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An acoustic system resolved into two characteristic waves

Example

Consider $U_{t} + A U_{x} = 0$ with $A = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ , base state $\overset{ˉ}{U} = (0, 0)^{T}$ , and Riemann data $U_{L} = (0, 0)^{T}$ , $U_{R} = (- ε, 0)^{T}$ . The eigenvalues are $- 1, 1$ , so the system is strictly hyperbolic and both fields are linearly degenerate. Use the constant fields $r_{-} = (1, - 1)^{T}$ and $r_{+} = (1, 1)^{T}$ . Choose $0 < ε < δ$ and $ε < 2 η$ , where $δ, η > 0$ are the local solver radii at $\overset{ˉ}{U}$ for these fields.

Facts & Assumptions

Given: The acoustic matrix, the two constant states chosen within the local solver radii, and the local solver for linearly-degenerate fields.

[F1]

Strictly hyperbolic conservation systems and characteristic families: characteristic waves have eigenvectors of the flux Jacobian, and a linear constant-coefficient system has constant eigenpairs.

[F2]

Local small-data Riemann solver near a strictly hyperbolic state: sufficiently close data are joined by the ordered local family waves; a linearly-degenerate family produces a contact.

[F3]

Local characteristic rarefaction curve and Mixed local wave curve for a genuinely nonlinear or linearly degenerate family: for a linearly-degenerate field, $Ψ_{i} (σ; U) = R_{i} (σ; U)$ , where $\partial_{σ} R_{i} = r_{i} (R_{i})$ and $R_{i} (0; U) = U$ .

Verification

1.1givenF1algebra

Direct multiplication gives eigenpairs $(- 1, r_{-} = (1, - 1)^{T})$ and $(1, r_{+} = (1, 1)^{T})$ . Since the system is linear, its two characteristic fields are linearly degenerate. The jump is $U_{R} - U_{L} = (- ε, 0)^{T} = - \frac{ε}{2} r_{-} - \frac{ε}{2} r_{+}$ .

2.1givenstep 1.1F2F3algebra

Since the chosen fields are constant, the initial-value equations in [F3] give $Ψ_{\pm} (σ; U) = U + σ r_{\pm}$ . Thus the coefficients $σ_{-} = σ_{+} = - ε / 2$ in step 1.1 are the solver's strengths for this normalization. They satisfy $∣ σ_{\pm} ∣ < η$ , and both data states lie in the prescribed radius- $δ$ neighborhood. After the first wave the intermediate state is $U_{M} = U_{L} - \frac{ε}{2} r_{-} = (- \frac{ε}{2}, \frac{ε}{2})^{T}$ ; after the second, $U_{M} - \frac{ε}{2} r_{+} = U_{R}$ . By uniqueness in [F2], these are the solver's ordered contacts. For the first jump $A (U_{M} - U_{L}) = (- 1) (U_{M} - U_{L})$ , and for the second $A (U_{R} - U_{M}) = (U_{R} - U_{M})$ , verifying Rankine–Hugoniot at the ordered speeds $- 1$ and $1$ .

3.1step 2.1F2∎

The profile equal to $U_{L}$ for $x / t < - 1$ , to $U_{M}$ for $- 1 < x / t < 1$ , and to $U_{R}$ for $x / t > 1$ is therefore the two-contact local Riemann solution.

Diagram

The horizontal axis is $x$ and the vertical axis is $t$ . Each ray from the origin has slope $d x / d t = λ$ ; the two constant wedges carry $U_{L}, U_{M}, U_{R}$ . This picture records the wave ordering only; the algebraic Rankine–Hugoniot checks above establish the solution.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

16 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Randall J. LeVeque, Numerical Methods for Conservation Laws (standard reference, not scraped)