Periodic two-velocity heat limit for well-prepared data

Define $j^{ϵ}=(c\mathrm{/}ϵ)(f_{+}^{ϵ}-f_{-}^{ϵ})$. Adding the two equations in [F1] gives ${\mathrm{\partial }}_t{\rho }^{ϵ}+{\mathrm{\partial }}_x{j}^{ϵ}=0$. Subtracting them and multiplying by $c\mathrm{/}ϵ$ gives ${ϵ}^2{\mathrm{\partial }}_t{j}^{ϵ}+2\lambda j^{ϵ}=-c^2{\mathrm{\partial }}_x{\rho }^{ϵ}$. The initial condition implies $j^{ϵ}(\cdot ,0)=0$.

Under [A1], take Fourier coefficients in $x$ in these two identities, with ${\kappa }_n$ and $\widehat{\cdot }(n)$ from [F1]; differentiate under the compact spatial integral using the smooth solution [F2], and integrate by parts in real and imaginary parts separately using [F4] after the measure-preserving rescaling $y=x\mathrm{/}L$. Then $y_n^{ϵ}(t):=\widehat{{\rho }^{ϵ}(t)}(n)$ satisfies ${ϵ}^2(y_n^{ϵ}{)}^{\mathrm{\prime }\mathrm{\prime }}+2\lambda (y_n^{ϵ}{)}^{\mathrm{\prime }}+c^2{\kappa }_n^2{y}_n^{ϵ}=0$, $y_n^{ϵ}(0)=\widehat{{\rho }_0}(n)$, and $(y_n^{ϵ}{)}^{\mathrm{\prime }}(0)=0$. The heat series in the Statement has coefficients $y_n(t)=\widehat{{\rho }_0}(n)e^{-D{\kappa }_n^{2}t}$.

Repeated integration by parts in the real and imaginary parts using [F4] gives ${\sum }_n(1+\mathrm{\mid }{\kappa }_n\mathrm{\mid }{)}^r\mathrm{\mid }\widehat{{\rho }_0}(n)\mathrm{\mid }<\mathrm{\infty }$ for every integer $r\ge 0$; convergence follows by grouping $2^j\le \mathrm{\mid }n\mathrm{\mid }<2^{j+1}$ after choosing more than $r+1$ derivatives. Therefore the heat series and every differentiated series converge uniformly on compact time intervals, and its coefficient equation $y_n^{\mathrm{\prime }}=-D{\kappa }_n^2{y}_n$ proves ${\mathrm{\partial }}_{t}u=D{\mathrm{\partial }}_x^{2}u$ with $u(0)={\rho }_0$. For any other smooth periodic solution $w$, integration by parts in its Fourier coefficients gives $\widehat{w}(n{)}^{\mathrm{\prime }}=-D{\kappa }_n^2\widehat{w}(n)$ with initial value $\widehat{{\rho }_0}(n)$. Multiplying by $e^{D{\kappa }_n^{2}t}$ shows that $e^{D{\kappa }_n^{2}t}\widehat{w}(n,t)$ has zero derivative, so $\widehat{w}(n,t)=\widehat{{\rho }_0}(n)e^{-D{\kappa }_n^{2}t}$ for every $n$ and $t$. Fourier completeness [F3] therefore gives $w(t)=u(t)$ as $L^2({\mathbb{T}}_L,d{m}_L)$ classes for each fixed $t$, hence almost everywhere. Both are continuous in $x$; if they differed at a point, their difference would be bounded away from zero on some nonempty open arc, which has positive normalized Lebesgue measure $d{m}_L$. This contradicts equality almost everywhere, so $w(t,x)=u(t,x)$ pointwise for every $t,x$.

