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A planar discontinuity and the space--time normal form of Rankine--Hugoniot

Example

Let n≥2, T>0, f∈C1(R;Rn), c∈R, a unit vector η∈Rn, and distinct states uL,uR∈R. Prescribe the Riemann datum u0(x)=uL for x⋅η<0 and u0(x)=uR for x⋅η>0, and for 0<t<T set u(t,x)=uL when x⋅η<ct and u(t,x)=uR when x⋅η>ct. This bounded piecewise-constant function has strong local L1 initial trace u0 and is a distributional weak solution of ut+div⁡xf(u)=0 exactly when (f(uR)−f(uL))⋅η=c (uR−uL). The plane interface Γ={(t,x):x⋅η=ct} has unit normal from the left side to the right side ν=(−c,η)/1+c2; hence the equivalent space--time normal equation is [u]νt+[f(u)]⋅νx=−c (uR−uL)+(f(uR)−f(uL))⋅η1+c2=0, where [u]=uR−uL and [f(u)]=f(uR)−f(uL). This direct plane calculation uses no division by the jump.

Facts & Assumptions

Given: n≥2, T>0, f∈C1(R;Rn), a unit vector η∈Rn, c∈R, distinct uL,uR, the piecewise constant function u above, and a test function φ∈Cc∞(Rn×(−∞,T)).

[F1]

The Cauchy weak identity is the integral over 0<t<T with its initial term (Distributional weak solutions of the Cauchy problem). For this profile the interior equation and the strong trace established in step 1.1 give that identity by a time cutoff and passage to t=0 (Scalar conservation laws, fluxes and Cauchy data).

[F3]

For every point of an open set and every neighbourhood of it there is a nonnegative smooth compactly supported test function, supported in that neighbourhood and positive at the point: take a finite product of rescaled translated copies of the bump of Explicit compactly supported smooth cutoffs.

Proof

technique · direct
1.1F1given

The initial trace. For a compact K⊆Rn and 0<t<T, the two definitions of u(t,x) and u0(x) differ exactly on {x∈K:x⋅η lies strictly between 0 and ct}, a set of measure at most CK∣t∣; hence ∫K∣u(t,x)−u0(x)∣ dx≤∣uR−uL∣ CK∣t∣→0 as t↓0. So u has the strong local L1 trace u0.

1.2F1F2given

The interface computation. Choose k with ηk≠0, and put γ(t,x′)=(ct−∑j≠kηjxj)/ηk. For a test supported in t>0, integrate first in xk on each side of xk=γ. FTC and the moving-endpoint rule give the weak pairing 1∣ηk∣∫Rn−1×(0,T)φ(t,x′,γ(t,x′))(c[u]−[f]⋅η) dx′ dt. For ηk>0 the lower side is the left state; for ηk<0 the lower side is the right state, which reverses the jump and converts ηk to ∣ηk∣. The endpoint derivatives are γt=c/ηk and γxj=−ηj/ηk.

2.1F1F2step 1.1step 1.2

The initial boundary. For a test meeting t=0, perform the same integration on δ<t<T. The bottom term is −∫u(δ,x)φ(δ,x) dx. By step 1.1 it converges to −∫u0φ(0,x) dx, cancelling the prescribed initial term. Thus the full Cauchy residual is the interface integral of step 1.2 over 0<t<T.

3.1F3step 2.1

Necessity. If c(uR−uL)−(f(uR)−f(uL))⋅η≠0, then it is nonzero on a small interface patch; by [F3] there is a nonnegative smooth compactly supported test function supported in a small space--time neighbourhood of a point of that patch and positive on the patch, and step 2.1 makes the weak residual for this test nonzero, contradicting the weak identity.

4.1step 1.1step 3.1F1∎

Sufficiency and normal form. Conversely, if (f(uR)−f(uL))⋅η=c(uR−uL), the interface bracket vanishes identically and step 2.1 shows that the weak identity holds for every test function; the trace was verified in step 1.1, so u is a distributional weak solution. Since η is a unit vector, ∇t,x(x⋅η−ct)=(−c,η) has length 1+c2, so the unit normal from the minus side is ν=(−c,η)/1+c2 and the condition becomes [u]νt+[f(u)]⋅νx=(−c[u]+[f(u)]⋅η)/1+c2=0. No division by the jump was used at any point.

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