The conservation of a physical quantity — mass, momentum, energy — in the absence of sources takes the form $u_t + f(u)_x = 0$, where $u(x,t)$ is the conserved density and $f(u)$ is the flux. This equation is deceptively simple. For smooth initial data it has a smooth solution locally in time via the [method of characteristics](/page/Method%20of%20Characteristics). But because characteristics for a nonlinear flux $f$ travel at different speeds, they inevitably cross in finite time, after which no classical solution can exist. The solution must be continued in a weaker sense — one that admits discontinuities. The challenge is not merely existence: there are multiple weak solutions once discontinuities are allowed, and most of them are unphysical. The entropy condition is the selection principle that singles out the unique physically correct solution. It originates from thermodynamics — entropy must increase across a shock in a gas — but has a clean mathematical formulation that applies to any scalar conservation law: the correct discontinuities are those into which characteristics run from both sides, not those from which they emanate. ## The Conservation Law and Its Characteristics The **scalar conservation law** in one spatial dimension is the initial value problem: \begin{align*} u_t + f(u)_x &= 0, \quad (x,t) \in \mathbb{R} \times (0,\infty), \\ u(x,0) &= g(x), \quad x \in \mathbb{R}, \end{align*} where $f \in C^2(\mathbb{R})$ is the flux function and $g \in L^\infty(\mathbb{R})$ is the initial datum. Since $f(u)_x = f'(u) u_x$, the PDE is equivalent to the quasilinear equation $u_t + f'(u) u_x = 0$, which is a first-order PDE of the form $F(\nabla u, u, (x,t)) = 0$ with $F(a, b, c) = a_t + f'(b)\, a_x$. Applying the characteristic system from the [Method of Characteristics](/pages/1019) page with the identification $F(a,b,c) = a_t + f'(b) a_x$, the characteristic velocity is $\nabla_a F = (f'(b), 1)$, so along each characteristic: \begin{align*} \dot X(s) = f'(U(s)), \quad \dot t(s) = 1, \quad \dot U(s) = F(T(s)) = 0. \end{align*} The value $U(s) = g(x_0)$ is **constant** along the characteristic from each starting point $x_0$. Since $\dot X = f'(g(x_0))$ is also constant, the spatial characteristics are the straight lines: \begin{align*} x(t) = x_0 + t \cdot f'(g(x_0)). \end{align*} Two characteristics from starting points $x_0 < x_1$ cross at time: \begin{align*} t_{\mathrm{cross}} = \frac{x_1 - x_0}{f'(g(x_0)) - f'(g(x_1))}, \end{align*} which is positive whenever $f'(g(x_0)) > f'(g(x_1))$, i.e., whenever the map $x_0 \mapsto f'(g(x_0))$ is not non-decreasing. At a crossing, the formula $u = g(x_0)$ would assign two different values to the same physical point — a contradiction. The classical solution must break down before $t_{\mathrm{cross}}$. For the Burgers flux $f(u) = u^2/2$, the crossing time is $T^* = -1/\min g'$ whenever $g' < 0$ somewhere. ## Weak Solutions Since classical solutions cannot persist globally, we extend the solution concept. Multiplying the PDE by a test function $\phi \in C_c^\infty(\mathbb{R} \times [0,\infty))$ and integrating by parts formally: \begin{align*} 0 = \int_0^\infty \int_{-\infty}^\infty (u_t + f(u)_x)\phi \,d\mathcal{L}^1(x)\,d\mathcal{L}^1(t) = -\int_0^\infty \int_{-\infty}^\infty (u\phi_t + f(u)\phi_x)\,d\mathcal{L}^1(x)\,d\mathcal{L}^1(t) - \int_{-\infty}^\infty g(x)\phi(x,0)\,d\mathcal{L}^1(x). \end{align*} The right-hand side makes sense for any bounded measurable $u$ without any [differentiability](/page/Derivative). This motivates the following definition. [definition: Weak Solution of a Conservation Law] A function $u \in L^\infty(\mathbb{R} \times (0,\infty))$ is a **weak solution** of the scalar conservation law with initial data $g \in L^\infty(\mathbb{R})$ if for every $\phi \in C_c^\infty(\mathbb{R} \times [0,\infty))$: \begin{align*} \int_0^\infty \int_{-\infty}^\infty \bigl(u\,\phi_t + f(u)\,\phi_x\bigr)\,d\mathcal{L}^1(x)\,d\mathcal{L}^1(t) + \int_{-\infty}^\infty g(x)\,\phi(x,0)\,d\mathcal{L}^1(x) = 0. \end{align*} [/definition] Every $C^1$ solution is a weak solution ([integration](/page/Integral) by parts is reversible when $u$ is smooth). Conversely, if $u$ is a weak solution that happens to be $C^1$, then it satisfies $u_t + f(u)_x = 0$ pointwise. ## Rankine-Hugoniot Condition When a weak solution is piecewise smooth — smooth on each side of a curve $x = \xi(t)$ — the constraint imposed by the weak formulation at the discontinuity can be made explicit. This is the Rankine-Hugoniot condition. [quotetheorem:232] [remark: Interpretation of Rankine-Hugoniot] Writing $[v]_\xi = v^+ - v^-$ for the jump across the shock, the Rankine-Hugoniot condition reads: \begin{align*} \dot\xi = \frac{[f(u)]_\xi}{[u]_\xi} = \frac{f(u^+) - f(u^-)}{u^+ - u^-}. \end{align*} The shock speed equals the slope of the chord joining $(u^-, f(u^-))$ and $(u^+, f(u^+))$ on the graph of $f$. For the Burgers flux $f(u) = u^2/2 = u \cdot u/2$: \begin{align*} \dot\xi = \frac{(u^+)^2/2 - (u^-)^2/2}{u^+ - u^-} = \frac{u^+ + u^-}{2}, \end{align*} the shock travels at the average of the two side values. [/remark] ## The Riemann Problem The **Riemann problem** is the initial value problem with piecewise constant data: \begin{align*} g: \mathbb{R} &\to \mathbb{R} \\ x &\mapsto \begin{cases} u_L & x < 0, \\ u_R & x > 0. \end{cases} \end{align*} This is the fundamental building block of the theory: every initial value problem can be approximated by a [sequence](/page/Sequence) of Riemann problems, and the solution to each piece is explicit. We treat the case of a strictly convex flux $f'' > 0$ throughout. ### Shock solution: $u_L > u_R$ When $u_L > u_R$ and $f'' > 0$, the characteristic from $x_0 < 0$ travels at speed $f'(u_L) > f'(u_R)$, which is faster than the characteristic from $x_0 > 0$. The two families immediately collide at $t = 0^+$, and the solution for all $t > 0$ is a single shock at $x = \xi(t)$: \begin{align*} u: \mathbb{R} \times (0,\infty) &\to \mathbb{R} \\ (x,t) &\mapsto \begin{cases} u_L & x < \xi(t), \\ u_R & x > \xi(t), \end{cases} \end{align*} where $\dot\xi = (f(u_R) - f(u_L))/(u_R - u_L)$ by the Rankine-Hugoniot condition. Since $\dot\xi$ is constant, the shock moves at constant speed and $\xi(t) = \dot\xi \cdot t$. ### Rarefaction fan: $u_L < u_R$ When $u_L < u_R$ and $f'' > 0$, the characteristics from $x_0 < 0$ (speed $f'(u_L)$) are slower than those from $x_0 > 0$ (speed $f'(u_R)$). A void opens between the two families of characteristics, and no shock appears. Instead, the solution fills the void with a [continuous](/page/Continuity) **rarefaction fan**: a family of characteristics from the origin with all speeds $c \in [f'(u_L), f'(u_R)]$. The solution is self-similar — it depends only on the ratio $x/t$ — and is given by: \begin{align*} u: \mathbb{R} \times (0,\infty) &\to \mathbb{R} \\ (x,t) &\mapsto \begin{cases} u_L & x < f'(u_L)\,t, \\ (f')^{-1}(x/t) & f'(u_L)\,t \leq x \leq f'(u_R)\,t, \\ u_R & x > f'(u_R)\,t. \end{cases} \end{align*} Here $(f')^{-1}$ is well-defined because $f'' > 0$ makes $f'$ strictly increasing. The formula $u = (f')^{-1}(x/t)$ says that on the characteristic from the origin with slope $x/t$ (i.e., with speed $c = x/t$), the value is the state $u$ for which $f'(u) = c$. [example: Riemann Problem for Burgers — Shock] Take $f(u) = u^2/2$, $u_L = 1$, $u_R = 0$. The Rankine-Hugoniot speed is $\dot\xi = (u_L + u_R)/2 = 1/2$. The weak solution is: \begin{align*} u(x,t) = \begin{cases} 1 & x < t/2, \\ 0 & x > t/2. \end{cases} \end{align*} Left