Kinetic relaxation, diffusive scaling and entropy dissipation

For each $n\in \mathbb{Z}$, set $v=c\mathrm{/}ϵ$, $\alpha =\lambda \mathrm{/}{ϵ}^2$, $M_n=\left(\begin{array}{cc}-i{\kappa }_{n}v-\alpha & \alpha \\ \alpha & i{\kappa }_{n}v-\alpha \end{array}\right)$, and $U_n(0)=({\widehat{f}}_{+,0}(n),{\widehat{f}}_{-,0}(n){)}^{\mathsf{T}}$. Realify $z\in {\mathbb{C}}^2$ by $\mathcal{R}z=(\mathrm{Re}{z}_1,\mathrm{Re}{z}_2,\mathrm{Im}{z}_1,\mathrm{Im}{z}_2)\in {\mathbb{R}}^4$, and for $M=P+iQ$ put $\mathcal{R}M=\left(\begin{array}{cc}P& -Q\\ Q& P\end{array}\right)$. Direct coordinate multiplication gives $\mathcal{R}(Mz)=(\mathcal{R}M)\mathcal{R}z$ and $\mathrm{\mid }\mathcal{R}z{\mathrm{\mid }}_2^2=\mathrm{\mid }{z}_1{\mathrm{\mid }}^2+\mathrm{\mid }{z}_2{\mathrm{\mid }}^2$. By [F6], ${\mathbb{R}}^4$ is Banach; [F7] applied to each row bounds every real $4\times 4$ matrix as an operator, and [F8]–[F9] specify the Euclidean and operator norms. By [F5], $\mathcal{R}{U}_n(t):=e^{t\mathcal{R}{M}_n}\mathcal{R}{U}_n(0)$ defines a solution of the realified mode IVP for all $t\ge 0$, and pulling back by $\mathcal{R}$ defines $U_n(t)$. The matrix-valued ODE supplier [F4] gives uniqueness of this vector solution on each compact interval $[0,T]$: embed any vector solution as the first column of a $4\times 4$ matrix solution with the other columns zero, and apply [F4]. For any smooth solution of the PDE, integration by parts in real and imaginary parts separately, using [F3], gives these same Fourier mode equations, so the constructed modes are the only possible ones. Define the candidate pair by the Fourier series $F^{ϵ}(t,x):={\sum }_{n\in \mathbb{Z}}{U}_n(t)e_n(x)$, with $e_n$ from [F1]. Since real initial data satisfy $U_{-n}(0)=\overline{U_n(0)}$ and $M_{-n}=\overline{M_n}$, uniqueness gives $U_{-n}(t)=\overline{U_n(t)}$, so this series is real. The skew-Hermitian transport part contributes zero to the derivative of $\mathrm{\mid }{U}_n{\mathrm{\mid }}^2$, and direct multiplication of the collision matrix gives $(\mathrm{\mid }{U}_n{\mathrm{\mid }}^2{)}^{\mathrm{\prime }}=-2\alpha \mathrm{\mid }{\widehat{f}}_{+}^{ϵ}(n,t)-{\widehat{f}}_{-}^{ϵ}(n,t){\mathrm{\mid }}^2\le 0$. For every integer $r\ge 0$ and $q>r+1$, periodicity and [F3] give $\widehat{{\mathrm{\partial }}_x^q{f}_{\pm ,0}}(n)=(i{\kappa }_n{)}^q{\widehat{f}}_{\pm ,0}(n)$ by repeated integration by parts; hence $\mathrm{\mid }{\widehat{f}}_{\pm ,0}(n)\mathrm{\mid }\le \mathrm{\mid }{\kappa }_n{\mathrm{\mid }}^{-q}{\int }_0^L\mathrm{\mid }{\mathrm{\partial }}_x^q{f}_{\pm ,0}(x)\mathrm{\mid }d{m}_L(x)$ for $n\mathrm{\ne }0$. Grouping $2^j\le \mathrm{\mid }n\mathrm{\mid }<2^{j+1}$ then shows ${\sum }_n(1+\mathrm{\mid }{\kappa }_n\mathrm{\mid }{)}^r\mathrm{\mid }{U}_n(0)\mathrm{\mid }<\mathrm{\infty }$. The mode bound $\mathrm{\mid }{U}_n(t)\mathrm{\mid }\le \mathrm{\mid }{U}_n(0)\mathrm{\mid }$ and $\mathrm{\mid }{M}_n\mathrm{\mid }=O(1+\mathrm{\mid }n\mathrm{\mid })$ imply absolute uniform convergence on every compact time interval of the Fourier series and of each differentiated series, since each space derivative contributes a factor $i{\kappa }_n$ and each time derivative a factor $M_n$. Fourier orthonormality [F2] gives the series coefficients $U_n$; their mode equations and the uniformly convergent differentiated series then verify the PDE termwise. At $t=0$ the series agrees with the prescribed smooth data by Fourier completeness [F2], first in $L^2$ and hence pointwise by continuity. Thus it is a global smooth solution of the Cauchy problem. Any other smooth solution has the same mode equations and initial coefficients; [A1] and [F2] make the solutions equal as $L^2({\mathbb{T}}_L,d{m}_L;{\mathbb{C}}^2)$ classes, hence almost everywhere. Both solutions are smooth and therefore continuous in $x$. If their difference were nonzero at some $x_0$, continuity would give a nonempty open arc on which its norm is bounded below by a positive constant; every nonempty open arc has positive normalized Lebesgue measure $d{m}_L$, contradicting equality almost everywhere. Thus they agree pointwise on ${\mathbb{T}}_L$.

