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Distributional weak solutions of the Cauchy problem are not unique

Statement

Take $n = 1$ , $f (u) = u^{2}$ and $u_{0} \equiv 0$ . For every $δ > 0$ define, for $t > 0$ , $u_{δ} (t, x) = {\begin{cases} 0, & x < - δ t, \\ - δ, & - δ t < x < 0, \\ + δ, & 0 < x < δ t, \\ 0, & x > δ t . \end{cases}$ This is a bounded distributional weak solution of $u_{t} + (u^{2})_{x} = 0$ on $Π_{T}$ with strong local $L^{1}$ initial trace $u_{0} = 0$ , and it is not identically zero. At its three jumps the speeds $s = [u^{2}] / [u]$ are, respectively, $- δ$ , $0$ and $+ δ$ , so each jump coefficient $[u^{2}] - s [u]$ in the weak equation vanishes. The middle stationary jump violates the Kruzhkov entropy inequality for $k = 0$ : with $η_{0} (u) = ∣ u ∣$ and $q_{0} (u) = sgn (u) u^{2}$ its entropy production is $[q_{0}] - 0 \cdot [η_{0}] = δ^{2} - (- δ^{2}) = 2 δ^{2} > 0$ , whereas the entropy inequality requires this coefficient to be nonpositive (Kruzhkov entropy solutions). Hence $u_{δ}$ is a weak solution but not an entropy solution, and the zero solution is a distinct weak solution with the same initial data: distributional weak solutions are not unique (Distributional weak solutions of the Cauchy problem, Piecewise smooth shocks and one-sided traces).

Facts & Assumptions

Given: $n = 1$ , $f (u) = u^{2}$ , $u_{0} \equiv 0$ , $T > 0$ , $δ > 0$ , and the piecewise constant function $u_{δ}$ above, whose jump rays are $Γ_{-} = {x = - δ t}$ , $Γ_{0} = {x = 0}$ and $Γ_{+} = {x = δ t}$ in $Π_{T}$ .

[F1]

On each of the four regions the function is constant and $f$ is smooth, so $u_{δ}$ solves the equation classically there; across a jump ray $x = s (t)$ of a piecewise $C^{1}$ weak solution of $u_{t} + (u^{2})_{x} = 0$ , the distributional identity holds iff the jump coefficient $[u^{2}] - s [u]$ vanishes, where $[\cdot]$ denotes the right minus left trace across the ray (Piecewise smooth shocks and one-sided traces, Distributional weak solutions of the Cauchy problem, Scalar conservation laws, fluxes and Cauchy data).

[F2]

The Kruzhkov entropy pair for $k = 0$ is $η_{0} (u) = ∣ u ∣$ , $q_{0} (u) = sgn (u) u^{2}$ with $sgn (0) = 0$ ; an entropy solution must satisfy $\partial_{t} η_{0} (u) + \partial_{x} q_{0} (u) \leq 0$ in $D^{'} (Π_{T})$ , so across a jump ray the entropy production coefficient $[q_{0}] - s [η_{0}]$ must be nonpositive (Kruzhkov entropy solutions).

[F3]

Basic computation with the explicit states and speeds: for the ray $x = - δ t$ the left state is $0$ , the right state is $- δ$ , and the rightward speed is $s = - δ$ ; for $x = 0$ the states are $- δ$ (left) and $+ δ$ (right) with $s = 0$ ; for $x = δ t$ the states are $+ δ$ (left) and $0$ (right) with $s = δ$ ; directly $[u^{2}] - s [u] = 0$ in all three cases.

Proof

technique · direct

1.1F1F3

Weak equation. The profile is constant on its four regions. At $x = - δ t$ the right-minus-left jumps are $[u] = - δ$ , $[u^{2}] = δ^{2}$ , and $s = - δ$ , so $[u^{2}] - s [u] = δ^{2} - (- δ) (- δ) = 0$ . At $x = 0$ , $[u] = 2 δ$ , $[u^{2}] = 0$ , and $s = 0$ . At $x = δ t$ , $[u] = - δ$ , $[u^{2}] = - δ^{2}$ , and $s = δ$ , again giving zero. To verify the distributional equation, integrate $u φ_{t} + u^{2} φ_{x}$ in each region using the moving-endpoint FTC formula: each interface contributes $(s [u] - [u^{2}]) \int φ (t, s t) d t$ , which vanishes.

1.2F1given

Initial trace. For every compact $K \subseteq R$ and $0 < t < δ_{0}$ , the set where $u_{δ} (t, \cdot)$ differs from $0$ is contained in $[- δ t, δ t]$ , so $\int_{K} ∣ u_{δ} (t, x) ∣ d x \leq 2 δ^{2} t \to 0$ as $t ↓ 0$ ; hence $u_{δ}$ has the strong local $L^{1}$ initial trace $0$ . The function is bounded, hence a distributional weak solution of the Cauchy problem with datum $u_{0} \equiv 0$ in the sense of [F1].

2.1F2step 1.2

Failure of the entropy condition. At the middle ray $x = 0$ the left and right states are $- δ$ and $+ δ$ . The entropy production coefficient is $[q_{0}] - s [η_{0}]$ with $[η_{0}] = ∣ + δ ∣ - ∣ - δ ∣ = 0$ , $s = 0$ , and $[q_{0}] = q_{0} (+ δ) - q_{0} (- δ) = δ^{2} - (- δ^{2}) = 2 δ^{2} > 0$ . By [F2] the required entropy inequality fails: the distribution $\partial_{t} η_{0} (u_{δ}) + \partial_{x} q_{0} (u_{δ})$ carries the positive coefficient $2 δ^{2}$ on the ray $x = 0$ .

3.1step 1.1step 1.2step 2.1∎

Non-uniqueness. By steps 1.1 and 1.2, both $u_{δ}$ and the zero function are bounded distributional weak solutions of the same Cauchy problem with initial datum $u_{0} \equiv 0$ ; they differ on a set of positive measure for every $δ > 0$ . By step 2.1, $u_{δ}$ is not a Kruzhkov entropy solution, so the non-uniqueness occurs strictly within the class of distributional weak solutions.

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Sources

G. A. Chechkin and A. Yu. Goritsky (translated by B. Andreianov), “S. N. Kruzhkov’s lectures on first-order quasilinear PDEs,” in Analytical and Numerical Aspects of PDEs, de Gruyter 2009, complete lecture-notes text (standard reference, not scraped)
Victor Ivrii, Partial Differential Equations, University of Toronto, current complete 415-page PDF (standard reference, not scraped)
S. N. Kruzhkov, “First order quasilinear equations in several independent variables,” Mat. USSR-Sbornik 10 (1970), 217–243, complete English translation (standard reference, not scraped)