For $n=0$, the mass mode is constant and equals the heat mode. Fix $n\mathrm{\ne }0$. For all sufficiently small $ϵ>0$ with $ϵc\mathrm{\mid }{\kappa }_n\mathrm{\mid }<\lambda$, set $s_n^{ϵ}=(-\lambda +\sqrt{{\lambda }^2-{ϵ}^2{c}^2{\kappa }_n^2})\mathrm{/}{ϵ}^2$ and $r_n^{ϵ}=(-\lambda -\sqrt{{\lambda }^2-{ϵ}^2{c}^2{\kappa }_n^2})\mathrm{/}{ϵ}^2$. The solution of the scalar mode equation is $y_n^{ϵ}(t)=\widehat{{\rho }_0}(n)(A_n^{ϵ}{e}^{s_n^{ϵ}t}+B_n^{ϵ}{e}^{r_n^{ϵ}t})$, where $A_n^{ϵ}=\frac{1}{2}(1+\lambda \mathrm{/}\sqrt{{\lambda }^2-{ϵ}^2{c}^2{\kappa }_n^2})$ and $B_n^{ϵ}=\frac{1}{2}(1-\lambda \mathrm{/}\sqrt{{\lambda }^2-{ϵ}^2{c}^2{\kappa }_n^2})$; these coefficients solve $A+B=1$ and $As+Br=0$. Rationalizing the slow root gives $s_n^{ϵ}=-c^2{\kappa }_n^2\mathrm{/}(\lambda +\sqrt{{\lambda }^2-{ϵ}^2{c}^2{\kappa }_n^2})\to -D{\kappa }_n^2$, while $r_n^{ϵ}\to -\mathrm{\infty }$, $A_n^{ϵ}\to 1$, and $B_n^{ϵ}\to 0$. Put $q=D{\kappa }_n^2\ge 0$; both $s_n^{ϵ}$ and $-q$ are nonpositive, so integrating the derivative of $s\mapsto e^{st}$ with [F4] gives $\mathrm{\mid }{e}^{s_n^{ϵ}t}-e^{-qt}\mathrm{\mid }\le t\mathrm{\mid }{s}_n^{ϵ}+q\mathrm{\mid }\le T\mathrm{\mid }{s}_n^{ϵ}+q\mathrm{\mid }$ on $[0,T]$. Since both exponentials have modulus at most one, $\mathrm{\mid }{A}_n^{ϵ}{e}^{s_n^{ϵ}t}+B_n^{ϵ}{e}^{r_n^{ϵ}t}-e^{-qt}\mathrm{\mid }\le \mathrm{\mid }{A}_n^{ϵ}-1\mathrm{\mid }+T\mathrm{\mid }{s}_n^{ϵ}+q\mathrm{\mid }+\mathrm{\mid }{B}_n^{ϵ}\mathrm{\mid }$. The right side tends to zero independently of $t\in [0,T]$, proving uniform convergence of this Fourier mode.

Let $P_{>K}$ denote the Fourier projection onto $\mathrm{\mid }n\mathrm{\mid }>K$, applied componentwise to pairs. The mode equations from step 2.1 are decoupled, so the projected pair solves the same system with initial pair $P_{>K}({\rho }_0\mathrm{/}2,{\rho }_0\mathrm{/}2)$; uniqueness in [F2] identifies it with that evolution. Applying [F2] gives $\mathrm{\parallel }{P}_{>K}(f_{+}^{ϵ},f_{-}^{ϵ})(t){\mathrm{\parallel }}_{2,L}\le \mathrm{\parallel }{P}_{>K}({\rho }_0\mathrm{/}2,{\rho }_0\mathrm{/}2){\mathrm{\parallel }}_{2,L}=\mathrm{\parallel }{P}_{>K}{\rho }_0{\mathrm{\parallel }}_{2,L}\mathrm{/}\sqrt{2}$. Since $\mathrm{\parallel }a+b{\mathrm{\parallel }}_{2,L}^2\le 2(\mathrm{\parallel }a{\mathrm{\parallel }}_{2,L}^2+\mathrm{\parallel }b{\mathrm{\parallel }}_{2,L}^2)$, it follows that $\mathrm{\parallel }{P}_{>K}{\rho }^{ϵ}(t){\mathrm{\parallel }}_{2,L}\le \mathrm{\parallel }{P}_{>K}{\rho }_0{\mathrm{\parallel }}_{2,L}$. The heat multiplier has modulus at most one, hence $\mathrm{\parallel }{P}_{>K}u(t){\mathrm{\parallel }}_{2,L}\le \mathrm{\parallel }{P}_{>K}{\rho }_0{\mathrm{\parallel }}_{2,L}$.

By Parseval [F3], $\mathrm{\parallel }{P}_{>K}{\rho }_0{\mathrm{\parallel }}_{2,L}\to 0$ as $K\to \mathrm{\infty }$. Given $\eta >0$, choose $K$ so that twice this tail is below $\eta \mathrm{/}2$. The finitely many modes $\mathrm{\mid }n\mathrm{\mid }\le K$ converge uniformly on $[0,T]$ by step 3.2, so for all sufficiently small $ϵ$ their combined $L^2$ error is below $\eta \mathrm{/}2$. Step 3.3 bounds the remaining density difference by twice the chosen tail, uniformly in $t$ and $ϵ$. Therefore ${\sup }_{0\le t\le T}\mathrm{\parallel }{\rho }^{ϵ}(t)-u(t){\mathrm{\parallel }}_{2,L}<\eta$ for all sufficiently small $ϵ$, proving the claim.