Regular-center TOV local existence and dependence

Write $P_c=[p_{-},p_{+}]$ and set $P=[p_{-}-\delta ,p_{+}+\delta ]$; the margin hypothesis puts $P$ in $I_p$, and [F6] makes it compact. Thus ${ϵ}_{\max }={\max }_Pϵ$ and $L_{ϵ}={\max }_P\mathrm{\mid }{ϵ}^{\mathrm{\prime }}\mathrm{\mid }$ exist and are finite. Put $V=K{ϵ}_{\max }\mathrm{/}3$, and first choose $R_1\le 1$ with $2R_1^{2}V\le 1\mathrm{/}2$. The set $Q=[-R_1,R_1]\times P\times [0,V]$ is closed and bounded in ${\mathbb{R}}^3$, hence compact by [F6]. On $Q$, the denominator of $H(r,x,y)=\frac{(ϵ(x)+x)(y+Kx)}{1-2r^{2}y}$ is at least $1\mathrm{/}2$, so $H,H_x,H_y$ are continuous there by [F5]; [F6] supplies finite bounds $C$ for $\mathrm{\mid }H\mathrm{\mid }$, $C_x$ for $\mathrm{\mid }{H}_x\mathrm{\mid }$, and $C_y$ for $\mathrm{\mid }{H}_y\mathrm{\mid }$. Set $C_1=C_x+K{L}_{ϵ}{C}_y\mathrm{/}3$. Choose $R_0\le R_1$ so that $C{R}_0^2\mathrm{/}2\le \delta$ and $C_1{R}_0^2\mathrm{/}2\le 1\mathrm{/}2$.

For each $p_c\in P_c$, let ${\mathcal{B}}_{p_c}$ be the closed sup-norm ball of radius $\delta$ about the constant function $p_c$ in $C([-R_0,R_0])$. It is complete: a sup-norm Cauchy sequence has pointwise limits in the closed real interval $P$, the convergence is uniform by the Cauchy bound, and its uniform limit is continuous. For $q\in {\mathcal{B}}_{p_c}$ define $v[q](r)=K{\int }_0^1{s}^2ϵ(q(rs))ds,({\mathcal{T}}_{p_c}q)(r)=p_c-{\int }_0^{r}tH(t,q(t),v[q](t))dt\mathrm{.}$ The function $ϵ\circ q$ is uniformly continuous on $[-R_0,R_0]$ by [F10]. Because $\mathrm{\mid }rs-\tilde{r}s\mathrm{\mid }\le \mathrm{\mid }r-\tilde{r}\mathrm{\mid }$ for $s\in [0,1]$, [F9] bounds $\mathrm{\mid }v[q](r)-v[q](\tilde{r})\mathrm{\mid }$ by $K\mathrm{/}3$ times ${\sup }_{s\in [0,1]}\mathrm{\mid }ϵ(q(rs))-ϵ(q(\tilde{r}s))\mathrm{\mid }$, which tends to zero; hence $v[q]$ is continuous. The integrand defining ${\mathcal{T}}_{p_c}q$ is then continuous, so [F3] makes that map continuous as well. By [F9] and [F8], $0\le v[q]\le K{ϵ}_{\max }\mathrm{/}3=V$; hence $1-2r^{2}v[q]\ge 1\mathrm{/}2$. The mean value theorem [F7], linearity and [F9] applied to the bounds $\mathrm{\mid }{s}^2(ϵ(q(rs))-ϵ(\tilde{q}(rs)))\mathrm{\mid }\le s^2{L}_{ϵ}\mathrm{\parallel }q-\tilde{q}{\mathrm{\parallel }}_{\mathrm{\infty }}$ give $\mathrm{\parallel }v[q]-v[\tilde{q}]{\mathrm{\parallel }}_{\mathrm{\infty }}\le K{L}_{ϵ}\mathrm{\parallel }q-\tilde{q}{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}3.$ Applying [F7] successively in the $x$ and $y$ variables gives $\mathrm{\mid }H(r,x,y)-H(r,\tilde{x},\tilde{y})\mathrm{\mid }\le C_x\mathrm{\mid }x-\tilde{x}\mathrm{\mid }+C_y\mathrm{\mid }y-\tilde{y}\mathrm{\mid }$ for state pairs in $P\times [0,V]$. For each endpoint $r$, [F9] bounds the absolute integral of $tH$ or of the difference $t(H-\tilde{H})$ by the corresponding constant times the integral of $\mathrm{\mid }t\mathrm{\mid }$ over the interval between $0$ and $r$. This last integral is $r^2\mathrm{/}2$ by [F8]; reversing endpoints changes only the integral sign, and the $r=0$ case is zero. Hence the bounds in step 1.1, [F9] and [F8] give $\mathrm{\parallel }{\mathcal{T}}_{p_c}q-p_c{\mathrm{\parallel }}_{\mathrm{\infty }}\le C{R}_0^2\mathrm{/}2\le \delta$ and $\mathrm{\parallel }{\mathcal{T}}_{p_c}q-{\mathcal{T}}_{p_c}\tilde{q}{\mathrm{\parallel }}_{\mathrm{\infty }}\le (C_1{R}_0^2\mathrm{/}2)\mathrm{\parallel }q-\tilde{q}{\mathrm{\parallel }}_{\mathrm{\infty }}\le \frac{1}{2}\mathrm{\parallel }q-\tilde{q}{\mathrm{\parallel }}_{\mathrm{\infty }}$. Thus ${\mathcal{T}}_{p_c}$ maps the ball to itself and is a contraction.

