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Choose transverse $x$ along the cusp from the line of cylinder contact, and $y$ across its narrow gap. Expanding the two circular boundaries for $x\ll a$ gives $y=\pm x^2/(2a)$, so the local full gap is $d(x)=x^2/a$. At the meniscus, $x\simeq w$, a semicircle fits across this gap with radius $d(w)/2=w^2/(2a)$. Thus the leading area and transverse curvature in the cusp flow between touching cylinders areThe meniscus area correction is $O(w^4/a^2)$ and is small relative to $w^3/a$. The curvature is stated as a positive suction magnitude; the liquid pressure relative to air is $p=-2\gamma a/w^2$. Axial curvature is negligible compared with this transverse curvature under the slow-variation approximation.

At fixed $x$, axial lubrication theory gives a local slit Poiseuille flow between rigid walls $y=\pm d/2$:Integrating across the cusp givesThe local meniscus edge correction occupies a transverse width small compared with $w$ and does not change this leading flux. Since $p_z=4\gamma a{w}^{-3}{w}_z$, the axial volume flux is $Q=-\gamma w^4{w}_z/(21\mu a^2)$. Area conservation ${\mathcal{A}}_t+Q_z=0$ now givesWriting $u=w^3$ also puts this in porous medium equation form $u_t=\gamma (u^{5/3}{)}_{zz}/(35\mu a)$, explaining the finite-support spreading profile.

Set $K=\gamma /(7\mu a)$. Fixed volume means $\int w^{3}dz=3aV$. If $w\propto t^{-\beta }$ and axial extent is proportional to $t^{\alpha }$, this conservation gives $\alpha =3\beta$. The evolution equation gives $3\beta +1=5\beta +2\alpha$, hence $\beta =1/8$, $\alpha =3/8$.

For the symmetric source-type similarity solution, write $w=t^{-1/8}f(\eta )$, $\eta =z/t^{3/8}$. The differential equation becomesIntegrate from the centre using zero flux and symmetry. This gives $K{f}^4{f}^{\prime }=-3\eta f^3/8$, so on the positive support $f{f}^{\prime }=-3\eta /(8K)$. With $f({\eta }_N)=0$,Therefore the fixed-volume capillary spreading in a cylindrical cusp profile isLet $B=21\mu a/(8\gamma t)$. Its volume isWith $s=\sin \theta$, the last integral is ${\int }_{-\pi /2}^{\pi /2}{\cos }^4\theta d\theta =3\pi /8$, equal to the supplied sine integral by symmetry. Consequently $V=\pi B^{3/2}{z}_N^4/(8a)$, yieldingThe other tip is at $-z_N$. The maximum width decays as $t^{-1/8}$. This is the leading slender outer solution: the ideal point injection and the steep microscopic nose require local descriptions outside its uniform lubrication range.

Finally, for vertical cylinders let $\rho$ denote the liquid density, and take the bath pressure at $z=0$ as the air-pressure reference. At rest the liquid pressure is $p(z)=-\rho gz$. Equating hydrostatic suction with the wetting-meniscus capillary pressure gives the hydrostatic rise in a cylindrical cusp:The cusp can therefore support an indefinitely high, increasingly narrow ideal wetting tail rather than a fixed terminal rise height. Large $z$ makes $w\ll a$ and strengthens the transverse cusp approximation, until microscopic physics limits the continuum description.