Sufficient conditions and direct methods

Second variation and the variational treatment of eigenvalue problems.

In this chapter 4

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The second variation
- The second variation and Legendre's condition
- Conjugate points and Jacobi's condition
Variational eigenvalue problems and direct methods
- Variational characterization of eigenvalues
- The Rayleigh-Ritz method

The second variation

Theorem

The second variation and Legendre's condition

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At a weak minimum of a functional the second variation is nonnegative in every direction, which forces F_{y'y'} >= 0 at every point of the curve (Legendre's condition); the unit also writes the second variation in the form used for sufficient conditions.

Prerequisites: The Euler-Lagrange equation, Differentiation under the integral sign, Taylor's theorem

DefinitionAdaptedAdapted from Gelfand & Fomin (1963), Ch. 5, §§24–25, pp. 99–101; Evans (2010), §8.1.3, p. 431

Let $F(x, y, z)$ be a real function on $[a, b] \times \mathbb{R} \times \mathbb{R}$ with continuous partial derivatives up to order two, and consider

J[y] = \int_a^b F\bigl(x, y(x), y'(x)\bigr)\, dx

on the admissible curves $y \in C^1[a, b]$ with $y(a) = A$ , $y(b) = B$ . The variations are the functions in

H = \{\, h \in C^1[a, b] : h(a) = h(b) = 0 \,\}, \qquad \|h\|_1 = \max_{[a,b]} |h| + \max_{[a,b]} |h'| .

An admissible $y$ is a weak minimum of $J$ if there is $r > 0$ with $J[\hat y] \ge J[y]$ for every admissible $\hat y$ with $\|\hat y - y\|_1 < r$ . "Weak" means that a competitor must be close to $y$ in both value and slope.

Throughout, $F_{y'y'}$ denotes $\partial^2 F / \partial z^2$ evaluated at $(x, y(x), y'(x))$ , and similarly $F_{yy'}$ and $F_{yy}$ . The second variation of $J$ at $y$ in the direction $h \in H$ is

\delta^2 J[y; h] = \frac{d^2}{d\varepsilon^2} J[y + \varepsilon h] \Big|_{\varepsilon = 0} = \int_a^b \bigl( F_{yy}\, h^2 + 2F_{yy'}\, h h' + F_{y'y'}\, h'^2 \bigr)\, dx .

Convention. Gelfand and Fomin define the second variation as the quadratic part of the increment $J[y+h] - J[y]$ , which is half of ours, so their coefficients $P$ and $Q$ are half of the ones below. Evans uses the second derivative, as we do.

TheoremAdaptedAdapted from Gelfand & Fomin (1963), Ch. 5, §24 Thm 1, p. 99, and §25, formula (10) and Legendre's theorem, pp. 102–103

Let $F$ have continuous second partial derivatives and let $y$ be admissible.

(Second-order necessary condition.) If $y$ is a weak minimum of $J$ , then $\delta^2 J[y; h] \ge 0$ for every $h \in H$ .
(Legendre's condition.) If $y$ is a weak minimum of $J$ , then $F_{y'y'}\bigl(x, y(x), y'(x)\bigr) \ge 0 \qquad \text{for every } x \in [a, b].$ For a weak maximum the inequality is reversed.
(Integrated form.) If in addition $F$ has continuous third partial derivatives and $y \in C^2[a, b]$ , then for every $h \in H$ $\delta^2 J[y; h] = \int_a^b \bigl( P\, h'^2 + Q\, h^2 \bigr)\, dx, \qquad P = F_{y'y'}, \quad Q = F_{yy} - \frac{d}{dx} F_{yy'} .$

Idea of the proofOriginal

Along the straight line $\varepsilon \mapsto y + \varepsilon h$ in function space, $J$ becomes an ordinary function of one variable with a minimum at $\varepsilon = 0$ , so its second derivative there cannot be negative. For Legendre's condition we pick a direction $h$ concentrated near one point: a bump of height $\delta$ and width $\delta$ . Its slope stays of size one while the bump itself is of size $\delta$ , so the three terms of $\delta^2 J$ scale like $\delta$ , $\delta^2$ and $\delta^3$ . The $h'^2$ term wins, and its sign is the sign of $F_{y'y'}$ at that point.

ProofAdaptedAdapted from Evans (2010), §8.1.3, pp. 431–432; Gelfand & Fomin (1963), Ch. 5, §25, Lemma, p. 103

Part 1. Fix $h \in H$ , $h \ne 0$ , and put $\varphi(\varepsilon) = J[y + \varepsilon h]$ .

$\varphi$ is twice continuously differentiable. The integrand $F(x, y + \varepsilon h, y' + \varepsilon h')$ has first and second $\varepsilon$ -derivatives that are continuous in $(x, \varepsilon)$ on $[a, b] \times [-1, 1]$ , because $F$ has continuous second derivatives and $y, y', h, h'$ are continuous. So differentiation under the integral sign applies twice, and $\varphi''(0)$ is the integral displayed in the definition.
$\varphi$ has a local minimum at $0$ . For $|\varepsilon| < r / \|h\|_1$ the curve $y + \varepsilon h$ is admissible (it has the right end values because $h(a) = h(b) = 0$ ) and lies within $r$ of $y$ in the norm $\|\cdot\|_1$ . The weak-minimum hypothesis gives $\varphi(\varepsilon) \ge \varphi(0)$ .
Conclusion. A local minimum forces $\varphi'(0) = 0$ . If $\varphi''(0)$ were negative, then by Taylor's theorem with the Lagrange remainder, $\varphi(\varepsilon) = \varphi(0) + \tfrac12 \varphi''(\theta\varepsilon)\, \varepsilon^2$ for some $\theta \in (0,1)$ , and continuity of $\varphi''$ would make this smaller than $\varphi(0)$ for small $\varepsilon \ne 0$ . Hence $\delta^2 J[y; h] = \varphi''(0) \ge 0$ .

Part 2. Write $P = F_{y'y'}$ , $R = F_{yy'}$ and $S = F_{yy}$ along the curve; these are continuous on $[a, b]$ . Let $M = \max_{[a,b]} (|R| + |S|)$ . Suppose, for a contradiction, that $P(x_0) < 0$ for some $x_0 \in [a, b]$ .

A short interval where $P$ is negative. By continuity of $P$ there are $\beta > 0$ , an interior point $x_1 \in (a, b)$ and $\delta_0 > 0$ with $[x_1 - \delta_0, x_1 + \delta_0] \subset (a, b)$ and $P \le -\beta$ on that interval. (If $x_0$ is an endpoint, move slightly inside; $P$ is still negative there.)
A rescaled bump. Let $g(s) = (1 - s^2)^2$ for $|s| \le 1$ and $g(s) = 0$ otherwise. Then $g \in C^1(\mathbb{R})$ , because $g$ and $g'$ both vanish at $s = \pm 1$ . For $0 < \delta \le \delta_0$ put $h_\delta(x) = \delta\, g\!\left( \frac{x - x_1}{\delta} \right), \qquad h_\delta'(x) = g'\!\left( \frac{x - x_1}{\delta} \right).$ So $h_\delta \in H$ , with support in $[x_1 - \delta, x_1 + \delta]$ .
Size of the three terms. Substitute $x = x_1 + \delta s$ , so $dx = \delta\, ds$ . With $G_1 = \int_{-1}^{1} g'(s)^2\, ds = \tfrac{256}{105}$ and the constant $G_2 = \int_{-1}^{1} \bigl( 2|g g'| + g^2 \bigr)\, ds$ , $\int_a^b P\, h_\delta'^2\, dx = \delta \int_{-1}^{1} P(x_1 + \delta s)\, g'(s)^2\, ds \le -\beta G_1\, \delta ,$ because $P \le -\beta$ on the support. Since $\delta \le 1$ may be assumed, $\delta^3 \le \delta^2$ and $\left| \int_a^b \bigl( 2R\, h_\delta h_\delta' + S\, h_\delta^2 \bigr)\, dx \right| \le M \bigl( 2\delta^2 \textstyle\int |g g'|\, ds + \delta^3 \int g^2\, ds \bigr) \le M G_2\, \delta^2 .$
Contradiction. Adding, $\delta^2 J[y; h_\delta] \le \delta\, (-\beta G_1 + M G_2\, \delta)$ , which is negative once $\delta < \beta G_1 / (M G_2)$ (any $\delta$ works if $M = 0$ ). This contradicts Part 1. Hence $P \ge 0$ on $[a, b]$ . For a weak maximum, apply this to $-J$ , whose minimum it is.

