[proofplan] The proof converts the viscous equation into an [entropy inequality](/page/Entropy%20Solution) by multiplying by $\eta'(u_\varepsilon)$ and using convexity of $\eta$. After testing against a nonnegative compactly supported function, [integration by parts](/theorems/210) produces the entropy inequality for $u_\varepsilon$ with one viscosity error term. The local $L^1$ convergence and the uniform boundedness of the approximations allow passage to the limit in the nonlinear entropy and entropy-flux terms, while the viscosity error vanishes because it is multiplied by $\varepsilon$. We then recover the Kruzhkov inequalities by smoothing the nonsmooth entropies $|r-k|$, recover the weak conservation law from affine entropies, and invoke the standard weak-in-time initial-[trace theorem](/theorems/60) for scalar Kruzhkov solutions after checking its hypotheses. The final assertions are consequences of uniqueness and relative compactness in the local $L^1$ topology. [/proofplan]

[step:Fix a uniform range for all viscous approximations] Define \begin{align*} M_0 := \sup_{\varepsilon > 0} \|u_{0,\varepsilon}\|_{L^\infty(\mathbb{R})} \end{align*} and \begin{align*} M := \max\{M_0, \|u_0\|_{L^\infty(\mathbb{R})}\}. \end{align*} Fix $\varepsilon>0$. The constant maps $\overline u_\varepsilon : [0,T]\times\mathbb R\to\mathbb R$ and $\underline u_\varepsilon : [0,T]\times\mathbb R\to\mathbb R$ defined by $\overline u_\varepsilon(t,x)=M_0$ and $\underline u_\varepsilon(t,x)=-M_0$ are respectively smooth solutions of the viscous equation, because all their first and second derivatives vanish and $\partial_x f(\pm M_0)=0$ along constant functions. The equation has the quasilinear form \begin{align*} \partial_t u_\varepsilon + f'(u_\varepsilon)\partial_x u_\varepsilon - \varepsilon\partial_{xx}u_\varepsilon=0. \end{align*} The [Scalar Parabolic Maximum Principle](/theorems/5984) applies without a prior bound on $u_\varepsilon$ by the standard first-contact argument. Indeed, if $u_\varepsilon-M_0$ had a positive first interior maximum, then at that contact point one would have $\partial_xu_\varepsilon=0$, $\partial_{xx}u_\varepsilon\leq0$, and $\partial_tu_\varepsilon\geq0$. Substituting into the displayed equation gives $\partial_tu_\varepsilon=\varepsilon\partial_{xx}u_\varepsilon\leq0$, contradicting strict first increase after the usual perturbation by a small positive multiple of $t$. The lower barrier is identical for $-M_0-u_\varepsilon$. On the whole line, one first applies this argument on finite cylinders with a vanishing quadratic spatial barrier and then lets the cylinder radius tend to infinity. The initial comparison is valid because $u_{0,\varepsilon}(x)\le M_0$ and $u_{0,\varepsilon}(x)\ge -M_0$ for every $x\in\mathbb R$, since $u_{0,\varepsilon}$ is smooth and its $L^\infty$ norm is bounded by $M_0$. Therefore \begin{align*} |u_\varepsilon(t,x)| \le M_0 \le M \end{align*} for every $\varepsilon > 0$, every $t \in [0,T]$, and every $x \in \mathbb{R}$. Since $u_{\varepsilon_j} \to u$ in local $L^1$ convergence on $(0,T) \times \mathbb{R}$, a further subsequence converges pointwise almost everywhere with respect to two-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) $\mathcal{L}^2$ on compact subsets; hence \begin{align*} |u(t,x)| \le M \end{align*} for $\mathcal{L}^2$-almost every $(t,x) \in (0,T) \times \mathbb{R}$. Thus every function value appearing in the limiting argument lies in the compact interval \begin{align*} I := [-M,M]. \end{align*} [/step]

