TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The classical $C^{2}$ fixed-end string: the sine series solves the wave equation with the given data, uniquely, with an explicit tail bound

Statement

Let $L > 0$ , $c > 0$ . Let $F : R \to R$ be odd and $2 L$ -periodic of class $C^{5}$ and let $G : R \to R$ be odd and $2 L$ -periodic of class $C^{4}$ (Sine coefficients, sine series and the odd periodic extension, $C^{k}$ Euclidean maps and diffeomorphisms), with $M_{5} : = \sup ∣ F^{(5)} ∣ < \infty$ , $M_{4} : = \sup ∣ G^{(4)} ∣ < \infty$ . Put

$a_{n} : = b_{n} (F), d_{n} : = \frac{b_{n} (G)}{ω_{n}}, ω_{n} : = \frac{c n π}{L} (n \geq 1),$

and define, for $(t, x) \in R^{2}$ ,

$\begin{matrix} u (t, x) : = \sum_{n \geq 1} [a_{n} \cos (ω_{n} t) + d_{n} \sin (ω_{n} t)] \sin (\frac{n π x}{L}) . & (S) \end{matrix}$

Then:

(i) the series (S) and the termwise series of $\partial_{t} u, \partial_{x} u, \partial_{t}^{2} u, \partial_{t} \partial_{x} u, \partial_{x}^{2} u$ converge absolutely and uniformly on every strip $[- T, T] \times [0, L]$ ; the sum $u$ is $C^{2}$ , is odd and $2 L$ -periodic in $x$ , satisfies $u_{t t} = c^{2} u_{x x}$ , obeys the fixed-end conditions $u (t, 0) = u (t, L) = 0$ , and has

$u (0, x) = F (x), u_{t} (0, x) = G (x) (x \in R) .$

(ii) (Uniqueness.) If $v$ is $C^{2}$ on $[0, T] \times [0, L]$ for some $T > 0$ , satisfies $v_{t t} = c^{2} v_{x x}$ there, $v (t, 0) = v (t, L) = 0$ for all $t$ , and $v (0, \cdot) = v_{t} (0, \cdot) = 0$ , then $v = 0$ on $[0, T] \times [0, L]$ .

(iii) (Tail bound.) For every $N \geq 1$ and every $(t, x) \in [- T, T] \times [0, L]$ ,

$∣ u (t, x) - \sum_{n \leq N} [a_{n} \cos (ω_{n} t) + d_{n} \sin (ω_{n} t)] \sin (\frac{n π x}{L}) ∣ \leq C (N + 1)^{- 4},$

where $C : = 2 (\frac{2 L^{5} M_{5}}{π^{5}} + \frac{2 L^{5} M_{4}}{c π^{5}})$ .

Facts & Assumptions

Given: $L, c > 0$ ; the odd periodic data $F, G$ with the stated regularities and bounds.

[F1]

Coefficient decay: $∣ a_{n} ∣ \leq \frac{2 L^{5} M_{5}}{π^{5} n^{5}}$ and $∣ b_{n} (G) ∣ \leq \frac{2 L^{4} M_{4}}{π^{4} n^{4}}$ , hence $∣ d_{n} ∣ \leq \frac{2 L^{5} M_{4}}{c π^{5} n^{5}}$ (Repeated integration by parts gives $∣ b_{n} (F) ∣ \leq 2 L^{5} M_{5} / (π^{5} n^{5})$ for odd $2 L$ -periodic $C^{5}$ functions and $∣ b_{n} (G) ∣ \leq 2 L^{4} M_{4} / (π^{4} n^{4})$ for $C^{4}$ ).

[F2]

The modes $\sin (n π x / L) \cos (ω_{n} t)$ and $\sin (n π x / L) \sin (ω_{n} t) / (ω_{n})$ solve $u_{t t} = c^{2} u_{x x}$ , vanish at $x = 0, L$ , and have initial data $(\sin (n π x / L), 0)$ and $(0, \sin (n π x / L))$ respectively (d'Alembert's formula solves the wave equation, is the unique $C^{2}$ solution, and the fixed-end sine modes solve it exactly(c), The derivatives of sine and cosine are cosine and minus sine, The zero sets of sine and cosine and the least positive common period 2 pi).

[F3]

Weierstrass majorant test: a series of functions majorised by a summable sequence converges absolutely and uniformly (The Weierstrass M-test gives absolute pointwise convergence and uniform convergence of a function series, A series of real-valued functions and its pointwise and uniform convergence through its partial sums).

[F4]

The rectangle differentiation lemma: uniform convergence of a $C^{2}$ sequence together with its first and second partial derivatives implies the limit is $C^{2}$ with those derivatives (Uniform convergence of a $C^{2}$ sequence with its first and second derivatives on a rectangle passes $C^{2}$ regularity and termwise differentiation to the limit).