Part 3. If $F$ has continuous third derivatives and $y \in C^2$ , the chain rule makes $R(x) = F_{yy'}(x, y(x), y'(x))$ continuously differentiable. Since $2hh' = (h^2)'$ and $h(a) = h(b) = 0$ , integration by parts gives

\int_a^b 2R\, h h'\, dx = \bigl[ R\, h^2 \bigr]_a^b - \int_a^b R'\, h^2\, dx = - \int_a^b R'\, h^2\, dx .

Substituting this into $\delta^2 J$ gives $\int_a^b \bigl( P h'^2 + (S - R') h^2 \bigr)\, dx$ , which is the claimed form with $Q = S - R'$ . $\square$

RemarkOriginal

Where the hypotheses are used.

Second derivatives of $F$ continuous: this makes $\varphi$ twice differentiable (Part 1, step 1) and $P$ continuous, so that a negative value at one point persists on an interval (Part 2, step 1). Part 2 never integrates by parts, so it needs no third derivatives; only Part 3 does.
Weak minimum: used only in Part 1, step 2, for the small multiples $\varepsilon h$ of one fixed direction. The bumps $h_\delta$ are not small in $\|\cdot\|_1$ , since their slopes stay of size one, but they are never compared with $y$ directly; only $\varepsilon h_\delta$ for small $\varepsilon$ is.
Fixed endpoints: $h(a) = h(b) = 0$ keeps $y + \varepsilon h$ admissible and removes the boundary term in Part 3.

Necessary, not sufficient. Even the strict inequality $F_{y'y'} > 0$ does not make an extremal a minimum. Take $J[y] = \int_0^{2\pi} (y'^2 - y^2)\, dx$ with $y(0) = y(2\pi) = 0$ . Here $F_{y'y'} = 2 > 0$ and $y \equiv 0$ is an extremal, yet $h = \sin(x/2)$ gives

\delta^2 J[0; h] = 2\int_0^{2\pi} \Bigl( \tfrac14 \cos^2 \tfrac{x}{2} - \sin^2 \tfrac{x}{2} \Bigr)\, dx = 2\Bigl( \frac{\pi}{4} - \pi \Bigr) = -\frac{3\pi}{2} < 0 .

Legendre's condition is a statement about single points, so it cannot detect that the interval is too long. That global information is what Jacobi's condition supplies.

ExampleOriginal

A double-well integrand. Let $J[y] = \int_0^1 (y'^2 - 1)^2\, dx$ with $y(0) = 0$ and $y(1) = \tfrac12$ .

The extremal. $F = (z^2 - 1)^2$ depends only on the slope, so the Euler–Lagrange equation says $F_z(y')$ is constant. The straight line $y_0(x) = x/2$ satisfies it and the boundary conditions, and $J[y_0] = (\tfrac14 - 1)^2 = \tfrac{9}{16}$ .

Legendre's test. $F_{zz} = 12 z^2 - 4$ , which at $z = \tfrac12$ equals $3 - 4 = -1 < 0$ . So $y_0$ is not a weak minimum. (Legendre's condition holds only for slopes with $|z| \ge 1/\sqrt3$ .)

A direct check. Take $h = \sin \pi x \in H$ and put $u = \varepsilon \pi \cos \pi x$ , so $(y_0 + \varepsilon h)' = \tfrac12 + u$ and the integrand is

\bigl( u^2 + u - \tfrac34 \bigr)^2 = u^4 + 2u^3 - \tfrac12 u^2 - \tfrac32 u + \tfrac{9}{16}.

Over $[0, 1]$ the averages of $\cos \pi x$ and $\cos^3 \pi x$ are $0$ , that of $\cos^2 \pi x$ is $\tfrac12$ and that of $\cos^4 \pi x$ is $\tfrac38$ . Therefore

J[y_0 + \varepsilon h] = \frac{9}{16} - \frac{\pi^2}{4}\, \varepsilon^2 + \frac{3\pi^4}{8}\, \varepsilon^4 .

The second derivative at $\varepsilon = 0$ is $-\pi^2/2$ , which agrees with the formula $\int_0^1 F_{y'y'} h'^2\, dx = -\pi^2 \int_0^1 \cos^2 \pi x\, dx = -\pi^2/2$ . The perturbed curve beats the line whenever $0 < |\varepsilon| < \sqrt6/(3\pi) \approx 0.2599$ . For instance $\varepsilon = 0.1$ gives $J \approx 0.5625 - 0.02467 + 0.00365 = 0.54148 < 0.5625$ .

ApplicationOriginal

Mechanics. For a Lagrangian $L = \tfrac12 m \dot q^2 - V(q)$ one has $L_{\dot q \dot q} = m > 0$ . So the action can never have a maximum at a nontrivial motion: Legendre's condition for a maximum fails at every point. A true path is either a minimum or a saddle of the action, and which one it is depends on the length of the time interval (see conjugate points).
Materials and phase transitions. Energies like the double well $(y'^2 - 1)^2$ model materials with two preferred strains, as in martensitic transformations. Where the imposed average strain lies in the region in which Legendre's condition fails, the material lowers its energy by forming fine mixtures of the two preferred strains (microstructure) instead of deforming uniformly. In several variables the same test becomes the Legendre–Hadamard (strong ellipticity) condition of elasticity; the scalar case is derived in Evans (Evans (2010), §8.1.3, p. 432).

Theorem · advanced

Conjugate points and Jacobi's condition

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Along an extremal with F_{y'y'} > 0, the second variation is positive definite exactly when the Jacobi equation has no conjugate point, which gives Jacobi's necessary condition and a set of sufficient conditions for a weak minimum.

Prerequisites: The second variation and Legendre's condition, Sturm-Liouville eigenvalue problems, Dependence on initial data and Gronwall's inequality, Structure of linear equations and the Wronskian, Taylor's theorem, Differentiation under the integral sign

DefinitionAdaptedAdapted from Gelfand & Fomin (1963), Ch. 5, §26, p. 106, and §27, Defs 1–2, p. 112

Let $J[y] = \int_a^b F(x, y, y')\, dx$ with $y(a) = A$ , $y(b) = B$ , where $F$ has continuous third partial derivatives, and let $y \in C^2[a, b]$ be an extremal. As in the second variation, put

P = F_{y'y'}, \qquad Q = F_{yy} - \frac{d}{dx} F_{yy'}

along $y$ , so that $\delta^2 J[y; h] = I[h]$ for $h \in H = \{ h \in C^1[a,b] : h(a) = h(b) = 0 \}$ , where

I[h] = \int_a^b \bigl( P\, h'^2 + Q\, h^2 \bigr)\, dx , \qquad I[g, k] = \int_a^b \bigl( P\, g' k' + Q\, g k \bigr)\, dx .

The same two integrals make sense, and define a symmetric bilinear form $I[g, k]$ with $I[h] = I[h, h]$ , for all continuous, piecewise $C^1$ functions on $[a, b]$ , since their derivatives are bounded with finitely many jumps. The proof of Part 2 uses $I$ on such functions.

Assume the strengthened Legendre condition $P > 0$ on $[a, b]$ .

The Jacobi equation of $J$ along $y$ is the Euler–Lagrange equation of $I$ :

-(P u')' + Q u = 0 .

We read it as the first-order linear system $u' = v / P$ , $v' = Q u$ for the pair $(u, v) = (u, P u')$ . Its coefficients are continuous because $P > 0$ , so by the theory of linear equations each initial value $(u(x_0), v(x_0))$ has exactly one solution on all of $[a, b]$ , and $u$ and $P u'$ are $C^1$ . Let $u_a$ be the solution with $u_a(a) = 0$ and $u_a'(a) = 1$ .

