem containing a parameter, such as

y'' + \lambda y = 0, \quad y(0) = 0, \quad y(L) = 0 .

The values of $λ$ for which a nontrivial solution exists are called eigenvalues, and the corresponding solutions are called eigenfunctions.

The function $y \equiv 0$ always satisfies a homogeneous problem of this kind, so the question is never whether a solution exists but whether a nonzero one does.

Find the eigenvalues and eigenfunctions of y'' + \lambda y = 0 with $y (0) = 0$ and $y (L) = 0$ , where $L > 0$ .

Solution

Three cases must be checked separately, according to the sign of $λ$ .

Case $λ < 0$ . Write $λ = - μ^{2}$ with $μ > 0$ . The general solution is $y = c_{1} e^{μ x} + c_{2} e^{- μ x}$ , and the conditions give

c_{1} + c_{2} = 0, c_{1} e^{μ L} + c_{2} e^{- μ L} = 0 .

Substituting $c_{2} = - c_{1}$ into the second gives $c_{1} (e^{μ L} - e^{- μ L}) = 0$ . Since $μ L > 0$ , the bracket is nonzero, so $c_{1} = c_{2} = 0$ . Only the trivial solution: no negative eigenvalues.

Case $λ = 0$ . Then y'' = 0, so $y = c_{1} + c_{2} x$ . The conditions give $c_{1} = 0$ and $c_{2} L = 0$ , hence $c_{2} = 0$ . Again only the trivial solution.

Case $λ > 0$ . Write $λ = μ^{2}$ with $μ > 0$ . The general solution is $y = c_{1} \cos μ x + c_{2} \sin μ x$ . The condition $y (0) = 0$ gives $c_{1} = 0$ , leaving $y = c_{2} \sin μ x$ . The condition $y (L) = 0$ then requires

c_{2} \sin μ L = 0 .

For a nontrivial solution we need $c_{2} \neq 0$ , hence $\sin μ L = 0$ , hence $μ L = n π$ for a positive integer $n$ .

Therefore the eigenvalues are

λ_{n} = {(\frac{n π}{L})}^{2}, n = 1, 2, 3, \dots

with eigenfunctions

y_{n} (x) = \sin \frac{n π x}{L},

each determined only up to a constant multiple.

These are the standing waves on a string of length $L$ fixed at both ends: $n = 1$ is the fundamental, $n = 2$ the first overtone, and so on. The same eigenvalue problem underlies Fourier series and the solution of the heat and wave equations by separation of variables.

Types of Boundary Conditions

Name	Form at the endpoint $a$	Physical reading
Dirichlet	$y (a) = α$	The value is held fixed, as with a clamped temperature
Neumann	y'(a) = \alpha	The flux is prescribed, as with an insulated end where y'(a) = 0
Robin (mixed)	\alpha y(a) + \beta y'(a) = \gamma	A relation between value and flux, as in convective cooling
Periodic	$y (a) = y (b)$ and y'(a) = y'(b)	The interval closes into a ring

A single problem may impose different types at its two ends, and doing so changes the eigenvalues.

Why the Solution Methods Differ

Numerical methods for initial value problems march forward: knowing the complete state at $x_{0}$ , a stepping method such as Euler's or a Runge-Kutta method advances the solution using only information already computed.

Boundary value problems permit no such march, since the data at the far boundary is not available at the start. They require iterative techniques that adjust the solution until every boundary condition is met.

The shooting method guesses the missing initial slope, solves the resulting initial value problem, checks the value reached at the far boundary, and revises the guess. The name is apt: it is artillery ranging, adjusting the elevation until the shell lands on target.
Finite difference methods discretize the whole interval at once, replacing derivatives by difference quotients and solving one large system of algebraic equations for all the grid values simultaneously.

Exercises

Determine whether y'' + 4y = 0 with $y (0) = 0$ and $y (π / 4) = 1$ has a unique solution, infinitely many, or none.

Solution

The general solution is $y = A \cos 2 x + B \sin 2 x$ . The condition $y (0) = 0$ gives $A = 0$ . At $x = π / 4$ we have $2 x = π / 2$ , so $\sin (π / 2) = 1$ and the second condition gives $B = 1$ .

