A Family of Explicit Solutions

The integral in Equation (84) contains a cubic polynomial. The nature of its three roots depends on the value of its discriminant $\Delta$ [#!WikiCubicEq!#]. For cubic polynomials of the form $ax^3+cx+d$ we have $\Delta=-4ac^3-27a^{2}d^{2}$. If $\Delta>0$ the polynomial has three distinct real roots, if $\Delta=0$ it has three real roots but two of them are equal, and if $\Delta<0$ it has one real and two complex roots which are conjugate to each other. In our case we have $a=-1$, $c=3$, $d=\eta$ and therefore $\Delta=27(4-\eta^{2})$.

Figure 1: Comparison between the dimensionless pressure $p(x)$ calculated analytically and numerically using the Runge-Kutta fourth-order algorithm for $\eta =2.0$, $x_{2}=5^{1/3}$ and $x_{M}=1.0$, resulting in $x_{1}=0.594881$ and $x_{\mu }=0.596494$.

Figure 2: The functions $\nu (x)$ and $\lambda (x)$ for $\eta =2.0$, $x_{2}=5^{1/3}$ and $x_{M}=1.0$. The shaded area indicates the matter region, to its right is the outer vacuum and to its left is the inner vacuum. Here we have $x_{1}=0.594881$ and $x_{\mu }=0.596494$.

Figure: The dimensionless pressure $p$ calculated numerically for $\eta=5.0$, $x_{2}=2.0$ and $x_{M}=1.0$. Here we have $x_{1}=1.24050$ and $x_{\mu}=1.03035$.

Figure: The functions $\nu (x)$ and $\lambda (x)$ for $\eta=5.0$, $x_{2}=2.0$ and $x_{M}=1.0$. The shaded area indicates the matter region, to its right is the outer vacuum and to its left is the inner vacuum. Here we have $x_{1}=1.24050$ and $x_{\mu}=1.03035$.

The value $\Delta=0$ corresponds to the case where the solution for $z(x)$ can be expressed in terms of elementary functions. Note that we have $\Delta=0$ when $\eta=\pm 2$, which corresponds to $x_{2}^{3}=\pm2+3x_{M}$. For $\eta=-2$ the polynomial in the integral in Equation (84) is non-positive for $x\ge 0$. Therefore, we must choose $\eta=2$. For this value of $\eta$ the polynomial is strictly positive in the interval $[0,2)$ and can be factored as

  $$2+3y-y^{3} = (2-y)(y+1)^{2}.$$ (86)

In this case we can express the integral in Equation (84) in terms of elementary functions. The calculation can be considerably simplified using a new integration variable $u$ defined by $u=\sqrt{y/(2-y)}$. The final result, up to an integration constant, is

$$\mathcal{I}(y) \equiv \int\,dy\, \frac{y^{5/2}}{(2-y)^{3/2}(y+1)^3}$$

$$= \frac{2y^{2}+15y+10}{18\,(y+1)^{2}}\sqrt{\frac{y}{2-y}} -\, \frac{5\sqrt{3}}{27} \arctan\!\left(\sqrt{\frac{3y}{2-y}}\,\right).$$ (87)

Thus, in terms of $\mathcal{I}(y)$ Equation (84) reads

  $$z(x) = \sqrt{\frac{2+3x-x^3}{x}} \left\{ \sqrt{\frac{x_{2}}{... ...2} \left[\rule{0em}{2.5ex}\mathcal{I}(x)-\mathcal{I}(x_{2})\right] \right\}.$$ (88)

Note that, in order to guarantee that the cubic polynomial for $\eta=2$ shown in Equation (86) is always positive, we need to have $y<2$. Therefore, since we already know that the polynomial is positive, the arguments of the square roots in Equation (87) are always positive.