Appendix: Resolution of the Painlevé Vector Field
1.1 The Affine Charts
1.1.1 Affine Chart \((u_{01}, v_{01})\)
The first affine chart is defined by the original coordinates
$$\begin{aligned} u_{01}= & {} u, \\ v_{01}= & {} v, \\ E= & {} -uv(u+v+2). \end{aligned}$$
1.1.2 Affine Chart \((u_{02}, v_{02})\)
The second affine chart is given by the following coordinates:
$$\begin{aligned} u_{02}= & {} \frac{1}{u}, \qquad v_{02}=\frac{v}{u}, \\ u= & {} \frac{1}{u_{02}}, \qquad v=\frac{v_{02}}{u_{02}}. \end{aligned}$$
The line at infinity is \(\mathcal {L}_0: u_{02}=0\).
The Painlevé vector field is given by
$$\begin{aligned} u_{02}'= & {} 1+2u_{02}+2v_{02}+\frac{1}{2z}(2\alpha _1u_{02}^2+u_{02}), \\ v_{02}'= & {} \frac{v_{02}}{u_{02}}(4u_{02}+3v_{02}+3)+\frac{1}{z}(-\alpha _2u_{02}+\alpha _1u_{02}v_{02}), \end{aligned}$$
which contain base points at
$$\begin{aligned} b_0\ :\ u_{02}=0,v_{02}=0 \quad \text {and}\quad b_1\ :\ u_{02}=0,v_{02}=-1. \end{aligned}$$
The energy is
1.1.3 Affine Chart \((u_{03}, v_{03})\)
We have the coordinates
$$\begin{aligned} u_{03}= & {} \frac{1}{v}, \qquad v_{03}=\frac{u}{v}, \\ u= & {} \frac{v_{03}}{u_{03}}, \qquad v=\frac{1}{u_{03}}, \end{aligned}$$
and the line at infinity is given by \(\mathcal {L}_0: u_{03}=0\).
The flow is given by
$$\begin{aligned}&u_{03}'=-1-2u_{03}-2v_{03}+\frac{1}{2z}(2\alpha _2u_{03}^2+u_{03}), \\&v_{03}'=-\frac{v_{03}}{u_{03}}(4u_{03}+3v_{03}+3)+\frac{1}{z}(-\alpha _1u_{03}+\alpha _2u_{03}v_{03}), \end{aligned}$$
which contains base points at
$$\begin{aligned} b_2\ :\ u_{03}=0,v_{03}=0 \end{aligned}$$
and \((u_{03}=0,v_{03}=-1)\), which is \(b_1\).
The energy is given by
$$\begin{aligned} E= & {} -\frac{v_{03} (1 + 2 u_{03} + v_{03})}{u_{03}^3}, \\ E'= & {} \frac{1}{2 u_{03}^3 z} \left( 4 \alpha _2 u_{03}^2 v_{03}+ 2 \alpha _2 u_{03} v_{03}^2+ 4 \alpha _1 u_{03}^2 + 3 v_{03}^2 \right. \\&\quad \left. +\,4(\alpha _1+\alpha _2+1)u_{03} v_{03} +\,2 \alpha _1 u_{03} + 3 v_{03} \right) . \end{aligned}$$
1.2 Resolution at Base Points \(b_0\), \(b_1\), \(b_2\)
1.2.1 Resolution at \(b_0\)
The first chart is given by the coordinate change:
$$\begin{aligned} u_{11}= & {} \frac{u_{02}}{v_{02}}=\frac{1}{v}, \qquad v_{11}=v_{02}=\frac{v}{u}, \\ u= & {} \frac{1}{u_{11}v_{11}}, \qquad v=\frac{1}{u_{11}}. \end{aligned}$$
The exceptional line is \(\mathcal {L}_1:v_{11}=0\). The preimage of line \(\mathcal {L}_0\) is visible in this chart and given by the equation \(u_{11}=0\).
The flow in this chart:
$$\begin{aligned} u_{11}'= & {} -\frac{1}{v_{11}}(v_{11}+2u_{11}v_{11}+2)+\frac{u_{11}}{2z}(1+2\alpha _2u_{11}), \\ v_{11}'= & {} \frac{1}{u_{11}}(3v_{11}+4u_{11}v_{11}+3)+\frac{u_{11}v_{11}}{z}(-\alpha _2+\alpha _1v_{11}), \end{aligned}$$
contains no new base points.
The energy is given by
The second chart is given by
$$\begin{aligned} u_{12}= & {} u_{02}=\frac{1}{u}, \qquad v_{12}=\frac{v_{02}}{u_{02}}=v, \\ u= & {} \frac{1}{u_{12}}, \qquad v=v_{12}. \end{aligned}$$
The exceptional line is \(\mathcal {L}_1:u_{12}=0\). The preimage of line \(\mathcal {L}_0\) is not visible in this chart.
