[proofplan] We prove that every vanishing-viscosity limit satisfying the parabolic bounds for the viscous approximations and the local viscous energy estimate satisfies the Kruzhkov entropy inequalities. For each constant state $k \in \mathbb{R}$, we apply the viscous equation to smooth convex approximations of the entropy $\eta_k(z)=|z-k|$ and to the corresponding entropy flux, which is later defined using the sign function. The viscosity produces a nonpositive dissipation term and a divergence term multiplied by $\varepsilon$; after testing against nonnegative compactly supported functions, the dissipation has the correct sign and the divergence term vanishes by the local energy bound. Passing to the limit in $L^1_{\mathrm{loc}}$ gives the open-time Kruzhkov entropy inequality, and the strengthened convergence on $[0,\infty)\times\mathbb R$ lets the same argument pass to half-space tests with the prescribed initial trace. [/proofplan] [step:Fix the measure notation and the viscous equation] Throughout the proof, $\mathcal{L}^1$ denotes one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$ and $\mathcal{L}^2$ denotes two-dimensional Lebesgue measure on $(0,\infty)\times\mathbb{R}$. For each $\varepsilon>0$, the function $u_\varepsilon:(0,\infty)\times\mathbb{R}\to\mathbb{R}$ is the stated classical solution of \begin{align*} \partial_t u_\varepsilon + \partial_x f(u_\varepsilon)=\varepsilon\partial_{xx}u_\varepsilon. \end{align*} [/step] [step:Define the Kruzhkov entropy pairs and their smooth convex approximations] Fix $k \in \mathbb{R}$. Define the sign function $\operatorname{sgn}:\mathbb{R}\to\{-1,0,1\}$ by $\operatorname{sgn}(r)=1$ for $r>0$, $\operatorname{sgn}(r)=0$ for $r=0$, and $\operatorname{sgn}(r)=-1$ for $r<0$. Define the Kruzhkov entropy function \begin{align*} \eta_k: \mathbb{R} \to [0,\infty), \qquad z \mapsto |z-k|. \end{align*} Define its entropy flux \begin{align*} q_k: \mathbb{R} \to \mathbb{R}, \qquad z \mapsto \operatorname{sgn}(z-k)(f(z)-f(k)). \end{align*} Let $(\eta_{k,\delta})_{\delta>0}$ be a family of functions \begin{align*} \eta_{k,\delta}: \mathbb{R} \to [0,\infty) \end{align*} such that $\eta_{k,\delta} \in C^2(\mathbb{R})$, each $\eta_{k,\delta}$ is convex, $\eta_{k,\delta} \to \eta_k$ locally uniformly on $\mathbb{R}$, and $\eta_{k,\delta}' \to \operatorname{sgn}(\cdot-k)$ pointwise on $\mathbb{R}\setminus\{k\}$ with $|\eta_{k,\delta}'|\le 1$. Define the associated smooth entropy flux \begin{align*} q_{k,\delta}: \mathbb{R} \to \mathbb{R}, \qquad z \mapsto \int_k^z \eta_{k,\delta}'(s) f'(s) \, d\mathcal{L}^1(s). \end{align*} Then $q_{k,\delta} \to q_k$ locally uniformly on $\mathbb{R}$, because $f'$ is continuous and locally bounded. [/step] [step:Derive the viscous entropy inequality for the smooth entropy pair] Fix $\varepsilon>0$ and $\delta>0$. Since $u_\varepsilon$ is classical, the chain rule gives \begin{align*} \partial_t \eta_{k,\delta}(u_\varepsilon) = \eta_{k,\delta}'(u_\varepsilon)\partial_t u_\varepsilon. \end{align*} The definition of $q_{k,\delta}$ gives $q_{k,\delta}'=\eta_{k,\delta}' f'$, hence \begin{align*} \partial_x