advanced34 min studyLesson 414 in Mathematics

First-order symmetric hyperbolic systems

Construct a positive energy from the system matrices and diagonalize a linear acoustic example into counterpropagating fields.

On this page

What you’ll understand

  • Identify positive time matrices and symmetric spatial matrices.
  • Derive the energy balance for a constant-coefficient symmetric system.
  • Compute acoustic characteristic variables and their propagation directions.

A vector-valued evolution can carry several wave families at once. Symmetric hyperbolic structure provides a robust way to control those families without assuming that one scalar wave formula applies componentwise. A positive time matrix defines an energy norm, and symmetric spatial matrices turn spatial derivatives into boundary fluxes. Linear acoustics supplies an explicit example in which pressure and velocity mix into two independently transported characteristic variables, with their own speeds and energy interpretation.

Positive energy is the essential structural resource

Consider A0Ut+∑jAjUxj=0A^0U_t+\sum_jA^jU_{x_j}=0, with real constant symmetric matrices and A0A^0 positive definite. The quadratic energy density is one half of UTA0UU^TA^0U. Positivity bounds the ordinary squared norm from above and below by fixed constants, making this an actual control of the state rather than a signed quantity that can cancel. Multiplication by UTU^T and symmetry give a local flux identity. No differentiation of eigenvectors is needed for this estimate.

On a periodic domain, or on the whole space with adequate decay, boundary fluxes cancel and total energy is constant. For smooth variable coefficients, derivatives of the matrices generate lower-order energy terms. Uniform positive definiteness and suitable bounded coefficient derivatives then support Grönwall-type estimates on a finite interval. Without uniform positivity, an energy may fail to control some state directions. Boundary-value problems also require conditions on incoming fields rather than arbitrary independent prescriptions for every component.

e=12UTA0U,Fj=12UTAjU,∂te+∑j∂xjFj=0e=\frac12U^TA^0U,\qquad F_j=\frac12U^TA^jU,\qquad\partial_te+\sum_j\partial_{x_j}F_j=0
This exact balance assumes constant symmetric matrices and no forcing or lower-order terms. Positive definiteness of A⁰ makes e comparable to |U|².↔ Scroll wide equations horizontally.

Directional eigenvalues are real in the energy metric

For a spatial direction, the propagation matrix is the inverse time matrix times the corresponding spatial matrix. It need not be symmetric in the ordinary Euclidean inner product, but it is similar to a real symmetric matrix obtained using the positive square root of the time matrix. Consequently its eigenvalues are real and it has a well-behaved characteristic basis relative to the energy metric. A uniform coefficient structure is what connects this algebra to stable evolution estimates.

A finite speed bound follows when those directional speeds are uniformly bounded over unit spatial directions and the region under consideration. Symmetric structure therefore supports a cone argument much like the scalar wave argument, with the appropriate maximal speed and lower-order estimates. It does not say that every component travels at the same speed or that an arbitrary coordinate norm is conserved. The energy norm and the boundary flux reveal which combinations are controlled and where information enters or leaves.

(A0)−1A(ξ)∼(A0)−1/2A(ξ)(A0)−1/2,A(ξ)=∑jξjAj(A^0)^{-1}A(\xi)\sim(A^0)^{-1/2}A(\xi)(A^0)^{-1/2},\qquad A(\xi)=\sum_j\xi_jA^j
The matrix on the right is symmetric, proving real directional speeds for the positive symmetric system. Uniform bounds matter for a common finite-speed estimate.↔ Scroll wide equations horizontally.

Pressure and velocity combine into acoustic invariants

For one-dimensional linear acoustics, let K be a positive bulk modulus and rho a positive density. The equations couple pressure rate to velocity gradient and momentum rate to pressure gradient. Dividing the pressure equation by K gives a symmetric form with time matrix diag(1/K,rho) and off-diagonal spatial matrix entries one. The energy is a sum of pressure and velocity contributions with their physical weights. The sound speed and impedance then emerge from the same coefficients.

Adding or subtracting impedance times velocity from pressure produces two characteristic variables. One moves right at the sound speed and the other left at that speed. A localized pressure perturbation with zero velocity generally excites both families, whereas a suitable relation between pressure and velocity can isolate one. This is a linear constant-medium decomposition; nonlinear gas dynamics or spatially changing media require different equations and may couple, reflect, or steepen those families.

pt+Kvx=0,ρvt+px=0,c=K/ρ,Z=ρc,(p±Zv)t±c(p±Zv)x=0p_t+Kv_x=0,\quad\rho v_t+p_x=0,\quad c=\sqrt{K/\rho},\quad Z=\rho c,\qquad (p\pm Zv)_t\pm c(p\pm Zv)_x=0
The plus variable travels right and the minus variable left. The signs connect each characteristic combination to its own transport equation.↔ Scroll wide equations horizontally.

Decomposing a mixed acoustic state

Take K=8 and rho=2. For a pressure p=3 and velocity v=0.25, compute the two characteristic variables and the pressure contributions of the right- and left-moving fields.

  1. The sound speed is √(8/2)=2 and impedance is rho·c=4. Thus w₊=p+Zv=4 and w₋=p−Zv=2.

  2. Recover pressure as (w₊+w₋)/2. The right-moving contribution is w₊/2=2 and the left-moving contribution is w₋/2=1.

  3. Their velocity contributions are 2/4=0.5 and −1/4=−0.25. They sum to the given velocity 0.25, while their pressures sum to three.

c=2,Z=4,w+=4,w−=2,(p,v)=(2,0.5)+(1,−0.25)c=2,\quad Z=4,\quad w_+=4,\quad w_-=2,\qquad(p,v)=(2,0.5)+(1,-0.25)
The first component pair is right-moving and the second left-moving in this constant acoustic medium.↔ Scroll wide equations horizontally.
The result

Both wave families are present, with unequal pressure amplitudes. The opposite signs in their velocity contributions determine the direction of travel.

Explore the model

Interactive lab

Interactive lab · vectors

State coordinates versus characteristic coordinates

Pause and Reset, then set angle 0° and magnitude 1. Regard the coordinates as nondimensional (p/Z,v). Decompose this state into r₊=(1,1) and r₋=(−1,1) on paper. For a fresh comparison, Reset and set angle 90° and magnitude 1, then repeat the decomposition.

State coordinates versus characteristic coordinatesPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. 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 forThe first state has coefficients (α₊,α₋)=(0.5,−0.5); the second has (0.5,0.5). The instrument shows coordinate projections, not transported acoustic fields; the characteristic-basis calculation is the learner's separate step.

Check your understanding

Which matrix property makes the symmetric-system quadratic density control the whole state?

Try it yourself

For K=18, rho=2, p=0, and v=1, find c, Z, the two characteristic variables, their pressure contributions, and the acoustic energy density.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use p=(w₊+w₋)/2 and v=(w₊−w₋)/(2Z). The density is p²/(2K)+rho·v²/2.

Compare with the solution

The speed is three and impedance six. Therefore w₊=6 and w₋=−6. Their pressure contributions are three and minus three, canceling to zero, while their velocity contributions are each one half, summing to one. The acoustic energy density is 0+2·1²/2=1. The negative left-moving pressure amplitude is a signed perturbation, not a negative energy.

Take it with you

  • Positive symmetric structure converts a coupled evolution into a controlled energy balance.
  • Directional speeds are real because the propagation matrix is similar to a symmetric matrix.
  • Acoustic characteristic variables mix pressure and velocity and travel in opposite directions.

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