Generator

upwind-identity

One function in the convection library, called 2 times across 2 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 5 claims it put to the test while drawing them, and where it stands against the rule this site is named for.

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.

At its defaults

Drawn even though every essay passes arguments — which on this site is every essay, at 100% of placements since the standard pass. A default nothing exercises is a trap for the next essay to call this with none, and this is the page where a default that has drifted from the figures around it becomes visible.

The upwind answer, against the two problems it might be answeringThe 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 1.3·10⁻¹⁶.00.250.50.75100.250.50.751xuexact at ε = 0.005exact at ε(1 + Pe)the upwind answeran exact answer to a different question‖upwind(ε) − central(ε(1+Pe))‖/‖·‖1.3·10⁻¹⁶distance to the problem it solves0.39distance to the problem posed71the added diffusion is h/2 = 0.01563, whatever ε isso refining removes 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.

eps: 0.005

The arguments are the ones The exact answer to a nearby problem passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

The upwind answer, against the two problems it might be answeringThe 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 1.3·10⁻¹⁶.00.250.50.75100.250.50.751xuexact at ε = 0.005exact at ε(1 + Pe)the upwind answeran exact answer to a different question‖upwind(ε) − central(ε(1+Pe))‖/‖·‖1.3·10⁻¹⁶distance to the problem it solves0.39distance to the problem posed71the added diffusion is h/2 = 0.01563, whatever ε isso refining removes 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.

What it checked while drawing

Every figure above asserted its own claims on the way to being drawn, and a claim that failed would have failed the build rather than drawn a wrong picture. Those assertions 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.

5 distinct claims across 2 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.

a diffusion coefficient the identity is worth drawing at

LU is for square matrices

the artificial diffusion is exactly h/2

the upwind answer is closer to the problem it solves than to the one posed

upwinding at ε is central differencing at ε(1 + Pe), n = 31 agree

Against the rule

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 52 of 99 generators — 37 print a residual and 15 are exempt with a published reason; 47 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.

Where it is called

Changing this generator changes every figure on this list. That is what makes the list worth publishing rather than keeping in a check script.

The whole library · All essays · What must fail