The upwind answer, against the two problems it might be answering

At its defaults it draws the upwind answer, against the two problems it might be answering. The upwind solution drawn against two exact solutions: the one for the ε that was asked for, whose layer is thin, and the one for ε(1 + Pe) = 0.02063, whose layer is 4.125 times thicker. It lies on the second. The two operators are identical entry for entry, to 1.3·10⁻¹⁶.

upwind-identity is one function in lib/figures/convection.js — convection — the stencil that is not symmetric, and the fourier arguments that assumed it was. Everything below came out of it during this build, at arguments taken from the essays rather than invented for this page. A figure here is the figure a reader meets in an essay, and if the generator changes, this page changes with it.

The upwind solution drawn against two exact solutions: the one for the ε that was asked for, whose layer is thin, and the one for ε(1 + Pe) = 0.02063, whose layer is 4.125 times thicker. It lies on the second. The two operators are identical entry for entry, to 1.3·10⁻¹⁶.

The upwind solution drawn against two exact solutions: the one for the ε that was asked for, whose layer is thin, and the one for ε(1 + Pe) = 0.006953, whose layer is 1.391 times thicker. It lies on the second. The two operators are identical entry for entry, to 0.

The upwind solution drawn against two exact solutions: the one for the ε that was asked for, whose layer is thin, and the one for ε(1 + Pe) = 0.02063, whose layer is 4.125 times thicker. It lies on the second. The two operators are identical entry for entry, to 1.3·10⁻¹⁶.

The upwind solution drawn against two exact solutions: the one for the ε that was asked for, whose layer is thin, and the one for ε(1 + Pe) = 0.1156, whose layer is 1.156 times thicker. It lies on the second. The two operators are identical entry for entry, to 0.

The upwind solution drawn against two exact solutions: the one for the ε that was asked for, whose layer is thin, and the one for ε(1 + Pe) = 0.03625, whose layer is 7.25 times thicker. It lies on the second. The two operators are identical entry for entry, to 2.6·10⁻¹⁷.

The upwind solution drawn against two exact solutions: the one for the ε that was asked for, whose layer is thin, and the one for ε(1 + Pe) = 0.01281, whose layer is 2.563 times thicker. It lies on the second. The two operators are identical entry for entry, to 2.6·10⁻¹⁷.

Every figure above checked its own claims on the way to being drawn, and a claim that failed would have stopped the picture rather than shipped a wrong one. Those checks used to leave no trace at all: a passing one returned true and the only evidence the figure had checked anything was that nothing crashed. The list below is what they actually said, collected by running this generator with an observer installed — not a description of what it is believed to check.

10 distinct claims across 6 sets of arguments, grouped below by shape — because most of them are one sentence with a different number in it, and how many separate times that sentence was put to the test is the informative part.

It draws a decomposition and prints its residual. It calls solve, and every figure above carries the badge — which residualcheck verifies by looking for it in the emitted SVG rather than by finding the call that builds one. A badge that is constructed and then left out of the body is the failure that check exists for.

Across the library: the rule bites on 217 of 397 generators — 199 print a residual and 18 are exempt with a published reason; 180 factorise nothing. Read from lib/residual-rule.js, which is the same body the gate enforces from, and the gate's last check fails the build if this page and it disagree about any generator.