Riccati comparison for scalar initial shape

$S_i$ satisfies the Riccati equation and is self-adjoint. [F1, F2, F3, given] On any interval where $Y_i$ is invertible, differentiating $S_i=Y_i^{\mathrm{\prime }}{Y}_i^{-1}$ with [F3] gives $S_i^{\mathrm{\prime }}=Y_i^{\mathrm{\prime }\mathrm{\prime }}{Y}_i^{-1}+Y_i^{\mathrm{\prime }}(-Y_i^{-1}{Y}_i^{\mathrm{\prime }}{Y}_i^{-1})=-R_i{Y}_i{Y}_i^{-1}-S_i^2=-R_i-S_i^2,$ the middle step using $Y_i^{\mathrm{\prime }\mathrm{\prime }}=-R_i{Y}_i$; this is the Riccati equation. For self-adjointness put $W_i:=Y_i^{\ast }{Y}_i^{\mathrm{\prime }}-(Y_i^{\mathrm{\prime }}{)}^{\ast }{Y}_i$. Differentiating and inserting the equation, $W_i^{\mathrm{\prime }}=(Y_i^{\mathrm{\prime }}{)}^{\ast }{Y}_i^{\mathrm{\prime }}+Y_i^{\ast }{Y}_i^{\mathrm{\prime }\mathrm{\prime }}-(Y_i^{\mathrm{\prime }\mathrm{\prime }}{)}^{\ast }{Y}_i-(Y_i^{\mathrm{\prime }}{)}^{\ast }{Y}_i^{\mathrm{\prime }}=Y_i^{\ast }(-R_i{Y}_i)+(R_i{Y}_i{)}^{\ast }{Y}_i=-Y_i^{\ast }{R}_i{Y}_i+Y_i^{\ast }{R}_i{Y}_i=0,$ because $R_i^{\ast }=R_i$ by [F2]. Hence $W_i$ is constant, and $W_i(0)=(\mathrm{id}{)}^{\ast }(\lambda \mathrm{id})-(\lambda \mathrm{id}{)}^{\ast }\mathrm{id}=0$, so $Y_i^{\ast }{Y}_i^{\mathrm{\prime }}=(Y_i^{\mathrm{\prime }}{)}^{\ast }{Y}_i$. Multiplying on the right by $Y_i^{-1}$ gives $Y_i^{\ast }{S}_i=(Y_i^{\mathrm{\prime }}{)}^{\ast }$, and multiplying on the left by $(Y_i^{\ast }{)}^{-1}$, so $S_i$ is self-adjoint.

The limit at zero. [F3, F7, given] The solutions are $C^2$ on $[0,T]$, so $Y_i(t)\to Y_i(0)={\mathrm{id}}_E$ and $Y_i^{\mathrm{\prime }}(t)\to Y_i^{\mathrm{\prime }}(0)=\lambda {\mathrm{id}}_E$ as $t\downarrow 0$. By [F3] inversion of matrices is continuous at the invertible point ${\mathrm{id}}_E$, so $Y_i(t{)}^{-1}\to {\mathrm{id}}_E$ and therefore $S_i(t)=Y_i^{\mathrm{\prime }}(t)Y_i(t{)}^{-1}\to \lambda {\mathrm{id}}_E$. In particular $U:=S_2-S_1$, defined on $(0,t_{\ast })$ with $t_{\ast }:=\min (T,t_1,t_2)$, extends to $t=0$ by $U(0):=0$ and is continuous there.

The scalar model. [F4, F5, given] Let $R_2=k{\mathrm{id}}_E$ and $f:={\mathrm{cs}}_k+\lambda {\mathrm{sn}}_k$. By [F5], $f^{\mathrm{\prime }\mathrm{\prime }}+kf=0$, $f(0)=1$ and $f^{\mathrm{\prime }}(0)=\lambda$, so $Y:=f{\mathrm{id}}_E$ satisfies $Y^{\mathrm{\prime }\mathrm{\prime }}+kY=(f^{\mathrm{\prime }\mathrm{\prime }}+kf){\mathrm{id}}_E=0$, $Y(0)={\mathrm{id}}_E$ and $Y^{\mathrm{\prime }}(0)=\lambda {\mathrm{id}}_E$. The second-order equation for $Y$ is the first-order linear system $Z^{\mathrm{\prime }}=CZ$ for $Z=(Y,Y^{\mathrm{\prime }})$ with the constant matrix $C=\left(\begin{array}{cc}0& 1\\ -k& 0\end{array}\right)$, so [F4] gives uniqueness of its solutions; hence the given $Y_2$ equals $f{\mathrm{id}}_E$, and $S_2=(f^{\mathrm{\prime }}\mathrm{/}f){\mathrm{id}}_E$ wherever $f\mathrm{\ne }0$.

