The viscous scalar Cauchy problem with smooth data has a global classical solution

Setup and kernel estimates. Put $M_0:=\mathrm{\parallel }{u}_0{\mathrm{\parallel }}_{\mathrm{\infty }}$, $R:=2M_0+1$ and $L:={\sup }_{\mathrm{\mid }s\mathrm{\mid }\le R}\mathrm{\mid }{f}^{\mathrm{\prime }}(s)\mathrm{\mid }<\mathrm{\infty }$. Since $f(0)=0$ and $f$ is $C^1$, the mean value theorem gives $\mathrm{\mid }f(s)\mathrm{\mid }\le L\mathrm{\mid }s\mathrm{\mid }$ for $\mathrm{\mid }s\mathrm{\mid }\le R$. By [F2] there are constants $C_1,C_2$ with $\mathrm{\parallel }\mathrm{\nabla }{H}_{\sigma }g{\mathrm{\parallel }}_p\le C_1{\sigma }^{-1\mathrm{/}2}\mathrm{\parallel }g{\mathrm{\parallel }}_p$ for $p\in \{1,\mathrm{\infty }\}$. The Gaussian first moment gives $\mathrm{\parallel }{H}_{\sigma }g-g{\mathrm{\parallel }}_{\mathrm{\infty }}\le C_2{\sigma }^{1\mathrm{/}2}\mathrm{\parallel }\mathrm{\nabla }g{\mathrm{\parallel }}_{\mathrm{\infty }}$ for $g\in C^1\cap L^{\mathrm{\infty }}$ with $\mathrm{\nabla }g$ bounded. Also, $\mathrm{\parallel }\mathrm{\nabla }{\mathrm{\Gamma }}_{\sigma }(\cdot +h)-\mathrm{\nabla }{\mathrm{\Gamma }}_{\sigma }{\mathrm{\parallel }}_1\le \min \{2C_1{\sigma }^{-1\mathrm{/}2},C_2\mathrm{\mid }h\mathrm{\mid }{\sigma }^{-1}\}$ and $\mathrm{\parallel }{\mathrm{\Gamma }}_{\sigma }(\cdot +h)-{\mathrm{\Gamma }}_{\sigma }{\mathrm{\parallel }}_1\le \min \{2,C_2\mathrm{\mid }h\mathrm{\mid }{\sigma }^{-1\mathrm{/}2}\}$ for every $h\in {\mathbb{R}}^n$ and $\sigma >0$. For the scaled kernel, $\mathrm{\parallel }{\mathrm{\partial }}_{\sigma }\mathrm{\nabla }{\mathrm{\Gamma }}_{\epsilon \sigma }{\mathrm{\parallel }}_1\le C_{\epsilon }{\sigma }^{-3\mathrm{/}2}$, so for $\mathrm{\Delta }>0$ and $\sigma >0$, $\mathrm{\parallel }\mathrm{\nabla }{\mathrm{\Gamma }}_{\epsilon (\sigma +\mathrm{\Delta })}-\mathrm{\nabla }{\mathrm{\Gamma }}_{\epsilon \sigma }{\mathrm{\parallel }}_1\le C_{\epsilon }\min \{{\sigma }^{-1\mathrm{/}2},\mathrm{\Delta }{\sigma }^{-3\mathrm{/}2}\}\mathrm{.}$

A local mild solution for bounded uniformly continuous $L^1$ data. Fix a starting datum $g\in BUC({\mathbb{R}}^n)\cap L^1({\mathbb{R}}^n)$ with $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\le M_0$. Choose ${\tau }_0\in (0,T]$ with $q:=2C_{1}L\sqrt{{\tau }_0\mathrm{/}\epsilon }\le \frac{1}{2}$ and let $X_{{\tau }_0}:=C([0,{\tau }_0];C_b({\mathbb{R}}^n))$ with the supremum norm (Closed subspaces of complete metric spaces are complete; the converse under countable choice). For $v\in X_{{\tau }_0}$ with $\mathrm{\parallel }v{\mathrm{\parallel }}_{X_{{\tau }_0}}\le R$ define $({\mathcal{T}}_{g}v)(t):=H_{\epsilon t}g-{\int }_0^t\mathrm{\nabla }{H}_{\epsilon (t-s)}\ast f(v(s))ds\mathrm{.