Eigenvalues and Eigenfunctions, differential equations lesson

The problem $y'' + y = f$ on $[0, \pi]$ with zero end values is resonant, because the homogeneous problem has the nonzero solution $\sin x$ . Now put a parameter into the equation and turn the question around. For which values of $\lambda$ does $-y'' = \lambda y$ , with $y' = dy/dx$ and given homogeneous boundary conditions, have a solution other than zero? This lesson finds those values for the four standard kinds of endpoint conditions, shows when the value $0$ is among them, and explains how to fix the free scale of the solutions.

Key terms

Eigenvalue problem

Definition Let $y' = dy/dx$ and let $\lambda$ be a real parameter. An eigenvalue problem on an interval $[a, b]$ consists of the equation

-y'' = \lambda y, \qquad a \le x \le b,

together with two homogeneous linear boundary conditions. The zero function always solves it. A number $\lambda$ is an eigenvalue if the problem has a solution $y$ that is not identically zero, and such a solution is an eigenfunction for $\lambda$ . For an eigenvalue $\lambda$ , the eigenfunctions together with the zero function form the eigenspace $E_\lambda$ , a vector space of dimension $1$ or $2$ . Equivalently, $\lambda$ is an eigenvalue exactly when the boundary-value problem $y'' + \lambda y = f$ with the same conditions is resonant. The minus sign in front of $y''$ is a convention that makes the eigenvalues of the most common problems nonnegative.

In plain words Without boundary conditions every value of $\lambda$ has nonzero solutions. The endpoint conditions are what make most values fail, and the eigenvalues are the special values whose solutions fit both ends.

Example On $[0, \pi]$ with $y(0) = y(\pi) = 0$ , the value $\lambda = 4$ is an eigenvalue with eigenfunction $\sin 2x$ . Indeed $(\sin 2x)'' = -4\sin 2x$ , so $-y'' = 4y$ , and $\sin 0 = \sin 2\pi = 0$ . The value $\lambda = 2$ is not an eigenvalue. Every solution of $-y'' = 2y$ has the form $A\cos(\sqrt{2}\,x) + B\sin(\sqrt{2}\,x)$ , the left condition gives $A = 0$ , and the right condition gives $B\sin(\sqrt{2}\,\pi) = 0$ . Since $\sqrt{2}$ is not an integer, $\sin(\sqrt{2}\,\pi) \ne 0$ , so $B = 0$ as well.

Dirichlet boundary conditions

Definition Dirichlet boundary conditions prescribe the values of $y$ at the endpoints. In an eigenvalue problem they are homogeneous, $y(a) = 0$ and $y(b) = 0$ . For

\begin{gathered} -y'' = \lambda y, \\[0.4em] y(0) = y(\pi) = 0, \end{gathered}

the eigenvalues and eigenfunctions are

\begin{gathered} \lambda_n = n^2, \qquad y_n = \sin nx, \\[0.4em] n = 1, 2, 3, \ldots, \end{gathered}

and $0$ is not an eigenvalue. The derivation treats the sign of $\lambda$ in three separate cases, so that no case is lost inside a formula with $\sqrt{\lambda}$ .

For $\lambda = 0$ the solutions are $y = A + Bx$ . The left condition gives $A = 0$ , and then $B\pi = 0$ gives $B = 0$ .
For $\lambda = -k^2$ with $k > 0$ the solutions are $y = A\cosh kx + B\sinh kx$ . The left condition gives $A = 0$ , and since $\sinh k\pi \ne 0$ the right condition gives $B = 0$ .
For $\lambda = k^2$ with $k > 0$ the solutions are $y = A\cos kx + B\sin kx$ . The left condition gives $A = 0$ , and a nonzero $B$ survives the right condition exactly when $\sin k\pi = 0$ , that is, when $k$ is a positive integer $n$ .

On a general interval $[0, L]$ the same argument gives $\lambda_n = (n\pi/L)^2$ and $y_n = \sin(n\pi x/L)$ for $n \ge 1$ .

In plain words A string held fixed at both ends can only take shapes that fit a whole number of half waves between the ends. Each extra half wave raises the eigenvalue, and a straight line or a hyperbolic shape cannot vanish at both ends unless it is zero.

