Three schemes at ε = 0.005, cell Péclet number 3.13
At its defaults it draws three schemes at ε = 0.005, cell péclet number 3.13. The computed solution of each scheme against position, with the exact solution drawn as a dashed line. The tuned scheme's values lie on it — the largest nodal difference is 2.4·10⁻¹⁷. Upwinding is smooth and 0.136 away at its worst. Central differencing oscillates and leaves the interval [0, 1] at 16 of the 31 points.
tuned-solution is one function in lib/figures/supg.js —
tuned diffusion — exact at the nodes, on the problem it was derived from. 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 computed solution of each scheme against position, with the exact solution drawn as a dashed line. The tuned scheme's values lie on it — the largest nodal difference is 2.4·10⁻¹⁷. Upwinding is smooth and 0.136 away at its worst. Central differencing oscillates and leaves the interval [0, 1] at 16 of the 31 points.
n: 31
The arguments are the ones Exact along one axis passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
The computed solution of each scheme against position, with the exact solution drawn as a dashed line. The tuned scheme's values lie on it — the largest nodal difference is 2.4·10⁻¹⁷. Upwinding is smooth and 0.136 away at its worst. Central differencing oscillates and leaves the interval [0, 1] at 16 of the 31 points.
eps: 0.005
The arguments are the ones The diffusion that makes the answer exact passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
The computed solution of each scheme against position, with the exact solution drawn as a dashed line. The tuned scheme's values lie on it — the largest nodal difference is 2.4·10⁻¹⁷. Upwinding is smooth and 0.136 away at its worst. Central differencing oscillates and leaves the interval [0, 1] at 16 of the 31 points.
eps: 0.2
The arguments are the ones The diffusion that makes the answer exact passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
The computed solution of each scheme against position, with the exact solution drawn as a dashed line. The tuned scheme's values lie on it — the largest nodal difference is 2.8·10⁻¹⁶. Upwinding is smooth and 0.025 away at its worst. Central differencing oscillates and leaves the interval [0, 1] at 0 of the 31 points.
eps: 0.1
The arguments are the ones The diffusion that makes the answer exact passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
The computed solution of each scheme against position, with the exact solution drawn as a dashed line. The tuned scheme's values lie on it — the largest nodal difference is 5.6·10⁻¹⁷. Upwinding is smooth and 0.051 away at its worst. Central differencing oscillates and leaves the interval [0, 1] at 0 of the 31 points.
eps: 0.05
The arguments are the ones The diffusion that makes the answer exact passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
The computed solution of each scheme against position, with the exact solution drawn as a dashed line. The tuned scheme's values lie on it — the largest nodal difference is 5.6·10⁻¹⁷. Upwinding is smooth and 0.092 away at its worst. Central differencing oscillates and leaves the interval [0, 1] at 0 of the 31 points.
What it checked while drawing
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.
6 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.
a diffusion the layer is resolvable near
a grid whose points can be told apart
and it stays inside [0, 1]
and the other two are not
LU is for square matrices
the tuned scheme is exact at the nodes
Against the rule
It draws a decomposition and prints its residual. It calls
solveScheme,
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.
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.
Exact along one axis
The tuned diffusion makes the answer exact at every node, and in two dimensions it holds at exactly one flow angle. Five degrees off the grid the relative error goes from 1.2·10⁻¹⁴ to 6.9, and by twenty degrees the scheme is worse than the upwinding it was built to improve on.
Iterating, instead of factorisingThe diffusion that makes the answer exact
Upwinding adds h/2 of artificial diffusion. Central differencing adds none. Add ε·ξ·Pe with ξ = coth(Pe) − 1/Pe and the computed solution is the exact one at every grid point, to 2.4·10⁻¹⁷ — at every Péclet number, on the problem it was derived from and on no other.