Exact regular constant-density solution with verified matching

Since $0<u<8\mathrm{/}9$, $a=\sqrt{1-u}>1\mathrm{/}3$. On $0\le r\le R$, $a\le B(r)\le 1$, so $b=B^2>0$ and $3a-B(r)\ge 3a-1>0$. Thus the lapse and pressure are finite and the lapse is positive; $p(R)=0$, $p(r)\ge 0$, and $p(0)={ϵ}_0(1-a)\mathrm{/}(3a-1)$.

To check sharpness, first take $8\mathrm{/}9\le u<1$. At $u=8\mathrm{/}9$, $a=1\mathrm{/}3$ and $B(0)=1=3a$, so the pressure denominator vanishes at the center while its numerator is positive, and $e^{\varphi (0)}=0$. If $8\mathrm{/}9<u<1$, then $a<1\mathrm{/}3$ and $a<3a<1=B(0)$, whereas $B(R)=a$; continuity gives $r_{\ast }\in (0,R)$ with $B(r_{\ast })=3a$, where the pressure denominator vanishes and the lapse is zero. If $u=1$, then $b(R)=0$; if $u>1$, $B(R)$ is not real. Thus no $u\ge 8\mathrm{/}9$ gives a regular positive branch, while step 1.1 proves that every $0<u<8\mathrm{/}9$ does.

The mass formula gives $M^{\mathrm{\prime }}=3M_R{r}^2\mathrm{/}{R}^3=4\pi {\rho }_0{r}^2=4\pi r^2{ϵ}_0\mathrm{/}{c}^2$. The functions $B,\varphi ,p$ are even and smooth near $r=0$ because $1-u{r}^2\mathrm{/}{R}^2>0$ and $3a-B>0$. Writing $b=1-r^{2}q$ with $q=u\mathrm{/}{R}^2$, [F2] gives a $C^2$ Cartesian center for the interior metric. The Einstein equations hold at the center by continuity of the curvature and source components from $r>0$.

For $0<r<R$, the chain, product, and quotient rules [F4], together with [F3], give $B^{\mathrm{\prime }}=-\frac{ur}{R^{2}B},{\varphi }^{\mathrm{\prime }}=\frac{ur}{R^{2}B(3a-B)},{\left(\frac{p}{{ϵ}_0}\right)}^{\mathrm{\prime }}=-\frac{2aur}{R^{2}B(3a-B{)}^2}\mathrm{.}$ Also $1+p\mathrm{/}{ϵ}_0=2a\mathrm{/}(3a-B)$, so $p^{\mathrm{\prime }}=-({ϵ}_0+p){\varphi }^{\mathrm{\prime }}$. Using $u=8\pi G{ϵ}_0{R}^2\mathrm{/}(3c^4)$ and $p\mathrm{/}{ϵ}_0=(B-a)\mathrm{/}(3a-B)$, direct simplification yields ${\varphi }^{\mathrm{\prime }}=\frac{4\pi Gr({ϵ}_0\mathrm{/}3+p)}{c^{4}b}=\frac{GM\mathrm{/}{c}^2+4\pi G{r}^{3}p\mathrm{/}{c}^4}{r^{2}b},$ where $M=M_R{r}^3\mathrm{/}{R}^3$. Substituting the displayed lapse into $p^{\mathrm{\prime }}=-({ϵ}_0+p){\varphi }^{\mathrm{\prime }}$ is the TOV pressure equation in [F1], so the interior mass, lapse, and pressure equations all hold.

For $r\ge R$, $M^{\mathrm{\prime }}=0$, $ϵ=p=0$, and $b=1-uR\mathrm{/}r>0$ since $u<1$. The chain and quotient rules [F4] give ${\varphi }^{\mathrm{\prime }}=\frac{1}{2}{b}^{\mathrm{\prime }}\mathrm{/}b=uR\mathrm{/}(2r^{2}b)=G{M}_R\mathrm{/}(c^2{r}^{2}b)$, so the vacuum mass and lapse equations hold and the pressure equation is $0=0$.

