Well-posedness, positivity and contraction for two-velocity transport

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 $\alpha =\lambda \mathrm{/}{ϵ}^2$.