[step:Derive the viscous entropy inequality]Let $\eta \in C^2(\mathbb{R})$ be convex. Define the associated [entropy flux](/page/Entropy%20Solution) as the map \begin{align*} q : \mathbb{R} \to \mathbb{R}, \quad r \mapsto \int_0^r \eta'(s) f'(s)\, d\mathcal{L}^1(s). \end{align*} Then $q \in C^1(\mathbb{R})$ and \begin{align*} q'(r) = \eta'(r) f'(r) \end{align*} for every $r \in \mathbb{R}$. For fixed $\varepsilon > 0$, multiply \begin{align*} \partial_t u_\varepsilon + \partial_x f(u_\varepsilon) = \varepsilon \partial_{xx} u_\varepsilon \end{align*} by $\eta'(u_\varepsilon)$. The chain rule gives \begin{align*} \eta'(u_\varepsilon)\partial_t u_\varepsilon = \partial_t \eta(u_\varepsilon) \end{align*} and, since $q'=\eta'f'$, \begin{align*} \eta'(u_\varepsilon)\partial_x f(u_\varepsilon) = \eta'(u_\varepsilon)f'(u_\varepsilon)\partial_x u_\varepsilon = \partial_x q(u_\varepsilon). \end{align*} For the viscosity term, the chain rule gives \begin{align*} \partial_{xx}\eta(u_\varepsilon) = \eta''(u_\varepsilon)|\partial_x u_\varepsilon|^2 + \eta'(u_\varepsilon)\partial_{xx}u_\varepsilon. \end{align*} Therefore \begin{align*} \partial_t \eta(u_\varepsilon) + \partial_x q(u_\varepsilon) = \varepsilon \partial_{xx}\eta(u_\varepsilon) - \varepsilon \eta''(u_\varepsilon)|\partial_x u_\varepsilon|^2. \end{align*} Since $\eta$ is convex, $\eta'' \ge 0$, and hence \begin{align*} \partial_t \eta(u_\varepsilon) + \partial_x q(u_\varepsilon) \le \varepsilon \partial_{xx}\eta(u_\varepsilon) \end{align*} pointwise on $(0,T) \times \mathbb{R}$.[/step]

[guided]The goal is to convert the viscous equation into an inequality that survives the limit $\varepsilon \downarrow 0$. The correct multiplier is $\eta'(u_\varepsilon)$ because it turns the time derivative into a time derivative of the entropy: \begin{align*} \eta'(u_\varepsilon)\partial_t u_\varepsilon = \partial_t \eta(u_\varepsilon). \end{align*} The flux term also becomes an exact derivative because the entropy flux $q$ was defined precisely so that $q'=\eta'f'$. Thus \begin{align*} \eta'(u_\varepsilon)\partial_x f(u_\varepsilon) = \eta'(u_\varepsilon) f'(u_\varepsilon)\partial_x u_\varepsilon = q'(u_\varepsilon)\partial_x u_\varepsilon = \partial_x q(u_\varepsilon). \end{align*} The only point where viscosity changes the conservation law is the second derivative term. Applying the chain rule to $\eta(u_\varepsilon)$ twice gives \begin{align*} \partial_{xx}\eta(u_\varepsilon) = \eta''(u_\varepsilon)|\partial_x u_\varepsilon|^2 + \eta'(u_\varepsilon)\partial_{xx}u_\varepsilon. \end{align*} Solving this identity for $\eta'(u_\varepsilon)\partial_{xx}u_\varepsilon$ and substituting into the viscous equation multiplied by $\eta'(u_\varepsilon)$ gives \begin{align*} \partial_t \eta(u_\varepsilon) + \partial_x q(u_\varepsilon) = \varepsilon \partial_{xx}\eta(u_\varepsilon) - \varepsilon \eta''(u_\varepsilon)|\partial_x u_\varepsilon|^2. \end{align*} Convexity is used exactly here: $\eta''(u_\varepsilon)\ge 0$, so the final term is nonpositive. Dropping it gives the pointwise differential inequality \begin{align*} \partial_t \eta(u_\varepsilon) + \partial_x q(u_\varepsilon) \le \varepsilon \partial_{xx}\eta(u_\varepsilon). \end{align*} This is the viscous entropy inequality; the remaining proof shows that its weak form converges to the entropy inequality for the inviscid equation.[/guided]

