Inviscid warp equations of a nearly Keplerian disk ($\Sigma R^2\Omega (W_t-i{\omega }_{1}W)=R^{-1}{G}_R,G_t-i{\omega }_{2}G=\frac{1}{4}P{R}^3\Omega W_R$)

Here $G$ is the complex internal horizontal torque, $P$ the vertically integrated pressure, and ${\omega }_1,{\omega }_2$ the approximate nodal precession and apsidal precession rates. In the exactly Keplerian case, local plane waves obey ${\omega }^2=k^{2}P/(4\Sigma )=k^2{c}_s^2/4$, so tilt disturbances propagate at $c_s/2$. The secular thin-disk approximation requires wavelengths large relative to the height while the local approximation requires them small relative to the orbital radius.

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 $x=4\sqrt{24|a|}{r}^{-5/4}/(5ϵ)$.