Iterate canonically from $q_0(r)=p_c$, $q_{j+1}={\mathcal{T}}_{p_c}{q}_j$. The differences satisfy $\mathrm{\parallel }{q}_{j+1}-q_j{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2^{-j}\mathrm{\parallel }{q}_1-q_0{\mathrm{\parallel }}_{\mathrm{\infty }}$, so the geometric tail estimate makes $(q_j)$ Cauchy in the complete ball. Its uniform limit $p$ is continuous and remains in the ball. The contraction estimate makes ${\mathcal{T}}_{p_c}$ continuous, hence ${\mathcal{T}}_{p_c}p=p$; any two fixed points have distance at most half their distance and therefore coincide. If $q$ is even, then $v[q]$ is even, $H(r,q(r),v[q](r))$ is even, and its product with $r$ is odd, so ${\mathcal{T}}_{p_c}q$ is even. Every iterate and hence $p$ is even; the fixed point equations give the two displayed integral identities.

The first FTC in [F3] makes $p$ continuously differentiable and yields $p^{\mathrm{\prime }}(r)=-rH(r,p(r),v[p](r))$, including $p^{\mathrm{\prime }}(0)=0$. The parameter-integral theorem [F3], together with the $C^k$ composition and algebra rules [F5], applied repeatedly on $[-R_0,R_0]\times [0,1]$, shows that whenever $p\in C^k$, $v[p]$ is $C^k$ because $(r,s)\mapsto s^2ϵ(p(rs))$ has continuous $r$-derivatives through order $k$. The denominator is bounded below by $1\mathrm{/}2$, so $H(r,p(r),v[p](r))$ is $C^k$; the integral equation then gives $p\in C^{k+1}$. Induction proves $p,v$ are smooth and even. Rewriting the $v$ integral as $r^{3}v(r)=K{\int }_0^r{t}^2ϵ(p(t))dt$, then applying the product rule and the first FTC gives $v^{\mathrm{\prime }}=(Kϵ(p)-3v)\mathrm{/}r$ for $r>0$; the pressure equation is the defining derivative of $p$.