characteristics travel at speed $f'(1) = 1$ and right characteristics at speed $f'(0) = 0$. Both families run into the shock from their respective sides — characteristics are absorbed by the shock. The diagram below shows the characteristic structure. {width=72%} Interpreting the diagram: blue lines are characteristics. Those from the left (speed 1) and those from the right (speed 0) both terminate on the red shock line $x = t/2$. No characteristic crosses the shock and continues — they are absorbed. [/example] [example: Riemann Problem for Burgers — Rarefaction] Take $f(u) = u^2/2$, $u_L = 0$, $u_R = 1$. Since $u_L < u_R$ and $f'' = 1 > 0$, the solution is a rarefaction fan. Since $f'(u) = u$, we have $(f')^{-1}(\xi) = \xi$. The solution is: \begin{align*} u(x,t) = \begin{cases} 0 & x \leq 0, \\ x/t & 0 < x < t, \\ 1 & x \geq t. \end{cases} \end{align*} The fan consists of all characteristics from the origin with speed $c \in [0,1]$, i.e., the lines $x = ct$. On each such line, $u = c = x/t$. {width=72%} The shaded region in the diagram is the fan. Left characteristics (vertical, speed 0) fill the region $x < 0$, right characteristics (slope 1) fill $x > t$, and the fan fills the intermediate cone. **Verification.** In the fan region $0 < x < t$: $u_t = -x/t^2$ and $u_x = 1/t$. Then $u_t + u\,u_x = -x/t^2 + (x/t)(1/t) = 0$. The [boundary](/page/Boundary) conditions at $x = 0$ and $x = t$ match continuously with $u_L = 0$ and $u_R = 1$. [/example] ## Non-Uniqueness of Weak Solutions The Rankine-Hugoniot condition is necessary but not sufficient to determine a unique physically meaningful solution. The following example shows that multiple distinct weak solutions can coexist. [example: Non-Uniqueness for Burgers] Take $f(u) = u^2/2$, $u_L = 0$, $u_R = 1$. We have already exhibited the rarefaction fan as one weak solution. But consider the shock: \begin{align*} v(x,t) = \begin{cases} 0 & x < t/2, \\ 1 & x > t/2. \end{cases} \end{align*} This satisfies the Rankine-Hugoniot condition: $\dot\xi = 1/2$ and $(f(1) - f(0))/(1-0) = 1/2$. ✓ It is therefore also a weak solution. Both $u$ (the rarefaction fan) and $v$ (the shock) are weak solutions with the same initial data. A selection criterion is needed to single out the physically correct one. The shock $v$ is non-physical: its characteristics emanate from the shock rather than impinging on it. Left characteristics (speed 0) run away from the shock to the left, and right characteristics (speed 1) run away to the right. Information is generated at the shock rather than absorbed by it — a violation of causality. [/example] ## The Entropy Condition ### Physical motivation: vanishing viscosity Consider the **viscous conservation law** $u^\varepsilon_t + f(u^\varepsilon)_x = \varepsilon\, u^\varepsilon_{xx}$ with the same initial data, where $\varepsilon > 0$ is a small viscosity. This parabolic equation has a unique smooth solution for each $\varepsilon > 0$. The physically correct weak solution of the inviscid law $u_t + f(u)_x = 0$ is the [limit](/page/Limit) $u = \lim_{\varepsilon \to 0} u^\varepsilon$, called the **vanishing viscosity limit**. The entropy condition is the intrinsic characterisation of this limit — it selects exactly those weak solutions that arise as vanishing viscosity limits, without reference to the regularised problem. ### Entropy-flux pairs and the entropy inequality The key idea is that for the viscous equation, every smooth convex function of $u^\varepsilon$ satisfies an inequality that survives the limit $\varepsilon \to 0$. This leads to the abstract entropy condition. [definition: Entropy-Flux Pair] A pair of $C^1$ [functions](/page/Function) $(\eta, q)$ with $\eta: \mathbb{R} \to \mathbb{R}$ and $q: \mathbb{R} \to \mathbb{R}$ is an **entropy-flux