The flow is
$$\begin{aligned} u_{12}'= & {} 1+2u_{12}+2u_{12}v_{12}+\frac{u_{12}}{2z}(1+2\alpha _1u_{12}), \\ v_{12}'= & {} \frac{v_{12}}{u_{12}}(2u_{12}+2+u_{12}v_{12})-\frac{1}{2z}(2\alpha _2+v_{12}). \end{aligned}$$
Both the vector field and the anticanonical pencil have base point at
$$\begin{aligned} b_3\ :\ u_{12}=0,v_{12}=0. \end{aligned}$$
The energy is given by
1.2.2 Resolution at \(b_1\)
The first chart is given by the coordinate change:
$$\begin{aligned} u_{21}= & {} \frac{u_{02}}{v_{02}+1}=\frac{1}{u+v}, \qquad v_{21}=v_{02}+1=\frac{u+v}{u}, \\ u= & {} \frac{1}{u_{21}v_{21}}, \qquad v=\frac{v_{21}-1}{u_{21}v_{21}}. \end{aligned}$$
The exceptional line is \(\mathcal {L}_2:v_{21}=0\). The preimage of the line \(\mathcal {L}_0\) is visible in this chart and given by the equation \(u_{21}=0\).
The flow is given by
$$\begin{aligned} u_{21}'= & {} \frac{(2u_{21}+1)(2-v_{21})}{v_{21}}+\frac{u_{21}}{2z}\big (2(\alpha _1+\alpha _2)u_{21}+1\big ), \\ v_{21}'= & {} \frac{(4u_{21}+3)(v_{21}-1)}{u_{21}}+\frac{u_{21}v_{21}}{z}(\alpha _1v_{21}-\alpha _1-\alpha _2), \end{aligned}$$
and contains a new base point at
$$\begin{aligned} b_4 :\ u_{21}=-\frac{1}{2},v_{21}=0. \end{aligned}$$
The energy is given by
$$\begin{aligned} E= & {} -\frac{( 2 u_{21} +1) (v_{21}-1)}{u_{21}^3 v_{21}^2}, \\ E'= & {} \frac{1}{2 u_{21}^3 v_{21}^2 z}\left( {-3} - 2(2+ \alpha _1 + \alpha _2) u_{21} + 3 v_{21} + 4 (1+ \alpha _2) u_{21} v_{21} \right. \\&\left. +\, 4( \alpha _2 - \alpha _1) u_{21}^2 v_{21} + 2 \alpha _1 u_{21} v_{21}^2 + 4 \alpha _1 u_{21}^2 v_{21}^2\right) . \end{aligned}$$
The second chart is given by
$$\begin{aligned} u_{22}= & {} u_{02}=\frac{1}{u}, \qquad v_{22}=\frac{v_{02}+1}{u_{02}}=u+v, \\ u= & {} \frac{1}{u_{22}}, \qquad v=v_{22}-\frac{1}{u_{22}}. \end{aligned}$$
The exceptional line is \(\mathcal {L}_2:u_{22}=0\). The preimage of line \(\mathcal {L}_0\) is not visible in this chart.
The flow is given by
$$\begin{aligned} u_{22}'= & {} -1+2u_{22}+2u_{22}v_{22}+\frac{u_{22}}{2z}(1+2\alpha _1u_{22}), \\ v_{22}'= & {} \frac{(v_{22}+2)(u_{22}v_{22}-2)}{u_{22}}-\frac{1}{2z}(v_{22}+2\alpha _1+2\alpha _2), \end{aligned}$$
and contains a base point \((u_{22}=0,v_{22}=-2)\), which is \(b_4\).
The energy is given by
1.2.3 Resolution at \(b_2\)
The first chart is given by
$$\begin{aligned} u_{31}= & {} \frac{u_{03}}{v_{03}}=\frac{1}{u}, \qquad v_{31}=v_{03}=\frac{u}{v}, \\ u= & {} \frac{1}{u_{31}}, \qquad v=\frac{1}{u_{31}v_{31}}. \end{aligned}$$
The exceptional line is \(\mathcal {L}_3:v_{31}=0\). The preimage of line \(\mathcal {L}_0\) is visible in this chart and given by the equation \(u_{31}=0\).
The flow
$$\begin{aligned} u_{31}'= & {} \frac{v_{31}+2u_{31}v_{31}+2}{v_{31}}+\frac{u_{31}}{2z}(1+2\alpha _1u_{31}), \\ v_{31}'= & {} -\frac{3+4u_{31}v_{31}+3v_{31}}{u_{31}}+\frac{u_{31}v_{31}}{z}(-\alpha _1+\alpha _2v_{31}), \end{aligned}$$
contains no base point.
The energy is given by
The second chart is given by
$$\begin{aligned} u_{32}= & {} u_{03}=\frac{1}{v}, \qquad v_{32}=\frac{v_{03}}{u_{03}}=u, \\ u= & {} v_{32}, \qquad v=\frac{1}{u_{32}}. \end{aligned}$$
The exceptional line is \(\mathcal {L}_3:u_{32}=0\). The preimage of line \(\mathcal {L}_0\) is not visible in this chart.