}$ By [F6], $t\mapsto H_{\epsilon t}g$ is continuous in the supremum norm at $t=0$; by [F1] the Duhamel term is continuous in $X_{{\tau }_0}$, since its integrand is norm continuous for $s<t$ and has the integrable majorant $C_1(\epsilon (t-s){)}^{-1\mathrm{/}2}\mathrm{\parallel }f(v(s)){\mathrm{\parallel }}_{\mathrm{\infty }}$. The bounds $\mathrm{\parallel }{H}_{\epsilon t}g{\mathrm{\parallel }}_{\mathrm{\infty }}\le M_0$ and $\mathrm{\parallel }f(v(s)){\mathrm{\parallel }}_{\mathrm{\infty }}\le LR$ give $\mathrm{\parallel }{\mathcal{T}}_{g}v{\mathrm{\parallel }}_{X_{{\tau }_0}}\le M_0+qR\le R,\mathrm{\parallel }{\mathcal{T}}_{g}u-{\mathcal{T}}_{g}v{\mathrm{\parallel }}_{X_{{\tau }_0}}\le q\mathrm{\parallel }u-v{\mathrm{\parallel }}_{X_{{\tau }_0}}\mathrm{.}$ Thus ${\mathcal{T}}_g$ is a contraction of the closed radius-$R$ ball; the Banach fixed point theorem gives a unique fixed point $u\in X_{{\tau }_0}$ satisfying $u(t)=H_{\epsilon t}g-{\int }_0^t\mathrm{\nabla }{H}_{\epsilon (t-s)}\ast f(u(s))ds$. The original datum $u_0\in C_c^{\mathrm{\infty }}$ satisfies these assumptions.

Hölder regularity on strips. Fix $0<\delta <{\tau }_0$ and let $t,t^{\mathrm{\prime }}\in [\delta ,{\tau }_0]$. Since $f(u)$ is bounded with $\mathrm{\parallel }f(u){\mathrm{\parallel }}_{\mathrm{\infty }}\le LR$, the mild identity and the kernel bounds of step 1.1 give, for every $\alpha \in (0,1)$, $\mathrm{\parallel }u(t,\cdot +h)-u(t,\cdot ){\mathrm{\parallel }}_{\mathrm{\infty }}\le C_{\delta }\mathrm{\mid }h{\mathrm{\mid }}^{\alpha },\mathrm{\parallel }u(t,\cdot )-u(t^{\mathrm{\prime }},\cdot ){\mathrm{\parallel }}_{\mathrm{\infty }}\le C_{\delta }\mathrm{\mid }t-t^{\mathrm{\prime }}{\mathrm{\mid }}^{1\mathrm{/}2}\mathrm{.}$ For the spatial estimate, $\mathrm{\parallel }\mathrm{\nabla }{H}_{\epsilon t}g{\mathrm{\parallel }}_{\mathrm{\infty }}\le C_{\epsilon ,\delta }\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$ on the strip. The Duhamel term is bounded by $LR{\int }_0^t\min \{C_{\epsilon }{\sigma }^{-1\mathrm{/}2},C_{\epsilon }\mathrm{\mid }h\mathrm{\mid }{\sigma }^{-1}\}d\sigma$, with $t\le {\tau }_0$; splitting at $\sigma =\mathrm{\mid }h{\mathrm{\mid }}^2$ gives at most $C_{\epsilon }(\mathrm{\mid }h\mathrm{\mid }+\mathrm{\mid }h\mathrm{\mid }{\log }_{+}({\tau }_0\mathrm{/}\mathrm{\mid }h{\mathrm{\mid }}^2))\le C_{\epsilon ,{\tau }_0,\alpha }\mathrm{\mid }h{\mathrm{\mid }}^{\alpha }$ for $0<\mathrm{\mid }h\mathrm{\mid }\le 1$, where ${\log }_{+}(r):=\max \{\log r,0\}$. Larger $\mathrm{\mid }h\mathrm{\mid }$ are covered by boundedness of $u$. For the temporal estimate assume $t=t^{\mathrm{\prime }}+\mathrm{\Delta }$ with $\mathrm{\Delta }>0$. The initial heat term is Lipschitz in $t$ on $[\delta ,{\tau }_0]$ by the positive-time estimate for ${\mathrm{\partial }}_t{H}_{\epsilon t}g=\epsilon \mathrm{\Delta }{H}_{\epsilon t}g$. On the common Duhamel interval, with $\sigma =t^{\mathrm{\prime }}-s$, the kernel difference is a time difference and satisfies $\mathrm{\parallel }\mathrm{\nabla }{\mathrm{\Gamma }}_{\epsilon (\sigma +\mathrm{\Delta })}-\mathrm{\nabla }{\mathrm{\Gamma }}_{\epsilon \sigma }{\mathrm{\parallel }}_1\le C_{\epsilon }\min \{{\sigma }^{-1\mathrm{/}2},\mathrm{\Delta }{\sigma }^{-3\mathrm{/}2}\}\mathrm{.