Let $b_{-}(r)=1-u{r}^2\mathrm{/}{R}^2$ on the interior side and $b_{+}(r)=1-uR\mathrm{/}r$ on the exterior side. Both formulas are smooth and positive on separate open neighborhoods of $R$. By [F3]–[F4], their negative one-half powers are smooth there. For a common value ${\mathrm{\ell }}_R$, define ${\mathrm{\ell }}_{\pm }(r)={\mathrm{\ell }}_R+{\int }_R^r{b}_{\pm }(s{)}^{-1\mathrm{/}2}ds$. By [F5] these functions have derivatives $d{\mathrm{\ell }}_{\pm }\mathrm{/}dr=b_{\pm }^{-1\mathrm{/}2}>0$; repeated differentiation of these identities makes ${\mathrm{\ell }}_{\pm }$ smooth. By [F6], after shrinking the neighborhoods, each has a smooth local inverse $r_{\pm }(\mathrm{\ell })$ satisfying $d{r}_{\pm }\mathrm{/}d\mathrm{\ell }=\sqrt{b_{\pm }(r_{\pm }(\mathrm{\ell }))}$. Use $\mathrm{\ell }<{\mathrm{\ell }}_R$ for the interior and $\mathrm{\ell }>{\mathrm{\ell }}_R$ for the exterior as the matching chart, so $r_{\pm }({\mathrm{\ell }}_R)=R$ and $d{r}_{\pm }\mathrm{/}d\mathrm{\ell }({\mathrm{\ell }}_R)=a$ on both sides. At $r=R$, [F1] gives $b_{-}(R)=b_{+}(R)=a^2$ and both lapse values equal $a$. The lapse derivatives with respect to $r$ agree by steps 2.3–2.4: $(e^{\varphi }{)}_{-}^{\mathrm{\prime }}(R)=u\mathrm{/}(2Ra)=(\sqrt{b_{+}}{)}^{\mathrm{\prime }}(R)$. Hence in the matching chart the coefficients $g_{00}=-e^{2\varphi }$ and $r_{\pm }(\mathrm{\ell }{)}^2\gamma$ have matching values and first derivatives, while $g_{\mathrm{\ell }\mathrm{\ell }}=1$ on both sides. Thus the metric is $C^1$ and piecewise $C^2$ in that chart. For a piecewise $C^2$ coefficient $f(\mathrm{\ell })$ with one-sided values and derivatives, define $Df[\psi ]:=-\int f(\mathrm{\ell }){\psi }^{\mathrm{\prime }}(\mathrm{\ell })d\mathrm{\ell }$ for $\psi \in C_c^{\mathrm{\infty }}$, and define ${\delta }_{{\mathrm{\ell }}_R}[\psi ]=\psi ({\mathrm{\ell }}_R)$ and ${\delta }_{{\mathrm{\ell }}_R}^{\mathrm{\prime }}[\psi ]=-{\psi }^{\mathrm{\prime }}({\mathrm{\ell }}_R)$. These functionals satisfy the stated test bounds: for nonempty compact $K\subset U$, choose a compact interval $[a,b]\subset U$ containing $K$ and let $M$ bound $\mathrm{\mid }f\mathrm{\mid }$ there on both sides of the interface. For $\mathrm{supp}\psi \subseteq K$, $\mathrm{\mid }\int f\psi \mathrm{\mid }\le (b-a)M{\sup }_K\mathrm{\mid }\psi \mathrm{\mid }$, $\mathrm{\mid }Df[\psi ]\mathrm{\mid }\le (b-a)M{\sup }_K\mathrm{\mid }{\psi }^{\mathrm{\prime }}\mathrm{\mid }$, and $\mathrm{\mid }{D}^{2}f[\psi ]\mathrm{\mid }=\mathrm{\mid }\int f{\psi }^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\mid }\le (b-a)M{\sup }_K\mathrm{\mid }{\psi }^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\mid }$; hence the orders $0,1,2$ suffice. Also $\mathrm{\mid }{\delta }_{{\mathrm{\ell }}_R}[\psi ]\mathrm{\mid }\le {\sup }_K\mathrm{\mid }\psi \mathrm{\mid }$ and $\mathrm{\mid }{\delta }_{{\mathrm{\ell }}_R}^{\mathrm{\prime }}[\psi ]\mathrm{\mid }\le {\sup }_K\mathrm{\mid }{\psi }^{\mathrm{\prime }}\mathrm{\mid }$, with both evaluations zero if ${\mathrm{\ell }}_R\mathrm{\notin }K$; empty $K$ admits only the zero test function. Linearity follows from these formulas. Integration by parts on the two sides gives $D^{2}f=f_{\mathrm{p}\mathrm{i}\mathrm{e}\mathrm{c}\mathrm{e}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}}^{\mathrm{\prime }\mathrm{\prime }}+[f^{\mathrm{\prime }}]{\delta }_{{\mathrm{\ell }}_R}+[f]{\delta }_{{\mathrm{\ell }}_R}^{\mathrm{\prime }}$, where $[f]=f({\mathrm{\ell }}_R+)-f({\mathrm{\ell }}_R-)$ and similarly for $[f^{\mathrm{\prime }}]$. The matched metric coefficients and first derivatives make both jumps zero. The Christoffel coefficients are formed from the metric, its inverse, and its first derivatives; since the metric is nondegenerate, they are continuous and piecewise $C^1$. With the curvature convention of [F1], its coordinate formula on the smooth sides is $R^a{}_{bcd}={\mathrm{\partial }}_c{\mathrm{\Gamma }}^a{}_{db}-{\mathrm{\partial }}_d{\mathrm{\Gamma }}^a{}_{cb}+{\mathrm{\Gamma }}^a{}_{ce}{\mathrm{\Gamma }}^e{}_{db}-{\mathrm{\Gamma }}^a{}_{de}{\mathrm{\Gamma }}^e{}_{cb}$; in this interface calculation take those derivatives in the distributional sense just defined. Since the Christoffel coefficients are continuous, the curvature formula produces no surface-supported term. The source fields are piecewise ordinary functions and likewise contain no delta-supported surface stress.