The flow is
$$\begin{aligned} u_{32}'= & {} -1-2u_{32}-2u_{32}v_{32}+\frac{u_{32}}{2z}(1+2\alpha _2u_{32}), \\ v_{32}'= & {} -\frac{v_{32}}{u_{32}}(2u_{32}+2+v_{32}u_{32})-\frac{1}{2z}(2\alpha _1+v_{32}). \end{aligned}$$
Both the vector field and the anticanonical pencil have a base point at
$$\begin{aligned} b_5\ :\ u_{32}=0,v_{32}=0. \end{aligned}$$
The energy is given by
1.3 Resolution at Points \(b_3\), \(b_4\), \(b_5\)
1.3.1 Resolution at \(b_3\)
The first chart is
$$\begin{aligned} u_{41}= & {} \frac{u_{12}}{v_{12}}=\frac{1}{uv}, \qquad v_{41}=v_{12}=v, \\ u= & {} \frac{1}{u_{41}v_{41}}, \qquad v=v_{41}, \end{aligned}$$
and the corresponding Jacobian is
$$\begin{aligned} J_{41}=\frac{\partial u_{41}}{\partial u}\frac{\partial v_{41}}{\partial v}-\frac{\partial u_{41}}{\partial v}\frac{\partial v_{41}}{\partial u} =-\frac{1}{u^2v}=-u_{41}^2v_{41}. \end{aligned}$$
The exceptional line is \(\mathcal {L}_4:v_{41}=0\). The preimage of line \(\mathcal {L}_1\) in this chart is \(u_{41}=0\). \(\mathcal {L}_0\) is not visible in this chart.
The flow is given by
$$\begin{aligned} u_{41}'= & {} -\frac{1}{v_{41}}+u_{41}v_{41}+\frac{u_{41} \left( \alpha _2+v_{41}+\alpha _1u_{41}v_{41}^2\right) }{zv_{41}}, \\ v_{41}'= & {} \frac{2}{u_{41}}+2v_{41}+v_{41}^2-\frac{1}{2z}(2\alpha _2+v_{41}), \end{aligned}$$
and contains a base point:
$$\begin{aligned} b_6\ :\ u_{41}=\frac{z}{\alpha _2},v_{41}=0. \end{aligned}$$
The energy and related quantities are
The second chart is given by
$$\begin{aligned} u_{42}= & {} u_{12}=\frac{1}{u}, \qquad v_{42}=\frac{v_{12}}{u_{12}}=uv, \\ u= & {} \frac{1}{u_{42}}, \qquad v=u_{42}v_{42}. \end{aligned}$$
The exceptional line is \(\mathcal {L}_4:u_{42}=0\). The preimages of \(\mathcal {L}_0\) and \(\mathcal {L}_1\) are not visible in this chart.
The flow is
$$\begin{aligned} \begin{aligned} u_{42}'&=1+2u_{42}+2u_{42}^2v_{42}+\frac{u_{42}}{2z}(1+2\alpha _1u_{42}), \\ v_{42}'&=\frac{v_{42}}{u_{42}}-u_{42}v_{42}^2-\frac{1}{zu_{42}} \left( \alpha _2+u_{42}v_{42}+\alpha _1u_{42}^2v_{42}\right) , \end{aligned} \end{aligned}$$
and contains a base point \((u_{42}=0,v_{42}=\frac{\alpha _2}{z})\), which is \(b_6\).
1.3.2 Resolution at \(b_4\)
The first chart is given by
$$\begin{aligned} u_{51}= & {} \frac{u_{21}+\frac{1}{2}}{v_{21}}=\frac{u(u+v+2)}{2(u+v)^2}, \qquad v_{51}=v_{21}=\frac{u+v}{u}, \\ u= & {} \frac{2}{v_{51} (2 u_{51} v_{51}-1)}, \qquad v=\frac{2 (v_{51}-1)}{v_{51} ( 2 u_{51} v_{51}-1)}. \end{aligned}$$
The exceptional line is \(\mathcal {L}_5:v_{51}=0\). The preimage of \(\mathcal {L}_2\) is not visible in this chart, while the preimage of \(\mathcal {L}_0\) is given by \(u_{51}v_{51}=\frac{1}{2}\).
The flow is given by
$$\begin{aligned} \begin{aligned} u_{51}'&=-\frac{2u_{51}(1+2u_{51}v_{51}(3v_{51}-4))}{v_{51}(2u_{51}v_{51}-1)} \\&+\frac{\left( 2u_{51}v_{51}-1\right) \left( \alpha _1+\alpha _2-1-4(\alpha _1+\alpha _2)u_{51}v_{51}+2\alpha _1u_{51}v_{51}^2\right) }{4zv_{51}}, \\ v_{51}'&=\frac{2(v_{51}-1)(4u_{51}v_{51}+1)}{2u_{51}v_{51}-1}+\frac{v_{51}(\alpha _1v_{51}-\alpha _1-\alpha _2)(2u_{51}v_{51}-1)}{2z}, \end{aligned} \end{aligned}$$
and contains a base point
$$\begin{aligned} b_7\ :\ u_{51}=\frac{1-\alpha _1-\alpha _2}{8z},v_{51}=0. \end{aligned}$$
The second chart is
$$\begin{aligned} u_{52}= & {} u_{21}+\frac{1}{2}=\frac{1}{u+v}+\frac{1}{2}, \qquad v_{52}=\frac{v_{21}}{u_{21}+\frac{1}{2}}=\frac{2(u+v)^2}{u(u+v+2)}, \\ u= & {} \frac{2}{(2 u_{52}-1) u_{52}v_{52}}, \qquad v=\frac{2 ( u_{52} v_{52}-1)}{( 2 u_{52}-1)u_{52} v_{52}}, \end{aligned}$$
which has Jacobian
$$\begin{aligned} J_{52}=\frac{\partial u_{52}}{\partial u}\frac{\partial v_{52}}{\partial v}-\frac{\partial u_{52}}{\partial v}\frac{\partial v_{52}}{\partial u} =-\frac{2}{u^2(u+v+2)}=-\frac{1}{8}u_{52}(2u_{52}-1)^3v_{52}^2. \end{aligned}$$
The exceptional line is \(\mathcal {L}_5:u_{52}=0\). In this chart, the preimage of \(\mathcal {L}_2\) is given by \(v_{52}=0\), and of \(\mathcal {L}_0\) by \(u_{52}=\frac{1}{2}\).
