Stationary bending warp of a Kerr disk ($W=W_{\infty }\cos (x_{\mathrm{in}}-x)/\cos x_{\mathrm{in}},x=4\sqrt{24a}{r}^{-5/4}/(5ϵ)$)

For $a>0$, $\Sigma \propto R^{-3/4}$, constant disk aspect ratio $ϵ=H/R$, and $r=R{c}^2/(GM)$, the inviscid steady disk warp equation reduces to $W_{xx}+W=0$ in the displayed coordinate. The vanishing inner torque gives $W_x(x_{\mathrm{in}})=0$; the outer tilt gives $W(0)=W_{\infty }$. The displayed solution holds when $\cos x_{\mathrm{in}}\ne 0$. If $x_{\mathrm{in}}=\pi$, it is $W=W_{\infty }\cos x$, with an aligned node at $r=2^{4/5}{r}_{\mathrm{in}}$ and opposite signed inner tilt. For negative spin the corresponding equation is hyperbolic, giving $W=W_{\infty }\cosh (x_{\mathrm{in}}-x)/\cosh x_{\mathrm{in}}$ with $x=4\sqrt{24|a|}{r}^{-5/4}/(5ϵ)$.