Put $F_0=(f_{+,0},f_{-,0})$. For $h\ge 0$, free transport $T_h(g_{+},g_{-}):=(g_{+}(\cdot -vh),g_{-}(\cdot +vh))$ preserves nonnegativity and pair norm. It preserves each component mass because a change of variable turns its integral into an integral of a smooth periodic function over another full period, which is unchanged by [F3]. The pointwise collision operator $C_h$ is the matrix $\frac{1}{2}\left(\begin{array}{cc}1+e^{-2\alpha h}& 1-e^{-2\alpha h}\\ 1-e^{-2\alpha h}& 1+e^{-2\alpha h}\end{array}\right)$. Its entries are nonnegative and its column sums are one. The orthonormal vectors $(1,1)\mathrm{/}\sqrt{2}$ and $(1,-1)\mathrm{/}\sqrt{2}$ have eigenvalues $1$ and $e^{-2\alpha h}\in [0,1]$, so $C_h$ preserves nonnegativity and total mass and contracts the pair norm pointwise, hence in pair $L^2$. Thus the alternating products $(T_{t\mathrm{/}m}{C}_{t\mathrm{/}m}{)}^m{F}_0$ are nonnegative when $F_0$ is and are $L^2$ contractions.

Fix a Fourier mode $k\in \mathbb{Z}$ and let $A_k=\mathrm{diag}(-i{\kappa }_{k}v,i{\kappa }_{k}v)$ and $B=\alpha \left(\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right)$. Under the realification of step 1.1, their generators are real $4\times 4$ matrices ${\tilde{A}}_k=\mathcal{R}(A_k)$ and $\tilde{B}=\mathcal{R}(B)$. Transport on this mode is $e^{{\tilde{A}}_{k}h}$, and collision is $e^{\tilde{B}h}$ because the eigenvectors and eigenvalues in step 1.2 diagonalize $B$. By [F6], ${\mathbb{R}}^4$ with its Euclidean norm is Banach; applying [F7] rowwise bounds every real matrix as an operator, and [F8]–[F9] give the Euclidean and operator norms used in [F5]. For $X\in \{{\tilde{A}}_k,\tilde{B},{\tilde{A}}_k+\tilde{B}\}$, [F5] gives $e^{Xh}=I+hX+R_X(h)$ with $\mathrm{\parallel }{R}_X(h)\mathrm{\parallel }\le K_X{h}^2$ on $0\le h\le t$, where $K_X:=\frac{1}{2}\mathrm{\parallel }X{\mathrm{\parallel }}^2{e}^{t\mathrm{\parallel }X\mathrm{\parallel }}$. Multiplying the first two expansions and subtracting the third shows $\mathrm{\parallel }{e}^{{\tilde{A}}_{k}h}{e}^{\tilde{B}h}-e^{({\tilde{A}}_k+\tilde{B})h}\mathrm{\parallel }\le C_{k,ϵ,t}{h}^2$, where one may take $C_{k,ϵ,t}=\mathrm{\parallel }{\tilde{A}}_k\mathrm{\parallel }\mathrm{\parallel }\tilde{B}\mathrm{\parallel }+K_{{\tilde{A}}_k}+K_{\tilde{B}}+K_{{\tilde{A}}_k+\tilde{B}}+t(\mathrm{\parallel }{\tilde{A}}_k\mathrm{\parallel }{K}_{\tilde{B}}+\mathrm{\parallel }\tilde{B}\mathrm{\parallel }{K}_{{\tilde{A}}_k})+t^2{K}_{{\tilde{A}}_k}{K}_{\tilde{B}}$. All factors are contractions: the split factors by step 1.2, and the exact mode evolution by the energy identity in step 1.1. Telescoping the $m$-fold products with $h=t\mathrm{/}m$ bounds the error by $m{C}_{k,ϵ,t}(t\mathrm{/}m{)}^2=C_{k,ϵ,t}{t}^2\mathrm{/}m$, so the split evolution converges to the exact solution in each fixed mode.