The flow is given by
$$\begin{aligned} \begin{aligned} u_{52}'&=-2u_{52}+\frac{4}{v_{52}}+\frac{(2u_{52}-1)(1+(\alpha _1+\alpha _2)(2u_{52}-1))}{4z}, \\ v_{52}'&=\frac{2\left( 1-8u_{52}+6u_{52}^2v_{52}\right) }{u_{52}(2u_{52}-1)} \\&\quad + \frac{v_{52}(2u_{52}-1)\left( \alpha _1+\alpha _2-1-4(\alpha _1+\alpha _2)u_{52}+2\alpha _1u_{52}^2v_{52}\right) }{4zu_{52}}, \end{aligned} \end{aligned}$$
which contains a base point \(\left( u_{52}=0, v_{52}={8z}/(1-\alpha _1-\alpha _2)\right) \), which is \(b_7\).
The energy and related quantities are
$$\begin{aligned} \begin{aligned} E&=-\frac{16 ( u_{52} v_{52}-1)}{u_{52} (2 u_{52}-1)^3 v_{52}^2}, \qquad EJ_{52}=2 ( u_{52} v_{52}-1), \\ E'&=\frac{4}{u_{52}^2 \left( 2 u_{52}-1\right) ^3 v_{52}^2 z} \left( (\alpha _1 + \alpha _2-1) - 2(2+ \alpha _1+ \alpha _2) u_{52} \right. \\&\qquad + (1-\alpha _1-\alpha _2) u_{52} v_{52} + 4(1+\alpha _1) u_{52}^2 v_{52} +4(\alpha _2-\alpha _1) u_{52}^3 v_{52} \\&\left. \qquad - 2 \alpha _1 u_{52}^3 v_{52}^2 + 4 \alpha _1 u_{52}^4 v_{52}^2\right) . \end{aligned} \end{aligned}$$
1.3.3 Resolution at \(b_5\)
The first chart is
$$\begin{aligned} u_{61}= & {} \frac{u_{32}}{v_{32}}=\frac{1}{uv}, \qquad v_{61}=v_{32}=u, \\ u= & {} v_{61}, \qquad v=\frac{1}{u_{61}v_{61}}, \\ J_{61}= & {} \frac{\partial u_{61}}{\partial u}\frac{\partial v_{61}}{\partial v}-\frac{\partial u_{61}}{\partial v}\frac{\partial v_{61}}{\partial u} =\frac{1}{uv^2}=u_{61}^2v_{61}. \end{aligned}$$
The exceptional line is \(\mathcal {L}_6:v_{61}=0\). In this chart, the preimage of \(\mathcal {L}_3\) is given by \(u_{61}=0\), and the preimage of \(\mathcal {L}_0\) is not visible.
The flow is
$$\begin{aligned} \begin{aligned} u_{61}'&=\frac{1-u_{61}v_{61}^2}{v_{61}}+\frac{u_{61}\left( v_{61}+\alpha _1+\alpha _2u_{61}v_{61}^2\right) }{zv_{61}}, \\ v_{61}'&=-\frac{2+u_{61}v_{61}(2+v_{61})}{u_{61}}-\frac{v_{61}+2\alpha _1}{2z}, \end{aligned} \end{aligned}$$
and contains a base point:
$$\begin{aligned} b_8\ :\ u_{61}=-\frac{z}{\alpha _1},v_{61}=0. \end{aligned}$$
The energy is given by
$$\begin{aligned} \begin{aligned} E&=-\frac{1 + 2 u_{61} v_{61} + u_{61} v_{61}^2}{u_{61}^2 v_{61}}, \qquad EJ_{61}=-\left( 1 + 2 u_{61} v_{61} + u_{61} v_{61}^2\right) , \\ E'&=\frac{1}{2 u_{61}^2 v_{61}^2 z} \left( 2 \alpha _1 + 3 v_{61} + 4 \alpha _1 u_{61} v_{61} + 4(1+\alpha _1+\alpha _2) u_{61} v_{61}^2 \right. \\&\left. \quad +\, 3 u_{61} v_{61}^3 + 4 \alpha _2 u_{61}^2 v_{61}^3 + 2 \alpha _2 u_{61}^2 v_{61}^4 \right) . \end{aligned} \end{aligned}$$
The second chart is
$$\begin{aligned} u_{62}= & {} u_{32}=\frac{1}{v}, \qquad v_{62}=\frac{v_{32}}{u_{32}}=uv, \\ u= & {} u_{62}v_{62}, \qquad v=\frac{1}{u_{62}}. \end{aligned}$$
The exceptional line is \(\mathcal {L}_6:u_{62}=0\). In this chart, the preimages of \(\mathcal {L}_3\) and \(\mathcal {L}_0\) are not visible.