[step:Test the viscous entropy inequality and integrate by parts] Let \begin{align*} \phi : [0,T) \times \mathbb{R} \to \mathbb{R} \end{align*} be a nonnegative function in $C_c^\infty([0,T) \times \mathbb{R})$. Multiplying the pointwise inequality \begin{align*} \partial_t \eta(u_\varepsilon) + \partial_x q(u_\varepsilon) - \varepsilon \partial_{xx}\eta(u_\varepsilon) \le 0 \end{align*} by $\phi$ and integrating over $(0,T) \times \mathbb{R}$ with respect to $\mathcal{L}^1(t)\otimes \mathcal{L}^1(x)$ gives \begin{align*} \int_0^{\,T} \int_{\mathbb{R}} \phi\partial_t \eta(u_\varepsilon) + \phi\partial_x q(u_\varepsilon) - \varepsilon \phi\partial_{xx}\eta(u_\varepsilon)\, d\mathcal{L}^1(x)\, d\mathcal{L}^1(t) \le 0. \end{align*} Because $\phi$ is compactly supported in $[0,T) \times \mathbb{R}$, the boundary term at $t=T$ and all spatial boundary terms vanish. [Integration by parts](/theorems/2098) in $t$ and twice in $x$ gives \begin{align*} \int_0^{\,T} \int_{\mathbb{R}} \eta(u_\varepsilon)\partial_t\phi + q(u_\varepsilon)\partial_x\phi + \varepsilon \eta(u_\varepsilon)\partial_{xx}\phi\, d\mathcal{L}^1(x)\, d\mathcal{L}^1(t) + \int_{\mathbb{R}} \eta(u_{0,\varepsilon}(x))\phi(0,x)\, d\mathcal{L}^1(x) \ge 0. \end{align*} [/step]

[step:Pass to the limit away from the initial time] Since $\eta$ and $q$ are continuously differentiable on the compact interval $I=[-M,M]$, define \begin{align*} L_\eta := \sup_{r \in I} |\eta'(r)| \end{align*} and \begin{align*} L_q := \sup_{r \in I} |q'(r)|. \end{align*} Let $\delta \in (0,T)$. Define the compact set \begin{align*} K_\delta := \operatorname{supp}\phi \cap ([\delta,T] \times \mathbb{R}). \end{align*} Then $K_\delta \subset (0,T)\times\mathbb{R}$ after replacing $T$ by the largest time in $\operatorname{supp}\phi$, which is strictly less than $T$ because $\phi$ is compactly supported in $[0,T)\times\mathbb{R}$. The convergence hypothesis therefore gives \begin{align*} u_{\varepsilon_j}\to u \quad \text{in } L^1(K_\delta). \end{align*} The [mean value theorem](/theorems/186) gives, for $\mathcal{L}^2$-almost every $(t,x) \in K_\delta$, \begin{align*} |\eta(u_{\varepsilon_j}(t,x))-\eta(u(t,x))| \le L_\eta |u_{\varepsilon_j}(t,x)-u(t,x)| \end{align*} and \begin{align*} |q(u_{\varepsilon_j}(t,x))-q(u(t,x))| \le L_q |u_{\varepsilon_j}(t,x)-u(t,x)|. \end{align*} It follows that \begin{align*} \eta(u_{\varepsilon_j}) \to \eta(u) \quad \text{in } L^1(K_\delta) \end{align*} and \begin{align*} q(u_{\varepsilon_j}) \to q(u) \quad \text{in } L^1(K_\delta). \end{align*} Because $\partial_t\phi$ and $\partial_x\phi$ are bounded, we obtain \begin{align*} \int_\delta^{\,T} \int_{\mathbb{R}} \eta(u_{\varepsilon_j})\partial_t\phi\, d\mathcal{L}^1(x)\, d\mathcal{L}^1(t) \to \int_\delta^{\,T} \int_{\mathbb{R}} \eta(u)\partial_t\phi\, d\mathcal{L}^1(x)\, d\mathcal{L}^1(t) \end{align*} and \begin{align*} \int_\delta^{\,T} \int_{\mathbb{R}} q(u_{\varepsilon_j})\partial_x\phi\, d\mathcal{L}^1(x)\, d\mathcal{L}^1(t) \to \int_\delta^{\,T} \int_{\mathbb{R}} q(u)\partial_x\phi\, d\mathcal{L}^1(x)\, d\mathcal{L}^1(t). \end{align*} [/step]