The zero mode in the equations of [F1] satisfies $({\widehat{f}}_{+}(0)+{\widehat{f}}_{-}(0){)}^{\mathrm{\prime }}=0$, since the switching terms cancel and the transport frequency is zero. As $M^{ϵ}(t)=L({\widehat{f}}_{+}^{ϵ}(0,t)+{\widehat{f}}_{-}^{ϵ}(0,t))$, this proves total-mass conservation.

For a finite set of modes, step 2.1 gives convergence of the split evolution to the exact solution. Let $P_{>K}$ be the Fourier projection onto $\mathrm{\mid }k\mathrm{\mid }>K$, applied componentwise. Its initial pair tail tends to zero by [A1] and [F2]. Every split product preserves modes and contracts this tail by step 1.2. The exact evolution also preserves modes; applying the mode energy identity in step 1.1 and summing by Parseval [F2] shows it contracts the same tail. Hence the split products converge in pair $L^2$ to the solution from step 1.1.

The nonnegative cone is closed in real $L^2$: if classes $g_m$ have nonnegative representatives almost everywhere and $g_m\to g$, then for representatives of $g$ and $g_m$ the inequality $g^{-}\le \mathrm{\mid }g-g_m\mathrm{\mid }$ holds almost everywhere, so $\mathrm{\parallel }{g}^{-}{\mathrm{\parallel }}_2\le \mathrm{\parallel }g-g_m{\mathrm{\parallel }}_2\to 0$. Applying this componentwise, the split products are nonnegative by step 1.2 and converge by step 3.1, so each component of the exact solution is nonnegative almost everywhere. Fix $t\ge 0$. Step 1.1 gives smooth, hence continuous, functions $x\mapsto f_{\pm }^{ϵ}(x,t)$. If either component were negative at some $x_0$, continuity would give a nonempty open arc on which it is bounded above by a negative constant. Such an arc has positive $d{m}_L$ measure, contradicting nonnegativity almost everywhere. Thus both components are nonnegative pointwise for every $t\ge 0$.

Under [A1], apply the mode identity in step 1.1 to the difference of two solutions; summing over $n$ by Parseval [F2] gives the stated contraction. Step 1.1 gives uniform convergence of the solution and time-derivative Fourier series, so termwise differentiation of Parseval's sum is valid. Applying the same sum to one solution gives $\frac{d}{dt}\mathrm{\parallel }{F}^{ϵ}(t){\mathrm{\parallel }}_{2,L}^2=-2\alpha \mathrm{\parallel }{f}_{+}^{ϵ}(t)-f_{-}^{ϵ}(t){\mathrm{\parallel }}_{2,L}^2$, which is the displayed dissipation identity because $\alpha =\lambda \mathrm{/}{ϵ}^2$.

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.