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Use the magnetic energy $E_B=(2{\mu }_0{)}^{-1}{\int }_V|\mathbf{B}{|}^{2}dV$. Vary the Euler potentials while keeping their boundary values fixed. The field variation isLet $\delta \mathbf{A}=\delta \alpha \nabla \beta -\delta \beta \nabla \alpha$. Integration by parts givesbecause $\delta \mathbf{A}=0$ on the boundary. ThereforeThe two interior variations are independent. At a smooth minimum their coefficients must vanish, so the curl is perpendicular to both potential gradients and hence parallel to $\mathbf{B}$ wherever $\mathbf{B}\ne 0$:This is a force-free magnetic field: the Lorentz force density vanishes. Taking the divergence also gives $\mathbf{B}\cdot \nabla \lambda =0$, so the proportionality factor is constant along a field line but need not be the same on different lines.

The fixed-label variation preserves the encoded field-line connection rather than allowing arbitrary reconnection during minimization. The argument proves a necessary condition for a smooth minimum, not that every force-free stationary field is a stable energy minimum. It also does not include gas pressure or gravity in the minimized functional.

Take $\mathbf{R}={\mathbf{r}}_2-{\mathbf{r}}_1$ and label each dumbbell once. Locally assume its probability density and number density vary negligibly over a connector length. The oriented connector crossing count follows geometrically: a connector crosses a small plane element precisely when its midpoint lies within a slab of normal thickness $|\mathbf{R}\cdot \mathbf{n}|$. For configurations in $d^{3}R$, the unsigned crossing number per unit area is thereforeThe expression without the absolute value is an oriented crossing count, or an unsigned count restricted to $\mathbf{R}\cdot \mathbf{n}\ge 0$. It cannot be an unrestricted nonnegative number: a connector $\mathbf{R}=-R\mathbf{n}$ gives a negative printed expression despite physically intersecting the plane.

To obtain traction, keep the force orientation as well. For $\mathbf{R}\cdot \mathbf{n}>0$, the spring force on the bead below the plane exerted by the bead above it is $H\mathbf{R}$. For the opposite orientation it is $-H\mathbf{R}$. Multiplying that force by the unsigned crossing count in either case givesIntegrating over all labelled configurations yieldsThere is no extra factor of two: each dumbbell was labelled and counted once, and the orientation-dependent force has already supplied the sign. This is the local continuum stress approximation; resolving appreciable spatial variation across a finite connector would require a nonlocal stress construction.

For a variation $y+\epsilon \eta$ with $\eta (a)=0$ but arbitrary $\eta (b)$, differentiation and integration by parts giveFirst choose variations supported inside the interval. The fundamental lemma of the calculus of variations gives the Euler-Lagrange equation. Then allow arbitrary endpoint variation to obtain the natural boundary conditions for a free endpoint:Because $F$ has no explicit dependence on $x$,Thus the Beltrami identity is $F-y^{\prime }{F}_{y^{\prime }}=k$, constant along an extremal.

For the brachistochrone problem, put $u=-y>0$, so conservation of energy gives the bead speed $v=\sqrt{2gu}$. Its travel time isThe Beltrami identity now gives $u(1+u^{\prime 2})=2R$ for a positive constant $R$. A descending cycloid parametrizes this relation:The free endpoint has $F_{u^{\prime }}=0$, hence $u^{\prime }(x_0)=0$. On the first descending arc this occurs at $\theta =\pi$. ThereforeAlong this arc $ds=2R\sin (\theta /2)d\theta$ and $v=2\sqrt{gR}\sin (\theta /2)$, soThe initial vertical tangent causes no divergent travel time: the quotient $ds/v$ has the finite constant limit $\sqrt{R/g}d\theta$.

To verify global minimality, take any admissible wire graph of finite travel time below the release level, and let $H>0$ be its maximum depth. For $0<u\le H$, the Cauchy-Schwarz inequality givesIndeed the two coefficients on the right, after multiplication by $\sqrt{u}$, form a unit vector. Since the path starts at depth zero and reaches depth $H$, the weighted total variation is at least the integral over all depths from $0$ to $H$. ConsequentlyThe last inequality is minimized at $H=2x_0/\pi$. Our cycloid attains equality throughout: its depth increases to $H$ and $u^{\prime }=\sqrt{H/u-1}$. Thus it achieves the genuine minimum over the wire graphs, not merely a stationary time. Reflection handles a negative horizontal endpoint coordinate by replacing $x_0$ with $|x_0|$.

Let the boundary split into a displacement part ${\Gamma }_D$ and a traction part ${\Gamma }_N$. The linearized stored elastic energy functional isIts corresponding total potential subtracts the work of fixed body and surface loads:The problem in the calculus of variations is to minimize $\Pi$ among admissible displacements with $u=u_D$ on ${\Gamma }_D$. For a stable quadratic elastic energy this is the usual energy minimization formulation. Its stationarity condition suffices to derive equilibrium.

Vary by $u+sv$, with $v=0$ on ${\Gamma }_D$. Major symmetry combines the two differentiated quadratic terms, givingIntegration by parts, first with test functions compactly supported in the interior, gives $-\mathrm{Div}{S}^{\mathrm{lin}}=f$. Then arbitrary traces on ${\Gamma }_N$ give the natural traction condition. The complete boundary value problem isFor a homogeneous hyperelastic material the elasticity tensor is constant. In isotropic linear elasticity the constitutive equation is $S^{\mathrm{lin}}=\lambda \mathrm{div}uI+\mu (Du+D{u}^T)$, giving the familiar Navier-Cauchy equation $-\mu \Delta u-(\lambda +\mu )\nabla \mathrm{div}u=f$.