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A stationary path that is not a local minimum: the quadratic model $L = \frac{1}{2} (v^{2} - 4 q^{2})$ on $[0, π]$

Example

Let $n = 1$ , $[a, b] = [0, π]$ , $U = R$ , and $L (q, v) = \frac{1}{2} (v^{2} - 4 q^{2})$ , a $C^{\infty}$ (hence $C^{2}$ ) Lagrangian on $R \times R \times R$ (A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations, $C^{k}$ Euclidean maps and diffeomorphisms). Let

$q_{*} (t) : = \sin 2 t, η (t) : = \sin t .$

Use fixed-endpoint admissible $C^{2}$ curves with $q (0) = q (π) = 0$ and the norm $∥ q ∥_{C^{2}} : = \max_{0 \leq j \leq 2} \sup_{t \in [0, π]} ∣ q^{(j)} (t) ∣$ . A local minimum means $S (q_{*}) \leq S (q)$ for all admissible $q$ in some ball about $q_{*}$ in this norm. Then $q_{*}$ is a stationary curve of the action $S (q) = \int_{0}^{π} \frac{1}{2} (q^{' 2} - 4 q^{2}) d t$ , and for the one-parameter family $q_{*} + ϵ η$ one has the exact identity

$S (q_{*} + ϵ η) = S (q_{*}) - \frac{3 π}{4} ϵ^{2} (ϵ \in R) .$

In particular $S (q_{*} + ϵ η) < S (q_{*})$ for every $ϵ \neq 0$ : the stationary curve is not a local minimum of the action. The model has no minimiser at all, since $S (λ η) = - \frac{3 π}{4} λ^{2} \to - \infty$ as $∣ λ ∣ \to \infty$ .

Facts & Assumptions

Given: The Lagrangian $L (q, v) = \frac{1}{2} (v^{2} - 4 q^{2})$ on $[0, π]$ , the curves $q_{*} (t) = \sin 2 t$ and $η (t) = \sin t$ , and the action $S$ of A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations.

[F1]

The functions $\sin$ and $\cos$ are differentiable on $R$ with $\sin^{'} = \cos$ , $\cos^{'} = - \sin$ , $\sin 0 = 0$ , $\cos 0 = 1$ (The derivatives of sine and cosine are cosine and minus sine, Sine and cosine defined by their real power series); the chain rule gives $q_{*}^{'} (t) = 2 \cos 2 t$ and $q_{*}^{''} (t) = - 4 \sin 2 t$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ). The addition formulas $\cos (2 t) = \cos^{2} t - \sin^{2} t$ , $\sin (2 t) = 2 \sin t \cos t$ and the Pythagorean identity $\cos^{2} t + \sin^{2} t = 1$ hold for all real $t$ (The addition formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine), and $\sin (2 π) = 0$ because $2 π$ is a period of $\sin$ (The zero sets of sine and cosine and the least positive common period 2 pi).

[F2]

The Euler–Lagrange expression of $L$ along a $C^{2}$ curve $q$ is $E_{L} (q) = - q^{''} - 4 q$ , because $L_{q} = - 4 q$ and $L_{v} = q^{'}$ ; stationarity of a $C^{2}$ curve for $S$ is equivalent to $E_{L} (q) = 0$ (A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations, Stationarity of the time-dependent action is equivalent to the vanishing of the Euler-Lagrange expression, The first variation of a time-dependent action on a finite interval: admissibility, the integral formula and continuity of the Euler-Lagrange coefficient).

[F3]

For a $C^{2}$ family $q_{ϵ}$ the first variation is the derivative at $ϵ = 0$ of $ϵ \mapsto S (q_{ϵ})$ ; the integral is linear and integrable functions are the continuous ones here (Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ , The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ , A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). The second fundamental theorem evaluates $\int_{0}^{π} \cos 2 t d t = \frac{1}{2} [\sin 2 t]_{0}^{π} = \frac{1}{2} (\sin 2 π - \sin 0) = 0$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

Verification

technique · direct; check the Euler–Lagrange equation and expand the quadratic action exactly

1.1F1F2given

The curve $q_{*}$ is $C^{2}$ on an open interval containing $[0, π]$ with $q_{*} (0) = q_{*} (π) = 0$ , and $E_{L} (q_{*}) = - q_{*}^{''} - 4 q_{*} = 4 \sin 2 t - 4 \sin 2 t = 0$ ; hence $q_{*}$ is stationary for $S$ by [F2].