Example The second mode $\sin 2x$ fits two half waves into $[0, \pi]$ and has eigenvalue $4$ . On $[0, 2]$ the first two eigenvalues are $\pi^2/4$ and $\pi^2$ , with eigenfunctions $\sin(\pi x/2)$ and $\sin \pi x$ . In general, doubling the length of the interval divides every eigenvalue by $4$ .

Neumann boundary conditions

Definition Neumann boundary conditions prescribe the derivative $y'$ at the endpoints, and in an eigenvalue problem they are $y'(a) = 0$ and $y'(b) = 0$ . For

\begin{gathered} -y'' = \lambda y, \\[0.4em] y'(0) = y'(\pi) = 0, \end{gathered}

the eigenvalues and eigenfunctions are

\begin{gathered} \lambda_n = n^2, \qquad y_n = \cos nx, \\[0.4em] n = 0, 1, 2, \ldots \end{gathered}

For $\lambda = k^2$ with $k > 0$ , the derivative of $A\cos kx + B\sin kx$ is $k(-A\sin kx + B\cos kx)$ , so the left condition gives $B = 0$ and the right condition requires $\sin k\pi = 0$ . For $\lambda = -k^2$ with $k > 0$ both coefficients vanish, because $\sinh k\pi \ne 0$ . For $\lambda = 0$ the solutions $A + Bx$ have derivative $B$ , so $B = 0$ and every constant survives. An eigenfunction with eigenvalue $0$ , that is, a nonzero solution of $y'' = 0$ that meets the boundary conditions, is called a zero mode. Here the zero mode is the constant function $1$ , and its eigenspace consists of the constant functions.

In plain words Insulated ends let no heat through, so the slope of the temperature profile is zero there. A constant profile already satisfies this, and in a heat problem on an insulated rod it carries the average temperature, which never changes. Copying the Dirichlet list, which starts at $n = 1$ , would lose this mode.

Example Check $\cos 2x$ directly. Its second derivative is $-4\cos 2x$ , so $-y'' = 4y$ , and its derivative $-2\sin 2x$ vanishes at $0$ and at $\pi$ . The constant $1$ satisfies $-y'' = 0 = 0 \cdot 1$ and has zero derivative everywhere, so it is an eigenfunction for $\lambda = 0$ . Under Dirichlet conditions the same constant fails $y(0) = 0$ , and there $0$ is not an eigenvalue at all.

Robin boundary conditions

Definition A Robin boundary condition prescribes a combination of the value and the derivative at an endpoint, $\alpha y + \beta y' = 0$ there with $\alpha \ne 0$ and $\beta \ne 0$ . A convenient form on $[0, \pi]$ is

\begin{gathered} y'(0) = h_0\,y(0), \\[0.4em] y'(\pi) = -h_\pi\,y(\pi), \end{gathered}

with constants $h_0$ and $h_\pi$ , where a zero coefficient gives back a Neumann condition at that end. The opposite signs come from the opposite outward directions at the two ends. With $h_0 > 0$ and $h_\pi > 0$ the conditions describe ends that lose heat to surroundings at temperature zero, at a rate proportional to their own temperature. For a real eigenfunction $y$ with eigenvalue $\lambda$ , multiplying $-y'' = \lambda y$ by $y$ and integrating by parts gives

\begin{gathered} \lambda\int_0^\pi y^2\,dx = \int_0^\pi (y')^2\,dx \\[0.4em] +\, h_0\,y(0)^2 + h_\pi\,y(\pi)^2. \end{gathered}

So if $h_0 \ge 0$ and $h_\pi \ge 0$ , every eigenvalue satisfies $\lambda \ge 0$ , and $\lambda = 0$ occurs only when $h_0 = h_\pi = 0$ , the Neumann case, because a zero right side forces $y$ to be a nonzero constant $c$ with $h_0c^2 = h_\pi c^2 = 0$ . With a negative coefficient the conclusion can fail. For example, $h_0 = -1$ and $h_\pi = 0$ give a negative eigenvalue near $-1.007$ .

In plain words A Robin end sits between a fixed end and an insulated one. With positive coefficients the boundary terms in the identity act like extra stiffness at the ends, and they push the eigenvalues above the Neumann ones.