The flow is
$$\begin{aligned} \begin{aligned} u_{62}'&=-1-2u_{62}-2u_{62}^2v_{62}+\frac{u_{62}(2\alpha _2u_{62}+1)}{2z}, \\ v_{62}'&=\frac{v_{62}(-1+u_{62}^2v_{62})}{u_{62}}-\frac{\alpha _1+u_{62}v_{62}(1+\alpha _2u_{62})}{zu_{62}}, \end{aligned} \end{aligned}$$
and contains a base point is \(u_{62}=0,v_{62}=-\,{\alpha _1}/{z}\), which is \(b_8\).
The energy is given by
$$\begin{aligned} \begin{aligned} E&=-\frac{v_{62}}{u_{62}} \left( 1 + 2 u_{62} + u_{62}^2 v_{62}\right) , \qquad EJ_{62}=-v_{62}\left( 1 + 2 u_{62} + u_{62}^2 v_{62}\right) , \\ E'&=\frac{1}{2 u_{62}^2 z} \left( 2 \alpha _1 + 4 \alpha _1 u_{62} + 3 u_{62} v_{62} + 4(1+\alpha _1+\alpha _2) u_{62}^2 v_{62} \right. \\&\left. \quad +\, 4 \alpha _2 u_{62}^3 v_{62} + 3 u_{62}^3 v_{62}^2 + 2 \alpha _2 u_{62}^4 v_{62}^2\right) . \end{aligned} \end{aligned}$$
1.4 Resolution at Points \(b_6\), \(b_7\), \(b_8\)
1.4.1 Resolution at \(b_6\)
The first chart is
$$\begin{aligned} u_{71}= & {} \frac{u_{42}}{v_{42}-\frac{\alpha _2}{z}}=\frac{z}{u(uvz-\alpha _2)}, \qquad v_{71}=v_{42}-\frac{\alpha _2}{z}=uv-\frac{\alpha _2}{z}, \\ u= & {} \frac{1}{u_{71} v_{71}}, \qquad v=u_{71} v_{71} \left( v_{71}+\frac{\alpha _2}{z} \right) , \end{aligned}$$
which gives the Jacobian
$$\begin{aligned} J_{71}&=\frac{\partial u_{71}}{\partial u}\frac{\partial v_{71}}{\partial v}-\frac{\partial u_{71}}{\partial v}\frac{\partial v_{71}}{\partial u} =\frac{z}{u(\alpha _2-uvz)}=-u_{71}, \\ J_{71}'&=-2u_{71}-3u_{71}^2v_{71}^2 -\frac{u_{71}}{2z}\big (3+4(\alpha _1+2\alpha _2)u_{71}v_{71}\big ) -\frac{(\alpha _1+\alpha _2)\alpha _2}{z^2}u_{71}^2. \end{aligned}$$
The exceptional line is \(\mathcal {L}_7:v_{71}=0\). In this chart, the preimage of \(\mathcal {L}_4\) is given by equation \(u_{71}=0\), while the preimages of \(\mathcal {L}_1\) and \(\mathcal {L}_0\) are not visible.
The flow is given by
$$\begin{aligned} \begin{aligned} u_{71}'&=2u_{71}+3u_{71}^2v_{71}^2 +\frac{u_{71}}{2z}\big (3+4(\alpha _1+2\alpha _2)u_{71}v_{71}\big ) +\frac{(\alpha _1+\alpha _2)\alpha _2}{z^2}u_{71}^2, \\ v_{71}'&= \frac{1}{u_{71}} - u_{71} v_{71}^3 -\frac{v_{71}}{z}\big (1 + (\alpha _1+2\alpha _2) u_{71} v_{71} \big ) -\frac{\alpha _2(\alpha _1+\alpha _2) u_{71} v_{71}}{z^2}, \end{aligned} \end{aligned}$$
and contains no base points.
The energy is given by
$$\begin{aligned} \begin{aligned} E&=-\frac{1 + 2 u_{71} v_{71} + u_{71}^2 v_{71}^3}{u_{71}}-\frac{\alpha _2}{u_{71} v_{71} z}\left( 1 + 2 u_{71} v_{71} + 2 u_{71}^2 v_{71}^3\right) -\frac{\alpha _2^2 }{z^2}u_{71} v_{71}, \\ EJ_{71}&=1 + 2 u_{71} v_{71} + u_{71}^2 v_{71}^3+ \frac{\alpha _2}{v_{71} z}\left( 1 + 2 u_{71} v_{71} + 2 u_{71}^2 v_{71}^3\right) +\frac{\alpha _2^2 }{z^2}u_{71}^2 v_{71}. \end{aligned} \end{aligned}$$
The second chart is
$$\begin{aligned} u_{72}= & {} u_{42}=\frac{1}{u}, \qquad v_{72}=\frac{v_{42}-\frac{\alpha _2}{z}}{u_{42}}=\frac{u(uvz-\alpha _2)}{z}, \\ u= & {} \frac{1}{u_{72}}, \qquad v=\frac{u_{72}}{z}(zu_{72}v_{72}+\alpha _2). \end{aligned}$$
In this chart, the exceptional line \(\mathcal {L}_7\) is given by equation \(u_{72}=0\), while the preimages of \(\mathcal {L}_4\), \(\mathcal {L}_1\), and \(\mathcal {L}_0\) are not visible.