2.1F1F2F3step 1.1algebra

For every $ϵ$ the curve $q_{*} + ϵ η$ is $C^{2}$ with zero endpoint values, and the integrand of its action is a quadratic polynomial in $ϵ$ : $\frac{1}{2} ((q_{*}^{'} + ϵ η^{'})^{2} - 4 (q_{*} + ϵ η)^{2}) = \frac{1}{2} (q_{*}^{' 2} - 4 q_{*}^{2}) + ϵ (q_{*}^{'} η^{'} - 4 q_{*} η) + \frac{ϵ^{2}}{2} (η^{' 2} - 4 η^{2}) .$ By [F3] the action is the sum of the integrals of the three summands, and the linear term is $ϵ δ S (q_{*}) [η] = 0$ by step 1.1 and [F2].

3.1F1F3step 2.1algebra

The quadratic term: with $η = \sin$ , $η^{'} = \cos$ and $η^{' 2} - 4 η^{2} = \cos^{2} t - 4 \sin^{2} t$ . By [F1], $\cos^{2} t = \frac{1}{2} (1 + \cos 2 t)$ and $\sin^{2} t = \frac{1}{2} (1 - \cos 2 t)$ , so $\cos^{2} t - 4 \sin^{2} t = \frac{1}{2} (1 + \cos 2 t) - 2 (1 - \cos 2 t) = - \frac{3}{2} + \frac{5}{2} \cos 2 t$ . Integrating over $[0, π]$ and using $\int_{0}^{π} \cos 2 t d t = 0$ from [F3] and $\int_{0}^{π} 1 d t = π$ gives $\int_{0}^{π} (η^{' 2} - 4 η^{2}) d t = - \frac{3 π}{2}$ . Substituting into step 2.1 gives $S (q_{*} + ϵ η) = S (q_{*}) - \frac{3 π}{4} ϵ^{2}$ for every real $ϵ$ .

4.1F1F2F3step 2.1step 3.1algebra

A positive direction is $ζ (t) : = \sin 3 t$ , which vanishes at both endpoints. The same expansion as step 2.1 has zero linear term by stationarity and gives $S (q_{*} + ϵ ζ) - S (q_{*}) = \frac{1}{2} ϵ^{2} \int_{0}^{π} (9 \cos^{2} 3 t - 4 \sin^{2} 3 t) d t = \frac{5 π}{4} ϵ^{2}$ . Use the double-angle formulas and $\int_{0}^{π} \cos 6 t d t = [\sin 6 t / 6]_{0}^{π} = 0$ , by periodicity from [F1]. Thus arbitrarily small nonzero variations in this direction increase the action, whereas step 3.1 decreases it in direction $η$ .

5.1step 3.1algebra∎

Since $∥ ϵ η ∥_{C^{2}} = ∣ ϵ ∣ ∥ η ∥_{C^{2}} \to 0$ and $- \frac{3 π}{4} ϵ^{2} < 0$ for every $ϵ \neq 0$ , arbitrarily small nonzero variations of the stationary curve strictly decrease the action; thus $q_{*}$ is not a local minimum. Moreover $S (λ η) = - \frac{3 π}{4} λ^{2}$ tends to $- \infty$ , so the model has no minimising curve at all.

Remarks

The example is a mathematical model with an indefinite quadratic Lagrangian; it shows that the Euler–Lagrange equation is a necessary condition only, and that "principle of least action" is a misnomer in general. The stationary curve here is a saddle in the infinite-dimensional sense realised by the decreasing family $q_{*} + ϵ \sin t$ and the increasing family $q_{*} + ϵ \sin 3 t$ of steps 3.1 and 4.1.
The zero curve $q = 0$ is also stationary for this model, and it is likewise not a local minimum, by the same computation with $q_{*}$ replaced by $0$ . Nothing here depends on the interpretation of $L$ as a physical energy; no positivity of the kinetic term is claimed.

A stationary path that is not a local minimum: the quadratic model $L = \frac{1}{2} (v^{2} - 4 q^{2})$ on $[0, π]$

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

A stationary path that is not a local minimum: the quadratic model L=12(v2−4q2)L=21​(v2−4q2) on [0,π][0,π]

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

A stationary path that is not a local minimum: the quadratic model $L = \frac{1}{2} (v^{2} - 4 q^{2})$ on $[0, π]$