[F5]

Orthogonality and completeness: $\int_{0}^{L} \sin (n π x / L) \sin (k π x / L) d x = (L / 2) δ_{n k}$ and the full-period identities for odd $2 L$ -periodic functions (Sine orthogonality on a half period and the full-period integrals of odd $2 L$ -periodic functions); a continuous one-periodic function with all character integrals zero vanishes, hence a continuous odd $2 L$ -periodic real function with all sine coefficients zero vanishes: rescale $x = 2 L y$ by Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ and use $e^{- i r} = \cos r - i \sin r$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ) (Fejer completeness: a continuous one-periodic function with all integrals against the characters vanishing is zero).

[F6]

For $N \geq 1$ and integers $m > N + 1$ , decreasing $x^{- 5}$ gives $\sum_{n = N + 2}^{m} n^{- 5} \leq \int_{N + 1}^{m} x^{- 5} d x \leq \frac{1}{4} (N + 1)^{- 4}$ . Passing to the limit of the nonnegative partial sums and adding $(N + 1)^{- 5}$ gives $\sum_{n > N} n^{- 5} \leq 2 (N + 1)^{- 4}$ , using only proper integrals (The integral test: for $f \geq 0$ nonincreasing on $[0, \infty)$ , $\sum_{k} f (k)$ converges if and only if the sequence $(\int_{0}^{N} f)_{N}$ is bounded, with $\int_{0}^{N} f \leq \sum_{k < N} f (k) \leq f (0) + \int_{0}^{N} f$ , For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ , The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ).

[F7]

For a $C^{1}$ function $H$ on a compact time-space rectangle, the time difference quotient $Q_{h} (x) : = (H (t + h, x) - H (t, x)) / h$ equals $h^{- 1} \int_{t}^{t + h} \partial_{t} H (s, x) d s$ by the fundamental theorem. Uniform continuity of $\partial_{t} H$ on the rectangle makes $Q_{h} \to \partial_{t} H (t, \cdot)$ uniformly in $x$ . These functions are continuous and integrable, so Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error gives convergence of their integrals and $\frac{d}{d t} \int_{0}^{L} H (t, x) d x = \int_{0}^{L} \partial_{t} H (t, x) d x$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous, Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error).

[F8]

For a $C^{2}$ function on an open neighbourhood of a point $(t, x)$ , the mixed partials $v_{t x}$ and $v_{x t}$ agree. The divided rectangular increment over $[t, t + h] \times [x, x + k]$ equals the averages of the two ordered mixed partials by applying the one-variable fundamental theorem in each order; continuity makes both averages tend to their point values (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion).

[F9]

For $C^{2}$ data, integration by parts in $x$ on $[0, L]$ gives $\int_{0}^{L} v_{x} v_{x t} d x = [v_{x} v_{t}]_{0}^{L} - \int_{0}^{L} v_{x x} v_{t} d x = - \int_{0}^{L} v_{x x} v_{t} d x$ when $v_{t} (t, 0) = v_{t} (t, L) = 0$ , using [F8] to identify $v_{x t} = \partial_{x} v_{t}$ (If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ , 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 integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ , 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 derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

Proof

technique · direct; majorise the differentiated series, identify the initial data by completeness, and prove uniqueness by the energy identity

1.1givenF1F2F3algebra

By [F1] the coefficient sequences satisfy $∣ a_{n} ∣ + ∣ d_{n} ∣ \leq C_{1} n^{- 5}$ with $C_{1} : = \frac{2 L^{5} M_{5}}{π^{5}} + \frac{2 L^{5} M_{4}}{c π^{5}}$ . Since $∣ \sin ∣ \leq 1$ , $∣ \cos ∣ \leq 1$ (Parity and the Pythagorean identity for sine and cosine) and $∣ \partial_{t}^{2} ∣ \leq ω_{n}^{2}$ , $∣ \partial_{t} \partial_{x} ∣ \leq ω_{n} (n π / L)$ , $∣ \partial_{x}^{2} ∣ \leq (n π / L)^{2}$ termwise (The derivatives of sine and cosine are cosine and minus sine), each of the five differentiated series is majorised termwise by a constant multiple of $n^{- 3}$ : $ω_{n}^{2} (∣ a_{n} ∣ + ∣ d_{n} ∣) \leq (\frac{c π}{L})^{2} C_{1} n^{- 3}, ω_{n} \frac{n π}{L} (∣ a_{n} ∣ + ∣ d_{n} ∣) \leq \frac{c π^{2}}{L^{2}} C_{1} n^{- 3}, (\frac{n π}{L})^{2} (∣ a_{n} ∣ + ∣ d_{n} ∣) \leq \frac{π^{2}}{L^{2}} C_{1} n^{- 3} .$ These majorants are independent of $t$ and $x$ , so [F3] gives absolute uniform convergence of the series and all five differentiated series on $R^{2}$ , in particular on every stated strip.