A point $c \in (a, b]$ is conjugate to $a$ if $u_a(c) = 0$ . Equivalently, some solution that is not identically zero vanishes at both $a$ and $c$ , since every solution vanishing at $a$ is a multiple of $u_a$ .

TheoremAdaptedAdapted from Gelfand & Fomin (1963), Ch. 5, §26 Thms 1–3, pp. 106–111, §27 (Jacobi's necessary condition), p. 112, and §28 (Theorem), p. 116

Let $P, Q$ be continuous on $[a, b]$ with $P > 0$ .

(Strong positivity.) If no point of $(a, b]$ is conjugate to $a$ , there is $\alpha > 0$ with
$I[h] \ge \alpha^2 \int_a^b h'^2\, dx \qquad \text{for all } h \in H .$
(Loss of positivity.) If $b$ is conjugate to $a$ , then $I[u_a] = 0$ with $u_a \in H$ , $u_a \ne 0$ . If some $c \in (a, b)$ is conjugate to $a$ , then $I[h] < 0$ for some $h \in H$ .

Together, 1 and 2 say: $I[h] > 0$ for all nonzero $h \in H$ if and only if no point of $(a, b]$ is conjugate to $a$ .
(Jacobi's necessary condition.) If the extremal $y$ is a weak minimum of $J$ and $P > 0$ on $[a, b]$ , then no point of the open interval $(a, b)$ is conjugate to $a$ .
(Sufficient conditions for a weak minimum.) If $y \in C^2[a, b]$ is an extremal, $P > 0$ on $[a, b]$ and no point of $(a, b]$ is conjugate to $a$ , then $y$ is a strict weak minimum: there is $r > 0$ such that $J[y + h] > J[y]$ for every nonzero $h \in H$ with $\|h\|_1 < r$ .

Idea of the proofOriginal

The sign of $I$ is settled by completing the square, an idea due to Legendre. If the Jacobi equation has a solution $u$ that stays positive on all of $[a, b]$ , the integrand $P h'^2 + Q h^2$ equals $P\,(h' - h u'/u)^2$ plus an exact derivative, so $I \ge 0$ , with equality only for $h = 0$ . Such a $u$ exists precisely when $u_a$ does not vanish again, which is the no-conjugate-point condition. Conversely, if $u_a$ returns to zero at an interior point $c$ , then $u_a$ cut off at $c$ gives $I = 0$ but has a corner at $c$ . Because of the corner, a small perturbation near $c$ makes $I$ negative, and rounding off the corner afterwards keeps it negative. For sufficiency, a second variation bounded below by $\alpha^2 \int h'^2$ dominates the Taylor remainder of $J$ for nearby curves.

Proof(Part 1: strong positivity)AdaptedAdapted from Gelfand & Fomin (1963), Ch. 5, §26 Thm 1 and footnote 9, pp. 106–108, and §28, pp. 116–117

Picone's identity. Let $\tilde P$ be continuous and positive, and let $u$ be a solution of $-(\tilde P u')' + Q u = 0$ (in the system sense) with $u > 0$ on $[a, b]$ . For $h \in H$ , expanding the derivative and using $(\tilde P u')' = Q u$ , $\Bigl( \frac{\tilde P u'}{u}\, h^2 \Bigr)' = Q h^2 + \frac{2 \tilde P u' h h'}{u} - \frac{\tilde P u'^2 h^2}{u^2}, \qquad\text{so}\qquad \tilde P h'^2 + Q h^2 = \tilde P \Bigl( h' - \frac{u'}{u} h \Bigr)^2 + \Bigl( \frac{\tilde P u'}{u}\, h^2 \Bigr)' .$ Integrating over $[a, b]$ , the exact derivative contributes nothing because $h(a) = h(b) = 0$ . Hence $\int_a^b (\tilde P h'^2 + Q h^2)\, dx = \int_a^b \tilde P\, (h' - h u'/u)^2\, dx \ge 0$ .
A positive solution of the Jacobi equation. Assume no point of $(a, b]$ is conjugate to $a$ . Then $u_a > 0$ on $(a, b]$ : it is positive just to the right of $a$ because $u_a'(a) = 1$ , and by the intermediate value theorem it cannot become negative without a zero. Let $u_1$ be the solution with $u_1(a) = 1$ , $(P u_1')(a) = 0$ . By continuity choose $\eta > 0$ with $u_a' \ge \tfrac12$ and $u_1 \ge \tfrac12$ on $[a, a + \eta]$ . Put $m = \min_{[a + \eta, b]} u_a > 0$ , $K = \max_{[a,b]} |u_1| \ge 1$ and $\epsilon = m / (2K)$ . Then $u = u_a + \epsilon u_1$ is positive on all of $[a, b]$ : on $[a, a + \eta]$ because $u_a \ge 0$ and $\epsilon u_1 \ge \epsilon/2$ , and on $[a + \eta, b]$ because $u \ge m - \epsilon K = m/2$ .
Perturbing $P$ . For $0 \le \alpha^2 \le \tfrac12 \min P$ , let $u^{(\alpha)}$ be the solution of $-((P - \alpha^2) w')' + Q w = 0$ with the same initial data as $u$ at $a$ : $w(a) = \epsilon$ , $((P - \alpha^2) w')(a) = P(a)$ . In system form the coefficients $1/(P - \alpha^2)$ and $Q$ depend continuously on $(x, \alpha)$ . By continuous dependence on parameters, $u^{(\alpha)} \to u^{(0)} = u$ uniformly on $[a, b]$ as $\alpha \to 0$ . Since $\min u > 0$ , there is $\alpha > 0$ with $u^{(\alpha)} > 0$ on $[a, b]$ .
Conclusion. Step 1 with $\tilde P = P - \alpha^2 > 0$ and the solution $u^{(\alpha)}$ gives $\int_a^b \bigl( (P - \alpha^2) h'^2 + Q h^2 \bigr)\, dx \ge 0$ , that is, $I[h] \ge \alpha^2 \int_a^b h'^2\, dx$ . $\square$

Proof(Part 2: loss of positivity)AdaptedAdapted from do Carmo (2016), §5-9, Lemma 2 and Thm 2, pp. 427–429

The Jacobi solution has zero energy between its zeros. Let $c \in (a, b]$ be conjugate to $a$ . Since $P u_a'$ is $C^1$ , integration by parts on $[a, c]$ gives $\int_a^c \bigl( P u_a'^2 + Q u_a^2 \bigr)\, dx = \bigl[ P u_a' u_a \bigr]_a^c + \int_a^c u_a \bigl( -(P u_a')' + Q u_a \bigr)\, dx = 0 ,$ because $u_a(a) = u_a(c) = 0$ and $u_a$ solves the Jacobi equation. If $c = b$ , then $u_a \in H$ , $u_a \ne 0$ and $I[u_a] = 0$ .
Cutting off at an interior conjugate point. Now let $c \in (a, b)$ . Define $\hat h = u_a$ on $[a, c]$ and $\hat h = 0$ on $[c, b]$ . It is continuous and $C^1$ on each piece, and $I[\hat h] = 0$ by step 1. Moreover $u_a'(c) \ne 0$ : otherwise $(u_a, P u_a')$ would vanish at $c$ , and uniqueness for the system would force $u_a \equiv 0$ . So $\hat h$ has a genuine corner at $c$ .
Pushing below zero. Take $w \in H$ with $w(c) = 1$ , for instance a $C^1$ bump supported in $(a, b)$ . Integrating by parts on $[a, c]$ exactly as in step 1, $I[\hat h, w] = \int_a^c \bigl( P u_a' w' + Q u_a w \bigr)\, dx = \bigl[ P u_a' w \bigr]_a^c = P(c)\, u_a'(c) \ne 0 .$ For real $t$ , bilinearity gives $I[\hat h - t w] = -2t\, P(c) u_a'(c) + t^2 I[w]$ . Choosing $t$ small, with the sign of $P(c) u_a'(c)$ , makes this negative. The function $g = \hat h - t w$ vanishes at $a$ and $b$ and is $C^1$ except at $c$ .
Rounding the corner (smoothing lemma). Let $g$ be continuous on $[a, b]$ , $C^1$ on $[a, c]$ and on $[c, b]$ , with $g(a) = g(b) = 0$ , and let $G = \sup |g'|$ . For small $\delta > 0$ let $k_\delta = g'$ outside $(c - \delta, c + \delta)$ and let $k_\delta$ be linear on that interval, joining $g'(c - \delta)$ to $g'(c + \delta)$ . Then $k_\delta$ is continuous, $|k_\delta| \le G$ , and $k_\delta$ differs from $g'$ only on a set of length $2\delta$ , so $\int_a^b |k_\delta - g'|\, dx \le 4G\delta$ . Put $g_\delta(x) = \int_a^x k_\delta\, dt - \frac{x - a}{b - a} \int_a^b k_\delta\, dt .$ Then $g_\delta \in H$ . Since $\int_a^b g'\, dx = g(b) - g(a) = 0$ , the correction term is at most $4G\delta$ in size, so $g_\delta \to g$ uniformly and $\int_a^b (g_\delta' - g')^2\, dx \to 0$ as $\delta \to 0$ . Because $P$ and $Q$ are bounded, the Cauchy–Schwarz inequality applied to $I[g] - I[g_\delta] = I[g - g_\delta,\, g + g_\delta]$ shows $I[g_\delta] \to I[g] < 0$ . So $I[g_\delta] < 0$ for small $\delta$ , with $g_\delta \in H$ . $\square$