pair** for the conservation law $u_t + f(u)_x = 0$ if: \begin{align*} q'(u) = \eta'(u)\,f'(u) \quad \text{for every } u \in \mathbb{R}. \end{align*} The function $\eta$ is called the **entropy** and $q$ is called the **entropy flux**. The pair is **convex** if additionally $\eta'' \geq 0$. [/definition] For any $C^1$ solution $u$ of the conservation law, the chain rule gives: \begin{align*} \partial_t \eta(u) + \partial_x q(u) = \eta'(u)\,u_t + q'(u)\,u_x = \eta'(u)(u_t + f'(u)\,u_x) = 0. \end{align*} So entropy is conserved exactly along smooth solutions. The crucial observation is that viscosity breaks this conservation — but only in one direction. [theorem: Entropy Inequality from Vanishing Viscosity] Let $f, \eta, q \in C^2(\mathbb{R})$ with $(\eta, q)$ a convex entropy-flux pair. Let $u^\varepsilon \in C^\infty(\mathbb{R} \times (0,\infty))$ be the smooth solution to the viscous conservation law $u^\varepsilon_t + f(u^\varepsilon)_x = \varepsilon\,u^\varepsilon_{xx}$. Then $u^\varepsilon$ satisfies: \begin{align*} \partial_t \eta(u^\varepsilon) + \partial_x q(u^\varepsilon) = \varepsilon\,\partial_x\bigl(\eta'(u^\varepsilon)\,u^\varepsilon_x\bigr) - \varepsilon\,\eta''(u^\varepsilon)\,(u^\varepsilon_x)^2. \end{align*} Since $\eta'' \geq 0$, the right-hand side satisfies: \begin{align*} \varepsilon\,\partial_x\bigl(\eta'(u^\varepsilon)\,u^\varepsilon_x\bigr) - \varepsilon\,\eta''(u^\varepsilon)\,(u^\varepsilon_x)^2 \leq \varepsilon\,\partial_x\bigl(\eta'(u^\varepsilon)\,u^\varepsilon_x\bigr). \end{align*} In the limit $\varepsilon \to 0$, if $u^\varepsilon \to u$ in a suitable sense, then for every non-negative [test function](/page/Test%20Function) $\phi \in C_c^\infty(\mathbb{R} \times (0,\infty))$: \begin{align*} \int_0^\infty \int_{-\infty}^\infty \bigl(\eta(u)\,\phi_t + q(u)\,\phi_x\bigr)\,d\mathcal{L}^1(x)\,d\mathcal{L}^1(t) \geq 0. \end{align*} [/theorem] [proof] **Step 1: Deriving the identity for $u^\varepsilon$.** Since $u^\varepsilon$ is smooth and satisfies $u^\varepsilon_t + f(u^\varepsilon)_x = \varepsilon\,u^\varepsilon_{xx}$, multiply both sides by $\eta'(u^\varepsilon)$: \begin{align*} \eta'(u^\varepsilon)\,u^\varepsilon_t + \eta'(u^\varepsilon)\,f'(u^\varepsilon)\,u^\varepsilon_x = \varepsilon\,\eta'(u^\varepsilon)\,u^\varepsilon_{xx}. \end{align*} The left side equals $\partial_t \eta(u^\varepsilon) + \partial_x q(u^\varepsilon)$ by the chain rule and the relation $q' = \eta' f'$. For the right side, we compute $\partial_x(\eta'(u^\varepsilon)\,u^\varepsilon_x)$ using the product rule: \begin{align*} \partial_x\bigl(\eta'(u^\varepsilon)\,u^\varepsilon_x\bigr) = \eta''(u^\varepsilon)\,(u^\varepsilon_x)^2 + \eta'(u^\varepsilon)\,u^\varepsilon_{xx}. \end{align*} Rearranging gives $\eta'(u^\varepsilon)\,u^\varepsilon_{xx} = \partial_x(\eta'(u^\varepsilon)\,u^\varepsilon_x) - \eta''(u^\varepsilon)(u^\varepsilon_x)^2$. Substituting: \begin{align*} \partial_t \eta(u^\varepsilon) + \partial_x q(u^\varepsilon) = \varepsilon\,\partial_x\bigl(\eta'(u^\varepsilon)\,u^\varepsilon_x\bigr) - \varepsilon\,\eta''(u^\varepsilon)\,(u^\varepsilon_x)^2. \end{align*} **Step 2: The sign.** Since $\eta'' \geq 0$ and $\varepsilon > 0$, the second term on the right is non-positive: \begin{align*} \partial_t \eta(u^\varepsilon) + \partial_x q(u^\varepsilon) \leq \varepsilon\,\partial_x\bigl(\eta'(u^\varepsilon)\,u^\varepsilon_x\bigr). \end{align*} **Step 3: Passing to the limit.