}$ Splitting its integral at $\sigma =\mathrm{\Delta }$ gives ${\int }_0^{t^{\mathrm{\prime }}}\mathrm{\parallel }\mathrm{\nabla }{\mathrm{\Gamma }}_{\epsilon (\sigma +\mathrm{\Delta })}-\mathrm{\nabla }{\mathrm{\Gamma }}_{\epsilon \sigma }{\mathrm{\parallel }}_{1}d\sigma \le C_{\epsilon }\sqrt{\mathrm{\Delta }}$. The remaining time slab has norm at most $LR{\int }_{t^{\mathrm{\prime }}}^t{C}_1(\epsilon (t-s){)}^{-1\mathrm{/}2}ds\le 2C_{1}LR\sqrt{\mathrm{\Delta }\mathrm{/}\epsilon }$. Thus both estimates hold with $C_{\delta }$ depending only on $\delta ,{\tau }_0,\epsilon ,n,\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }},LR$, and $u$ is continuous in $(t,x)$ for $t>0$.

$L^1$ continuity and identification of the limits. For the arbitrary starting datum $g\in BUC\cap L^1$ of step 2.1, take Picard iterates $u^{(0)}:=g$, $u^{(k+1)}:={\mathcal{T}}_g{u}^{(k)}$. By [F6], $u^{(0)}\in C([0,{\tau }_0];L^1)$. If $u^{(k)}\in C([0,{\tau }_0];L^1)$, the integral defining $u^{(k+1)}$ is defined by norm limits of Riemann sums on truncated intervals $s\le t-\eta$. Completeness supplies these limits, and the bound on the omitted interval is at most $2C_{1}L\mathrm{\parallel }{u}^{(k)}{\mathrm{\parallel }}_{C_t{L}^1}\sqrt{\eta \mathrm{/}\epsilon }$, which tends to zero. Its integrand has integrable majorant $C_1(\epsilon s{)}^{-1\mathrm{/}2}L\mathrm{\parallel }{u}^{(k)}{\mathrm{\parallel }}_{C_t{L}^1}$, so $u^{(k+1)}$ is $L^1$-continuous and $\mathrm{\parallel }{u}^{(k+1)}{\mathrm{\parallel }}_{C_t{L}^1}\le \mathrm{\parallel }g{\mathrm{\parallel }}_1+q\mathrm{\parallel }{u}^{(k)}{\mathrm{\parallel }}_{C_t{L}^1}\le 2\mathrm{\parallel }g{\mathrm{\parallel }}_1$. Successive differences obey $\mathrm{\parallel }{u}^{(k+1)}-u^{(k)}{\mathrm{\parallel }}_{C_t{L}^1}\le q\mathrm{\parallel }{u}^{(k)}-u^{(k-1)}{\mathrm{\parallel }}_{C_t{L}^1}$, so the iterates converge in the complete space $C([0,{\tau }_0];L^1)$ (Riesz-Fischer completeness of $L^p$ for $1\le p\le \mathrm{\infty }$) to some $w$. They also converge uniformly on $[0,{\tau }_0]\times {\mathbb{R}}^n$ to the fixed point $u$ of step 2.1. On every bounded ball $B_m$, this uniform convergence implies convergence to $u$ in $C([0,{\tau }_0];L^1(B_m))$, while the global $L^1$ convergence gives convergence to $w$ in the same local space. Uniqueness of the local $L^1$ limit yields $u=w$ a.e. on every $B_m$, hence a.e. on ${\mathbb{R}}^n$; thus $u\in C([0,{\tau }_0];L^1)$ and $\mathrm{\parallel }u(t){\mathrm{\parallel }}_1\le 2\mathrm{\parallel }g{\mathrm{\parallel }}_1$.