Proof(Parts 3 and 4: Jacobi's condition and sufficiency)AdaptedAdapted from Gelfand & Fomin (1963), Ch. 5, §27, p. 112, and §28, pp. 116–117

Part 3. If $y$ is a weak minimum, the second-order necessary condition gives $I[h] = \delta^2 J[y; h] \ge 0$ for every $h \in H$ . A point of $(a, b)$ conjugate to $a$ would produce $h \in H$ with $I[h] < 0$ by Part 2. So there is none.

Part 4. Let $\alpha$ be as in Part 1, and fix a nonzero $h \in H$ with $\|h\|_1 \le \rho \le 1$ , where $\rho$ is chosen below. Put $\varphi(t) = J[y + t h]$ .

Taylor expansion. As in the second-variation unit, $\varphi$ is $C^2$ (we may differentiate under the integral twice), with $\varphi''(t)$ given by the second-variation integral at the curve $y + t h$ . Taylor's formula with integral remainder gives $J[y + h] - J[y] = \varphi'(0) + \int_0^1 (1 - t)\, \varphi''(t)\, dt = \tfrac12\, I[h] + \int_0^1 (1 - t) \bigl( \varphi''(t) - \varphi''(0) \bigr)\, dt .$ Here $\varphi'(0) = 0$ because $y$ is an extremal, and $\varphi''(0) = I[h]$ .
Controlling the remainder. The second derivatives $F_{yy}, F_{yy'}, F_{y'y'}$ are uniformly continuous on the compact set $\{ (x, y(x) + s, y'(x) + \sigma) : x \in [a,b],\ |s|, |\sigma| \le 1 \}$ . Let $\omega(\rho)$ be the largest change of any of them when $|s|, |\sigma| \le \rho$ ; then $\omega(\rho) \to 0$ as $\rho \to 0$ . For $t \in [0, 1]$ , $|\varphi''(t) - \varphi''(0)| \le \omega(\rho) \int_a^b (h^2 + 2|h h'| + h'^2)\, dx \le 2\omega(\rho) \int_a^b (h^2 + h'^2)\, dx$ .
A Poincaré inequality. Since $h(a) = 0$ , Cauchy–Schwarz gives $h(x)^2 = \bigl( \int_a^x h' \bigr)^2 \le (x - a) \int_a^b h'^2$ , and integrating, $\int_a^b h^2 \le \tfrac12 (b - a)^2 \int_a^b h'^2$ . With $C = 1 + \tfrac12 (b - a)^2$ and $\int_0^1 (1 - t)\, dt = \tfrac12$ , steps 1 and 2 and Part 1 give $J[y + h] - J[y] \ge \Bigl( \frac{\alpha^2}{2} - C\, \omega(\rho) \Bigr) \int_a^b h'^2\, dx .$
Conclusion. Choose $\rho \le 1$ with $C\,\omega(\rho) < \alpha^2/2$ . A nonzero $h \in H$ has $\int h'^2 > 0$ (otherwise $h$ would be constant, hence $0$ ), so $J[y + h] > J[y]$ whenever $0 < \|h\|_1 < \rho$ . $\square$

RemarkOriginal

Where the hypotheses are used.

$P > 0$ on the closed interval. It turns the Jacobi equation into a regular linear system, so that $u_a$ exists and has isolated zeros with $u_a' \ne 0$ there (Part 2, step 2). It also makes the square in Picone's identity nonnegative, and it leaves room to subtract $\alpha^2$ (Part 1, step 3). Where $P$ vanishes the Jacobi equation is singular, and conjugate points need a separate analysis.
The endpoint $b$ . Sufficiency asks that $b$ itself not be conjugate; necessity only excludes the open interval. The gap is real. For $J[y] = \int_0^\pi (y'^2 - y^2)\, dx$ with $y(0) = y(\pi) = 0$ , the Jacobi equation is $-2u'' - 2u = 0$ , so $u_a = \sin x$ and $b = \pi$ is conjugate. Here $y \equiv 0$ is a minimum, because $J[h] = \tfrac12 I[h] = \int_0^\pi (h'^2 - h^2)\, dx \ge 0$ for $h \in H$ . To see this, apply Picone's identity (Part 1, step 1) with $\tilde P = 1$ and $u = \sin x$ on $[\varepsilon, \pi - \varepsilon]$ : the integral there is $\int (h' - h \cot x)^2\, dx \ge 0$ plus the boundary values of $h^2 \cot x$ , and these tend to $0$ as $\varepsilon \to 0$ because $|h(x)| \le \max|h'| \cdot \min(x, \pi - x)$ . (This is the statement that the lowest Dirichlet eigenvalue of $-u''$ on $[0, \pi]$ is $1$ ; see the variational characterization of eigenvalues.) The minimum is not a strict one: $J[\varepsilon \sin x] = 0 = J[0]$ . On a longer interval $[0, \ell]$ with $\ell > \pi$ , the conjugate point $\pi$ is interior, and indeed $J[\sin(\pi x/\ell)] = \tfrac{\ell}{2}\bigl( \pi^2/\ell^2 - 1 \bigr) < 0$ .
Extremal. Used only to remove $\varphi'(0)$ in Part 4.
Weak, not strong. Part 4 compares $y$ only with curves whose slopes are close to $y'$ . A minimum against all curves close in value alone (a strong minimum) needs further conditions, such as Weierstrass's condition, which this unit does not treat.

Conjugate points and neighbouring extremals. Suppose $y(x; s)$ is a smooth family of extremals through the point $(a, A)$ , with $y(x; 0) = y(x)$ . Differentiating the Euler–Lagrange equation in $s$ at $s = 0$ shows that $\partial_s y(x; 0)$ solves the Jacobi equation and vanishes at $a$ . So a conjugate point is where extremals leaving $(a, A)$ with slightly different slopes come back, to first order, to meet $y$ (Gelfand & Fomin (1963), §27, pp. 112–115). A point $c$ is conjugate to $a$ exactly when $0$ is an eigenvalue of the Sturm–Liouville problem $-(P u')' + Q u = \mu u$ on $[a, c]$ with $u(a) = u(c) = 0$ .

ExampleOriginal

When does a catenary minimize? Take $F = y\sqrt{1 + y'^2}$ . Up to a factor $2\pi$ , $J$ is then the area of the surface swept out by rotating the curve about the $x$ -axis (a soap film between two coaxial rings). Consider the arc of $y = \cosh x$ starting at $a = -1$ .

Extremal. $F_y = \sqrt{1 + y'^2} = \cosh x$ and $F_{y'} = y y'/\sqrt{1 + y'^2} = \sinh x$ , whose derivative is $\cosh x = F_y$ . So $y = \cosh x$ satisfies the Euler–Lagrange equation.