Unique solution: $y = \sin 2 x$ .

Determine whether y'' + 4y = 0 with $y (0) = 0$ and $y (π / 2) = 1$ has a solution.

Solution

Again $A = 0$ , leaving $y = B \sin 2 x$ . At $x = π / 2$ we have $2 x = π$ , so $\sin π = 0$ and the condition reads $0 = 1$ .

No solution. The endpoint $π / 2$ is a zero of $\sin 2 x$ , so every candidate solution is forced to vanish there and can never equal $1$ .

Note that changing the required value to $0$ would give infinitely many solutions instead, since $B$ would then be unconstrained. The boundary between "none" and "infinitely many" is exactly this.

Find the eigenvalues and eigenfunctions of y'' + \lambda y = 0 with y'(0) = 0 and y'(\pi) = 0, two Neumann conditions.

Solution

For $λ = μ^{2} > 0$ , the general solution is $y = c_{1} \cos μ x + c_{2} \sin μ x$ , with y' = -c_1\mu\sin\mu x + c_2\mu\cos\mu x. The condition y'(0) = 0 gives $c_{2} = 0$ , leaving $y = c_{1} \cos μ x$ . Then y'(\pi) = -c_1\mu\sin\mu\pi = 0 requires $\sin μ π = 0$ , so $μ = n$ for a positive integer $n$ .

Eigenvalues $λ_{n} = n^{2}$ with eigenfunctions $y_{n} = \cos n x$ , for $n = 1, 2, 3, \dots$

The case $λ = 0$ must also be checked, and here it succeeds: $y = c_{1} + c_{2} x$ has y' = c_2, and both conditions give $c_{2} = 0$ , leaving the nonzero constant solution $y = 1$ . So $λ_{0} = 0$ is also an eigenvalue, with eigenfunction $y_{0} = 1$ .

Unlike the Dirichlet problem, the Neumann problem admits $λ = 0$ , which corresponds physically to a rod with both ends insulated settling at a uniform temperature.

Frequently Asked Questions

Why can a boundary value problem have no solution?

Because the two conditions are not independent. The equation controls what happens between the endpoints, so once the solution leaves the first boundary its arrival value at the second is largely determined. If the second condition demands something the equation cannot deliver, nothing satisfies both.

Why can it have infinitely many solutions?

When some nonzero homogeneous solution happens to satisfy both boundary conditions, any multiple of it can be added freely. In case (b) above, $\sin x$ vanishes at both $0$ and $π$ , so the coefficient $A$ is completely unconstrained.

Is there a test for which case will occur?

Yes. Solve the associated homogeneous boundary value problem, the one with both boundary values set to zero. If only the trivial solution exists, the original problem has exactly one solution for any prescribed boundary values. If a nontrivial solution exists, the original has either infinitely many or none, depending on a compatibility condition. This is the Fredholm alternative.

What are eigenvalues and eigenfunctions?

For a problem containing a parameter $λ$ , the eigenvalues are the values of $λ$ admitting a nonzero solution, and the eigenfunctions are those solutions. They are the resonant configurations of the system: standing waves on a string, vibration modes of a beam, energy levels of a quantum particle in a box.

Why does the eigenvalue calculation split into three cases?

Because the form of the general solution depends on the sign of $λ$ : negative gives real exponentials, zero gives a linear function, positive gives sines and cosines. Only the oscillatory case can vanish at two separate points without being identically zero, which is why the eigenvalues turn out to be positive here.

How are BVPs solved in practice?

Analytically when the equation is simple, as in the examples above. Numerically, either by the shooting method, which guesses the missing initial data and iterates, or by finite differences, which discretize the whole interval and solve one large linear system. Neither can simply march forward the way an initial value solver does.

Quick Reference

Recalling the Distinction

Existence and Uniqueness Fail

Eigenvalues and Eigenfunctions

Types of Boundary Conditions

Why the Solution Methods Differ

Exercises

Frequently Asked Questions

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