The transport equation for $U$. [F2, step 1.1, given] On $(0,t_{\ast })$ both $S_1$ and $S_2$ are defined, and step 1.1 gives $U^{\mathrm{\prime }}=S_2^{\mathrm{\prime }}-S_1^{\mathrm{\prime }}=(-S_2^2-R_2)-(-S_1^2-R_1)=S_1^2-S_2^2+(R_1-R_2)\mathrm{.}$ Put $X:=-\frac{1}{2}(S_1+S_2)$, self-adjoint by [F2] and step 1.1, and $S:=R_1-R_2\ge 0$. The mixed terms collapse, $XU+UX=-\frac{1}{2}(S_1+S_2)(S_2-S_1)-\frac{1}{2}(S_2-S_1)(S_1+S_2)=S_1^2-S_2^2,$ so that $U^{\mathrm{\prime }}=XU+UX+S,U(0)=0,S\ge 0.$

Positivity of $U$: $S_1\le S_2$ on $(0,t_{\ast })$. [F4, F6, F7, step 1.2, step 2.1] Fix $\tau \in (0,t_{\ast })$. The linear matrix initial value problem $g^{\mathrm{\prime }}=Xg$, $g(0)={\mathrm{id}}_E$, has a unique solution $g$ on the compact interval $[0,\tau ]$ by [F4], and every $g(s)$ is invertible by [F4]. For $v\in E$ and $s\in (0,\tau )$ define $w(s):=g(s{)}^{-1}U(s)(g(s{)}^{-1}{)}^{\ast };$ since $U(s)\to 0$ and $g(s{)}^{-1}\to {\mathrm{id}}_E$ as $s\downarrow 0$ by step 1.2, $w$ extends continuously to $s=0$ with $w(0)=0$. Using $(g^{-1}{)}^{\mathrm{\prime }}=-g^{-1}X$ from [F3] and $X^{\ast }=X$, $w^{\mathrm{\prime }}=-g^{-1}XU(g^{-1}{)}^{\ast }+g^{-1}(XU+UX+S)(g^{-1}{)}^{\ast }-g^{-1}UX(g^{-1}{)}^{\ast }=g^{-1}S(g^{-1}{)}^{\ast },$ which is positive semidefinite at every $s$, because $⟨g^{-1}S(g^{-1}{)}^{\ast }v,v⟩=⟨S(g^{-1}{)}^{\ast }v,(g^{-1}{)}^{\ast }v⟩\ge 0$ by $S\ge 0$ and [F7]. Therefore the continuous real function $s\mapsto ⟨w^{\mathrm{\prime }}(s)v,v⟩$ is nonnegative on $[0,\tau ]$, and [F6] gives $⟨w(\tau )v,v⟩=⟨w(0)v,v⟩+{\int }_0^{\tau }⟨w^{\mathrm{\prime }}(s)v,v⟩ds\ge 0.$ As $v$ was arbitrary, $w(\tau )\ge 0$; since $U(\tau )=g(\tau )w(\tau )g(\tau {)}^{\ast }$ is a congruence by the invertible $g(\tau )$, $⟨U(\tau )v,v⟩=⟨w(\tau )g(\tau {)}^{\ast }v,g(\tau {)}^{\ast }v⟩\ge 0$ for all $v$, that is $U(\tau )\ge 0$. Thus $S_1\le S_2$ on $(0,t_{\ast })$.

