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A Riemann shock and a rarefaction of the scalar model: explicit weak solutions, entropy coefficients and initial trace

Physical scope

A worked scalar-model example of shock and rarefaction weak solutions: the Riemann data 2 and 1 with the entropy coefficients of the shock computed on the interval (1,2), and the rarefaction fan of the data 1 and 2 checked for continuity, the classical equation inside the fan and convergence of the initial trace. The scalar conservation law u_t + (u^2/2)_x = 0 is a dimensionless idealized one-dimensional model; its scalar variable is not identified with a particular physical density or velocity. The example asserts no derivation from three-dimensional fluid mechanics and reports no measurement.

Example

Assume ACω (The Axiom of Countable Choice (ACω)) for the cited Lebesgue weak/entropy formulation.

Let F(u)=12u2 and use the weak and entropy formulations of Weak and entropy solutions of the scalar conservation law ut+(u2/2)x=0, with the scalar field u read as a dimensionless model variable. The coordinates t,x, the states uL,uR and flux F(u) are all dimensionless; no assignment to a measured density, velocity, length or time is made. Then:

  1. (Shock.) For the Riemann data uL=2, uR=1 the piecewise constant field u(t,x)=2 for x<32t and u(t,x)=1 for x>32t (t>0), taking the right-state value on the shock line, is a weak solution of ut+(F(u))x=0: its interface residual is −s[u]+[F(u)]=−32⋅(1−2)+12(12−22)=32−32=0 with s=32, so the weak identity (W) is satisfied for every test function by the split integration over the two half-planes.
  2. (Entropy of the shock.) For every k∈(1,2) the interface coefficient of the pair (ηk,qk)=(∣u−k∣,sign⁡(u−k)(u2−k2)/2) is s[ηk]−[qk]=−(k−2)(k−1)=(2−k)(k−1)>0, so the shock satisfies the entropy inequality with strictly positive interface production on this range of k; the integrated inequality is strict only for nonnegative tests with ∫0∞φ(t,st)dt>0; for k∉[1,2] the coefficient vanishes and the inequality holds with equality.
  3. (Rarefaction.) For the Riemann data uL=1, uR=2 the field u(t,x)=1 for x≤t, u(t,x)=x/t for t<x<2t and u(t,x)=2 for x≥2t (t>0) is continuous on (0,∞)×R, satisfies ut+(F(u))x=0 classically inside the fan, and has initial trace f: ∫R∣u(t,x)−f(x)∣ dx≤32t→0 as t↓0.
  4. (Weak kinks are unavoidable here.) The shock profile is discontinuous across x=32t and the rarefaction is not differentiable across x=t and x=2t; the classical derivation of the transport and balance identities of the continuum page does not apply at such sets, which is exactly why the weak formulation of the model is used. The admissibility of the shock and the equality case of the rarefaction are those of Riemann shock and rarefaction waves for ut+(u2/2)x=0 are weak solutions, and the shock is an entropy solution.

Facts & Assumptions

Given: Countable choice. the flux F(u)=u2/2; the Riemann data uL=2,uR=1 and uL=1,uR=2; the two explicit fields.

[F1]

Weak and entropy solutions (Weak and entropy solutions of the scalar conservation law ut+(u2/2)x=0): the weak identity (W) and the entropy inequality (E) for convex entropy pairs with q′=uη′, and the family (ηk,qk) with ηk(u)=∣u−k∣, qk(u)=sign⁡(u−k)(u2−k2)/2.

[F2]

Riemann theorem (Riemann shock and rarefaction waves for ut+(u2/2)x=0 are weak solutions, and the shock is an entropy solution): for uL>uR the shock with speed s=(uL+uR)/2 is a weak solution and satisfies every entropy inequality of the family (ηk,qk) with coefficient s[ηk]−[qk]=−(k−uL)(k−uR)≥0 for uR≤k≤uL, and coefficient 0 outside that interval; for uL<uR the rarefaction fan is continuous, is a weak solution and satisfies the entropy inequalities with equality.

[F3]

Integration (Integral over a measurable subset, The Lebesgue integral is linear on L1(μ)): the integral of a bounded measurable function over a set of finite measure, and additivity over the regions of the piecewise description.

Verification

technique · direct; compute the Rankine-Hugoniot residual and the entropy coefficients for the chosen data, and check the fan solution pointwise and its initial trace by explicit integration
1.1givenF1F2algebra

Shock residual: with uL=2,uR=1 the speed is s=(uL+uR)/2=3/2; [u]=uR−uL=−1 and [F(u)]=(uR2−uL2)/2=(1−4)/2=−3/2, so −s[u]+[F(u)]=3/2−3/2=0; this is the interface coefficient of the weak identity (W) when the field is integrated piecewise, so (W) holds for every test function.

1.2givenF1F2algebra

Rarefaction equation and continuity: inside the fan u=x/t gives ut=−x/t2, ux=1/t and u ux=x/t2, so ut+uux=0; at x=t the value is 1=uL and at x=2t it is 2=uR, matching the constant pieces, so u is continuous on the open upper half-plane.

2.1step 1.1F1F2algebra

Shock entropy: by [F2], for k∈[1,2] the coefficient is s[ηk]−[qk]=−(k−2)(k−1)=(2−k)(k−1)≥0, strictly positive for 1<k<2 because k−2<0 and k−1>0, and zero at k=1,2. For k<1 both states exceed k, so [ηk]=[u]=−1 and [qk]=[F]=−3/2, giving s[ηk]−[qk]=(3/2)(−1)−(−3/2)=0. For k>2 both states are below k, so [ηk]=−[u]=1 and [qk]=−[F]=3/2, again giving zero. Thus (E) holds for every nonnegative test. For 1<k<2 it is strict exactly when the test has positive integral on the shock line; zero or off-interface tests give equality, and for k∉[1,2] every test gives equality.

2.2step 1.2F2F3algebra

Initial trace: for 0<t one has u(t,x)≠f(x) only on the set where the fan and the constants differ, and ∫∣u−f∣≤∫0t∣1−2∣ dx+∫t2t∣x/t−2∣ dx=t+∫t2t(2−x/t) dx=t+t/2=3t/2→0 as t↓0, by additivity and evaluation of the elementary integrals [F3]; moreover the entropy inequality of the fan holds with equality by [F2].

3.1step 1.1step 1.2F1∎

Regularity reading: the shock is discontinuous and the fan fails to be differentiable across x=t and x=2t; hence the classical smooth-motion hypotheses of the balance and transport items of this page are not met on these sets, and the weak formulation of [F1] is the correct setting for these solutions. This is a statement about the models, not an empirical claim.

Remarks

  • Physical reading. The dimensionless scalar field u is a model variable of an idealized one-dimensional transport problem; it is not claimed to be the velocity or density of a three-dimensional compressible fluid, and no derivation of the scalar equation from the momentum and mass balances of this page is asserted. The example shows that a scalar continuum model can have nondifferentiable weak solutions whose admissibility is controlled by entropy inequalities, in contrast with the smooth motions required by the transport and balance theorems of this page.

  • Admissibility is not automatic. Both the shock and the fan are weak solutions; only the entropy inequalities select the physically admissible one among weak solutions of this model family. The computed coefficients show the shock is admissible for the stated data and the fan is the equality case. No statement is made about entropy solutions for general initial data or for systems of conservation laws.

Depends on

Dependency tree · 2 levels

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