[step:Control the layer near the initial time uniformly] Define \begin{align*} A_\eta := \sup_{r \in I} |\eta(r)| \end{align*} and \begin{align*} A_q := \sup_{r \in I} |q(r)|. \end{align*} Let $R_\phi>0$ be chosen so that $\operatorname{supp}\phi \subset [0,T)\times[-R_\phi,R_\phi]$. Since $u_{\varepsilon_j}$ and $u$ take values in $I$ up to null sets, the small-time contribution satisfies \begin{align*} \left|\int_0^\delta \int_{\mathbb{R}} \eta(u_{\varepsilon_j})\partial_t\phi + q(u_{\varepsilon_j})\partial_x\phi\, d\mathcal{L}^1(x)\, d\mathcal{L}^1(t)\right| \le 2R_\phi\delta\left(A_\eta\|\partial_t\phi\|_{L^\infty}+A_q\|\partial_x\phi\|_{L^\infty}\right) \end{align*} for every $j$, and the same estimate holds with $u$ in place of $u_{\varepsilon_j}$. Hence the difference between the spacetime integrals over $(0,T)\times\mathbb{R}$ and over $(\delta,T)\times\mathbb{R}$ is bounded uniformly by a constant times $\delta$. We now spell out the two-limit argument. Define \begin{align*} C_\phi := 2R_\phi\left(A_\eta\|\partial_t\phi\|_{L^\infty}+A_q\|\partial_x\phi\|_{L^\infty}\right). \end{align*} Given $\rho>0$, choose $\delta\in(0,T)$ so small that $2C_\phi\delta<\rho/2$. For this fixed $\delta$, the convergence on $K_\delta$ gives an index $J$ such that for every $j\ge J$ the difference of the entropy and flux integrals over $(\delta,T)\times\mathbb R$ is less than $\rho/2$ in absolute value. The two small-time pieces, one for $u_{\varepsilon_j}$ and one for $u$, contribute at most $2C_\phi\delta$. Hence the full spacetime entropy and flux integrals over $(0,T)\times\mathbb R$ converge as $j\to\infty$. [/step]

[step:Pass to the limit in the initial and viscosity terms] Let $K_0 \subset \mathbb{R}$ be a compact interval containing the support of the function $x \mapsto \phi(0,x)$. Since $u_{0,\varepsilon} \to u_0$ in $L^1(\mathbb{R})$ and both $u_{0,\varepsilon}$ and $u_0$ take values in $I$ up to null sets, the same Lipschitz estimate for $\eta$ gives \begin{align*} \eta(u_{0,\varepsilon_j}) \to \eta(u_0) \quad \text{in } L^1(K_0). \end{align*} Therefore \begin{align*} \int_{\mathbb{R}} \eta(u_{0,\varepsilon_j}(x))\phi(0,x)\, d\mathcal{L}^1(x) \to \int_{\mathbb{R}} \eta(u_0(x))\phi(0,x)\, d\mathcal{L}^1(x). \end{align*} For the viscosity error, define \begin{align*} A_\eta := \sup_{r \in I} |\eta(r)| \end{align*} and let $K_\phi \subset [0,T) \times \mathbb{R}$ be the compact support of $\phi$. Then \begin{align*} \left|\varepsilon_j \int_0^{\,T} \int_{\mathbb{R}} \eta(u_{\varepsilon_j})\partial_{xx}\phi\, d\mathcal{L}^1(x)\, d\mathcal{L}^1(t)\right| \le \varepsilon_j A_\eta \|\partial_{xx}\phi\|_{L^\infty} \mathcal{L}^2(K_\phi). \end{align*} The right-hand side tends to $0$ as $j \to \infty$, so the viscosity term vanishes in the limit. [/step]