The flow
$$\begin{aligned} \begin{aligned} u_{72}'&=\frac{u_{72}}{2z}\big (1 + 2 (\alpha _1 + 2 \alpha _2) u_{72}\big ) + 1 + 2 u_{72} + 2 u_{72}^3 v_{72}, \\ v_{72}'&=-\frac{2 \alpha _2(\alpha _1+\alpha _2) + 4(\alpha _1+2 \alpha _2) u_{72} v_{72} z + v_{72} z \left( 3 + 4 z + 6 u_{72}^2 v_{72} z\right) }{2 z^2}, \end{aligned} \end{aligned}$$
contains no base points.
1.4.2 Resolution at \(b_7\)
The first chart is
$$\begin{aligned} \begin{aligned} u_{81}&=\frac{u_{51}-\frac{1-\alpha _1-\alpha _2}{8z}}{v_{51}} \\&=\frac{u}{8 (u + v)^3 z}\big ((\alpha _1 + \alpha _2-1) v^2 + (\alpha _1 + \alpha _2 -1 + 4 z)u^2 \\&\quad + 2(\alpha _1 + \alpha _2 -1 + 2 z)uv + 8 u z \big ), \\ v_{81}=&v_{51}=\frac{u+v}{u}, \\ u&=\frac{8 z}{v_{81} ( 8 u_{81} v_{81}^2 z-( \alpha _1 + \alpha _2-1) v_{81} - 4 z )}, \\ v&=\frac{8 (v_{81}-1) z}{v_{81} ( 8 u_{81} v_{81}^2 z-( \alpha _1 + \alpha _2-1) v_{81} - 4 z )}. \end{aligned} \end{aligned}$$
In this chart, the exceptional line \(\mathcal {L}_8\) is given by equation \(v_{81}=0\), while the preimages of \(\mathcal {L}_5\) and \(\mathcal {L}_2\) are not visible.
The flow is given by
$$\begin{aligned} \begin{aligned} u_{81}'&= \left( \frac{1 - 4 \alpha _1 - 4 \alpha _2}{2 z}-10\right) u_{81} -\frac{\alpha _1 (\alpha _1 + \alpha _2-1)^2}{64 z^3}v_{81}-\frac{\alpha _1 (\alpha _1 + \alpha _2-1)}{16 z^2} \\&\quad +\left( \frac{\alpha _1}{z}-\frac{5 ( \alpha _1 + \alpha _2-1) (\alpha _1 + \alpha _2)}{8 z^2} \right) u_{81}v_{81} +\frac{(\alpha _1 + \alpha _2-1)^2 (\alpha _1 + \alpha _2)}{32 z^3} \\&\quad +\frac{3 \alpha _1 (\alpha _1 + \alpha _2-1)}{8 z^2} u_{81}v_{81}^2 +\frac{3(\alpha _1+\alpha _2)}{z}u_{81}^2v_{81}^2-\frac{2\alpha _1}{z}u_{81}^2v_{81}^3 \\&\quad + \frac{3}{16z^2(8 u_{81} v_{81}^2 z-(\alpha _1 + \alpha _2-1) v_{81} - 4 z)} \times \big ({-(\alpha _1 + \alpha _2-1)^3} \\&\quad -4 ( \alpha _1 + \alpha _2-1)^2 z -32 (3(\alpha _1 + \alpha _2-1) + 8 z) z^2u_{81} \\&\quad +8 ( \alpha _1 + \alpha _2-1)( \alpha _1 + \alpha _2-1+ 4z)z u_{81} v_{81} +512 z^3 u_{81}^2 v_{81} \big ), \\ v_{81}'&=-\,4 + 4 v_{81} - \frac{\alpha _1+\alpha _2}{z}u_{81}v_{81}^3+\frac{\alpha _1}{z}u_{81}v_{81}^4 +\frac{(\alpha _1+\alpha _2-1)(\alpha _1+\alpha _2)}{8z^2}v_{81}^2 \\&\quad -\frac{\alpha _1(\alpha _1+\alpha _2-1)}{8z^2}v_{81}^3-\frac{\alpha _1}{2z}v_{81}^2 +\frac{\alpha _1+\alpha _2}{2z}v_{81} \\&\quad +\frac{24z(v_{81}-1)}{8 u_{81} v_{81}^2 z-(\alpha _1 + \alpha _2-1) v_{81} - 4 z }. \end{aligned} \end{aligned}$$
There are no new base points.
