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✓ 4 results · all verified · 2 also independently AI-judged

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Hyperbolic DG and summation-by-parts methods — Examples

1 · Prerequisites

2 · Summary

These examples calculate the inflow energy supply for a one-cell upwind DG state, compare central and upwind periodic flux dissipation, and verify the sign of a two-node SBP inflow penalty. The underintegrated Burgers collocation example gives an exact positive quadratic-entropy rate despite satisfying the SBP matrix identity; it illustrates why nonlinear entropy stability needs the stated flux-differencing and quadrature conditions.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Underintegrated Burgers collocation can increase quadratic entropy

Statement refuted

The SBP matrix identity alone guarantees that a naively collocated strong-form discretization of a nonlinear conservation law cannot increase its discrete quadratic entropy.

Witness Data

Use one reference interval $[- 1, 1]$ with three Legendre–Gauss–Lobatto nodes $(- 1, 0, 1)$ , weights $H = diag (1 / 3, 4 / 3, 1 / 3)$ , derivative matrix $D = (\begin{array}{ccc} - 3 / 2 & 2 & - 1 / 2 \\ - 1 / 2 & 0 & 1 / 2 \\ 1 / 2 & - 2 & 3 / 2 \end{array}),$ and boundary matrix $B = diag (- 1, 0, 1)$ . For periodic Burgers flux $f (u) = u^{2} / 2$ , define the naive collocated strong form $\dot{U} + D f (U) + H^{- 1} B ({\hat{f}}_{\partial} - f (U_{\partial})) = 0,$ where ${\hat{f}}_{\partial} = (f (U_{0}) + f (U_{2})) / 2$ is used at both periodic endpoints. Take $U = (- 3, - 1, - 2)^{T}$ .

Facts & Assumptions

Given: The three-node nodal vector, positive diagonal quadrature norm, exact SBP identity, pointwise Burgers flux, central periodic boundary flux, and quadratic entropy $E (U) = \frac{1}{2} U^{T} H U$ .

[F1]

Diagonal-norm summation-by-parts element data: SBP data satisfy $D 1 = 0$ and $H D + D^{T} H = B$ with positive diagonal $H$ .

Verification

1.1givenF1algebra

Direct multiplication gives $D 1 = 0$ and $H D + D^{T} H = diag (- 1, 0, 1) = B$ , so the derivative and norm satisfy the stated SBP identity. The three-point Gauss–Lobatto rule is exact through degree three, while both the quadratic entropy integrand $u_{h}^{2}$ and cubic entropy flux $u_{h}^{3}$ exceed that degree for a quadratic polynomial; the pointwise collocation therefore underintegrates these nonlinear products.

2.1givenstep 1.1algebra

At the chosen state, $f (U) = (9 / 2, 1 / 2, 2)^{T}$ and ${\hat{f}}_{\partial} = 13 / 4$ . Hence $D f (U) = (- 27 / 4, - 5 / 4, 17 / 4)^{T}$ and $H^{- 1} B ({\hat{f}}_{\partial} - f (U_{\partial})) = (15 / 4, 0, 15 / 4)^{T}$ . The semidiscrete residual is therefore $(- 3, - 5 / 4, 8)^{T}$ , so $\dot{U} = (3, 5 / 4, - 8)^{T}$ .

3.1givenstep 2.1algebra∎

The discrete entropy rate is $\dot{E} = U^{T} H \dot{U} = (- 3) (1 / 3) (3) + (- 1) (4 / 3) (5 / 4) + (- 2) (1 / 3) (- 8) = 2 / 3 > 0$ . This exact periodic state contradicts the asserted universal nonincrease, despite the SBP identity; it does not contradict the entropy-stable flux-differencing theorem, which uses a different volume flux.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

Central and upwind fluxes have different periodic dissipation

Example

On two equal periodic interval cells of length $h$ , consider degree-zero DG for $u_{t} + a u_{x} = 0$ with $a > 0$ and cell states $u_{1}, u_{2}$ . Compare the central flux ${\hat{f}}_{c} (a_{0}, b_{0}) = a (a_{0} + b_{0}) / 2$ with the upwind flux ${\hat{f}}_{u} (a_{0}, b_{0}) = a a_{0}$ .

Facts & Assumptions

Given: The degree-zero weak DG balance with a single shared numerical flux at each oriented face and periodic indexing $u_{0} = u_{2}$ , $u_{3} = u_{1}$ .