The equation in distributions. Testing the mild identity and using Fubini (the bound $(t-s{)}^{-1\mathrm{/}2}$ is integrable on each finite time triangle) gives ${\int }_{\mathrm{\Pi }}(u{\phi }_t+f(u)\cdot \mathrm{\nabla }\phi )+\int g\phi (\cdot ,0)=-\epsilon {\int }_{\mathrm{\Pi }}u\mathrm{\Delta }\phi \mathrm{.}$ Indeed the initial heat term pairs as $-\int g\phi (\cdot ,0)$ against ${\phi }_t+\epsilon \mathrm{\Delta }\phi$, and the divergence Duhamel term pairs as $-\int f(u)\cdot \mathrm{\nabla }\phi$ against the same expression, by integrating ${\mathrm{\partial }}_t{H}_{\epsilon (t-s)}=\epsilon \mathrm{\Delta }{H}_{\epsilon (t-s)}$ in $t$. Thus $u_t+\mathrm{div}f(u)=\epsilon \mathrm{\Delta }u$ distributionally, with datum $g$.

Hölder continuity and boundedness of the spatial gradient. Fix $0<{\delta }_0<\delta <{\tau }_0$ and use the mild identity restarted at ${\delta }_0$: $u(t)=H_{\epsilon (t-{\delta }_0)}u({\delta }_0)-\mathrm{div}W(t),W(t):={\int }_{{\delta }_0}^t{H}_{\epsilon (t-s)}f(u(s))ds\mathrm{.}$ By step 2.2, $f(u)$ is spatially $C^{\alpha }$ and temporally $C^{\alpha \mathrm{/}2}$ on each positive strip (the $C^{1\mathrm{/}2}$ bound implies the weaker $C^{\alpha \mathrm{/}2}$ bound). Componentwise, differentiation of the divergence heat potential gives ${\mathrm{\partial }}_{k}u(t,x)={\mathrm{\partial }}_k{H}_{\epsilon (t-{\delta }_0)}u({\delta }_0,x)-{\sum }_i{\int }_0^{t-{\delta }_0}{\int }_{{\mathbb{R}}^n}{D}_{ki}{\mathrm{\Gamma }}_{\epsilon \sigma }(z)(f_i(u(t-\sigma ,x-z))-f_i(u(t,x)))dzd\sigma ,$ where the frozen value is subtracted using $\int D_{ki}{\mathrm{\Gamma }}_{\epsilon \sigma }=0$. Gaussian scaling gives $\mathrm{\parallel }{D}^2{\mathrm{\Gamma }}_{\epsilon \sigma }{\mathrm{\parallel }}_1\le C_{\epsilon }{\sigma }^{-1}$, $\mathrm{\parallel }\mathrm{\nabla }{D}^2{\mathrm{\Gamma }}_{\epsilon \sigma }{\mathrm{\parallel }}_1\le C_{\epsilon }{\sigma }^{-3\mathrm{/}2}$, and $\mathrm{\parallel }{\mathrm{\partial }}_{\sigma }{D}^2{\mathrm{\Gamma }}_{\epsilon \sigma }{\mathrm{\parallel }}_1\le C_{\epsilon }{\sigma }^{-2}$. With the parabolic $C^{\alpha ,\alpha \mathrm{/}2}$ modulus of $f(u)$, a spatial increment $h$ is bounded after splitting at $\sigma =\mathrm{\mid }h{\mathrm{\mid }}^2$ by $C{\int }_0^{\mathrm{\mid }h{\mathrm{\mid }}^2}{\sigma }^{-1+\alpha \mathrm{/}2}d\sigma +C\mathrm{\mid }h\mathrm{\mid }{\int }_{\mathrm{\mid }h{\mathrm{\mid }}^2}^{{\tau }_0-{\delta }_0}{\sigma }^{-3\mathrm{/}2+\alpha \mathrm{/}2}d\sigma \le C\mathrm{\mid }h{\mathrm{\mid }}^{\alpha },$ and a time increment $\mathrm{\Delta }=\mathrm{\mid }t-t^{\mathrm{\prime }}\mathrm{\mid }$ is bounded after splitting at $\sigma =\mathrm{\Delta }$ by $C{\int }_0^{\mathrm{\Delta }}{\sigma }^{-1+\alpha \mathrm{/}2}d\sigma +C\mathrm{\Delta }{\int }_{\mathrm{\Delta }}^{{\tau }_0-{\delta }_0}{\sigma }^{-2+\alpha \mathrm{/}2}d\sigma \le C{\mathrm{\Delta }}^{\alpha \mathrm{/}2}\mathrm{.}$ If either split point exceeds the finite integration horizon, the same bounds follow by increasing $C$. The heat initial term is smooth with bounded derivatives on $t\ge \delta$. Thus for each $0<\alpha <1$, $[\mathrm{\nabla }u{]}_{C_x^{\alpha }({\mathbb{R}}^n\times [\delta ,{\tau }_0))}+[\mathrm{\nabla }u{]}_{C_t^{\alpha \mathrm{/}2}({\mathbb{R}}^n\times [\delta ,{\tau }_0))}\le C_{{\delta }_0,\delta }\mathrm{.