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The length of a curve is $L(\alpha )={\int }_a^b\Vert {\alpha }^{\prime }(t)\Vert dt$. Its arc length parameter is $s(t)={\int }_a^t\Vert {\alpha }^{\prime }(u)\Vert du$. Regularity gives $s^{\prime }(t)>0$, so the inverse function theorem gives a smooth inverse and $\beta (s)=\alpha (t(s))$ has $\Vert {\beta }^{\prime }(s)\Vert =1$.

Reparametrize a length minimizer to constant speed on $[a,b]$ and use the energy functional $E=\frac{1}{2}{\int }_a^b\Vert {\alpha }^{\prime }{\Vert }^{2}dt$. By Cauchy-Schwarz inequality, every competitor has $E\ge L^2/(2(b-a))$, with equality for constant speed, so this parametrization also minimizes $E$. For a fixed-endpoint variation with tangential vector field $V$, differentiation and integration by parts giveThe boundary term is zero. Arbitrary smooth tangential $V$ of compact interior support are realizable by the stated variation fact. Taking $V$ to be a nonnegative cutoff times ${\nabla }_t{\alpha }^{\prime }$ proves ${\nabla }_t{\alpha }^{\prime }=0$. Hence a length minimizer is a geodesic after constant-speed reparametrization. The original parameter need not be affine: arbitrary varying-speed parametrizations preserve length but have nonzero covariant acceleration. This qualification is necessary for the literal wording.

For distinct nonantipodal $p,q$ on the punctured unit sphere, there is a unique shorter great circle arc, of length $d=\arccos (p\cdot q)<\pi$. A minimizing curve exists exactly when that shorter arc avoids the removed north pole. If it avoids the pole it attains the spherical lower bound. If it passes through the pole, arbitrarily small smooth detours have lengths tending to $d$, but equality would force the unique shorter arc, which is unavailable. Thus the infimum is not attained; the longer great circle arc is not a substitute minimizer.

For antipodal $p,q$ in the punctured sphere, one can choose a great circle semicircle avoiding the north pole; it attains length $\pi$. Thus every admissible antipodal pair has a minimizer. If identical endpoints are included despite the earlier distinctness stipulation, the infimum is zero, but no smooth regular curve attains it; only a nonregular constant curve does.

Put $\eta =\psi -\varphi$. The common Dirichlet boundary condition gives $\eta =0$ on $S$. Expanding the energy functional,Integration by parts and the field equation make the cross term zero:Consequently

For $m>0$, equality holds exactly when $\eta =0$. For $m=0$, equality holds exactly when $\nabla \eta =0$, so $\eta$ is constant on the connected region; its zero boundary value then again forces $\eta =0$. The minimizing property therefore proves uniqueness for the Dirichlet problem in both cases:

Choose the Sobolev space $V$ encoding the homogeneous essential boundary conditions, setafter the appropriate integration by parts, and define the energy functionalIts first variation is $J^{\prime }(u)v=a(u,v)-\ell (v)$, so its stationary points are exactly the solutions of the weak formulationIf $a$ is bounded, symmetric, and coercive and $\ell \in V^{\prime }$, the Lax-Milgram theorem supplies a unique weak solution $u$. Moreover, for every $w\in V$,unless $w=0$. Thus $J$ is strictly convex, and $u$ is its unique global minimizer. This proves existence and uniqueness of the minimizer and of the weak solution simultaneously.

For real $f\in L^2({\mathbb{R}}^3)$, this energy functional on $H^1({\mathbb{R}}^3)$ is coercive and attains its minimum by the direct method in the calculus of variations. The nonnegative potential term is weakly lower semicontinuous by local Sobolev compactness gives lower semicontinuity of a nonnegative integral. Its Euler-Lagrange equation is $-\Delta u+u+\sin u=f$ in the weak solution sense. The extra linear restoring term screens the Sine-Gordon equation nonlinearity. The scalar potential $V(s)=s^2/2+1-\cos s$ is convex because $V^{\prime \prime }(s)=1+\cos s\ge 0$; hence any weak critical point is a minimizer. The elliptic regularity estimate for $-\Delta +1$, together with $|\sin u|\le |u|$, gives $u\in H^2$. The Sobolev inequality then puts $u$ in $W^{1,6}$, so Morrey's inequality and uniformly continuous integrable functions vanish at infinity give a continuous representative tending to zero.