Concept

Convection diffusion — where it appears

An equation with a transport term and a diffusion term, whose solution develops a boundary layer when the first dominates. It is the smallest problem on which the symmetric arguments in the iterative field stop applying, and the boundary layer it develops is often thinner than one grid cell.

Named by 12 essays across 2 fields — each of them below, with the objects they name alongside it.

10⁻¹110¹10²-45-35-25-15-551525s, the smallest of the three interpolation pointsrightmost pole of the reduced modelunstable above this linebalanced truncation, order 3exact, and unusablefull system's pole-3.5placements swept19unstable models4worst pole24their interpolation1.3·10⁻¹⁴balanced truncation-0.84the conditions all holdand the model cannot be run

A model that cannot be run

A stable system, reduced by matching its transfer function at three points exactly, comes back with a pole in the right half plane at four of nineteen placements — and matches at all three points to 1.3·10⁻¹⁴ while doing it. The construction did what it promised.

reduction · Reduced stability
10⁻¹110¹10²s, the smallest interpolation pointrightmost pole of the reduced model−10−10+1+10unstable above zeronumerical range edge, -0.986system's abscissa, -3.49two-sided, order 3one-sided, order 3one-sided, order 6Pe 10, nineteen placementssystem's rightmost eigenvalue-3.5edge of the numerical range-0.99two-sided, order 3: unstable placements4one-sided, order 3: unstable placements0one-sided, order 6: unstable placements0signed logarithmic axisshaded: poles in the right half plane

Half the conditions and a certificate

A two-sided interpolatory reduction of a stable convection–diffusion system came back unstable at four placements of nineteen. The one-sided reduction built from the same kind of solves came back stable at all 152 placements measured across eight Péclet numbers, and not by luck: every pole of a Galerkin model lies inside the numerical range of the operator, whose edge here is exactly the diffusion term's largest eigenvalue, −0.986 at Péclet 10. The price is the slope conditions, and spent as six one-sided shifts instead of three two-sided ones it beats the stable two-sided model at seven placements of ten.

reduction · Reduced stability
00.250.50.75100.250.50.751xuexactcentral differencesupwindthe oscillation is exact‖Ax − b‖/‖b‖ for the central answer4.6·10⁻¹⁸values outside [0, 1]16worst excursion0.52the dashed lines are 0 and 1, which the equation guaranteesno solver was involved

The stencil that is not symmetric

Past a cell Péclet number of exactly one — measured by bisection at 1.0000000000000002 — the central-difference solution of a convection–diffusion problem oscillates from point to point and leaves the interval the equation guarantees, at 16 of 31 grid points. It is the exact solution of its own linear system, to 4.6·10⁻¹⁸. No solver was involved.

iterative · Convection
edge of the numerical rangespectrum, -3.49zero37.9-0.986-3.496.59unstable one-sided models, of nineteen placements051015-3-2-1.5-1-0.500.511.522.53cond 1.4·10⁶cond 112t, the rescaling exponent — the k-th state is multiplied by γ to the power tk; t = 1 symmetrisesorder 1order 2order 3order 6the transfer function is the same at every tonly the inner product changes

A certificate written in coordinates

Every one-sided reduced model of the convection–diffusion system was stable, because its poles cannot leave the numerical range and the range sat in the left half plane. Rescale the state by a diagonal matrix and nothing about the system changes — not its eigenvalues, not its transfer function, not one pole of any two-sided model — but the numerical range moves. At Péclet 40 a rescaling with condition number 260 pushes its edge to +6.6 and ten placements of nineteen return unstable one-sided models. Projecting in the inner product of a Lyapunov solution puts the certificate back in any coordinates.

reduction · Reduced stability
00.250.50.75100.51xuexact & tunedupwindcentralone of these is exacttuned, worst nodal error2.4·10⁻¹⁷upwind, worst nodal error0.14central, points outside [0, 1]16exact at every nodeand only at the nodes

The 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.

iterative · Convection
Péclet 100, orders 1, 2, 3, 6nodal edge at 1,000, outflow8.9weighted edge at 1,000-0.097unstable nodal models7110¹10²10³-20246810largest cell over smallestright edge of the numerical rangenodal, small cells at the outflownodal, small cells at the inflowcell-size weightedabove the dashed line the certificate is goneit is a sufficient condition, not a failure

The inner product the mesh already computed

A Galerkin reduced model is certified stable when the operator's numerical range sits in the left half plane, and the certificate belongs to the coordinates. The coordinates that broke it before were a constructed rescaling. Grade a convection–diffusion mesh towards its outflow boundary layer — the grading anyone resolving the layer would choose — and nodal values break it too: the range's edge is past zero at a ratio of ten and reaches +8.9 at a thousand, and at Péclet 100 seven reduced models come back unstable. Weight the projection by the cell sizes the discretisation already computed and every model at every grading is stable, with the range's edge back at −0.098.