}$ These are seminorm bounds; in particular they establish continuity of $\mathrm{\nabla }u$ before any absolute bound is used. The absolute gradient bound follows from boundedness of $u$. If $\mathrm{\nabla }u(t,x)\mathrm{\ne }0$, put $e=\mathrm{\nabla }u(t,x)\mathrm{/}\mathrm{\mid }\mathrm{\nabla }u(t,x)\mathrm{\mid }$. Along the unit segment $x+se$, $0\le s\le 1$, the fundamental theorem of calculus and the spatial seminorm $M_{\delta }:=[\mathrm{\nabla }u{]}_{C_x^{\alpha }}$ give $\mathrm{\mid }\mathrm{\nabla }u(t,x)\mathrm{\mid }\le \mathrm{\mid }u(t,x+e)-u(t,x)\mathrm{\mid }+{\int }_0^1\mathrm{\mid }\mathrm{\nabla }u(t,x+se)-\mathrm{\nabla }u(t,x)\mathrm{\mid }ds\le 2R+\frac{M_{\delta }}{1+\alpha }\mathrm{.}$ Thus $\mathrm{\nabla }u$ is bounded on the positive-time strip. Since $f^{\mathrm{\prime }\mathrm{\prime }}$ is bounded on $[-R,R]$, $f^{\mathrm{\prime }}(u)$ has the same parabolic Hölder regularity as $u$ there; hence $h:=-f^{\mathrm{\prime }}(u)\cdot \mathrm{\nabla }u$ is bounded and belongs to $C^{\alpha ,\alpha \mathrm{/}2}({\mathbb{R}}^n\times [\delta ,{\tau }_0))$.

Heat-potential cancellation and $C^{1,2}$. Fix $0<\delta <{\tau }_0$, use the bounds of step 3.3 on a slightly larger positive strip, and restart the heat equation at $\delta$. Then $u(t)=H_{\epsilon (t-\delta )}u(\delta )+w(t),w(t,x):={\int }_{\delta }^t{H}_{\epsilon (t-s)}h(s)(x)ds\mathrm{.}$ For $\sigma >0$, ${\int }_{{\mathbb{R}}^n}{D}_{ij}{\mathrm{\Gamma }}_{\epsilon \sigma }(z)dz=0$, so cancellation gives $D_{ij}w(t,x)={\int }_0^{t-\delta }{\int }_{{\mathbb{R}}^n}{D}_{ij}{\mathrm{\Gamma }}_{\epsilon \sigma }(z)(h(t-\sigma ,x-z)-h(t,x))dzd\sigma \mathrm{.}$ The parabolic Hölder bound from step 3.3 yields $\mathrm{\mid }h(t-\sigma ,x-z)-h(t,x)\mathrm{\mid }\le C(\mathrm{\mid }z{\mathrm{\mid }}^{\alpha }+{\sigma }^{\alpha \mathrm{/}2})$. Gaussian scaling therefore gives ${\int }_{{\mathbb{R}}^n}\mathrm{\mid }{D}^2{\mathrm{\Gamma }}_{\epsilon \sigma }(z)\mathrm{\mid }(\mathrm{\mid }z{\mathrm{\mid }}^{\alpha }+{\sigma }^{\alpha \mathrm{/}2})dz\le C_{\epsilon ,n,\alpha }{\sigma }^{-1+\alpha \mathrm{/}2},$ which is integrable at $\sigma =0$. Truncating the heat-potential integral at $\sigma =\eta >0$, differentiating, and letting $\eta \downarrow 0$ with this integrable dominator shows that $D_{ij}w$ exists and is continuous; the heat-kernel identity also gives $w_t-\epsilon \mathrm{\Delta }w=h$ in distributions. Since $h$ and $\mathrm{\Delta }w$ are continuous, this identity gives a continuous classical time derivative. The first term $H_{\epsilon (t-\delta )}u(\delta )$ is smooth for $t>\delta$, so $u\in C^{1,2}({\mathbb{R}}^n\times (\delta ,{\tau }_0))$ and satisfies $u_t=\epsilon \mathrm{\Delta }u-f^{\mathrm{\prime }}(u)\cdot \mathrm{\nabla }u$, hence $u_t+{\mathrm{div}}_{x}f(u)=\epsilon \mathrm{\Delta }u$ pointwise. Since $\delta >0$ is arbitrary, $u\in C^{1,2}({\mathbb{R}}^n\times (0,{\tau }_0))$.