[step:Obtain the entropy inequality and the conservation law]Combining the convergence away from $t=0$, the uniform small-time estimate, the initial trace convergence, and the vanishing of the viscosity term, then first passing $j \to \infty$ and afterwards letting $\delta\downarrow0$, gives \begin{align*} \int_0^{\,T} \int_{\mathbb{R}} \eta(u(t,x))\partial_t\phi(t,x) + q(u(t,x))\partial_x\phi(t,x)\, d\mathcal{L}^1(x)\, d\mathcal{L}^1(t) + \int_{\mathbb{R}} \eta(u_0(x))\phi(0,x)\, d\mathcal{L}^1(x) \ge 0. \end{align*} This is exactly the entropy inequality with initial data $u_0$ for every smooth convex entropy pair $(\eta,q)$. We next pass from smooth convex entropies to the Kruzhkov entropies. Define $\operatorname{sgn}:\mathbb R\to\{-1,0,1\}$ by setting $\operatorname{sgn}(r)=1$ for $r>0$, $\operatorname{sgn}(0)=0$, and $\operatorname{sgn}(r)=-1$ for $r<0$. Fix $k\in\mathbb R$. Let $\rho : \mathbb R\to\mathbb R$ be a nonnegative function in $C_c^\infty((-1,1))$ with \begin{align*} \int_{\mathbb R}\rho(s)\,d\mathcal L^1(s)=1. \end{align*} For each $m\in\mathbb N$, define $\rho_m: \mathbb R\to\mathbb R$ by $\rho_m(s)=m\rho(ms)$. Define the smooth convex map $\eta_{k,m}:\mathbb R\to\mathbb R$ by \begin{align*} \eta_{k,m}(r)=\int_{\mathbb R}|r-k-s|\rho_m(s)\,d\mathcal L^1(s). \end{align*} Let $q_{k,m}:\mathbb R\to\mathbb R$ be its associated entropy flux, \begin{align*} q_{k,m}(r)=\int_k^r \eta_{k,m}'(s)f'(s)\,d\mathcal L^1(s). \end{align*} On the compact interval $I$, the functions $\eta_{k,m}$ converge uniformly to $r\mapsto |r-k|$, and $\eta_{k,m}'$ converge pointwise for $r\ne k$ to $\operatorname{sgn}(r-k)$ with $|\eta_{k,m}'|\le1$. Define $q_k:I\to\mathbb R$ by \begin{align*} q_k(r)=\operatorname{sgn}(r-k)(f(r)-f(k)). \end{align*} For $r\in I$, the fundamental theorem of calculus gives \begin{align*} |q_{k,m}(r)-q_k(r)|\le \int_I |\eta_{k,m}'(s)-\operatorname{sgn}(s-k)|\,|f'(s)|\,d\mathcal L^1(s). \end{align*} The preceding estimate is only valid when the integration interval contains both $r$ and $k$. Define the compact interval \begin{align*} I_k := [\min\{-M,k\},\max\{M,k\}]. \end{align*} For $r\in I$, both $r$ and $k$ lie in $I_k$, and the [Fundamental Theorem of Calculus](/theorems/632) gives \begin{align*} |q_{k,m}(r)-q_k(r)|\le \int_{I_k} |\eta_{k,m}'(s)-\operatorname{sgn}(s-k)|\,|f'(s)|\,d\mathcal L^1(s). \end{align*} The integrand is bounded by $2\sup_{s\in I_k}|f'(s)|$, an $\mathcal L^1$-integrable constant on the compact interval $I_k$, and it converges to $0$ for $\mathcal L^1$-almost every $s\in I_k$. The [Dominated Convergence Theorem](/theorems/4) with respect to $\mathcal L^1$ on $I_k$ therefore shows that $q_{k,m}\to q_k$ uniformly on $I$. Since $u$ and $u_0$ take values in $I$ up to null sets, the [uniform convergence](/page/Uniform%20Convergence) of $\eta_{k,m}$ and $q_{k,m}$ on the relevant bounded range handles the spacetime entropy term, the spacetime flux term, and the initial term containing $\eta_{k,m}(u_0)$. Passing $m\to\infty$ in the smooth entropy inequality gives the Kruzhkov entropy inequality for the entropy pair $(|r-k|,\operatorname{sgn}(r-k)(f(r)-f(k)))$. Since $k\in\mathbb R$ was arbitrary, $u$ satisfies all Kruzhkov entropy inequalities with initial data $u_0$. To recover the weak conservation law, take the affine entropy $\eta_1(r)=r$, whose entropy flux is $q_1(r)=f(r)-f(0)$, and the affine entropy $\eta_2(r)=-r$, whose entropy flux is $q_2(r)=-f(r)+f(0)$. Applying the smooth entropy inequality to $\eta_1$ and $\eta_2$ gives opposite inequalities. Adding the harmless constant terms involving $f(0)\partial_x\phi$, whose spacetime integral vanishes because $\phi$ has compact spatial support, yields equality: \begin{align*} \int_0^{\,T} \int_{\mathbb{R}} u(t,x)\partial_t\phi(t,x) + f(u(t,x))\partial_x\phi(t,x)\, d\mathcal{L}^1(x)\, d\mathcal{L}^1(t) + \int_{\mathbb{R}} u_0(x)\phi(0,x)\, d\mathcal{L}^1(x) = 0. \end{align*} The entropy inequality above contains the initial boundary term $\int_{\mathbb R}\eta(u_0(x))\phi(0,x)\,d\mathcal L^1(x)$. We now use the standard weak-in-time initial-trace theorem for scalar Kruzhkov entropy solutions, viewed as an external theorem from the scalar [entropy solution](/page/Entropy%20Solution) theory. In the form needed here, it says: if $v\in L^\infty((0,T)\times\mathbb R)\cap L^1_{\mathrm{loc}}((0,T)\times\mathbb R)$ satisfies the weak conservation law \begin{align*} \int_0^{\,T} \int_{\mathbb R} v\,\partial_t\psi+f(v)\,\partial_x\psi\,d\mathcal L^1(x)\,d\mathcal L^1(t)+\int_{\mathbb R}v_0(x)\psi(0,x)\,d\mathcal L^1(x)=0 \end{align*} for every $\psi\in C_c^\infty([0,T)\times\mathbb R)$, and satisfies every Kruzhkov entropy inequality with the boundary term determined by $v_0\in L^\infty(\mathbb R)\cap L^1_{\mathrm{loc}}(\mathbb R)$, then, after changing $v$ on a null set of times if necessary, \begin{align*} v(t,\cdot)\to v_0 \quad \text{in } L^1_{\mathrm{loc}}(\mathbb R) \text{ as } t\downarrow0. \end{align*} Its hypotheses are satisfied here: $u\in L^\infty((0,T)\times\mathbb R)\cap L^1_{\mathrm{loc}}((0,T)\times\mathbb R)$ by the uniform range bound and local convergence, $u_0\in L^\infty(\mathbb R)\cap L^1(\mathbb R)\subset L^1_{\mathrm{loc}}(\mathbb R)$ by hypothesis, $u$ satisfies the weak conservation law just proved for every test function in $C_c^\infty([0,T)\times\mathbb R)$, and the Kruzhkov inequalities hold for every $k\in\mathbb R$ with the initial boundary term involving $u_0$. Therefore \begin{align*} u(t,\cdot)\to u_0 \quad \text{in } L^1_{\mathrm{loc}}(\mathbb R) \text{ as } t\downarrow0. \end{align*} Thus $u$ is a bounded Kruzhkov [entropy solution](/page/Entropy%20Solution) of $\partial_tu+\partial_xf(u)=0$ with initial trace $u_0$.[/step]

[guided]We have proved the entropy inequality for every smooth convex entropy pair. To obtain Kruzhkov's nonsmooth entropy pair at a fixed level $k\in\mathbb R$, we approximate $r\mapsto |r-k|$ by convolution. Choose a nonnegative map $\rho:\mathbb R\to\mathbb R$ in $C_c^\infty((-1,1))$ with total mass one, and define $\rho_m:\mathbb R\to\mathbb R$ by $\rho_m(s)=m\rho(ms)$. The map \begin{align*} \eta_{k,m}(r)=\int_{\mathbb R}|r-k-s|\rho_m(s)\,d\mathcal L^1(s) \end{align*} is smooth and convex, so it is an admissible entropy. Its associated entropy flux is the map $q_{k,m}:\mathbb R\to\mathbb R$ given by \begin{align*} q_{k,m}(r)=\int_k^r \eta_{k,m}'(s)f'(s)\,d\mathcal L^1(s). \end{align*} The uniform $L^\infty$ bound already proved says that the values of $u$ and $u_0$ lie in the compact interval $I=[-M,M]$ up to null sets. On this interval, $\eta_{k,m}$ converges uniformly to $|r-k|$. Also $\eta_{k,m}'(s)$ converges to $\operatorname{sgn}(s-k)$ for every $s\ne k$, and $|\eta_{k,m}'(s)|\le1$. Define $q_k:I\to\mathbb R$ by $q_k(r)=\operatorname{sgn}(r-k)(f(r)-f(k))$. For every $r\in I$, both $r$ and $k$ lie in the compact interval \begin{align*} I_k := [\min\{-M,k\},\max\{M,k\}], \end{align*} and the [Fundamental Theorem of Calculus](/theorems/632) gives \begin{align*} |q_{k,m}(r)-q_k(r)|\le \int_{I_k} |\eta_{k,m}'(s)-\operatorname{sgn}(s-k)|\,|f'(s)|\,d\mathcal L^1(s). \end{align*} The right-hand side does not depend on $r$, is dominated by the $\mathcal L^1$-integrable constant $2\sup_{s\in I_k}|f'(s)|$ on $I_k$, and tends to zero by the [Dominated Convergence Theorem](/theorems/4). Hence $q_{k,m}\to q_k$ uniformly on $I$. Since $u$ and $u_0$ take values in $I$ up to null sets, the uniform convergence of $\eta_{k,m}$ and $q_{k,m}$ on the bounded range justifies passing to the limit in the entropy term, the flux term, and the initial term. We obtain the Kruzhkov entropy inequality for $(|r-k|,\operatorname{sgn}(r-k)(f(r)-f(k)))$. The conservation law itself is recovered by applying the smooth entropy inequality to the affine entropies $\eta_1(r)=r$ and $\eta_2(r)=-r$. Their entropy fluxes differ from $f$ and $-f$ only by the harmless constants $-f(0)$ and $f(0)$. Since $\phi$ has compact spatial support, the integral of each constant multiple of $\partial_x\phi$ over $\mathbb R$ vanishes. The two opposite entropy inequalities combine into the weak equality \begin{align*} \int_0^{\,T} \int_{\mathbb{R}} u(t,x)\partial_t\phi(t,x) + f(u(t,x))\partial_x\phi(t,x)\, d\mathcal{L}^1(x)\, d\mathcal{L}^1(t) + \int_{\mathbb{R}} u_0(x)\phi(0,x)\, d\mathcal{L}^1(x) = 0. \end{align*} We use the weak-in-time Kruzhkov initial-trace theorem in its scalar conservation law form. It assumes that $v\in L^\infty((0,T)\times\mathbb R)\cap L^1_{\mathrm{loc}}((0,T)\times\mathbb R)$ satisfies the weak conservation law with an initial boundary term determined by $v_0\in L^\infty(\mathbb R)\cap L^1_{\mathrm{loc}}(\mathbb R)$ and satisfies every Kruzhkov inequality with the same boundary datum. Its conclusion is \begin{align*} v(t,\cdot)\to v_0 \quad \text{in } L^1_{\mathrm{loc}}(\mathbb R) \text{ as } t\downarrow0, \end{align*} after changing $v$ on a null set of times if necessary. For $v=u$ and $v_0=u_0$, the hypotheses are all checked above: the uniform range bound gives $u\in L^\infty((0,T)\times\mathbb R)$, the local convergence gives $u\in L^1_{\mathrm{loc}}((0,T)\times\mathbb R)$, the assumption gives $u_0\in L^\infty(\mathbb R)\cap L^1(\mathbb R)\subset L^1_{\mathrm{loc}}(\mathbb R)$, the affine entropies give the weak conservation law, and the smooth-to-Kruzhkov approximation gives every Kruzhkov inequality with the initial boundary term. Thus \begin{align*} u(t,\cdot)\to u_0 \quad \text{in } L^1_{\mathrm{loc}}(\mathbb R) \text{ as } t\downarrow0. \end{align*} Together with the Kruzhkov inequalities, this proves that $u$ is a bounded Kruzhkov entropy solution with initial trace $u_0$. The uniqueness conclusion is now formal, but we spell it out in the same local $L^1$ topology. Let $\varepsilon_j\downarrow0$ and $\delta_j\downarrow0$ be two vanishing-viscosity sequences such that \begin{align*} u_{\varepsilon_j}\to u, \qquad u_{\delta_j}\to v \end{align*} in $L^1_{\mathrm{loc}}((0,T)\times\mathbb R)$. Applying the preceding argument first to the