** Multiply by a non-negative test function $\phi \in C_c^\infty(\mathbb{R} \times (0,\infty))$ and integrate over $\mathbb{R} \times (0,\infty)$. Integrating by parts on the left converts $\partial_t$ and $\partial_x$ onto $\phi$, and integrating by parts on the right: \begin{align*} -\int_0^\infty \int_{-\infty}^\infty \bigl(\eta(u^\varepsilon)\,\phi_t + q(u^\varepsilon)\,\phi_x\bigr)\,d\mathcal{L}^1(x)\,d\mathcal{L}^1(t) \leq -\varepsilon \int_0^\infty \int_{-\infty}^\infty \eta'(u^\varepsilon)\,u^\varepsilon_x\,\phi_x\,d\mathcal{L}^1(x)\,d\mathcal{L}^1(t). \end{align*} As $\varepsilon \to 0$, the right side vanishes (since $u^\varepsilon \to u$ in $L^\infty$ implies $u^\varepsilon_x$ is bounded in $L^2$ in a suitable sense and $\varepsilon \to 0$). Passing to the limit: \begin{align*} -\int_0^\infty \int_{-\infty}^\infty \bigl(\eta(u)\,\phi_t + q(u)\,\phi_x\bigr)\,d\mathcal{L}^1(x)\,d\mathcal{L}^1(t) \leq 0, \end{align*} which is the [distributional](/page/Distribution) form of the entropy inequality. [/proof] This motivates the central definition. [definition: Entropy Solution] A weak solution $u \in L^\infty(\mathbb{R} \times (0,\infty))$ of the conservation law with initial data $g \in L^\infty(\mathbb{R})$ is an **[entropy solution](/page/Entropy%20Solution)** if for every convex entropy-flux pair $(\eta, q)$ and every non-negative $\phi \in C_c^\infty(\mathbb{R} \times (0,\infty))$: \begin{align*} \int_0^\infty \int_{-\infty}^\infty \bigl(\eta(u)\,\phi_t + q(u)\,\phi_x\bigr)\,d\mathcal{L}^1(x)\,d\mathcal{L}^1(t) \geq 0. \end{align*} [/definition] [remark: Thermodynamic Interpretation] In gas dynamics, the state variable $u$ encodes density, velocity, and specific internal energy. The physical thermodynamic entropy $s = s(u)$ satisfies the second law: it cannot decrease across a shock. In the mathematical framework, the physical entropy corresponds to choosing $\eta = -s$ (with the sign reversed, since the second law $s \nearrow$ becomes $\eta \searrow$). The entropy inequality $\partial_t \eta(u) + \partial_x q(u) \leq 0$ then encodes exactly the second law of thermodynamics: physical entropy is produced at shocks, never destroyed. For a gas with internal energy $e$ and pressure $p = p(\rho, e)$, the Gibbs relation $T\,ds = de + p\,d(1/\rho)$ determines the entropy flux $q$, and the inequality $\dot\xi[\eta]_\xi \geq [q]_\xi$ at a shock is the Clausius inequality of classical thermodynamics. The mathematical framework thus abstracts the physically essential property — irreversibility — into a clean distributional inequality. [/remark] ### Entropy inequality at shocks: recovering Lax and Oleinik The entropy inequality has a concrete consequence for piecewise smooth solutions: it imposes a jump condition at each shock that turns out to be equivalent to the Oleinik entropy condition. Let $u$ be a piecewise smooth entropy solution with a shock at $x = \xi(t)$, smooth on either side. Applying the divergence theorem to the vector field $(\eta(u)\phi, q(u)\phi)$ on each half-domain $\Omega^\pm$ (exactly as in the proof of the Rankine-Hugoniot condition), the entropy inequality at the shock reduces to the pointwise condition: \begin{align*} \dot\xi\,[\eta(u)]_\xi \geq [q(u)]_\xi, \end{align*} i.e., $\dot\xi(\eta(u^+) - \eta(u^-)) \geq q(u^+) - q(u^-)$. This must hold for every convex entropy-flux pair $(\eta, q)$. Taking $\eta(u) = u$ and $\eta(u) = -u$ (both linear, hence convex with $\eta'' = 0$) recovers the Rankine-Hugoniot condition $\dot\xi\,[u]_\xi = [f(u)]_\xi$. For strictly convex $\eta$, the inequality provides genuine additional information. Choosing the Kružkov entropies $\eta_k(u) = |u - k|$ with entropy flux $q_k(u) = \mathrm{sgn}(u-k)(f(u) - f(k))$ for each $k \in \mathbb{R}$, and passing through the calculation, one recovers the Oleinik entropy condition for general flux, and the Lax condition when $f'' > 0$. [example: Entropy Inequality Rejects the Non-Physical Shock] Return to $f(u) = u^2/2$, $u_L = 0$, $u_R = 1$, and the non-physical shock $v(x,t) = \mathbb{1}_{x > t/2}$. Test the entropy inequality with the convex pair $\eta(u) = u^2/2$, $q(u) = u^3/3$ (which satisfies $q' = \eta' f'$ since $u^2 = u \cdot u$). At the shock, $u^- = 0$, $u^+ = 1$, $\dot\xi = 1/2$: \begin{align*} \dot\xi\,[\eta]_\xi &= \frac{1}{2}\left(\frac{1}{2} - 0\right) = \frac{1}{4}, \\ [q]_\xi &= \frac{1}{3} - 0 = \frac{1}{3}. \end{align*} The entropy inequality requires $\dot\xi[\eta]_\xi \geq [q]_\xi$, i.e., $1/4 \geq 1/3$ — false. The entropy inequality is violated, confirming that $v$ is not an entropy solution. For the physical shock with $u^- = 1$, $u^+ = 0$: $\dot\xi[\eta]_\xi = (1/2)(0 - 1/2) = -1/4$ and $[q]_\xi = 0 - 1/3 = -1/3$. The condition $-1/4 \geq -1/3$ holds. ✓ [/example] ### The Lax entropy condition For a strictly convex flux $f'' > 0$, the entropy condition simplifies to a clean geometric formulation in terms of characteristics. [definition: Lax Entropy Condition] Let $f \in C^2(\mathbb{R})$ with $f'' > 0$. A piecewise smooth weak solution with a shock at $x = \xi(t)$ satisfies the **Lax entropy condition** at $\xi(t)$ if: \begin{align*} f'(u^-(t)) \geq \dot\xi(t) \geq f'(u^+(t)). \end{align*} A weak solution satisfying this condition at every shock is a **Lax entropy solution**. [/definition] The Lax condition says the characteristic speeds on both sides of the shock straddle the shock speed: left-side characteristics (speed $f'(u^-)$) are faster than the shock, so they catch up and are absorbed; right-side characteristics (speed $f'(u^+)$) are slower, so the shock overtakes them. Since $f'' > 0$ and $f'$ is strictly increasing, $f'(u^-) \geq \dot\xi \geq f'(u^+)$ is equivalent to $u^- \geq u^+$ — the shock is **compressive**. The shock in the previous example for $u_L = 0$, $u_R = 1$ has $f'(u^-) = 0$ and $f'(u^+) = 1$, giving $0 \geq 1/2 \geq 1$ — false. So the non-physical shock correctly fails the Lax condition. The rarefaction fan has no shock, so no condition is required, and it is the entropy solution. For the physical shock with $u_L = 1$, $u_R = 0$: $f'(u^-) = 1 \geq 1/2 \geq 0 = f'(u^+)$. ✓ {width=72%} The diagram illustrates the Lax condition geometrically: left characteristics (arrows running right, speed $f'(u^-) = 1$) and right characteristics (arrows running up, speed $f'(u^+) = 0$) both terminate on the shock. The shock speed $\dot\xi = 1/2$ lies strictly between the two characteristic speeds. ### The Oleinik entropy condition For a general flux $f$ that need not be convex, the Lax condition is insufficient (characteristic speeds may not be monotone in $u$). The **Oleinik entropy condition** extends the entropy criterion to arbitrary smooth fluxes. [definition: Oleinik Entropy Condition] Let $f \in C^1(\mathbb{R})$. A piecewise smooth weak solution with a shock at $x = \xi(t)$ and side values $u^\pm = u^\pm(t)$ satisfies the **Oleinik entropy condition** if for every $u$ strictly between $u^-(t)$ and $u^+(t)$: \begin{align*} \frac{f(u) - f(u^-(t))}{u - u^-(t)} \geq \dot\xi(t) \geq \frac{f(u) - f(u^+(t))}{u - u^+(t)}. \end{align*} [/definition] [remark: Geometric Meaning of Oleinik] The Oleinik condition is a chord condition on the graph of $f$. Consider the chord from $(u^-, f(u^-))$ to $(u^+, f(u^+))$, which has slope $\dot\xi$ by Rankine-Hugoniot. The condition requires: for every point $(u, f(u))$ on the graph strictly between $u^-$ and $u^+$, the slope of the chord from $(u^-, f(u^-))$ to $(u, f(u))$ is at least $\dot\xi$, and the slope from $(u, f(u))$ to $(u^+, f(u^+))$ is at most $\dot\xi$. Geometrically: if $u^- > u^+$, the graph of $f$ on $[u^+, u^-]$ must lie below the chord; if $u^- < u^+$, above the chord. For $f'' > 0$, this is equivalent to the Lax condition. [/remark] ## Existence and Uniqueness for Convex Flux For a uniformly convex flux, the theory is complete: there is a unique entropy solution, and it is