The range bound by a barrier. Let $a<b$ in $[0,{\tau }_0)$ and put $M_a:={\sup }_{x}u(a,x)$. Write $B:=f^{\mathrm{\prime }}(u(t,x))$, which is bounded on $[a,b]\times {\mathbb{R}}^n$ by $L$. For ${\delta }_1>0$ and $\lambda >0$ set $\mathrm{\Phi }(t,x):={\delta }_1{e}^{\lambda (t-a)}(1+\mathrm{\mid }x{\mathrm{\mid }}^2)$; then, using steps 4.1 and [F4], $({\mathrm{\partial }}_t-\epsilon \mathrm{\Delta }+B\cdot \mathrm{\nabla })\mathrm{\Phi }={\delta }_1{e}^{\lambda (t-a)}(\lambda (1+\mathrm{\mid }x{\mathrm{\mid }}^2)-2\epsilon n+2B\cdot x)>0$ for $\lambda$ large enough, because $\mathrm{\mid }B\mathrm{\mid }\le L$ and $-\epsilon \mathrm{\Delta }(1+\mathrm{\mid }x{\mathrm{\mid }}^2)=-2\epsilon n$. The function $z:=u-M_a-\mathrm{\Phi }$ is negative at $t=a$ and, by step 2.1, $\mathrm{\mid }u\mathrm{\mid }\le R$ while $\mathrm{\Phi }\ge {\delta }_1(1+\mathrm{\mid }x{\mathrm{\mid }}^2)$, so $z$ is negative on the lateral boundary of every sufficiently large ball intersected with $[a,b]$; if $z$ had a positive maximum in the closed cylinder, then at that point $z_t\ge 0$, $Dz=0$, $\mathrm{\Delta }z\le 0$ (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line), contradicting $({\mathrm{\partial }}_t-\epsilon \mathrm{\Delta }+B\cdot \mathrm{\nabla })z<0$ there. Letting ${\delta }_1\downarrow 0$ gives $u\le M_a$ on $[a,b]$; applying the same argument to $-u$, whose flux is $\tilde{f}(s)=-f(-s)$ and whose range bound is the same, gives $u\ge m_a:={\inf }_{x}u(a,x)$. In particular, with $a=0$, ${\sup }_{x}u(t,x)\le {\sup }_x{u}_0$ and ${\inf }_{x}u(t,x)\ge {\inf }_x{u}_0$ on $[0,{\tau }_0)$, the range of $u$ is contained in $I_0:=[\inf u_0,\sup u_0]$.

Global continuation and conclusion. For every $t_1>0$, the positive-strip spatial modulus of step 2.2 makes $u(t_1,\cdot )$ bounded and uniformly continuous, step 3.1 gives $u(t_1)\in L^1$, and step 5.1 places its values in $I_0$. Thus $g:=u(t_1)$ is an admissible $BUC\cap L^1$ starting datum for the local fixed-point and Picard iteration of steps 2.1 and 3.1. The strong heat continuity required for this restart is [F6]: Gaussian approximate-identity convergence gives $H_{\epsilon t}g\to g$ in the supremum norm for BUC data and in $L^1$ for $L^1$ data. With the same global constants $R,L$, the local solution on $[t_1,t_1+{\tau }_0)$ agrees with $u$ at $t_1$; local uniqueness in step 2.1 makes the pieces agree on overlaps, and $\mathrm{\parallel }u(t){\mathrm{\parallel }}_{\mathrm{\infty }}\le \max \{\mathrm{\mid }\inf u_0\mathrm{\mid },\mathrm{\mid }\sup u_0\mathrm{\mid }\}\le R$ throughout. Iterating finitely many times covers $[0,T]$, and the same local uniqueness shows that two solutions obtained with different terminal times $T,T^{\mathrm{\prime }}$ agree on their common interval. The range bounds of step 5.1 hold on all of $[0,T]$; $L^1$ continuity holds by step 3.1 and $C^{1,2}$ regularity on ${\mathbb{R}}^n\times (0,T)$ by step 4.1. This is the asserted global mild solution.