Coefficients. Along the curve $1 + y'^2 = \cosh^2 x$ , and

P = \frac{y}{(1 + y'^2)^{3/2}} = \frac{1}{\cosh^2 x} > 0, \qquad F_{yy} = 0, \qquad F_{yy'} = \frac{y'}{\sqrt{1 + y'^2}} = \tanh x, \qquad Q = -\frac{1}{\cosh^2 x}.

So the Jacobi equation is $\bigl( h'/\cosh^2 x \bigr)' + h/\cosh^2 x = 0$ .

Two solutions from two families of catenaries. Translating, $\cosh(x - d)$ , and rescaling, $s \cosh(x/s)$ , give families of extremals. Their $d$ - and $s$ -derivatives at $d = 0$ , $s = 1$ suggest $h_1 = \sinh x$ and $h_2 = \cosh x - x \sinh x$ . Direct substitution confirms both. For $h_1$ : $h_1'/\cosh^2 x = 1/\cosh x$ , whose derivative $-\tanh x/\cosh x$ cancels $h_1/\cosh^2 x = \tanh x/\cosh x$ . For $h_2$ : $h_2' = -x\cosh x$ , and $(-x/\cosh x)' = -1/\cosh x + x \tanh x/\cosh x$ cancels $h_2/\cosh^2 x = 1/\cosh x - x\tanh x/\cosh x$ . Their Wronskian is $h_1 h_2' - h_1' h_2 = -\cosh^2 x \ne 0$ , so they are independent.

The conjugate point. The solution with $u(a) = 0$ , $u'(a) = 1$ is

u_a(x) = \frac{h_1(x)\, h_2(a) - h_2(x)\, h_1(a)}{\cosh^2 a},

since the numerator has derivative $\cosh^2 a$ at $x = a$ . For $a < 0$ we have $u_a(0) = -\sinh a/\cosh^2 a > 0$ . For $x \ne 0$ , dividing by $\sinh a \sinh x$ shows that $u_a(x) = 0$ exactly when

\coth x - x = \coth a - a .

The function $\coth t - t$ has derivative $-1/\sinh^2 t - 1 < 0$ , so it is strictly decreasing on $(-\infty, 0)$ and on $(0, \infty)$ . On $(a, 0)$ it stays below its value at $a$ , so there is no zero there. On $(0, \infty)$ it falls from $+\infty$ to $-\infty$ , so there is exactly one zero. For $a = -1$ , $\coth(-1) + 1 \approx -0.313035$ , and solving numerically gives the conjugate point $c \approx 1.433605$ .

Verdict. Write $c = 1.433605\ldots$ for the conjugate point found above. By Part 4, the arc of $\cosh x$ over $[-1, b]$ is a strict weak minimum of the area when $b < c$ . By Part 3, it is not even a weak minimum when $b > c$ . As a check, the tangent to $\cosh x$ at $t$ meets the $x$ -axis at $t - \coth t$ . This is $0.313035$ both for $t = -1$ and for $t = c$ , so the tangents at a point and at its conjugate point meet on the axis, a classical property of the catenary.

ApplicationOriginal

Least action is only stationary action. For a harmonic oscillator, $L = \tfrac12 m(\dot q^2 - \omega^2 q^2)$ , the Jacobi equation is the equation of motion itself, and $u_a = \sin(\omega(t - a))/\omega$ . The first conjugate time is half a period, $\pi/\omega$ . A true motion minimizes the action over a time interval shorter than $\pi/\omega$ but is a saddle over a longer one.
Geodesics. On a sphere the point conjugate to $M$ along a great circle is the antipode of $M$ (Gelfand & Fomin (1963), §27, p. 115), so a great-circle arc longer than a half-circle is not even a local minimum of length. In optics and gravitational lensing, conjugate points are the foci and caustics where neighbouring rays cross.
Counting instabilities. For an extremal with $P > 0$ , the number of points conjugate to $a$ in $(a, b)$ equals the number of negative eigenvalues of the second variation (Morse index theorem, stated here without proof). Engineers use this to count unstable buckling modes, and physicists to count unstable fluctuation modes about a classical solution.

Variational eigenvalue problems and direct methods

Theorem

Variational characterization of eigenvalues

Adapted

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The eigenvalues of a regular Sturm-Liouville problem are the successive minima of the Rayleigh quotient, and each one is also a min-max over subspaces, which gives upper bounds and comparison theorems without solving the equation.

Prerequisites: Isoperimetric problems, Sturm-Liouville eigenvalue problems, Eigenfunction expansions

DefinitionAdaptedAdapted from Strauss (2008), §11.1, p. 300, and §11.4, p. 314

Consider the regular Sturm–Liouville problem with Dirichlet conditions

-(p u')' + q u = \lambda\, w\, u \quad \text{on } [a, b], \qquad u(a) = u(b) = 0,

where $p \in C^1[a, b]$ , $q, w \in C[a, b]$ , and $p > 0$ , $w > 0$ on $[a, b]$ . Write $(u, v)_w = \int_a^b w\, u\, v\, dx$ and $\|u\|_w^2 = (u, u)_w$ . By Sturm–Liouville theory and the expansion theorem, the eigenvalues are real and simple and form a sequence $\lambda_1 < \lambda_2 < \cdots \to \infty$ . Their eigenfunctions $\varphi_n$ can be chosen real, in $C^2$ , with $(\varphi_m, \varphi_n)_w = \delta_{mn}$ , and complete: every continuous $u$ satisfies Parseval's identity $\sum_n (u, \varphi_n)_w^2 = \|u\|_w^2$ .

A trial function is a continuous, piecewise $C^1$ function $u$ on $[a, b]$ with $u(a) = u(b) = 0$ and $u \not\equiv 0$ . Let $\mathcal{A}$ be the set of trial functions together with $0$ ; it is a vector space. The energy and the Rayleigh quotient are

D[u, v] = \int_a^b \bigl( p\, u' v' + q\, u v \bigr)\, dx, \qquad D[u] = D[u, u], \qquad \mathcal{R}[u] = \frac{D[u]}{\|u\|_w^2} .

TheoremAdaptedAdapted from Strauss (2008), §11.1 Thms 1–2, pp. 301–303, and §11.2 Thm 1 (minimax principle), pp. 307–308

Under the hypotheses above:

(Ground state.) $\mathcal{R}[u] \ge \lambda_1$ for every trial function $u$ , with equality exactly when $u$ is a multiple of $\varphi_1$ . So $\lambda_1 = \min \mathcal{R}$ .
(Higher eigenvalues.) If $u$ is a trial function with $(u, \varphi_k)_w = 0$ for $k = 1, \dots, n - 1$ , then $\mathcal{R}[u] \ge \lambda_n$ , with equality exactly when $u$ is a multiple of $\varphi_n$ .
(Courant–Fischer min-max.) For each $n$ , $\lambda_n = \min_{\substack{V \subset \mathcal{A} \\ \dim V = n}} \ \max_{u \in V \setminus \{0\}} \mathcal{R}[u],$ and the minimum is attained at $V = \operatorname{span}\{\varphi_1, \dots, \varphi_n\}$ . Equivalently, $\lambda_n$ is the largest value, over all choices of $n - 1$ continuous functions $v_1, \dots, v_{n-1}$ on $[a, b]$ , of the infimum of $\mathcal{R}$ over trial functions $w$ -orthogonal to all $v_k$ . The largest value is attained at $v_k = \varphi_k$ , where the infimum is a minimum (this form is stated here without proof).

Idea of the proofAdaptedAdapted from Gelfand & Fomin (1963), §41.1, p. 198; Riley, Hobson & Bence (2006), §22.6, pp. 791–792, and §22.7, p. 793

Minimizing the energy $D[u]$ subject to $\|u\|_w^2 = 1$ is an isoperimetric problem. Its Euler–Lagrange equation with multiplier $\lambda$ is exactly $-(p u')' + q u = \lambda w u$ , and multiplying by $u$ and integrating shows that the minimum value equals the multiplier. So a minimizer, if one exists, is an eigenfunction, and the minimum is an eigenvalue. To avoid the question of existence, the proof goes the other way. We expand an arbitrary trial function in eigenfunctions, with coefficients $c_n$ . Then the energy is at least $\sum \lambda_n c_n^2$ while the norm is $\sum c_n^2$ , so the quotient is at least a weighted average of the eigenvalues, and hence at least the smallest one.