1.2givenF7F8F9algebra

(Uniqueness.) Let $v$ be as in (ii) and put $E (t) : = \frac{1}{2} \int_{0}^{L} (v_{t} (t, x)^{2} + c^{2} v_{x} (t, x)^{2}) d x$ , which is finite for each $t$ because the integrand is continuous on the compact interval (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion). By [F7] applied to this integrand, $E$ is differentiable and its derivative is the integral of its time derivative. Using [F9], $E^{'} (t) = \int_{0}^{L} (v_{t} v_{t t} + c^{2} v_{x} v_{x t}) d x = \int_{0}^{L} v_{t} (v_{t t} - c^{2} v_{x x}) d x = 0,$ where the boundary term vanishes because $\partial_{t} v (t, 0) = \partial_{t} v (t, L) = 0$ : the endpoint conditions $v (\cdot, 0) = v (\cdot, L) = 0$ hold identically in $t$ , so their time derivatives vanish. The integral-evaluation theorem The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ gives $E (t) - E (0) = \int_{0}^{t} E^{'} (s) d s = 0$ , so $E$ is constant; $E (0) = 0$ because $v (0, \cdot) = v_{t} (0, \cdot) = 0$ ; so $v_{t} (t, x) = v_{x} (t, x) = 0$ for all $(t, x)$ by continuity and nonnegativity of the integrand. Thus $v$ is constant on the connected rectangle and, vanishing at $t = 0$ , is identically $0$ . This is (ii).

2.1step 1.1F2F4algebra

Let $u_{N}$ be the $N$ -th partial sum of (S). Fix any compact rectangle in $R^{2}$ and translate its coordinates to the form required by [F4]. Step 1.1 gives uniform convergence there of the partial sums and all first and second partial-derivative series, so [F4] makes the limit $u$ $C^{2}$ on the rectangle interior with those termwise derivatives. Such rectangles cover $R^{2}$ . Since every summand satisfies $u_{n, t t} = c^{2} u_{n, x x}$ by [F2] and the differentiated series converge uniformly, $u_{t t} = c^{2} u_{x x}$ . Also every summand vanishes at $x = 0$ and $x = L$ , so $u (t, 0) = u (t, L) = 0$ , and each term is odd and $2 L$ -periodic in $x$ , so $u$ is too.

2.2step 1.1F1F3F5algebra

(Initial data.) Let $R : = F - \sum_{n} a_{n} \sin (n π x / L)$ ; uniform convergence from step 1.1 makes $R$ continuous, odd and $2 L$ -periodic. For each $k \geq 1$ , the uniform integral-error bound and [F5] give $\int_{0}^{L} R (x) \sin (k π x / L) d x = \frac{L}{2} b_{k} (F) - \frac{L}{2} a_{k} = 0$ . Rescaling $x = 2 L y$ and using the exponential identity in [F5], the character integral of $R (2 L \cdot)$ at positive $k$ is $- \frac{i}{L} \int_{0}^{L} R (x) \sin (k π x / L) d x = 0$ ; its cosine part vanishes by oddness. The $k = 0$ integral vanishes by the mean identity of [F5], and negative $k$ gives the opposite sine sign with the same zero integral. Thus every character integral vanishes and Fejer completeness [F5] yields $R = 0$ . For $G$ , its coefficient series converges uniformly by the $n^{- 4}$ bound [F1], so the same argument gives $G = \sum_{n} b_{n} (G) \sin (n π x / L)$ . Evaluating (S) and its uniformly convergent time-derivative series at $t = 0$ gives $u (0, x) = F (x)$ and $u_{t} (0, x) = G (x)$ for all real $x$ .

3.1step 1.1F1F6algebra∎

(Tail bound.) For fixed $(t, x)$ the tail of (S) is bounded in absolute value by $\sum_{n > N} (∣ a_{n} ∣ + ∣ d_{n} ∣) \leq C_{1} \sum_{n > N} n^{- 5} \leq 2 C_{1} (N + 1)^{- 4} = C (N + 1)^{- 4}$ by [F1], the triangle inequality and [F6]. This is (iii).

Remarks

The theorem solves the fixed-end wave problem for $C^{5}$ / $C^{4}$ odd periodic data with an explicit uniform tail bound; it does not treat merely continuous or $L^{2}$ data, non-periodic compatibility conditions beyond the odd extension, or the $L^{2}$ convergence of the full trigonometric system. The uniqueness clause is stated for $C^{2}$ solutions on a strip and uses only the energy identity.
The exact constants in (iii) inherit the constants of the coefficient decay; they are not asserted to be optimal, and no numerical error is claimed. The initial-data convergence is uniform on $R$ because the same convergence mode produced the series.

The classical $C^{2}$ fixed-end string: the sine series solves the wave equation with the given data, uniquely, with an explicit tail bound

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

The classical C2C2 fixed-end string: the sine series solves the wave equation with the given data, uniquely, with an explicit tail bound

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

The classical $C^{2}$ fixed-end string: the sine series solves the wave equation with the given data, uniquely, with an explicit tail bound