[F1]

Upwind discontinuous Galerkin discretization of constant transport: the semidiscrete cell balance is $h {\dot{u}}_{j} = - ({\hat{f}}_{j + 1 / 2} - {\hat{f}}_{j - 1 / 2})$ for a degree-zero test function.

Verification

1.1givenF1algebra

For the central flux, both faces of cell 1 see the same pair $(u_{1}, u_{2})$ and hence the same flux; likewise for cell 2. Thus ${\dot{u}}_{1} = {\dot{u}}_{2} = 0$ and $E = \frac{h}{2} (u_{1}^{2} + u_{2}^{2})$ has $\dot{E} = 0$ .

2.1F1step 1.1algebra∎

For the upwind flux, the two cell equations are ${\dot{u}}_{1} = - \frac{a}{h} (u_{1} - u_{2})$ and ${\dot{u}}_{2} = - \frac{a}{h} (u_{2} - u_{1})$ . Therefore $\dot{E} = h (u_{1} {\dot{u}}_{1} + u_{2} {\dot{u}}_{2}) = - a (u_{1} - u_{2})^{2}$ . In particular, at $(u_{1}, u_{2}) = (1, - 1)$ the central rate is zero and the upwind rate is $- 4 a < 0$ .

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

A constant upwind DG state with inflow data on one interval

Example

Take the one-cell interval $K = (0, 1)$ , degree $p = 0$ , speed $a > 0$ , constant cell state $U (t)$ , and inflow value $g (t)$ at $x = 0$ . The outflow trace at $x = 1$ is $U (t)$ .

Facts & Assumptions

Given: The one-dimensional upwind DG weak form on $(0, 1)$ , with $β = a > 0$ , cell length one, and left inflow flux $- a g$ .

[F1]

Upwind discontinuous Galerkin discretization of constant transport: for a degree-zero test function the volume derivative term vanishes, and the two endpoint fluxes enter with their outward signs.

Verification

1.1givenF1algebra

Use the constant test $v_{h} = 1$ in the DG equation. The cell mass is $U$ , the outflow contribution is $a U$ , and the inflow contribution is $- a g$ , so $\dot{U} + a U - a g = 0$ and $\dot{U} = a (g - U)$ .

2.1givenstep 1.1algebra∎

For $E (t) = \frac{1}{2} U (t)^{2}$ , multiply the ODE by $U$ to get $\dot{E} = a U (g - U) = \frac{a}{2} g^{2} - \frac{a}{2} U^{2} - \frac{a}{2} (U - g)^{2}$ . Rearranging displays the outflow energy, mismatch dissipation at inflow, and energy supplied by the boundary datum.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

The SBP inflow SAT produces the boundary energy term

Example

On the interval $[0, 1]$ , use two endpoint nodes, $H = \frac{1}{2} I$ , $D = (\begin{array}{cc} - 1 & 1 \\ - 1 & 1 \end{array})$ , left and right selectors $R_{L} = (1, 0)$ and $R_{R} = (0, 1)$ , and $B = diag (- 1, 1)$ . For $u_{t} + a u_{x} = 0$ with $a > 0$ , inflow value $g (t)$ at $x = 0$ , and upwind outflow at $x = 1$ , the SBP-SAT ODE is $H \dot{U} + a H D U + a e_{L} (U_{L} - g) = 0.$

Facts & Assumptions

Given: The two-node SBP data, the upwind inflow SAT sign, and the discrete energy $E = \frac{1}{2} U^{T} H U$ .

[F1]

Upwind SBP-SAT discretization of constant transport: the inflow SAT in the strong form is $- a e_{L} (g - U_{L}) = a e_{L} (U_{L} - g)$ and the outflow SAT is zero.

[F2]

Upwind SBP-SAT transport satisfies the semidiscrete L2 energy identity: the SBP identity converts the volume derivative term into the signed endpoint quadratic boundary term.

Verification

1.1givenF2algebra

The matrices satisfy $H D = (\begin{array}{cc} - 1 / 2 & 1 / 2 \\ - 1 / 2 & 1 / 2 \end{array})$ and $H D + D^{T} H = B$ , so $U^{T} H D U = \frac{1}{2} (U_{R}^{2} - U_{L}^{2})$ .

2.1givenF1step 1.1algebra∎

Multiply the ODE in [F1] by $U^{T}$ . The inflow penalty contributes $a U_{L} (U_{L} - g)$ , and step 1.1 gives $\dot{E} + \frac{a}{2} U_{R}^{2} + \frac{a}{2} (U_{L} - g)^{2} = \frac{a}{2} g^{2}$ . Thus the penalty has the sign required by the upwind energy identity.