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The factor ${\eta }_{xx}/(1+{\eta }_x^2{)}^{3/2}$ is the signed curvature of the graph, so the pressure jump is surface tension times curvature, the planar Young–Laplace equation.

Take total velocity potentials ${\Phi }_{-}={\varphi }_{-}$ and ${\Phi }_{+}=Ux+{\varphi }_{+}$. Each disturbance potential satisfies the Laplace equation. The interface moves with the fluid: for $F=y-\eta (x,t)$, the material equation $F_t+v\cdot \nabla F=0$ gives ${\eta }_t+u{\eta }_x=v_y$. Keeping only first-order terms and evaluating at $y=0$ yieldsThe unsteady Bernoulli equation is ${\Phi }_t+|\nabla \Phi {|}^2/2+p/\rho =C(t)$, with any common gravitational contribution canceling across this equal-density interface. Subtracting the basic flow and linearizing gives $p_{-}^{\prime }=-\rho {\varphi }_{-,t}$ and $p_{+}^{\prime }=-\rho ({\varphi }_{+,t}+U{\varphi }_{+,x})$. The linearized pressure jump is $T{\eta }_{xx}$. HenceUse complex modes with real parts understood:The decaying vertical factors solve the Laplace equation in their respective half-planes. The kinematic conditions give $B_{-}=-i\omega A/k$ and $B_{+}=i(\omega -kU)A/k$. Substitution in the dynamic condition givesWhen the discriminant is nonnegative, the phase speeds are $U/2\pm \sqrt{Tk/(2\rho )-U^2/4}$. They are two dispersive capillary-wave branches advected by the mean velocity $U/2$.

When $\boxed{U^2>2Tk/\rho }$, the frequencies are complex conjugates and one mode grows exponentially, with growth rate $k\sqrt{U^2/4-Tk/(2\rho )}$. This is Kelvin-Helmholtz instability. Surface tension stabilizes sufficiently short waves, while for any nonzero shear, sufficiently long waves are unstable in this gravity-free equal-density model. A real cosine with real frequency describes the stable branches; the unstable case is represented by the real part of the corresponding growing complex mode.

Orient $\partial B$ counterclockwise, and define positive circulation by that orientation. The outward normal from the body satisfies $nds=(dy,-dx)$. Pressure acts inward, soThe steady Bernoulli equation for this constant-density irrotational flow is $p+\rho |v{|}^2/2=C$. The constant has zero integral around a closed curve. With complex potential of a two-dimensional potential flow $w=\varphi +i\psi$, let $W=w^{\prime }=u-iv$. The no-penetration condition makes the boundary a streamline, so $dw=Wdz$ is real along it. Thus $Wdz=\overline{W}d\overline{z}$, and $|W{|}^{2}d\overline{z}=W^{2}dz$. Substitution proves Blasius force theorem:At infinity the Laurent expansion has the formThere is no source term because the impermeable body has zero net flux. The circulation term follows from $\oint Wdz=\Gamma$. The coefficient of $z^{-1}$ in $W^2$ is $\Gamma (U-iV)/(\pi i)$. Analyticity in the fluid lets the body contour be enlarged to a large counterclockwise circle, so the residue theorem gives $\oint W^{2}dz=2\Gamma (U-iV)$. HenceIts component along $(U,V)$ is zero: the drag is zero. The lift magnitude is $\boxed{\rho |\Gamma |\sqrt{U^2+V^2}}$. With the stated positive counterclockwise circulation, the signed lift along the left normal $(-V,U)/\sqrt{U^2+V^2}$ is $-\rho \Gamma \sqrt{U^2+V^2}$. Reversing the circulation convention reverses this signed formula. This is the Kutta–Joukowski theorem with its orientation made explicit.

Differentiating orthonormality gives ${\dot{e}}_i\cdot e_j=-e_i\cdot {\dot{e}}_j$. With $\omega =\frac{1}{2}{\sum }_i{e}_i\times {\dot{e}}_i$, the vector triple product yieldsThe body coordinates $x_i$ of a fixed material point are constant. Differentiating its position $R_G+{\sum }_i{x}_i{e}_i$ therefore givesThis identifies $\omega$ as the angular velocity of the rigid body.

Because $G$ is the center of mass, $\int \rho rdV=0$, so the translational contribution to the angular momentum about $G$ vanishes. The remaining vector triple product givesThis symmetric matrix is the inertia tensor about $G$. The angular-momentum balance about the center of mass is $dL/dt={\tau }_{\mathrm{ext},G}$ in an inertial frame. With no external forces, that torque is zero, so the inertial vector $L$ is constant. Its components in rotating body axes need not individually be constant.

For the tool, let the disc lie at $z=0$ and the rod extend to $z=l=\sqrt{3}a$. The total mass is $3M$ and its center is at $z_G=l/3$. Axial symmetry makes the inertia tensor diagonal with $I_x=I_y$.

The standard disc moments are $M{R}^2/2$ about its normal and $M{R}^2/4$ about a diameter. A thin rod has moment $m{l}^2/12$ about a transverse axis through its center. The parallel axis theorem adds $m{d}^2$ for a parallel shift of distance $d$. ThereforeandThe rod contributes zero about its own longitudinal axis. Thus $\boxed{J=2M{a}^2{I}_3}$, an isotropic inertia tensor even though the shape is not spherical. This gives a nonspherical rigid body with isotropic inertia. Consequently $L=2M{a}^2\omega$. Conservation of $L$ in torque-free motion proves the angular velocity is constant for every initial orientation and spin.