Example Take an insulated left end and a cooling right end, $y'(0) = 0$ and $y'(\pi) = -y(\pi)$ , so $h_0 = 0$ and $h_\pi = 1$ . The identity rules out $\lambda \le 0$ . For $\lambda = k^2$ with $k > 0$ the left condition leaves multiples of $\cos kx$ , and the right condition reads $-k\sin k\pi = -\cos k\pi$ , which is equivalent to

k\tan(k\pi) = 1.

On each interval $(n, n + \tfrac{1}{2})$ with $n = 0, 1, 2, \ldots$ the left side increases from $0$ to $\infty$ , and at every other $k > 0$ it is at most $0$ or undefined, so there is exactly one root in each of these intervals and no other root. The first two roots are $k_1 \approx 0.3834$ and $k_2 \approx 1.2187$ . The eigenvalues $\lambda_1 \approx 0.1470$ and $\lambda_2 \approx 1.4853$ are not squares of integers, and they lie above the first two Neumann eigenvalues $0$ and $1$ .

Periodic boundary conditions

Definition Periodic boundary conditions on $[0, 2\pi]$ are

\begin{gathered} y(0) = y(2\pi), \\[0.4em] y'(0) = y'(2\pi). \end{gathered}

They relate the two ends to each other instead of constraining each end on its own. For $-y'' = \lambda y$ the eigenvalues are $n^2$ for every integer $n \ge 0$ . The zero mode is the constant $1$ , with a one-dimensional eigenspace. For each $n \ge 1$ the eigenspace is two-dimensional,

E_{n^2} = \operatorname{span}\{\cos nx, \sin nx\}.

For $\lambda = k^2$ with $k > 0$ , the two conditions on $A\cos kx + B\sin kx$ form a linear system for $A$ and $B$ with determinant $k(2 - 2\cos 2\pi k)$ . It vanishes exactly when $k$ is an integer, and then both conditions hold for every $A$ and $B$ . For $\lambda = -k^2$ with $k > 0$ the corresponding determinant is $k(2 - 2\cosh 2\pi k) < 0$ , so only the zero solution remains, and for $\lambda = 0$ the condition $y(0) = y(2\pi)$ removes the term $Bx$ and leaves the constants.

In plain words Periodic conditions describe a ring rather than a rod with two ends. A wave can sit anywhere around the ring, so for each $n \ge 1$ the cosine shape, the sine shape and every shifted copy $\sin(n(x - c))$ fit. That is why one eigenvalue carries two independent eigenfunctions here, while each Dirichlet or Neumann eigenvalue on an interval has only one.

Example For $n = 1$ the functions $\cos x$ and $\sin x$ both satisfy $-y'' = y$ , both take equal values at $0$ and $2\pi$ , and both have equal derivatives there. Neither is a multiple of the other, so $\lambda = 1$ has a two-dimensional eigenspace. Keep the same equation and interval but impose Dirichlet conditions, and the eigenvalues become $n^2/4$ , each with the single eigenfunction $\sin(nx/2)$ . Then $\lambda = 1$ belongs to $\sin x$ alone, and $\cos x$ is excluded because $\cos 0 = 1$ .

Normalization of eigenfunctions

Definition Every nonzero multiple of an eigenfunction is again an eigenfunction for the same eigenvalue, so the problem does not fix its scale. With the inner product and norm

\begin{gathered} \langle f, g \rangle = \int_a^b f g\,dx, \\[0.4em] \lVert f \rVert = \sqrt{\langle f, f \rangle}, \end{gathered}

an eigenfunction $y$ is normalized when $\lVert y \rVert = 1$ , and any eigenfunction can be normalized by dividing it by its norm, $\phi = y/\lVert y \rVert$ . In a one-dimensional eigenspace the result is fixed only up to sign, because $-\phi$ is normalized as well. In a two-dimensional eigenspace normalization alone does not pick a direction, and one usually chooses an orthonormal pair, two normalized eigenfunctions $\phi$ and $\psi$ with $\langle \phi, \psi \rangle = 0$ .

In plain words An eigenfunction is a shape, and normalizing it fixes its size so that coefficients computed from it mean something definite. The convention has to be stated once and then used consistently.

Example On $[0, \pi]$ ,

\int_0^\pi \sin^2 2x\,dx = \frac{\pi}{2},

so the normalized second Dirichlet mode is

\phi_2 = \sqrt{2/\pi}\,\sin 2x.