The second chart is
$$\begin{aligned} u_{82}= & {} u_{51}-\frac{1-\alpha _1-\alpha _2}{8z}=\frac{u(u+v+2)}{2(u+v)^2}-\frac{1-\alpha _1-\alpha _2}{8z}, \\ v_{82}= & {} \frac{v_{51}}{u_{51}-\frac{1-\alpha _1-\alpha _2}{8z}}= \frac{u+v}{u}\left( \frac{u(u+v+2)}{2(u+v)^2}-\frac{1-\alpha _1-\alpha _2}{8z}\right) ^{-1}, \\ u= & {} \frac{-8 u_{82}^2 z}{v_{82} (4 u_{82} z + v_{82} ( \alpha _1 + \alpha _2 -1 - 8 u_{82} z))}, \\ v= & {} \frac{8 u_{82} (u_{82} - v_{82}) z}{v_{82} (4 u_{82} z + v_{82} ( \alpha _1 + \alpha _2 -1 - 8 u_{82} z))}. \end{aligned}$$
In this chart, the exceptional line \(\mathcal {L}_8\) is given by equation \(u_{82}=0\), and the preimage of \(\mathcal {L}_5\) by \(v_{82}=0\). The preimage of \(\mathcal {L}_2\) is not visible.
The Jacobian is
$$\begin{aligned} \begin{aligned} J_{82}&= \frac{v_{82} (4 u_{82} z + v_{82} (\alpha _1 + \alpha _2-1 - 8 u_{82} z))^3}{512 u_{82}^3 z^3}, \end{aligned} \end{aligned}$$
while the derivative of the Jacobian is
$$\begin{aligned} \begin{aligned} J_{82}'&=\frac{\partial J_{82}}{u_{82}}u_{82}'+\frac{\partial J_{82}}{v_{82}}v_{82}'+\frac{\partial J_{82}}{z} \\&= -\,\frac{(4 u_{82} z + v_{82} ( \alpha _1 + \alpha _2-1 - 8 u_{82} z))^2}{512 u_{82}^4 z^3}\times \left( 3 ( \alpha _1 + \alpha _2-1) v_{82}^2 u_{82}' \right. \\&\quad -\, 4u_{82}(u_{82} z + v_{82} ( \alpha _1 + \alpha _2 -1- 8 u_{82} z)) v_{82}' \\&\left. \quad +\,3 (\alpha _1 + \alpha _2-1) v_{82}^2 \right) . \end{aligned} \end{aligned}$$
The flow is given by
$$\begin{aligned} u_{82}'= & {} \frac{8}{v_{82}}-6u_{82} -\frac{(\alpha _1 + \alpha _2-1)^2}{64 z^3} u_{82} v_{82} (\alpha _1u_{82} v_{82}-2 \alpha _1 -2 \alpha _2) \\&-\,\frac{1}{4 z}\big (3(1 - \alpha _1 - \alpha _2) +2(3\alpha _1+3\alpha _2-1) u_{82} - 2 \alpha _1 u_{82}^2 v_{82} \\&-\, 8 (\alpha _1+\alpha _2) u_{82}^3 v_{82} + 4 \alpha _1 u_{82}^4 v_{82}^2\big ) \\&+\,\frac{\alpha _1 + \alpha _2-1}{16 z^2} \big (3( \alpha _1 + \alpha _2-1) - \alpha _1 u_{82} v_{82} - 8 (\alpha _1+\alpha _2) u_{82}^2 v_{82} \\&+\, 4 \alpha _1 u_{82}^3 v_{82}^2\big ) \\&+\,3\frac{32 z^2 + v_{82} ( \alpha _1 + \alpha _2 -1+ 4 z) ( \alpha _1 + \alpha _2-1 - 8 u_{82} z)}{4 v_{82} z (8 u_{82}^2 v_{82} z-( \alpha _1 + \alpha _2-1) u_{82} v_{82} - 4 z )}, \\ v_{82}'= & {} 10v_{82} +\frac{( \alpha _1 + \alpha _2-1)^2 v_{82}^2 ( \alpha _1 u_{82} v_{82}-2 \alpha _2 -2 \alpha _1)}{64 z^3} \\&+\,\frac{( \alpha _1 + \alpha _2-1) v_{82}^2 (\alpha _1+10\left( \alpha _1+ \alpha _2\right) u_{82} - 6\alpha _1 u_{82}^2 v_{82})}{16 z^2} \\&+\,\frac{v_{82}}{2z} \big ( 4 \alpha _1 + 4 \alpha _2-1 - 2 \alpha _1 u_{82} v_{82} - 6 (\alpha _1+\alpha _2) u_{82}^2 v_{82} + 4 \alpha _1 u_{82}^3 v_{82}^2\big ) \\&+\, \frac{3 v_{82}}{16 z^2 ((1 - \alpha _1 - \alpha _2) u_{82} v_{82} - 4 z + 8 u_{82}^2 v_{82} z)} \\&\quad \times \, \big ( ( \alpha _1 + \alpha _2-1)^3 v_{82} -4 (\alpha _1 + \alpha _2-1)^2 (2 u_{82}-1) v_{82} z \\&\quad -\,32 ( \alpha _1 + \alpha _2-1) (u_{82} v_{82}-3) z^2 +256(1-2u_{82})z^3 \big ). \end{aligned}$$
There are no new base points.
