ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)

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.

Finite sine sums admit a backward Dirichlet heat solution

Example

For $T > 0$ and a finite terminal sine sum $g (x) = \sum_{k = 1}^{N} a_{k} \sin (k x)$ on $(0, π)$ , the function $u (x, t) = \sum_{k = 1}^{N} a_{k} e^{k^{2} (T - t)} \sin (k x) (0 \leq t \leq T)$ solves $u_{t} - u_{x x} = 0$ , vanishes at $x = 0$ and $x = π$ , and has $u (x, T) = g (x)$ . Every such finite terminal datum therefore has a classical backward extension, although its mode amplification grows without bound as $k$ increases.

Facts & Assumptions

Given: $T > 0$ , a finite integer $N \geq 1$ , real coefficients $a_{1}, \dots, a_{N}$ , and the terminal sum $g (x) = \sum_{k = 1}^{N} a_{k} \sin (k x)$ on $(0, π)$ .

[F1]

$(\sin x)^{'} = \cos x$ and $(\cos x)^{'} = - \sin x$ (The derivatives of sine and cosine are cosine and minus sine), so differentiating twice gives $\partial_{x}^{2} \sin (k x) = - k^{2} \sin (k x)$ by the chain rule The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ .

[F2]

The exponential satisfies $\frac{d}{d t} e^{k^{2} (T - t)} = - k^{2} e^{k^{2} (T - t)}$ (The exponential function is smooth and $(\exp)^{'} = \exp$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

[F3]

Finite sums and scalar multiples of differentiable functions are differentiable with the expected derivatives (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[F4]

$\sin 0 = 0$ and $\sin (k π) = 0$ for every integer $k$ (The zero sets of sine and cosine and the least positive common period 2 pi), and the cylinder vocabulary is that of Parabolic cylinder and parabolic boundary.

Verification

Given: $T > 0$ , $N \geq 1$ , real $a_{1}, \dots, a_{N}$ , the terminal sum $g$ , and $u (x, t) = \sum_{k = 1}^{N} a_{k} e^{k^{2} (T - t)} \sin (k x)$ .

1.1F1F2given

For each $k$ the summand $u_{k} (x, t) = a_{k} e^{k^{2} (T - t)} \sin (k x)$ satisfies $\partial_{t} u_{k} = - k^{2} u_{k}$ by [F2] and $\partial_{x}^{2} u_{k} = - k^{2} u_{k}$ by [F1], so $\partial_{t} u_{k} - \partial_{x}^{2} u_{k} = 0$ on $(0, π) \times (0, T]$ .

2.1step 1.1F3F4given

Since $u$ is a finite sum of the summands of step 1.1, [F3] gives $\partial_{t} u = \sum_{k} \partial_{t} u_{k}$ and $\partial_{x}^{2} u = \sum_{k} \partial_{x}^{2} u_{k}$ , so $u_{t} - u_{x x} = \sum_{k} (\partial_{t} u_{k} - \partial_{x}^{2} u_{k}) = 0$ ; moreover $u (0, t) = \sum_{k} a_{k} \sin 0 = 0$ and $u (π, t) = \sum_{k} a_{k} \sin (k π) = 0$ for every $t$ by [F4], while $u (x, T) = \sum_{k} a_{k} \sin (k x) = g (x)$ because $e^{0} = 1$ ; the sum is smooth because it has finitely many smooth summands.

3.1step 2.1given∎

Step 2.1 exhibits, for every finite terminal sine sum $g$ , the classical backward solution $u$ on the closed rectangle, so existence is unconditional for finite data; the $k$ -th summand carries the factor $e^{k^{2} (T - t)}$ , which equals $e^{k^{2} T}$ at $t = 0$ and grows without bound as $k$ increases, so no uniform amplification bound over all $k$ is claimed, in agreement with the example's final sentence and the unboundedness of the backward solution map.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

36 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Per Kristen Jakobsen, An Introduction to Partial Differential Equations (2019) (standard reference, not scraped)
Gerald Teschl, Partial Differential Equations: From Classical to Modern (2025 archived author manuscript, AMS Graduate Studies in Mathematics) (standard reference, not scraped)