given by an explicit variational formula. [definition: Uniformly Convex Flux] A function $f \in C^2(\mathbb{R})$ is **uniformly convex** if there exists $\theta > 0$ such that $f''(u) \geq \theta$ for every $u \in \mathbb{R}$. [/definition] [definition: Legendre Transform] For a uniformly convex function $f \in C^2(\mathbb{R})$, the **Legendre transform** of $f$ is: \begin{align*} L: \mathbb{R} &\to \mathbb{R} \\ q &\mapsto \sup_{p \in \mathbb{R}} \bigl(p\,q - f(p)\bigr). \end{align*} Since $f$ is uniformly convex, $L$ is well-defined, finite, and uniformly convex with $L'' \geq 1/\|f''\|_{L^\infty}$. The supremum is achieved at $p = (f')^{-1}(q)$, giving $L(q) = q\,(f')^{-1}(q) - f((f')^{-1}(q))$. [/definition] [theorem: Existence and Uniqueness of the Entropy Solution] Let $f \in C^2(\mathbb{R})$ be uniformly convex and let $g \in L^\infty(\mathbb{R})$. Define: \begin{align*} G: \mathbb{R} &\to \mathbb{R} \\ y &\mapsto \int_0^y g(s)\,d\mathcal{L}^1(s). \end{align*} For each $(x,t) \in \mathbb{R} \times (0,\infty)$, let $y^*(x,t)$ denote a minimiser of the function: \begin{align*} \Phi(\,\cdot\,;\, x,t): \mathbb{R} &\to \mathbb{R} \\ y &\mapsto t\, L\!\left(\frac{x - y}{t}\right) + G(y). \end{align*} Then the **Lax-Oleinik formula**: \begin{align*} u: \mathbb{R} \times (0,\infty) &\to \mathbb{R} \\ (x,t) &\mapsto (f')^{-1}\!\left(\frac{x - y^*(x,t)}{t}\right) \end{align*} defines the unique entropy solution of $u_t + f(u)_x = 0$ with initial data $g$. This solution satisfies the Oleinik entropy condition everywhere. [/theorem] The minimiser $y^*(x,t)$ is the point where the backward characteristic from $(x,t)$ meets the initial line $\{t = 0\}$: it is the foot of the optimal path carrying mass from initial position $y$ to final position $x$ in time $t$. When two candidate paths achieve the same minimum, a shock occurs at $(x,t)$. ## Further Examples [example: Shock Merging in Burgers Equation] Consider Burgers' equation $f(u) = u^2/2$ with piecewise constant initial data: \begin{align*} g: \mathbb{R} &\to \mathbb{R} \\ x &\mapsto \begin{cases} 1 & x < 0, \\ 0 & 0 < x < 1, \\ -1 & x > 1. \end{cases} \end{align*} Two shocks form immediately at $t = 0$: shock $S_1$ at $x = 0$ separating $u_L = 1$ from $u_R = 0$, and shock $S_2$ at $x = 1$ separating $u_L = 0$ from $u_R = -1$. Rankine-Hugoniot for each: $\dot\xi_1 = (0^2/2 - 1^2/2)/(0 - 1) = 1/2$ and $\dot\xi_2 = ((-1)^2/2 - 0^2/2)/(-1 - 0) = -1/2$. So $S_1$ moves right at speed $1/2$ and $S_2$ moves left at speed $1/2$. The shocks approach each other and meet at position $x = 1/2$ at time $t = 1$. For $t < 1$: the solution is $u = 1$ for $x < t/2$, $u = 0$ for $t/2 < x < 1 - t/2$, $u = -1$ for $x > 1 - t/2$. At $t = 1$: the shocks merge. For $t > 1$: a single shock with $u_L = 1$ and $u_R = -1$ forms, moving at speed $\dot\xi = ((-1)^2/2 - 1^2/2)/(-1 - 1) = 0$. The merged shock is stationary at $x = 1/2$ for all $t > 1$. Lax check for the merged shock: $f'(u^-) = f'(1) = 1 \geq 0 \geq -1 = f'(u^+)$. ✓ [/example] [example: Non-Convex Flux and the Oleinik Condition] Consider the flux $f(u) = u^3/3$ (not globally convex: $f''(u) = 2u$, which changes sign at $u = 0$) with Riemann data $u_L = -1$, $u_R = 1$. Rankine-Hugoniot: $\dot\xi = ((1)^3/3 - (-1)^3/3)/(1 - (-1)) = (2/3)/2 = 1/3$. Check the Oleinik condition for $u \in (-1, 1)$: \begin{align*} \frac{f(u) - f(u^-)}{u - u^-} = \frac{u^3/3 - (-1/3)}{u + 1} = \frac{u^3 + 1}{3(u+1)} = \frac{(u+1)(u^2 - u + 1)}{3(u+1)} = \frac{u^2 - u + 1}{3}. \end{align*} The minimum of $u^2 - u + 1$ over $\mathbb{R}$ occurs at $u = 1/2$, giving $1/4 - 1/2 + 1 = 3/4$. So $(f(u) - f(u^-))/(u - u^-)$ achieves a minimum of $3/4 \cdot 1/3 = 1/4 < 1/3 = \dot\xi$. The left Oleinik inequality fails: this shock is not an entropy solution. The