The less-curved tensor has no earlier singular time. [F3, F7, step 3.1, given] Before either first singular time, $\det Y_i>0$, since it starts at one and cannot change sign without vanishing. Jacobi's determinant formula (The determinant differential is $D\det (A)[H]=\mathrm{tr}(\mathrm{adj}(A)H)$ at every matrix, and Jacobi's formula holds on the invertible locus) gives ${(\log \frac{\det Y_2}{\det Y_1})}^{\mathrm{\prime }}=\mathrm{tr}(S_2-S_1)\ge 0.$ The trace is nonnegative because it is the sum of $⟨(S_2-S_1)e_j,e_j⟩\ge 0$ in a finite orthonormal basis. The determinant ratio starts at one, so $\det Y_2\ge \det Y_1>0$ on this interval. If $t_2<t_1$ and $t_2\le T$, continuity as $t\uparrow t_2$ would give $0=\det Y_2(t_2)\ge \det Y_1(t_2)>0$, a contradiction. Thus $Y_2$ is invertible for all $t\le T$ with $t<t_1$, and step 3.1 extends $S_1\le S_2$ to any included nonsingular endpoint by continuity. This also proves $t_1\le t_2$ when $t_1\le T$. If $t_1=+\mathrm{\infty }$, there is no singular time of $Y_2$ in $[0,T]$.

The logarithmic norm inequality. [F2, step 3.1, step 1.3, given] For $u=0$ the norm inequality is immediate. Fix $u\in E\setminus \{0\}$ and put $J(t):=Y_1(t)u$ for $t\in (0,t_1)$. Since $Y_1(t)$ is invertible there, $J(t)\mathrm{\ne }0$ and $J^{\mathrm{\prime }}(t)=Y_1^{\mathrm{\prime }}(t)u=S_1(t)Y_1(t)u=S_1(t)J(t)\mathrm{.}$ On the initial interval where $t<t_{\ast }:=\min (T,t_1,t_2)$ and $f>0$ (which holds near $0$), the function $\log (\mathrm{\mid }J(t)\mathrm{\mid }\mathrm{/}(f(t)\mathrm{\mid }u\mathrm{\mid }))$ is defined and $(\log \frac{\mathrm{\mid }J\mathrm{\mid }}{f\mathrm{\mid }u\mathrm{\mid }}{)}^{\mathrm{\prime }}=\frac{⟨S_{1}J,J⟩}{\mathrm{\mid }J{\mathrm{\mid }}^2}-\frac{f^{\mathrm{\prime }}}{f}\le \frac{⟨S_{2}J,J⟩}{\mathrm{\mid }J{\mathrm{\mid }}^2}-\frac{f^{\mathrm{\prime }}}{f}=0,$ using $S_1\le S_2$ from step 4.1 and $S_2=(f^{\mathrm{\prime }}\mathrm{/}f)\mathrm{id}$ from step 1.3. Moreover $\frac{\mathrm{\mid }J(t)\mathrm{\mid }}{f(t)\mathrm{\mid }u\mathrm{\mid }}⟶\frac{\mathrm{\mid }{Y}_1(0)u\mathrm{\mid }}{f(0)\mathrm{\mid }u\mathrm{\mid }}=1(t\downarrow 0)$ by continuity of $Y_1$ and $f$ (step 1.2 and [F5]). Hence $\mathrm{\mid }{Y}_1(t)u\mathrm{\mid }\le f(t)\mathrm{\mid }u\mathrm{\mid }$ on this initial interval.

The model stays positive on the compared interval; conclusion. [step 4.1, step 4.2, given] If $f$ has a zero in $(0,T]$ before $t_1$, let $t_0$ be its first such zero. Then $f>0$ on $[0,t_0)$ and step 4.2 applies there, so for every $u\in E$, $\mathrm{\mid }{Y}_1(t)u\mathrm{\mid }\le f(t)\mathrm{\mid }u\mathrm{\mid }⟶0(t\uparrow t_0)\mathrm{.}$ Continuity forces $Y_1(t_0)=0$, contradicting its invertibility. Thus $f>0$ on $[0,T]$ wherever $t<t_1$, and $Y_2=f{\mathrm{id}}_E$ is invertible there. Steps 3.1 and 4.2 give $S_1\le S_2$ and $\mathrm{\mid }{Y}_1(t)u\mathrm{\mid }\le f(t)\mathrm{\mid }u\mathrm{\mid }$ for every $u\in E$ and $t\in (0,T]$ with $t<t_1$. If $t_1\le T$, continuity extends the norm inequality to $t=t_1$, so $t_1\le t_2$. If $t_1>T$, then $Y_2$ is nonsingular on all of $[0,T]$, so $t_2=+\mathrm{\infty }$ by definition. Step 4.1 proves the singular-time comparison for arbitrary $R_2$, and the scalar argument here proves the additional norm conclusion.