q_{k,\delta}(u_\varepsilon) = \eta_{k,\delta}'(u_\varepsilon) f'(u_\varepsilon)\partial_x u_\varepsilon = \eta_{k,\delta}'(u_\varepsilon)\partial_x f(u_\varepsilon). \end{align*} Multiplying the viscous equation by $\eta_{k,\delta}'(u_\varepsilon)$ therefore yields \begin{align*} \partial_t \eta_{k,\delta}(u_\varepsilon) + \partial_x q_{k,\delta}(u_\varepsilon) = \varepsilon \eta_{k,\delta}'(u_\varepsilon)\partial_{xx}u_\varepsilon. \end{align*} Using the product rule, \begin{align*} \eta_{k,\delta}'(u_\varepsilon)\partial_{xx}u_\varepsilon = \partial_x\bigl(\eta_{k,\delta}'(u_\varepsilon)\partial_x u_\varepsilon\bigr) - \eta_{k,\delta}''(u_\varepsilon)|\partial_x u_\varepsilon|^2. \end{align*} Since $\eta_{k,\delta}$ is convex, $\eta_{k,\delta}'' \ge 0$. Thus, in the sense of distributions on $(0,\infty)\times\mathbb{R}$, \begin{align*} \partial_t \eta_{k,\delta}(u_\varepsilon) + \partial_x q_{k,\delta}(u_\varepsilon) \le \varepsilon \partial_x\bigl(\eta_{k,\delta}'(u_\varepsilon)\partial_x u_\varepsilon\bigr). \end{align*} [guided] The purpose of introducing $\eta_{k,\delta}$ is to make the formal Kruzhkov calculation legitimate at the point $z=k$, where $|z-k|$ is not differentiable. For fixed $\delta>0$, the function $\eta_{k,\delta}$ is $C^2$, so we may apply the ordinary chain rule to the classical solution $u_\varepsilon$: \begin{align*} \partial_t \eta_{k,\delta}(u_\varepsilon) = \eta_{k,\delta}'(u_\varepsilon)\partial_t u_\varepsilon. \end{align*} The entropy flux is chosen so that its derivative exactly matches the flux term after multiplication by $\eta_{k,\delta}'$. Namely, by definition, \begin{align*} q_{k,\delta}'(z)=\eta_{k,\delta}'(z)f'(z) \end{align*} for every $z \in \mathbb{R}$. Applying the chain rule in the spatial variable gives \begin{align*} \partial_x q_{k,\delta}(u_\varepsilon)=q_{k,\delta}'(u_\varepsilon)\partial_x u_\varepsilon=\eta_{k,\delta}'(u_\varepsilon)f'(u_\varepsilon)\partial_xu_\varepsilon. \end{align*} Since $\partial_x f(u_\varepsilon)=f'(u_\varepsilon)\partial_xu_\varepsilon$, multiplying the viscous equation by $\eta_{k,\delta}'(u_\varepsilon)$ gives \begin{align*} \partial_t \eta_{k,\delta}(u_\varepsilon)+\partial_x q_{k,\delta}(u_\varepsilon)=\varepsilon \eta_{k,\delta}'(u_\varepsilon)\partial_{xx}u_\varepsilon. \end{align*} The right-hand side is where viscosity selects the [entropy solution](/page/Entropy%20Solution). We rewrite it by the product rule: \begin{align*} \eta_{k,\delta}'(u_\varepsilon)\partial_{xx}u_\varepsilon=\partial_x\bigl(\eta_{k,\delta}'(u_\varepsilon)\partial_xu_\varepsilon\bigr)-\eta_{k,\delta}''(u_\varepsilon)|\partial_xu_\varepsilon|^2. \end{align*} The second term is nonpositive after multiplication by $\varepsilon$, because $\eta_{k,\delta}''\ge 0$ by convexity. Dropping this nonpositive term gives the distributional inequality \begin{align*} \partial_t \eta_{k,\delta}(u_\varepsilon)+\partial_x q_{k,\delta}(u_\varepsilon)\le \varepsilon \partial_x\bigl(\eta_{k,\delta}'(u_\varepsilon)\partial_xu_\varepsilon\bigr). \end{align*} [/guided] [/step] [step:Test against a nonnegative compactly supported function and remove the viscous divergence] Let \begin{align*} \varphi: (0,\infty)\times\mathbb{R} \to [0,\infty) \end{align*} be a function in $C_c^\infty((0,\infty)\times\mathbb{R})$. Multiplying the preceding distributional inequality by $\varphi$ and integrating by parts gives \begin{align*} \int_{(0,\infty)\times\mathbb{R}} \eta_{k,\delta}(u_\varepsilon)\partial_t\varphi + q_{k,\delta}(u_\varepsilon)\partial_x\varphi \, d\mathcal{L}^2(t,x) \ge \varepsilon \int_{(0,\infty)\times\mathbb{R}} \eta_{k,\delta}'(u_\varepsilon)\partial_xu_\varepsilon \partial_x\varphi \, d\mathcal{L}^2(t,x). \end{align*} Let $K=\operatorname{supp}\varphi$ denote the closed support of $\varphi$. Since $|\eta_{k,\delta}'|\le 1$, the [Cauchy-Schwarz inequality](/theorems/432) gives \begin{align*} \left|\varepsilon \int_K \eta_{k,\delta}'(u_\varepsilon)\partial_xu_\varepsilon \partial_x\varphi \, d\mathcal{L}^2(t,x)\right| \le \varepsilon^{1/2}\left(\varepsilon \int_K |\partial_xu_\varepsilon|^2 \, d\mathcal{L}^2(t,x)\right)^{1/2}\left(\int_K |\partial_x\varphi|^2 \, d\mathcal{L}^2(t,x)\right)^{1/2}. \end{align*} By the local viscous energy hypothesis in the theorem statement, applied to the compact set $K\subset [0,\infty)\times\mathbb R$, there is a constant $C_K>0$ such that \begin{align*} \varepsilon \int_K |\partial_xu_\varepsilon|^2 \, d\mathcal{L}^2(t,x)\le C_K \end{align*} for all sufficiently small $\varepsilon$. Therefore the preceding bound is at most \begin{align*} \varepsilon^{1/2} C_K^{1/2}\left(\int_K |\partial_x\varphi|^2 \, d\mathcal{L}^2(t,x)\right)^{1/2}, \end{align*} which tends to $0$ as $\varepsilon \to 0$, uniformly in $\delta$. Hence \begin{align*} \liminf_{\varepsilon\to 0}\int_{(0,\infty)\times\mathbb{R}} \eta_{k,\delta}(u_\varepsilon)\partial_t\varphi + q_{k,\delta}(u_\varepsilon)\partial_x\varphi \, d\mathcal{L}^2(t,x) \ge 0. \end{align*} [/step] [step:Pass first to the inviscid limit and then to the Kruzhkov entropy] Let $K_0\subset (0,\infty)\times\mathbb{R}$ be compact. Choose $T>0$ such that $K_0\subset (0,T)\times\mathbb{R}$. By the uniform parabolic bound assumed in the theorem statement, there is a constant $M>0$, independent of $\varepsilon$, such that $|u_\varepsilon|\le M$ on $K_0$ for all sufficiently small $\varepsilon$. Since $u_\varepsilon\to u$ in $L^1(K_0)$, a subsequence converges to $u$ $\mathcal{L}^2$-a.e. on $K_0$, and hence $|u|\le M$ $\mathcal{L}^2$-a.e. on $K_0$. Because the compact set $K_0\subset (0,\infty)\times\mathbb R$ was arbitrary, this proves $u\in L^\infty_{\mathrm{loc}}((0,\infty)\times\mathbb R)$. Therefore the derived uniform $L^\infty$ bound on $u_\varepsilon$ and the convergence $u_\varepsilon\to u$ in $L^1_{\mathrm{loc}}((0,\infty)\times\mathbb{R})$ imply \begin{align*} \eta_{k,\delta}(u_\varepsilon)\to \eta_{k,\delta}(u) \end{align*} and \begin{align*} q_{k,\delta}(u_\varepsilon)\to q_{k,\delta}(u) \end{align*} in $L^1_{\mathrm{loc}}((0,\infty)\times\mathbb{R})$, because $\eta_{k,\delta}$ and $q_{k,\delta}$ are Lipschitz on every bounded interval containing the ranges of the functions. Therefore, \begin{align*} \int_{(0,\infty)\times\mathbb{R}} \eta_{k,\delta}(u)\partial_t\varphi + q_{k,\delta}(u)\partial_x\varphi \, d\mathcal{L}^2(t,x) \ge 0. \end{align*} Now let $\delta\to 0$. Since $u$ is locally bounded and $\eta_{k,\delta}\to\eta_k$, $q_{k,\delta}\to q_k$ locally uniformly on bounded intervals, dominated convergence gives \begin{align*} \int_{(0,\infty)\times\mathbb{R}} |u-k|\partial_t\varphi + \operatorname{sgn}(u-k)(f(u)-f(k))\partial_x\varphi \, d\mathcal{L}^2(t,x) \ge 0. \end{align*} This is the Kruzhkov entropy inequality for the constant state $k$ and the nonnegative [test function](/page/Test%20Function) $\varphi$. [guided] The limiting step has two separate passages, and each uses a different compactness fact. First fix $\delta>0$ and keep the smooth entropy pair $(\eta_{k,\delta},q_{k,\delta})$. Let $K_0\subset (0,\infty)\times\mathbb R$ be compact and choose $T>0$ with $K_0\subset (0,T)\times\mathbb R$. The uniform parabolic bound in the theorem statement gives a constant $M>0$, independent of $\varepsilon$, such that $|u_\varepsilon|\le M$ on $K_0$ for all sufficiently small $\varepsilon$. Since $u_\varepsilon\to u$ in $L^1(K_0)$, a subsequence converges to $u$ $\mathcal L^2$-a.e. on $K_0$, so $|u|\le M$ $\mathcal L^2$-a.e. on $K_0$. As $K_0$ was arbitrary, $u\in L^\infty_{\mathrm{loc}}((0,\infty)\times\mathbb R)$. For this fixed $\delta$, the functions $\eta_{k,\delta}$ and $q_{k,\delta}$ are Lipschitz on the bounded interval $[-M,M]$. Therefore $u_\varepsilon\to u$ in $L^1(K_0)$ implies \begin{align*} \eta_{k,\delta}(u_\varepsilon)\to \eta_{k,\delta}(u) \end{align*} and \begin{align*} q_{k,\delta}(u_\varepsilon)\to q_{k,\delta}(u) \end{align*} in $L^1(K_0)$. Applying this on the compact support of $\varphi$ allows passage to the limit $\varepsilon\to0$ in the tested entropy inequality, while the viscous divergence term has already been shown to vanish. Hence \begin{align*} \int_{(0,\infty)\times\mathbb{R}} \eta_{k,\delta}(u)\partial_t\varphi + q_{k,\delta}(u)\partial_x\varphi \, d\mathcal{L}^2(t,x) \ge 0. \end{align*} Now send $\delta\to0$. On the essential range of $u$ over $\operatorname{supp}\varphi$, the convergence $\eta_{k,\delta}\to\eta_k$ and $q_{k,\delta}\to q_k$ is uniform, so dominated convergence applies. This gives \begin{align*} \int_{(0,\infty)\times\mathbb{R}} |u-k|\partial_t\varphi + \operatorname{sgn}(u-k)(f(u)-f(k))\partial_x\varphi \, d\mathcal{L}^2(t,x) \ge 0. \end{align*} This is precisely the Kruzhkov entropy inequality on open time for the constant state $k$ and the chosen nonnegative test function $\varphi$. [/guided] [/step] [step:Pass the viscous entropy inequality up to the initial boundary] It remains to recover the initial datum in the entropy formulation. Fix the same constant state $k\in\mathbb R$ as above, and let \begin{align*} \Psi: [0,\infty)\times\mathbb{R}\to[0,\infty) \end{align*} be a function in $C_c^\infty([0,\infty)\times\mathbb{R})$. Repeating the smooth entropy calculation for $\eta_{k,\delta}$ on the half-space and integrating by parts in time gives \begin{align*} \int_{(0,\infty)\times\mathbb{R}} \eta_{k,\delta}(u_\varepsilon)\partial_t\Psi + q_{k,\delta}(u_\varepsilon)\partial_x\Psi \, d\mathcal{L}^2(t,x) +\int_{\mathbb{R}}\eta_{k,\delta}(u_{0,\varepsilon}(x))\Psi(0,x)\,d\mathcal L^1(x) \geq \varepsilon\int_{(0,\infty)\times\mathbb{R}}\eta_{k,\delta}'(u_\varepsilon)\partial_xu_\varepsilon\partial_x\Psi\,d\mathcal L^2(t,x). \end{align*} The boundary term has the displayed positive sign because $\Psi$ is supported in $t\geq0$ and the time [integration by parts](/theorems/210) contributes the initial trace at $t=0$. Let $K=\operatorname{supp}\Psi$. Since $|\eta_{k,\delta}'|\leq1$, the same Cauchy-Schwarz estimate used for open-time tests gives \begin{align*} \left|\varepsilon\int_{(0,\infty)\times\mathbb{R}}\eta_{k,\delta}'(u_\varepsilon)\partial_xu_\varepsilon\partial_x\Psi\,d\mathcal L^2(t,x)\right| \leq \varepsilon^{1/2}\left(\varepsilon\int_K|\partial_xu_\varepsilon|^2\,d\mathcal L^2(t,x)\right)^{1/2} \left(\int_K|\partial_x\Psi|^2\,d\mathcal L^2(t,x)\right)^{1/2}. \end{align*} The local viscous energy estimate applies to this compact set $K\subset[0,\infty)\times\mathbb R$, so the right-hand side tends to $0$ as $\varepsilon\to0$, uniformly in $\delta$. For fixed $\delta>0$, the uniform parabolic bound on $\operatorname{supp}\Psi\cap((0,\infty)\times\mathbb R)$ and the local convergence $u_\varepsilon\to u$ imply convergence of the two interior entropy terms exactly as in the open-time step. Also, since $\eta_{k,\delta}$ is Lipschitz with Lipschitz constant at most $1$ and $u_{0,\varepsilon}\to u_0$ in $L^1(\mathbb R)$, \begin{align*} \eta_{k,\delta}(u_{0,\varepsilon})\to\eta_{k,\delta}(u_0) \end{align*} in $L^1(\mathbb R)$. Passing first $\varepsilon\to0$ and then $\delta\to0$ therefore gives \begin{align*} \int_{(0,\infty)\times\mathbb{R}} |u-k|\partial_t\Psi + \operatorname{sgn}(u-k)(f(u)-f(k))\partial_x\Psi \, d\mathcal{L}^2(t,x) + \int_{\mathbb{R}} |u_0(x)-k|\Psi(0,x) \, d\mathcal{L}^1(x)\ge0. \end{align*} This is the Kruzhkov entropy inequality with the initial boundary term. [/step] [step:Conclude that the limit is the Kruzhkov entropy solution] The inviscid-limit step proves that $u\in L^\infty_{\mathrm{loc}}((0,\infty)\times\mathbb R)$. The open-time entropy step proves that for every $k\in\mathbb R$ and every nonnegative $\varphi\in C_c^\infty((0,\infty)\times\mathbb R)$, \begin{align*} \int_{(0,\infty)\times\mathbb{R}} |u-k|\partial_t\varphi+\operatorname{sgn}(u-k)(f(u)-f(k))\partial_x\varphi \, d\mathcal{L}^2(t,x)\ge 0. \end{align*} The boundary entropy step proves the initial-data formulation: for every $k\in\mathbb R$ and every nonnegative $\Psi\in C_c^\infty([0,\infty)\times\mathbb R)$, \begin{align*} \int_{(0,\infty)\times\mathbb{R}} |u-k|\partial_t\Psi + \operatorname{sgn}(u-k)(f(u)-f(k))\partial_x\Psi \, d\mathcal{L}^2(t,x) + \int_{\mathbb{R}} |u_0(x)-k|\Psi(0,x) \, d\mathcal{L}^1(x)\ge0. \end{align*} In this proof, a Kruzhkov entropy solution with initial data $u_0$ means a locally essentially bounded function satisfying this boundary entropy inequality for every constant state $k\in\mathbb R$ and every nonnegative half-space test function $\Psi$. The preceding paragraphs establish exactly these requirements. Hence $u$ is the Kruzhkov entropy solution with initial data $u_0$. [/step]