ProofAdaptedAdapted from Strauss (2008), §11.3, proof of Thm 2, pp. 311–313, and §11.2, proof of Thm 1, p. 308

Energy against an eigenfunction. For $u \in \mathcal{A}$ and every $n$ , $D[u, \varphi_n] = \lambda_n\, (u, \varphi_n)_w .$ Integrate $\int p\, u' \varphi_n'$ by parts on each interval where $u$ is $C^1$ . The interior boundary terms cancel because $u$ and $p \varphi_n'$ are continuous, and the end terms vanish because $u(a) = u(b) = 0$ . This leaves $D[u, \varphi_n] = \int_a^b u\, \bigl( -(p \varphi_n')' + q \varphi_n \bigr)\, dx = \lambda_n \int_a^b w\, u\, \varphi_n\, dx$ . In particular $D[\varphi_m, \varphi_n] = \lambda_n \delta_{mn}$ .
Making the energy nonnegative. Pick $\kappa \ge 0$ with $q + \kappa w \ge 0$ on $[a, b]$ , possible because $w$ has a positive minimum. Then $D_\kappa[u] = D[u] + \kappa \|u\|_w^2 = \int (p u'^2 + (q + \kappa w) u^2)\, dx \ge 0$ for all $u \in \mathcal{A}$ , since $p > 0$ . Step 1 holds for $D_\kappa$ with $\lambda_n + \kappa$ in place of $\lambda_n$ .
The energy dominates partial sums. For a trial function $u$ let $c_n = (u, \varphi_n)_w$ and $u_N = \sum_{n \le N} c_n \varphi_n$ , which lies in $\mathcal{A}$ . Expanding by bilinearity and using step 1 with orthonormality, $0 \le D_\kappa[u - u_N] = D_\kappa[u] - 2\sum_{n \le N} c_n (\lambda_n + \kappa) c_n + \sum_{n \le N} (\lambda_n + \kappa) c_n^2 = D_\kappa[u] - \sum_{n \le N} (\lambda_n + \kappa)\, c_n^2 .$
Parts 1 and 2. Suppose $c_k = 0$ for $k < n$ (no condition when $n = 1$ ). Since $\lambda_k \ge \lambda_n$ for $k \ge n$ , step 3 gives $D_\kappa[u] \ge (\lambda_n + \kappa) \sum_{k \le N} c_k^2$ . Letting $N \to \infty$ and using Parseval, $D_\kappa[u] \ge (\lambda_n + \kappa) \|u\|_w^2$ , that is, $\mathcal{R}[u] \ge \lambda_n$ . The value at $\varphi_n$ is $D[\varphi_n] = \lambda_n$ by step 1, so the bound is attained.
Equality. If $\mathcal{R}[u] = \lambda_n$ , then splitting off the gap $\lambda_{n+1} - \lambda_n > 0$ in step 3 gives $(\lambda_n + \kappa)\|u\|_w^2 \ge (\lambda_n + \kappa) \sum_{k \le N} c_k^2 + (\lambda_{n+1} - \lambda_n) \sum_{n < k \le N} c_k^2 .$ As $N \to \infty$ the first sum tends to $\|u\|_w^2$ , so $c_k = 0$ for all $k > n$ . By Parseval $\|u - c_n \varphi_n\|_w^2 = \sum_{k \ne n} c_k^2 = 0$ . Since $w > 0$ and both functions are continuous, $u = c_n \varphi_n$ . Simplicity of the eigenvalues, which makes the gap positive, is used here.
Part 3, lower bound. Let $V \subset \mathcal{A}$ have dimension $n$ . The $n - 1$ linear conditions $(u, \varphi_k)_w = 0$ , $k < n$ , on the $n$ -dimensional space $V$ have a nonzero solution $u$ , and Part 2 gives $\max_V \mathcal{R} \ge \mathcal{R}[u] \ge \lambda_n$ . The maximum exists: since $w > 0$ and the elements of $V$ are continuous, $\|\cdot\|_w$ is a norm on $V$ , and $\mathcal{R}$ is unchanged by scaling, so it suffices to maximize the continuous function $D$ on the unit sphere of $V$ , which is compact because $V$ is finite-dimensional.
Part 3, attainment. For $u = \sum_{k \le n} d_k \varphi_k \ne 0$ , step 1 gives $D[u] = \sum_{k \le n} \lambda_k d_k^2$ and $\|u\|_w^2 = \sum_{k \le n} d_k^2$ . So $\mathcal{R}[u] \le \lambda_n$ , with equality at $u = \varphi_n$ . Hence $\max_V \mathcal{R} = \lambda_n$ for $V = \operatorname{span}\{\varphi_1, \dots, \varphi_n\}$ . $\square$

RemarkOriginal

Where the hypotheses are used.

Boundary conditions. They kill the end terms in step 1. If a trial function violates them, the bound fails. For $-u'' = \lambda u$ on $[0, \pi]$ with $u(0) = u(\pi) = 0$ we have $\lambda_1 = 1$ , but the constant $u = 1$ , which is not a trial function, gives $\mathcal{R} = 0$ .
Completeness (Parseval) turns the partial-sum inequality of step 3 into a statement about $u$ itself (steps 4–5). Without it, a trial function with a component outside the span of the eigenfunctions would escape the argument.
$p > 0$ and $w > 0$ : $p > 0$ makes $D_\kappa$ nonnegative (step 2), and $w > 0$ makes $(\cdot,\cdot)_w$ an inner product, so that zero norm forces zero function (step 5). They also underlie the regular Sturm–Liouville theory quoted in the definition.
Simplicity of the eigenvalues is used only for the equality cases.

Other boundary conditions. For a Robin condition $p u' = \sigma u$ at $b$ (with $\sigma$ a constant), the Rayleigh quotient acquires the boundary term $-\sigma u(b)^2$ in $D$ , and trial functions need not satisfy any condition at $b$ . For the Neumann condition $u'(b) = 0$ , trial functions are simply free at $b$ : the condition is a natural boundary condition, produced by the minimization itself (Strauss (2008), §11.3 Thm 1, p. 310, and §11.4, p. 315).

Comparison. Since every $\lambda_n$ is a min-max of $\mathcal{R}$ , a change that does not lower $\mathcal{R}[u]$ for any trial function does not lower any eigenvalue. This applies to increasing $p$ or $q$ , and to shrinking the interval (a trial function on a subinterval, extended by zero, is a trial function on the larger one). Decreasing $w$ raises $\mathcal{R}[u]$ only when $D[u] > 0$ . So it does not lower any eigenvalue when $q \ge 0$ , because then $D[u] \ge \int_a^b p\, u'^2\, dx > 0$ for every trial function ( $u' \not\equiv 0$ , since $u$ vanishes at $a$ and is not identically zero). Without $q \ge 0$ this fails: $-u'' - 100 u = \lambda w u$ on $[0, \pi]$ with Dirichlet conditions has eigenfunctions $\sin nx$ , with $\lambda_1 = -99$ for $w = 1$ but $\lambda_1 = -198$ for $w = \tfrac12$ . The min-max gives only non-strict inequalities in all these comparisons.

ExampleOriginal

A string with linearly increasing density. Take $-u'' = \lambda (1 + x) u$ on $[0, 1]$ with $u(0) = u(1) = 0$ , so $p = 1$ , $q = 0$ and $w = 1 + x$ .

Two-sided bounds from min-max. For every trial function, $\int u^2 \le \int (1 + x) u^2 \le 2 \int u^2$ , so

\tfrac12\, \mathcal{R}_1[u] \le \mathcal{R}[u] \le \mathcal{R}_1[u],

where $\mathcal{R}_1$ is the quotient for $-u'' = \lambda u$ , whose eigenvalues are $n^2 \pi^2$ . Taking the min-max of each side gives

\tfrac12\, n^2 \pi^2 \le \lambda_n \le n^2 \pi^2, \qquad \text{so} \qquad 4.935 \le \lambda_1 \le 9.870 .

