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Two distinct oscillator solutions with the same zero endpoints at $T = π$ , and their common action value

Example

Take $m = 1$ , $k = 1$ , hence $ω = 1$ , and $T = π$ in The undamped oscillator: Dirichlet non-uniqueness at $T = n π / ω$ despite unique initial-value solutions, and consider the two curves

$x_{1} (t) = \sin t, x_{2} (t) = 2 \sin t (t \in [0, π]) .$

Both are $C^{\infty}$ solutions of $x^{''} + x = 0$ on $[0, π]$ with $x (0) = x (π) = 0$ , and they are distinct; they have different initial velocities $1$ and $2$ . On the chart $L (q, v) = \frac{1}{2} (v^{2} - q^{2})$ over $[0, π] \times R$ (A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations) both are stationary curves of the action, and their action values agree:

$S (x_{1}) = S (x_{2}) = 0.$

This is the concrete $T = π$ case of the non-uniqueness in The undamped oscillator: Dirichlet non-uniqueness at $T = n π / ω$ despite unique initial-value solutions and of the signed-action checks of An explicit saddle of the oscillator action at $T = 2 π$ : the two variations $\sin (t / 2)$ and $\sin (3 t / 2)$ ; all numbers are exact.

Facts & Assumptions

Given: The model $x^{''} + x = 0$ on $[0, π]$ , its curves $x_{1} (t) = \sin t$ and $x_{2} (t) = 2 \sin t$ , and the chart $L (q, v) = \frac{1}{2} (v^{2} - q^{2})$ with its action $S$ .

[L1]

The non-uniqueness proposition (The undamped oscillator: Dirichlet non-uniqueness at $T = n π / ω$ despite unique initial-value solutions): for $m = k = 1$ and $T = π = n π / ω$ with $n = 1$ , the set of $C^{2}$ solutions of $x^{''} + x = 0$ on $[0, π]$ with $x (0) = x (π) = 0$ is exactly ${A \sin t : A \in R}$ , and each such curve is a stationary curve of the action of the chart $L (q, v) = \frac{1}{2} (v^{2} - q^{2})$ (Stationarity of the time-dependent action is equivalent to the vanishing of the Euler-Lagrange expression, A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations).

[L2]

Trigonometric calculus and integrals (The derivatives of sine and cosine are cosine and minus sine, Sine and cosine defined by their real power series, The addition formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine, The zero sets of sine and cosine and the least positive common period 2 pi, The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , 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): $\sin$ and $\cos$ are differentiable with $\sin^{'} = \cos$ , $\cos^{'} = - \sin$ , the zeros of $\sin$ are the integer multiples of $π$ and in particular $\sin 0 = \sin π = 0$ ; the double-angle formula gives $\cos^{2} t - \sin^{2} t = \cos 2 t$ ; the fundamental theorem evaluates $\int_{0}^{π} \cos 2 t d t = \frac{1}{2} [\sin 2 t]_{0}^{π} = 0$ because $\sin 2 π = 0$ ; and continuous integrands on $[0, π]$ are integrable.

Verification

Proof technique: direct; verify the two solutions and their endpoint data, note their distinctness, and evaluate the action along the family.

1.1givenL2algebra

The two solutions: by [L2], $x_{1}^{''} = - \sin t = - x_{1}$ and $x_{2}^{''} = - 2 \sin t = - x_{2}$ , so both curves solve $x^{''} + x = 0$ on $[0, π]$ ; and $x_{1} (0) = x_{1} (π) = 0$ , $x_{2} (0) = x_{2} (π) = 0$ by the zero set of $\sin$ .

2.1step 1.1givenL2algebra

Distinctness with equal boundary data: $x_{2} - x_{1} = \sin t \neq 0$ (at $t = π / 2$ the value is $1$ ), so the solutions are distinct, while both take the value $0$ at $0$ and at $π$ ; their initial velocities are $x_{1}^{'} (0) = 1$ and $x_{2}^{'} (0) = 2$ , so they are different initial-value solutions, in accordance with the initial-value uniqueness clause of [L1].

3.1step 1.1step 2.1L1L2algebra

Action along the family: for $A \in R$ , using $L (A \sin t, A \cos t) = \frac{A^{2}}{2} (\cos^{2} t - \sin^{2} t)$ and linearity of the integral, $S (A \sin) = \frac{A^{2}}{2} \int_{0}^{π} (\cos^{2} t - \sin^{2} t) d t = \frac{A^{2}}{2} \int_{0}^{π} \cos 2 t d t = \frac{A^{2}}{4} [\sin 2 t]_{0}^{π} = 0$ by [L2]; in particular $S (x_{1}) = S (x_{2}) = 0$ .

4.1step 1.1step 2.1step 3.1L1∎

Reading: by [L1] the curves $x_{1}, x_{2}$ are stationary for the action of the chart and share the endpoint data $(0, 0)$ ; they are distinct and have the same action value $0$ , exhibiting in one explicit computation the failure of uniqueness of the fixed-endpoint problem at $T = π$ , with no claim about any other duration.

Two distinct oscillator solutions with the same zero endpoints at $T = π$ , and their common action value

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources

Two distinct oscillator solutions with the same zero endpoints at T=πT=π, and their common action value

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources

Two distinct oscillator solutions with the same zero endpoints at $T = π$ , and their common action value