Discontinuous Galerkin fluxes and stability
Join elementwise weak equations through oriented traces and derive a sign-consistent semidiscrete energy balance.
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The question for today
What you’ll understand
- Write the cellwise weak advection equation with correctly oriented boundary fluxes.
- Derive upwind energy dissipation using an explicit jump convention.
- Separate exact-quadrature semidiscrete stability from fully discrete and nonlinear guarantees.
A discontinuous Galerkin approximation uses a polynomial independently on each element. Neighboring traces need not agree, so the numerical flux is the mechanism that couples elements and determines stability. A sign convention is essential: changing the definition of a jump without changing the flux can turn intended dissipation into growth. The scalar periodic advection model gives a complete energy identity that can be checked cell by cell.
Integrate by parts on each cell
For constant a and periodic one-dimensional advection, choose a polynomial trial and test space separately on every cell. Multiply u_t+a u_x=0 by a test polynomial, integrate over one cell and integrate the spatial derivative by parts. The right face contributes numerical flux times the cell’s right test trace, while the left face contributes its negative times the cell’s left test trace. These opposite signs are outward-normal orientation in one dimension, not arbitrary choices of notation.
At an interface let u_minus be the trace from the left element and u_plus the trace from the right. Define the jump as u_plus−u_minus and the average as their half sum. The upwind flux is a times the average minus one half the absolute speed times this jump. For positive a it reduces to a u_minus; for negative a it reduces to a u_plus. The same single flux is inserted into both neighboring weak equations, with their opposite boundary signs, ensuring local conservation.
The jump penalty has a definite energy sign
Take the test function equal to the numerical solution on each element and sum. The integrated volume terms contribute differences of squared traces. Pairing them with the two contributions from each shared flux leaves one half the absolute speed times the squared jump, with a negative sign on the energy derivative. Thus the exact semidiscrete flow dissipates the sum of element L2 energies. For a central flux, the jump penalty is absent and the periodic scalar semidiscrete energy is conserved instead.
The proof does not require continuity of the approximation and does not identify jumps with physical shocks. Even a smooth exact solution can have small discrete jumps because adjacent element polynomials are represented separately. The energy identity also does not say that arbitrary nonlinear systems are stable under the same algebra: nonlinear entropy variables, compatible fluxes and quadrature become relevant. Exact integration, or a verified discrete summation-by-parts substitute, is part of the stated scalar argument.
Mass matrices and time stepping carry their own conditions
In coefficient form, the cellwise weak equations produce a mass matrix times the derivative plus a flux-coupled spatial operator. The natural energy is the mass-matrix norm, not necessarily an unweighted coefficient norm. A change of polynomial basis changes coefficient coordinates without changing the represented L2 energy when mass integration is consistent. Replacing the exact mass matrix by a quadrature approximation requires checking the corresponding discrete energy relation rather than copying the continuous-integration proof unchanged.
A fully discrete method must still satisfy an appropriate time-step and time-integrator stability condition. The p=0 upwind case reduces to a finite-volume method, but its Courant constant cannot be assigned unchanged to all higher polynomial degrees and Runge–Kutta choices. Central-flux semidiscrete advection is skew in the mass norm, while Forward Euler can amplify its nonzero imaginary modes. Therefore an elementwise energy inequality is an important spatial property, not an unconditional guarantee for every implemented space-time DG method.
Work it through
The p=0 method makes the energy loss explicit
Use three periodic cells of width 1/3, piecewise constant values(1,2,0), and speed a=1. Compute the semidiscrete DG energy and its instantaneous derivative under upwind fluxes.
For piecewise constants, the mass norm gives energy E=(h/2)sum U_i²=(1/6)(1+4+0)=5/6. The right-minus-left jumps around the periodic interfaces are 1,−2,1, whose squared sum is six.
The jump formula gives E′=−(1/2)·6=−3. Directly, the upwind semidiscrete derivatives are−3(1−0)=−3,−3(2−1)=−3, and−3(0−2)=6.
Differentiate E directly: E′=h sum U_iU_i′=(1/3)(−3−6+0)=−3. The sum of coefficient derivatives is zero, so the conserved average and dissipating energy are compatible.
Energy is 5/6 and its instantaneous derivative is−3. This verifies the jump sign and normalization without asserting a fully discrete time-step bound.
The living notebook
Explore the model
Interactive lab · function
The upwind penalty is quadratic in the jump
Pause and Reset, then set coefficient a = 0.50 and position x = 1. Interpret x as one interface jump for unit-speed advection: y=0.5 is the magnitude of its energy dissipation contribution. Reset and set both a = 0.50 and x = 2: predict y=2, a fourfold penalty for a doubled jump. The instrument evaluates a quadratic, not DG polynomial traces, a mass matrix or time evolution; the actual energy derivative carries the negative of this penalty.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forWith the correct jump convention, the upwind correction contributes a nonnegative quadratic penalty and a nonpositive energy derivative.
Pause & reason
Check your understanding
Your turn
Try it yourself
For two periodic piecewise constant cells of width 1/2, values(1,−1) and speed a=2, compute the semidiscrete energy, derivative and conserved mean.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
There are two periodic interfaces. Count both jumps, then confirm with the two upwind coefficient derivatives.
Compare with the solution
Energy is . The two jumps are−2 and 2, giving squared sum eight. Therefore E′=−|2|·8/2=−8. The upwind derivatives are−(2/h)(1−(−1))=−8 and−(2/h)(−1−1)=8. Direct differentiation gives h[1·(−8)+(−1)·8]=−8, agreeing with the jump identity. The spatial integral is h(1−1)=0 and its derivative is h(−8+8)=0. A time-discrete scheme must still be chosen to respect the appropriate stability restriction for these rates.
Keep the idea
Take it with you
- Element boundary orientation and jump convention fix the flux signs.
- Upwind traces dissipate the mass-norm energy through squared jumps in the periodic scalar model.
- Quadrature and time integration need their own stability checks beyond the semidiscrete identity.
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