A sharper upper bound from one trial function. With $u = \sin \pi x$ : $\int_0^1 u'^2\, dx = \pi^2/2$ , and since $\int_0^1 x \sin^2 \pi x\, dx = \tfrac14$ , the weighted norm is $\int_0^1 (1 + x) \sin^2 \pi x\, dx = \tfrac12 + \tfrac14 = \tfrac34$ . Hence

\lambda_1 \le \mathcal{R}[\sin \pi x] = \frac{\pi^2/2}{3/4} = \frac{2\pi^2}{3} \approx 6.5797 .

The exact value is $\lambda_1 \approx 6.5484$ (found by shooting: integrate from $x = 0$ with $u(0) = 0$ , $u'(0) = 1$ and adjust $\lambda$ until $u(1) = 0$ ). The bound is $0.5\%$ high.

A warning about higher eigenvalues. The same computation with $\sin 2\pi x$ gives $\mathcal{R} = (2\pi^2)/(3/4) = 8\pi^2/3 \approx 26.319$ , while $\lambda_2 \approx 26.465$ . The quotient is below $\lambda_2$ . There is no contradiction: $\sin 2\pi x$ is not $w$ -orthogonal to $\varphi_1$ , so Part 2 does not apply, and Part 1 only guarantees $26.319 \ge \lambda_1$ . Upper bounds for $\lambda_2$ need a two-dimensional space of trial functions, as in the Rayleigh–Ritz method.

ApplicationOriginal

Vibrations. For a string or a membrane the eigenvalues are squared frequencies, $\omega_n^2 = \lambda_n$ (up to the tension). Part 1 is Rayleigh's principle: the squared fundamental frequency $\omega_1^2$ is the minimum, over admissible shapes, of the ratio of potential energy to mass-weighted kinetic energy. When $q \ge 0$ (for instance a string on an elastic foundation), comparison says that stiffening a structure does not lower any frequency, and adding mass does not raise any.
Quantum mechanics. For $H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V(x)$ in a box, Part 1 states that the ground-state energy is the minimum of $(\psi, H\psi)/(\psi, \psi)$ . Any normalized guess bounds it from above. Part 3 extends this to excited states without knowledge of the lower eigenfunctions.
Heat flow and diffusion. The slowest decay rate of a solution of $w\, u_t = (p u_x)_x - q u$ with fixed ends is $\lambda_1$ . Min-max bounds on $\lambda_1$ therefore bound the time a rod, or a diffusing population, takes to relax.

Method

The Rayleigh-Ritz method

Adapted

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Restricting the Rayleigh quotient to the span of N trial functions turns a Sturm-Liouville eigenvalue problem into a small matrix problem K c = Lambda M c whose eigenvalues are guaranteed upper bounds for the first N eigenvalues and decrease to them as the trial space grows.

Prerequisites: Variational characterization of eigenvalues, Generalized eigenproblems and normal modes

DefinitionAdaptedAdapted from Strauss (2008), §11.2, pp. 305–307

Use the setting of the variational characterization: the problem $-(p u')' + q u = \lambda w u$ on $[a, b]$ with $u(a) = u(b) = 0$ , the energy $D[u, v] = \int_a^b (p u' v' + q u v)\, dx$ , the weighted inner product $(u, v)_w$ , the space $\mathcal{A}$ of continuous, piecewise $C^1$ functions vanishing at both ends, and eigenvalues $\lambda_1 < \lambda_2 < \cdots$ .

Choose $N$ linearly independent trial functions $\psi_1, \dots, \psi_N \in \mathcal{A}$ and let $V_N$ be their span. The Ritz matrices are the $N \times N$ symmetric matrices

K_{ij} = D[\psi_i, \psi_j] = \int_a^b \bigl( p\, \psi_i' \psi_j' + q\, \psi_i \psi_j \bigr)\, dx, \qquad M_{ij} = (\psi_i, \psi_j)_w = \int_a^b w\, \psi_i \psi_j\, dx .

$M$ is the Gram matrix of independent functions in the inner product $(\cdot, \cdot)_w$ , hence positive definite. For $u = \sum_i c_i \psi_i$ ,

\mathcal{R}[u] = \frac{c^{\mathsf T} K c}{c^{\mathsf T} M c} .

The Ritz values $\Lambda_1^{(N)} \le \cdots \le \Lambda_N^{(N)}$ are the roots of $\det(K - \Lambda M) = 0$ , that is, the eigenvalues of the pencil $K c = \Lambda M c$ . If $c^{(k)}$ is an eigenvector for $\Lambda_k^{(N)}$ , then $\sum_i c^{(k)}_i \psi_i$ is the $k$ -th Ritz function.

MethodAdaptedAdapted from Stakgold & Holst (2011), Thm 6.5.3 and Remark 4, pp. 398–400; Gelfand & Fomin (1963), §40.1, Theorem, p. 196

(Upper bounds.) $\lambda_k \le \Lambda_k^{(N)}$ for $k = 1, \dots, N$ . In particular $\Lambda_1^{(N)} = \min_{V_N \setminus \{0\}} \mathcal{R}$ is an upper bound for $\lambda_1$ , with equality exactly when $\varphi_1 \in V_N$ .
(Monotonicity.) If $V_N \subset V_{N+1}$ , then $\Lambda_k^{(N+1)} \le \Lambda_k^{(N)}$ for $k \le N$ .
(Convergence, statement.) Let $\psi_1, \psi_2, \dots$ be an infinite sequence in $\mathcal{A}$ that is complete in energy: for every $u \in \mathcal{A}$ and every $\epsilon > 0$ there are $N$ and $g \in V_N$ with $\int_a^b \bigl( (u' - g')^2 + (u - g)^2 \bigr)\, dx < \epsilon$ . Then $\Lambda_k^{(N)} \downarrow \lambda_k$ as $N \to \infty$ , for each fixed $k$ .

Idea of the proofOriginal

A Ritz value is a min-max of the Rayleigh quotient taken over subspaces of $V_N$ only. The true eigenvalue is the same min-max over subspaces of all of $\mathcal{A}$ . Minimizing over fewer subspaces can only give a larger answer, so every Ritz value lies above the eigenvalue it approximates. Enlarging $V_N$ lets the min-max see more subspaces and pushes the Ritz values down. If the trial spaces eventually approximate everything, they approximate the true eigenfunctions, and the Ritz values close in.

ProofAdaptedAdapted from Strauss (2008), §11.2, proof of Thm 1, pp. 307–308; Stakgold & Holst (2011), proof of Thm 6.5.3, p. 399; Gelfand & Fomin (1963), §40.1, proof of the Theorem, p. 196

