Put $\tau =T/\rho >0$ and take the positive-frequency branch for $k>0$. The phase velocity and group velocity areSince the square root is increasing, minimizing $c_p$ is equivalent to minimizing $g/k+\tau k$. Its derivative is $-g/k^2+\tau$, which changes sign from negative to positive atThe phase speed diverges at both positive endpoints $k\downarrow 0$ and $k\to \infty$, so this stationary point is its global minimum. At $k_c$, $\tau k_c^2=g$, givingThe relative speed is especially transparent:A narrow wave packet moves with $c_g$, while its crests move with $c_p$. For $k<k_c$, $c_p>c_g$: crests move forward through the packet, from its rear towards its front. For $k>k_c$, $c_p<c_g$: crests drift backwards relative to the packet, although both absolute velocities are positive. At $k_c$ the leading-order relative drift vanishes. This is the minimum phase speed of a gravity-capillary wave.

Use the complex potential of a two-dimensional potential flow convention $W=\varphi +i\psi$, for which $W^{\prime }(z)=u-iv$. A positive circulation is counterclockwise. The two point vortices and the two-dimensional point sink giveBranch choices change additive constants, not the local velocity field. A vortex is transported by the nonsingular velocity due to the other vortex and the sink; its own singular term is omitted. At $z=c$ this givesThe other vortex follows $-c$ by symmetry. Put $c=r{e}^{i\theta }$, with $r(0)=a$ and $\theta (0)=0$. Multiplication by $\bar{c}$ givesFor $Q>0$, integration yieldsEquivalently, with $s=1-Qt/(\pi a^2)$,The curves obey $\theta =(\Gamma /(2Q))\log (a/r)$, so they are logarithmic spirals. At $t=\pi a^2/Q$, both vortices reach the sink at the origin. For nonzero circulation they make infinitely many turns in that finite time; for zero circulation they approach radially. The point-vortex evolution becomes singular at capture, so the formula does not define a regular continuation through the event.

Take independent smooth variations $y+\epsilon \eta$ and $z+\epsilon \zeta$. Differentiating the functional and using integration by parts gives its first variationFirst choose variations vanishing at the endpoints. To use the stated form of the fundamental lemma of the calculus of variations, take $\eta =(x-a)(b-x)\chi$ with arbitrary smooth $\chi$, and similarly for $\zeta$. The lemma makes $(x-a)(b-x)(F_y-d{F}_{y^{\prime }}/dx)$ vanish, and hence makes the unweighted coefficient vanish in the interior; continuity extends this to the endpoints. Applying the same argument to $z$ gives the Euler-Lagrange equationsIf either endpoint value of $y$ or $z$ is free, the corresponding variation there is arbitrary. The remaining boundary term gives the natural boundary conditions for a free endpoint: $F_{y^{\prime }}=0$ for a free $y$ value and $F_{z^{\prime }}=0$ for a free $z$ value at that endpoint. With both coordinates free at both endpoints, all four conditions apply.

For the specified integrand these equations reduce toand the natural conditions are $y^{\prime }=z^{\prime }=0$ at each free endpoint. Eliminating $z$ givesIts four independent modes give the full general solutionAt zero, $y^{\prime }(0)=C+2D$ and $z^{\prime }(0)=2C+D$. Vanishing forces $C=D=0$, givingAt $x=\pi /6$ the two derivative conditions become $A/2+\sqrt{3}B=0$ and $A+\sqrt{3}B/2=0$. Their coefficient determinant is nonzero, so $A=B=0$. Thus the unique stationary extremal with all four natural boundary conditions is $y=z=0$. The word extremal here means a stationary point of the functional; the indefinite integrand does not require this point to be a minimum.