reduction · Reduced stability
Péclet 100centred nodal edge at 1,0008.9upwind nodal edge at 1,0006.8upwind weighted edge at 1,000-0.45110¹10²10³-202468largest cell over smallestright edge of the numerical rangecentred, nodalupwind, nodalcentred, weightedupwind, weightedabove the dashed line the certificate is goneupwinding lowers the line and does not cross it

The smaller cell downstream

Nodal values on a convection–diffusion mesh graded towards its outflow layer put the numerical range in the right half plane, and Galerkin reduced models then fail. A code that grades towards the layer usually also upwinds, and upwinding adds diffusion. It does not give the certificate back: at Péclet 100 and a grading of a thousand the range's edge falls from 8.94 to 6.78 and stays positive, and 18 reduced models fail where 23 did. Each row of the upwind stencil's symmetric part sums to half the difference between the inverse widths of the cell after the node and the cell before it, so wherever the cells shrink downstream the dissipation upwinding adds is outweighed by the coupling to the smaller cell next door.

reduction · Reduced stability
05101520253035404510⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²angle between the flow and the grid (degrees)worst nodal errortunedcentralupwindexact, and then nottuned, on the axis2.4·10⁻¹⁷tuned, five degrees off0.0079tuned at 45°0.066upwind at 45°0.0019fifteen orders of magnitude for five degreesand the worst of the three by forty-five

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.

iterative · Convection
worst nodal error, and nodes outside the interval the equation guarantees0°, streamline only2.39·10⁻¹⁷ · 0 outside0°, with crosswind5.25·10⁻¹⁷ · 0 outside15°, streamline only0.032 · 18 outside15°, with crosswind0.0145 · 9 outside30°, streamline only0.0571 · 50 outside30°, with crosswind0.0247 · 11 outside45°, streamline only0.0661 · 48 outside45°, with crosswind0.0281 · 0 outsidewhat the crosswind term buyserror ratio at 0°0.45error ratio at 15°2.2error ratio at 30°2.3error ratio at 45°2.3free where the scheme was exactand half the error everywhere else

The direction the diffusion does not go

Streamline diffusion adds τbbᵀ, a rank-one tensor that annihilates every direction across the flow. That is the design. The price is 18, 50 and 48 nodes where the computed solution leaves the interval the equation guarantees — and half a coefficient of crosswind diffusion halves the error at every angle while costing exactly nothing where the scheme was exact.

iterative · Convection
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))‖/‖·‖0distance to the problem it solves0.026distance to the problem posed0.36the added diffusion is h/2 = 0.001953, whatever ε isso refining removes it

A different equation on every grid

Upwinding is the exact discretisation of a convection–diffusion problem with diffusion ε + h/2, entry for entry, at a relative difference of between 0 and 1.26·10⁻¹⁶ on every mesh from 15 points to 511. The equation it is exact for is chosen by the mesh and not by ε — the added diffusion is 0.01563 on a 31-point grid whether ε is 0.2 or 0.001.

iterative · Convection
10²10⁻⁴10⁻³10⁻²10⁻¹grid points nworst nodal errorupwindtunedcentralthe same tuning, another problemtuned ÷ central at n = 3149tuned ÷ central at n = 12796central's error at the finest grid10·10⁻⁵exact on the problem it was derived fromand harmful on the one beside it

A parameter that is also a price

ξ = coth(Pe) − 1/Pe is the fraction of h/2 that makes a boundary-layer solution exact at every node. On a problem with no layer in it, the error the same scheme commits is ξ times upwinding's — 0.2511 against a ξ of 0.2504, 0.7461 against 0.7448 — so the number that buys the exactness is also the invoice.

iterative · Convection
00.7853981.57082.356193.1415900.250.50.751frequency θdamping |g(θ)|the oscillatory half →⅓ — the symmetric optimumno convectionwith convectionits imaginary parta modulus, not a valuesmoothing factor at this ω0.71best over every ω0.71the symmetric operator's, at ω = 2/30.33the imaginary part does not depend on ωso no ω removes it

A smoother that stops being one

Weighted Jacobi's smoothing factor on the convection–diffusion operator is a function of the cell Péclet number and nothing else — identical to eight digits at five grid sizes at matched Pe. It is 0.3335 at Pe = 0.016, exactly 1/√2 at Pe = 1, and 5.2190 at Pe = 7.8, where the sweep amplifies the modes it exists to remove.

iterative · Multigrid

Named alongside it

The objects these essays reach for when they reach for this one.

Peclet numberArtificial diffusionBoundary layerM-matrixNon-normalityReduced stabilityTransfer functionUpwind differencingGalerkin projectionNumerical rangePetrov–GalerkinRational krylov

All concepts