The same $R_0$ and contraction factor work for every central value in $P_c$. Subtracting the fixed-point equations for $p_a,p_b$ gives $\mathrm{\parallel }{p}_a-p_b{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\mid }a-b\mathrm{\mid }+\frac{1}{2}\mathrm{\parallel }{p}_a-p_b{\mathrm{\parallel }}_{\mathrm{\infty }},$ hence $\mathrm{\parallel }{p}_a-p_b{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2\mathrm{\mid }a-b\mathrm{\mid }$. The integral formula for $v$, [F9] and [F8] give $\mathrm{\parallel }{v}_a-v_b{\mathrm{\parallel }}_{\mathrm{\infty }}\le K{L}_{ϵ}\mathrm{\parallel }{p}_a-p_b{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}3$. The lapse integrand $J(r,x,y)=r(y+Kx)\mathrm{/}(1-2r^{2}y)$ and its $x,y$ derivatives are continuous and bounded on the same compact set $Q$ by [F6]. Applying [F7] successively in $x,y$ bounds its state increment by a constant times $\mathrm{\mid }{p}_a-p_b\mathrm{\mid }+\mathrm{\mid }{v}_a-v_b\mathrm{\mid }$; integrating over $0\le r\le R_0$ with [F9] yields $\mathrm{\parallel }{\varphi }_a-{\varphi }_b{\mathrm{\parallel }}_{\mathrm{\infty }}\le C_J{R}_0(\mathrm{\parallel }{p}_a-p_b{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }{v}_a-v_b{\mathrm{\parallel }}_{\mathrm{\infty }})$ for a finite $C_J$. Thus the solution map is continuous in the sup norm. If another regular-center solution stays in the same pressure ball, integrating its mass and pressure equations yields the same two fixed-point equations; uniqueness in step 3.1 proves it is this solution.

Set $b=1-2r^{2}v$ and define $\varphi$ by the displayed integral. Then $b\ge 1\mathrm{/}2$, $\varphi$ is smooth and even, and ${\varphi }^{\mathrm{\prime }}(0)=0$. The product rule and the equation for $v$ give $M^{\mathrm{\prime }}=4\pi r^2ϵ(p)\mathrm{/}{c}^2$ from $M=c^2{r}^{3}v\mathrm{/}G$; also $GM\mathrm{/}{c}^2=r^{3}v$, so ${\varphi }^{\mathrm{\prime }}=r(v+Kp)\mathrm{/}b$ is the lapse equation in [F1]. Apply [F2] with $q=2v$ and $h=q\mathrm{/}b$ to obtain the $C^2$ Cartesian metric extension. Its center coefficients are $g_{00}(0)=-e^{2{\varphi }_c}$ and $g_{ij}(0)={\delta }_{ij}$, so it is Lorentzian at the center.

For $r>0$, let $U=\{(r,p,v):r>0,p\in I_p,1-2r^{2}v>0\}$ and define the vector field $F(r,p,v)=(-r\frac{(ϵ(p)+p)(v+Kp)}{1-2r^{2}v},\frac{Kϵ(p)-3v}{r})\mathrm{.}$ By [F5], $F$ is $C^1$ on the open set $U$. Around each point of $U$ choose a closed time-state cylinder $[r_{-},r_{+}]\times \bar{B}((p_{\ast },v_{\ast }),\eta )\subset U$. It is a closed bounded subset of ${\mathbb{R}}^3$, hence compact by [F6]. The state derivative of $F$ is continuous there, and its operator norm is continuous, so [F6] gives a finite uniform bound $L$. For each fixed $r\in [r_{-},r_{+}]$, the state ball is convex and open; the mean-value inequality [thm-mean-value-inequality-for-total-derivatives] gives $\mathrm{\parallel }F(r,z)-F(r,\tilde{z}){\mathrm{\parallel }}_2\le L\mathrm{\parallel }z-\tilde{z}{\mathrm{\parallel }}_2$ on it. Thus $F$ is locally Lipschitz in the state variable, locally uniformly in time, as required by [F4]. If the solution graph remains in a compact subset of $U$ near a finite endpoint, [F4] extends $(p,v)$ past that endpoint. The integral defining $\varphi$ then also extends. This is only the stated compact-interior continuation criterion; it gives no global surface or stability claim.