The pencil. Since $K$ is symmetric and $M$ is symmetric positive definite, the generalized eigenvalue problem $K c = \Lambda M c$ has real eigenvalues $\Lambda_1^{(N)} \le \dots \le \Lambda_N^{(N)}$ and eigenvectors $e_1, \dots, e_N$ with $e_i^{\mathsf T} M e_j = \delta_{ij}$ .
Part 1. Fix $k \le N$ and let $W_k$ be the set of functions $\sum_i c_i \psi_i$ with $c \in \operatorname{span}\{e_1, \dots, e_k\}$ . Because the $\psi_i$ are independent, $W_k$ is a $k$ -dimensional subspace of $\mathcal{A}$ . Writing $c = \sum_{j \le k} d_j e_j$ gives $c^{\mathsf T} K c = \sum_{j \le k} \Lambda_j^{(N)} d_j^2$ and $c^{\mathsf T} M c = \sum_{j \le k} d_j^2$ , so $\max_{W_k} \mathcal{R} = \Lambda_k^{(N)}$ . The min-max characterization (Part 3 of the variational-characterization unit) gives $\lambda_k \le \max_{W_k} \mathcal{R} = \Lambda_k^{(N)}$ . For $k = 1$ , the case of equality is the equality case of $\mathcal{R} \ge \lambda_1$ : the minimum over $V_N$ equals $\lambda_1$ exactly when some multiple of $\varphi_1$ lies in $V_N$ .
Part 2. The proof of the min-max formula, repeated word for word for the pencil, shows $\Lambda_k^{(N)} = \min \max_{W} \mathcal{R}$ over $k$ -dimensional subspaces $W \subset V_N$ . Every such $W$ is also a subspace of $V_{N+1}$ , so the minimum over subspaces of $V_{N+1}$ is no larger.
Part 3 for $k = 1$ . By Part 2, $\Lambda_1^{(N)}$ decreases, and it is bounded below by $\lambda_1$ . Since $p, q, w$ are bounded and $\|\varphi_1\|_w = 1$ , the numerator and denominator of $\mathcal{R}$ depend continuously on $u$ with respect to $\int \bigl( (u')^2 + u^2 \bigr)$ , so $\mathcal{R}$ is continuous at $\varphi_1$ in that sense. Given $\epsilon > 0$ , completeness in energy gives $g \in V_N$ so close to $\varphi_1$ that $\mathcal{R}[g] < \lambda_1 + \epsilon$ . Then $\lambda_1 \le \Lambda_1^{(N')} \le \mathcal{R}[g] < \lambda_1 + \epsilon$ for all $N' \ge N$ . The case $k \ge 2$ uses the same idea with $k$ -dimensional subspaces close to $\operatorname{span}\{\varphi_1, \dots, \varphi_k\}$ , and we only state it. $\square$

MethodAdaptedAdapted from Strauss (2008), §11.2, pp. 305–307; Stakgold & Holst (2011), §6.5, Remark 6, p. 400

Write the problem in Sturm–Liouville form. Identify $p$ , $q$ , $w$ and the boundary conditions, and write down $D$ and $\mathcal{R}$ . For Robin or Neumann conditions, use the modified quotient described in the variational-characterization unit.
Choose trial functions. They must satisfy the Dirichlet (essential) conditions; Neumann and Robin conditions are natural and need not be imposed. Use independent functions shaped like the modes you expect. Polynomials times a factor that vanishes at the ends, sines, or piecewise-linear "hat" functions are standard. Prefer nearly orthogonal families: raw monomials make $M$ badly conditioned when $N$ is large.
Assemble $K_{ij} = D[\psi_i, \psi_j]$ and $M_{ij} = (\psi_i, \psi_j)_w$ , exactly or by accurate quadrature.
Solve $K c = \Lambda M c$ , for example with a symmetric generalized eigensolver (scipy.linalg.eigh(K, M)), or by hand from $\det(K - \Lambda M) = 0$ when $N \le 2$ .
Read off the bounds $\lambda_k \le \Lambda_k^{(N)}$ and the Ritz functions $\sum_i c^{(k)}_i \psi_i$ as approximate eigenfunctions.
Refine. Add trial functions so that the spaces are nested, and watch $\Lambda_k^{(N)}$ decrease. Stop when successive values agree to the accuracy you need. The method gives only upper bounds; a lower bound needs a different argument, such as comparison through the min-max principle.

RemarkAdaptedAdapted from Riley, Hobson & Bence (2006), §22.7, p. 793; Arfken, Weber & Harris (2013), §22.4, p. 1119

Why few trial functions give good eigenvalues. Write a trial function as $u = \varphi_1 + e$ with $(e, \varphi_1)_w = 0$ . By step 1 of the variational-characterization proof, $D[\varphi_1, e] = \lambda_1 (\varphi_1, e)_w = 0$ . Hence $D[u] = \lambda_1 + D[e]$ and $\|u\|_w^2 = 1 + \|e\|_w^2$ , and exactly

\mathcal{R}[u] - \lambda_1 = \frac{D[e] - \lambda_1 \|e\|_w^2}{1 + \|e\|_w^2} .

The numerator is nonnegative by the variational characterization, since $e$ is orthogonal to $\varphi_1$ . The error in the eigenvalue is quadratic in the error $e$ of the eigenfunction. For example, if $\|e\|_w = 0.1$ , the eigenvalue error is $0.01\,(\mathcal{R}[e] - \lambda_1)/1.01$ , about one hundredth of the gap between $\mathcal{R}[e]$ and $\lambda_1$ . The Ritz functions themselves converge more slowly than the Ritz values.

ExampleOriginal

Variable stiffness. Estimate the first two eigenvalues of

-\bigl( (1 + x)\, u' \bigr)' = \lambda u \quad \text{on } [0, 1], \qquad u(0) = u(1) = 0,

so $p = 1 + x$ , $q = 0$ , $w = 1$ . Take $\psi_1 = x(1 - x)$ and $\psi_2 = x^2(1 - x)$ , which vanish at both ends, so $\psi_1' = 1 - 2x$ and $\psi_2' = 2x - 3x^2$ .

One trial function. $K_{11} = \int_0^1 (1 + x)(1 - 2x)^2\, dx = \tfrac13 + \tfrac16 = \tfrac12$ and $M_{11} = \int_0^1 x^2 (1 - x)^2\, dx = \tfrac{1}{30}$ . So $\lambda_1 \le \Lambda_1^{(1)} = 15$ .

Two trial functions. The remaining integrals are polynomial:

K = \begin{pmatrix} \tfrac12 & \tfrac{17}{60} \\[2pt] \tfrac{17}{60} & \tfrac{7}{30} \end{pmatrix}, \qquad M = \begin{pmatrix} \tfrac{1}{30} & \tfrac{1}{60} \\[2pt] \tfrac{1}{60} & \tfrac{1}{105} \end{pmatrix}.

Expanding the determinant,

\det(K - \Lambda M) = \frac{\Lambda^2}{25200} - \frac{13\Lambda}{4200} + \frac{131}{3600} = \frac{\Lambda^2 - 78\Lambda + 917}{25200},

so $\Lambda^{(2)}_{1,2} = 39 \mp 2\sqrt{151} \approx 14.4236$ and $63.5764$ .

Exact values. With $t = 1 + x$ the equation becomes $(t u_t)_t + \lambda u = 0$ , solved by $J_0(2\sqrt{\lambda t})$ and $Y_0(2\sqrt{\lambda t})$ . The boundary conditions at $t = 1$ and $t = 2$ give the eigenvalue condition

J_0(2\sqrt{\lambda})\, Y_0(2\sqrt{2\lambda}) - J_0(2\sqrt{2\lambda})\, Y_0(2\sqrt{\lambda}) = 0,

whose first two roots are $\lambda_1 \approx 14.33767$ and $\lambda_2 \approx 57.48028$ . Adding $x^3(1 - x)$ and $x^4(1 - x)$ to the basis, computed the same way, gives the table below.

$N$	$\Lambda_1^{(N)}$	$\Lambda_2^{(N)}$
1	15	—
2	14.4236	63.5764
3	14.3467	59.7992
4	14.33773	57.7910
exact	14.33767	57.48028

Every entry lies above the exact value and decreases with $N$ , as Parts 1 and 2 require. The lowest Ritz value is already within $0.6\%$ at $N = 2$ , while the second is $10.6\%$ high. This is typical: with a fixed basis, the higher Ritz values are less accurate.

ApplicationOriginal

Quantum chemistry. Expanding a wavefunction in a finite basis of atomic orbitals and diagonalizing the Hamiltonian matrix is the Rayleigh–Ritz method with $M$ the overlap matrix. Part 1 guarantees that each computed level lies above the true one (the Hylleraas–Undheim–MacDonald theorem for excited states), so a larger basis can only improve the energies.
Structural dynamics and buckling. Engineers estimate natural frequencies and critical buckling loads of beams, plates and frames from a few assumed mode shapes. The guaranteed upper bound on a frequency means the estimate errs on a known side.
Finite elements. With piecewise-linear hat functions as trial functions, the Ritz matrices become the sparse stiffness and mass matrices of the finite element method. The upper-bound and convergence statements of this unit then apply to finite-element eigenvalues computed on nested meshes.