Some New Results for $N \geq 4$

The Painlevé Condition is invariant under arbitrary transformations $\phi=F(\theta),$ since,

$$\Omega=\sum_{i=1}^N\phi_{x_i}^4\sum_{k>j}J_{x_j,x_k}\left(\fr... ...phi_{x_k}}{\phi_{x_i }}, \frac{\phi_{x_j}}{\phi_{x_i}}\right)=0$$ (20)

where, $J_{x,y} \left(U,V\right) = U_x V_y - U_y V_x.$

There is a Quadratic Solution $\phi\left(\hat x \right)=\hat x\cdot A\hat x$ where

$$A=\left(\begin{array}{cccc} a_{11} & \mp 1 & \xi & \lambda \... ...1} & 1 \\ \lambda & \eta & 1 & -a_{22} \end{array} \right)$$ (21)

$$a_{11}=\frac 12 (\lambda - 1/\lambda)\xi \pm (\lambda+1/\lambda)\eta,$$

$$a_{22}=\frac 12 (\lambda + 1/\lambda)\xi \pm (\lambda-1/\lambda)\eta,$$

$$\lambda^2=-\frac{(\xi\mp\eta)^2}{4+(\xi\pm\eta)^2}.$$

The Painlevé Condition is identical to

$$\Omega=\nabla\phi\cdot \left(-\left(\nabla\nabla^t\phi\right)... ...elta\phi\nabla\nabla^t\phi + \Omega_2(\phi)I\right)\nabla\phi=0$$ (22)

where,

$${\displaystyle tr}\gr3=\Delta\phi=\lambda_1+\cdots+\lambda_N,$$

$$\Omega_2(\phi)=\frac 12\left({\displaystyle tr}\left(\gr3^2\right) - (\Delta \phi)^2\right),$$

$$\Omega_2=-\lambda_1\lambda_2 - \cdots - \lambda_{N-1}\lambda_N.$$

The Characteristic Equation for $\dr1$ is:

$$\gr3^N-\Omega_1\gr3^{N-1} -\Omega_2\gr3^{N-2} - \cdots \cdots -\Omega_N I=0.$$ (23)

where

$$\Omega_j=(-1)^{j+1}(\lambda_1\ldots\lambda_j + \cdots + and \lambda_{N-j+1}\ldots\lambda_N)$$

$$j\Omega_j={\displaystyle tr}\left(\gr3^j\right) -\sum_{k=1}^{j-1}\Omega_k{\displaystyle tr}\left(\gr3^{j-k}\right)$$ (24)

The definition [4] of the Legendre Transformation implies:

$${\gr3^2-\Delta\phi\gr3-\Omega_2I}=\Omega_N(\phi)\times$$

$$\hr3^{N-2}-\Omega_1(w) \hr3^{N-3} - \cdots - \Omega_{N-3}(w)\hr3$$ (25)

since [4] $\gr3\hr3=I$ and

$$\Omega_j(w)=-\Omega_{N-j}(\phi)/\Omega_N(\phi).$$

Therefore,

$$\m2=0$$

implies the Legendre Transform

$${\hat \xi}\cdot(\hr3^{N-2} -\Delta w\hr3^{N-3} - \cdots - \Omega_{N-3}\hr3){\hat \xi}=0$$ (26)

Let

$${d\over ds} = {\hat \xi}\cdot{\nabla_{\hat \xi}}.$$

For N=3, as found previously,

$${\hat \xi}^t\hr3{\hat \xi}= w_{ss} - w_s=0$$

$$w=w_0+w_1$$

$${d\over ds}w_0=0,{d\over ds}w_1=w_1$$

For N=4

$${\hat \xi}^t\left(\hr3^2 - \Delta w \hr3\right){\hat \xi}=$$ (27)

$$\nabla(w_s-w)\cdot\nabla(w_s-w)-(w_{ss}-w_s)\Delta w=0$$ (28)

Let $w$ be homogeneous of degree $m$, i.e. $w_s=mw$. Then,

$$(m-1)^2\:\nabla w\cdot\nabla w = m(m-1)\;w\Delta w$$

The substitution, with $m \neq 0,1$,

$$w=\Theta^m,$$

represents the solution as a Spherical Harmonics of degree one,

$$\Delta\Theta=0,$$

$$\Theta_s=\Theta.$$

From Hobson [5], the Spherical Harmonics of degree one can be found from the Spherical Harmonics of degree zero, $V=V(\theta,\phi,\beta)$. We define Spherical Harmonics of degree zero with $N=4$ by

$$x=r\sin\theta \cos\phi \,\, y=r\sin\theta \sin\phi,$$

$$z=r\cos\theta \cos\beta\,\, t=r\cos\theta \sin\beta ,$$

$${\partial\over\partial\phi}\left({\cos\theta\over\sin\theta}V... ...r\partial\theta}\left({\sin\theta\cos\theta}V_\theta\right) =0.$$

The change of variable,

$$\psi=\frac14\log\tan\theta,$$

obtains

$$e^{-4\psi}\left(V_{\phi\phi}+V_{\psi\psi}\right) + e^{4\psi}\left(V_{\beta\beta} + V_{\psi\psi}\right)=0.$$

After a Fourier transform in $(\phi,\beta),$ find

$$V_{\psi\psi}=\left({t^2+s^2\over 2} + {t^2-s^2\over 2}\tanh 4\psi\right)V.$$