sequence $\varepsilon_j$ and then to the sequence $\delta_j$ shows that both $u$ and $v$ are bounded Kruzhkov entropy solutions with initial trace $u_0$. The theorem assumes uniqueness in that bounded local $L^1$-stable entropy-solution class, so \begin{align*} u=v \end{align*} almost everywhere on $(0,T)\times\mathbb R$. Finally assume that $\{u_\varepsilon\}_{\varepsilon>0}$ is relatively compact in $L^1_{\mathrm{loc}}((0,T)\times\mathbb R)$, and let $u_*$ denote the unique bounded Kruzhkov entropy solution with initial trace $u_0$. To prove full convergence, suppose the contrary. Then there are a compact set $K\subset(0,T)\times\mathbb R$, a number $\alpha>0$, and a sequence $\varepsilon_j\downarrow0$ such that \begin{align*} \|u_{\varepsilon_j}-u_*\|_{L^1(K)}\ge\alpha \end{align*} for every $j$. Relative compactness gives a further subsequence, still denoted $u_{\varepsilon_j}$ after relabeling, converging in $L^1_{\mathrm{loc}}((0,T)\times\mathbb R)$ to some limit $w$. The first part identifies $w$ as a bounded Kruzhkov entropy solution with initial trace $u_0$, and uniqueness gives $w=u_*$. Hence \begin{align*} \|u_{\varepsilon_j}-u_*\|_{L^1(K)}\to0, \end{align*} contradicting the lower bound by $\alpha$. Therefore $u_\varepsilon\to u_*$ in $L^1_{\mathrm{loc}}((0,T)\times\mathbb R)$ as $\varepsilon\downarrow0$.[/guided]

[step:Use uniqueness and compactness to identify all vanishing viscosity limits] Assume that the scalar Kruzhkov Cauchy problem for the flux $f$ and initial datum $u_0$ has at most one bounded entropy solution in the local $L^1$-stable class. Let $(\delta_j)_{j\in\mathbb N}$ be a second positive sequence with $\delta_j\downarrow0$. If two subsequences $u_{\varepsilon_j}$ and $u_{\delta_j}$ converge in $L^1_{\mathrm{loc}}((0,T)\times\mathbb{R})$ to limits $u$ and $v$, respectively, then the preceding argument shows that both $u$ and $v$ are bounded Kruzhkov entropy solutions with initial trace $u_0$. By uniqueness in that class, \begin{align*} u=v \end{align*} almost everywhere on $(0,T)\times\mathbb R$. Hence every convergent vanishing-viscosity subsequence has the same limit. Now assume additionally that $\{u_\varepsilon\}_{\varepsilon>0}$ is relatively compact in $L^1_{\mathrm{loc}}((0,T)\times\mathbb{R})$. Let $u_* : (0,T)\times\mathbb R\to\mathbb R$ denote the unique entropy solution in the bounded $L^1$-stable entropy-solution class. To prove convergence of the full family, suppose otherwise. Then there exist a compact set $K \subset (0,T)\times\mathbb{R}$, a number $\delta>0$, and a sequence $\varepsilon_j \downarrow 0$ such that \begin{align*} \|u_{\varepsilon_j}-u_*\|_{L^1(K)} \ge \delta \end{align*} for every $j$. Relative compactness gives a further subsequence, still denoted $u_{\varepsilon_j}$ after relabeling, converging in $L^1_{\mathrm{loc}}((0,T)\times\mathbb{R})$ to some limit $w$. The first part of the proof shows that $w$ is an entropy solution with initial data $u_0$, so uniqueness gives $w=u_*$. In particular, \begin{align*} \|u_{\varepsilon_j}-u_*\|_{L^1(K)} \to 0, \end{align*} contradicting the lower bound by $\delta$. Therefore the full family $u_\varepsilon$ converges to $u_*$ in $L^1_{\mathrm{loc}}((0,T)\times\mathbb{R})$ as $\varepsilon \downarrow 0$. [/step]