Distinct Dirichlet modes are orthogonal, for example $\int_0^\pi \sin x\sin 2x\,dx = 0$ , so the functions $\sqrt{2/\pi}\,\sin nx$ with $n \ge 1$ form an orthonormal list. The Neumann modes $\cos nx$ with $n \ge 1$ normalize with the same factor, but the zero mode needs a different one. Since $\int_0^\pi 1\,dx = \pi$ , the normalized zero mode is $1/\sqrt{\pi}$ . On $[0, 2\pi]$ the periodic pair $\cos nx/\sqrt{\pi}$ and $\sin nx/\sqrt{\pi}$ is orthonormal, and so is a rotated pair such as $(\cos x + \sin x)/\sqrt{2\pi}$ and $(\cos x - \sin x)/\sqrt{2\pi}$ .

Check yourself

Verify that $\cos 3x$ is a Neumann eigenfunction on $[0, \pi]$ , give its eigenvalue, and normalize it.

Show the answer

The second derivative is $-9\cos 3x$ , so $-y'' = 9y$ and the eigenvalue is $9$ . The derivative $-3\sin 3x$ vanishes at $0$ and at $\pi$ , because $\sin 3\pi = 0$ . Since

\int_0^\pi \cos^2 3x\,dx = \frac{\pi}{2},

the normalized eigenfunction is $\sqrt{2/\pi}\,\cos 3x$ , or its negative.

For which of the following problems is $\lambda = 0$ an eigenvalue of $-y'' = \lambda y$ ? Dirichlet conditions on $[0, \pi]$ , Neumann conditions on $[0, \pi]$ , the Robin conditions $y'(0) = y(0)$ and $y'(\pi) = -y(\pi)$ on $[0, \pi]$ , and periodic conditions on $[0, 2\pi]$ . Give the zero mode whenever it exists.

Show the answer

At $\lambda = 0$ every solution has the form $y = A + Bx$ , with derivative $B$ .

Dirichlet. The conditions give $A = 0$ and $A + B\pi = 0$ , so only the zero function remains, and $0$ is not an eigenvalue.
Neumann. Both conditions give $B = 0$ , so every constant works, and the zero mode is $1$ .
Robin. Here $h_0 = h_\pi = 1$ , and the conditions read $B = A$ and $B = -(A + B\pi)$ . Substituting the first into the second gives $(2 + \pi)A = 0$ , so $A = B = 0$ and $0$ is not an eigenvalue, as the identity on this page predicts.
Periodic. The condition $y(0) = y(2\pi)$ gives $2\pi B = 0$ , the derivative condition then holds, and the zero mode is again $1$ .

Find all eigenvalues and eigenfunctions of $-y'' = \lambda y$ on $[0, \pi]$ with $y(0) = 0$ and $y'(\pi) = 0$ .

Show the answer

Treat the three signs of $\lambda$ separately. For $\lambda = 0$ the left condition leaves $y = Bx$ , and $y'(\pi) = B = 0$ . For $\lambda = -k^2$ with $k > 0$ it leaves $y = B\sinh kx$ , and $y'(\pi) = Bk\cosh k\pi$ is zero only for $B = 0$ . For $\lambda = k^2$ with $k > 0$ it leaves $y = B\sin kx$ , and $y'(\pi) = Bk\cos k\pi$ vanishes with $B \ne 0$ exactly when $\cos k\pi = 0$ , that is, $k = n - \tfrac{1}{2}$ for a positive integer $n$ . So for $n = 1, 2, 3, \ldots$

\begin{gathered} \lambda_n = \left(n - \tfrac{1}{2}\right)^2, \\[0.4em] y_n = \sin\big((n - \tfrac{1}{2})\,x\big), \end{gathered}

which gives the eigenvalues $\tfrac{1}{4}$ , $\tfrac{9}{4}$ , $\tfrac{25}{4}$ and so on. Neither the Dirichlet list nor the Neumann list applies here, and $0$ is not an eigenvalue.

Eigenvalues and Eigenfunctions

Key terms

Eigenvalue problem

Dirichlet boundary conditions

Neumann boundary conditions

Robin boundary conditions

Periodic boundary conditions

Normalization of eigenfunctions

Check yourself

How this lesson connects

Builds on

Used later in