correct entropy solution for $u_L < u_R$ with a non-monotone flux requires a **composite wave**: a combination of rarefaction fans and shocks chosen so that the Oleinik condition is satisfied everywhere. Specifically, one constructs the lower convex envelope of $f$ on $[u_L, u_R]$. The graph of $f(u) = u^3/3$ on $[-1,1]$ has an inflection at $u=0$, and the correct entropy solution mixes a shock at some intermediate state with rarefaction regions, determined by the convex hull construction. [/example] ## Problems [problem] Let $f(u) = u^2/2$ (Burgers) and consider the initial datum: \begin{align*} g: \mathbb{R} &\to \mathbb{R} \\ x &\mapsto \begin{cases} 0 & x < 0, \\ 1 - x & 0 \leq x \leq 1, \\ 0 & x > 1. \end{cases} \end{align*} 1. Write down the characteristic curve from each $x_0 \in \mathbb{R}$. 2. Find the time $T^*$ at which the first characteristic crossing occurs and the location $x^*$ at which it occurs. 3. For $t < T^*$, write down the classical solution $u(x,t)$ in each region separated by characteristics. 4. Determine the Rankine-Hugoniot shock speed just after $t = T^*$ and verify the Lax entropy condition. [/problem] [solution] **Part 1: Characteristics.** The characteristic from $x_0$ carries value $g(x_0)$ and satisfies $x(t) = x_0 + t\,g(x_0) = x_0 + t\,f'(g(x_0))$ (since $f'(u) = u$): \begin{align*} x(t) = \begin{cases} x_0 & x_0 < 0 \text{ (speed 0)}, \\ x_0 + t(1-x_0) & 0 \leq x_0 \leq 1 \text{ (speed } 1-x_0\text{)}, \\ x_0 & x_0 > 1 \text{ (speed 0)}. \end{cases} \end{align*} **Part 2: First crossing time.** Within the interval $[0,1]$, the characteristic from $x_0$ has speed $1 - x_0$, which is decreasing in $x_0$. Two characteristics from $x_0 < x_1$ in $[0,1]$ cross at: \begin{align*} x_0 + t(1-x_0) = x_1 + t(1-x_1) \implies t(x_1 - x_0) = x_1 - x_0 \implies t = 1. \end{align*} This is independent of $x_0, x_1 \in [0,1]$: all characteristics from the decreasing portion of $g$ meet simultaneously at $t = 1$. To find the crossing location, take $x_0 \to x_1$ in $(0,1)$: $x^* = x_0 + 1\cdot(1-x_0) = 1$ for any $x_0 \in [0,1]$. So all interior characteristics converge to $x^* = 1$ at $T^* = 1$. **Part 3: Classical solution for $t < 1$.** We invert $x = x_0 + t(1-x_0) = x_0(1-t) + t$ for $x_0 \in [0,1]$: $x_0 = (x-t)/(1-t)$. This is valid for $x_0 \in [0,1]$, i.e., $t \leq x \leq 1$. The value is $u = g(x_0) = 1 - x_0 = 1 - (x-t)/(1-t) = (1-x)/(1-t)$. The full solution for $0 < t < 1$: \begin{align*} u(x,t) = \begin{cases} 0 & x \leq t, \\ \dfrac{1-x}{1-t} & t \leq x \leq 1, \\ 0 & x \geq 1. \end{cases} \end{align*} **Part 4: Shock after $t = 1$.** At $t = 1$, the gradient $|u_x| = 1/(1-t) \to \infty$, confirming the gradient catastrophe. For $t > 1$, a shock forms at $\xi(t)$ separating $u^- = 0$ (left, from the region $x_0 < 0$) from $u^+ = 0$ (right, from the region $x_0 > 1$). But $u^- = u^+ = 0$, so there is no jump — the solution returns to $u \equiv 0$ instantaneously at $t = 1$, and the shock collapses immediately. This is consistent: the total mass $\int g\,d\mathcal{L}^1 = \int_0^1 (1-x)\,d\mathcal{L}^1(x) = 1/2$ is conserved, and for $t > 1$ it has all been swept to the single point $x = 1$. The Rankine-Hugoniot shock at $t = 1^+$ has $u^- = u^+ = 0$ and $\dot\xi = 0$. The Lax condition $f'(0) = 0 \geq 0 \geq 0 = f'(0)$ holds trivially. [/solution] ## References 1. L. C. Evans, *Partial Differential Equations*, 2nd ed., AMS (2010). §3.4. 2. B. Dacorogna, *Introduction to the [Calculus of Variations](/page/Calculus%20of%20Variations)*, 3rd ed., Imperial College Press (2014). 3. C. M. Dafermos, *Hyperbolic Conservation Laws in Continuum Physics*, 4th ed., Springer (2016). 4. P. D. Lax, *Hyperbolic Systems of Conservation Laws and the Mathematical Theory of Shock Waves*, SIAM (1973).