The energy is given by
$$\begin{aligned} \begin{aligned} E&=-\frac{128 ( u_{82} v_{82}-1) z^2 (1 - \alpha _1 - \alpha _2 + 8 u_{82} z)}{u_{82} v_{82} (8 u_{82}^2 v_{82} z-( \alpha _1 + \alpha _2-1) u_{82} v_{82} - 4 z)^3}, \\ EJ_{82}&=-\frac{( u_{82} v_{82}-1) (1 - \alpha _1 - \alpha _2 + 8 u_{82} z)}{4 u_{82}^4 z (8 u_{82}^2 v_{82} z-( \alpha _1 + \alpha _2-1) u_{82} v_{82} - 4 z)^3} \\&\qquad \times \big (4 u_{82} z + v_{82} ( \alpha _1 + \alpha _2 -1 - 8 u_{82} z)\big )^3. \end{aligned} \end{aligned}$$
1.4.3 Resolution at \(b_8\)
The first chart is
$$\begin{aligned} u_{91}= & {} \frac{u_{62}}{v_{62}+\frac{\alpha _1}{z}}=\frac{z}{v(uvz+\alpha _1)}, \qquad v_{91}=v_{62}+\frac{\alpha _1}{z}=uv+\frac{\alpha _1}{z}, \\ u= & {} u_{91}v_{91}\left( v_{91}-\frac{\alpha _1}{z}\right) , \qquad v=\frac{1}{u_{91}v_{91}}, \\ J_{91}= & {} \frac{\partial u_{91}}{\partial u}\frac{\partial v_{91}}{\partial v}-\frac{\partial u_{91}}{\partial v}\frac{\partial v_{91}}{\partial u} =\frac{z}{v(\alpha _1 + u v z)}=u_{91}, \\ J_{91}'= & {} -2 u_{91} - 3 u_{91}^2 v_{91}^2 - \frac{\alpha _1(\alpha _1+ \alpha _2) u_{91}^2}{z^2} + \frac{3 u_{91}}{2 z} +\frac{2(2\alpha _1+\alpha _2)u_{91}^2 v_{91}}{z}. \end{aligned}$$
The exceptional line is \(\mathcal {L}_9:v_{91}=0\). In this chart, the preimage of \(\mathcal {L}_6\) is given by equation \(u_{91}=0\), while the preimages of \(\mathcal {L}_3\) and \(\mathcal {L}_0\) are not visible.
The flow is
$$\begin{aligned} \begin{aligned} u_{91}'&=-2 u_{91} - 3 u_{91}^2 v_{91}^2 - \frac{\alpha _1(\alpha _1+ \alpha _2) u_{91}^2}{z^2} + \frac{3 u_{91}}{2 z} +\frac{2(2\alpha _1+\alpha _2)u_{91}^2 v_{91}}{z}, \\ v_{91}'&=-\frac{1}{u_{91}} + u_{91} v_{91}^3 + \frac{\alpha _1(\alpha _1+\alpha _2) u_{91} v_{91}}{z^2} - \frac{v_{91}}{z} - \frac{(2 \alpha _1+\alpha _2) u_{91} v_{91}^2}{z}, \end{aligned} \end{aligned}$$
and contains no base points.
Energy:
$$\begin{aligned} \begin{aligned} E&=\frac{(v_{91} z-\alpha _1 ) (\alpha _1 u_{91}^2 v_{91}^2 - z - 2 u_{91} v_{91} z - u_{91}^2 v_{91}^3 z)}{u_{91} v_{91} z^2}, \\ EJ_{91}&=-1 - 2 u_{91} v_{91} - u_{91}^2 v_{91}^3 - \frac{\alpha _1^2 u_{91}^2 v_{91}}{z^2} + \frac{ 2 \alpha _1 u_{91}}{z} + \frac{\alpha _1}{v_{91} z} + \frac{2 \alpha _1 u_{91}^2 v_{91}^2}{z}. \end{aligned} \end{aligned}$$
The second chart is
$$\begin{aligned} u_{92}= & {} u_{62}=\frac{1}{v}, \qquad v_{92}=\frac{v_{62}+\frac{\alpha _1}{z}}{u_{62}}=\frac{v(uvz+\alpha _1)}{z}, \\ u= & {} u_{92}^2v_{92}-\frac{\alpha _1}{z}u_{92}, \qquad v=\frac{1}{u_{92}}. \end{aligned}$$
The exceptional line is \(\mathcal {L}_9:u_{92}=0\). In this chart, the preimages of \(\mathcal {L}_6\), \(\mathcal {L}_3\) and \(\mathcal {L}_0\) are not visible.
The flow is given by
$$\begin{aligned} \begin{aligned} u_{92}'&=-1 - 2 u_{92} - 2 u_{92}^3 v_{92} + \frac{u_{92}}{2 z} + \frac{(2 \alpha _1+\alpha _2) u_{92}^2}{z}, \\ v_{92}'&=2 v_{92} + 3 u_{92}^2 v_{92}^2 + \frac{\alpha _1(\alpha _1+\alpha _2)}{z^2} -\frac{ 3 v_{92}}{2 z} - \frac{ 2(2 \alpha _1+\alpha _2) u_{92} v_{92}}{z}. \end{aligned} \end{aligned}$$
and contains no base points.
