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For a symmetric bounded bilinear form $B$ that is strictly positive on nonzero vectors, and a bounded linear functional $\ell$, set $I(v)=B(v,v)-2\ell (v)$. The first variation is $DI(v)[w]=2(B(v,w)-\ell (w))$. A weak solution $u$ of $B(u,w)=\ell (w)$ satisfies $I(u+w)-I(u)=B(w,w)$, so it is the unique minimizer. Conversely, any minimizer solves the weak equation. A coercive bilinear form gives existence by the Lax-Milgram theorem; strict positivity alone does not.

An integral constraint need not admit a normal Lagrange multiplier unless its first variation is nonzero. For the quadratic normalization in the title, the variation in direction $y$ has derivative $2\int w{y}^2=2$, so the constraint is always regular at an admissible function. With free endpoint values this justifies the Euler-Lagrange equation for $F-\lambda w{y}^2$ and its natural endpoint conditions. For $F=p{y}^{\prime 2}-q{y}^2$, multiplication of the differential equation by $y$ gives $\lambda =J[y]$.

For an ideal soap film with equal constant surface tension $\gamma$ on both faces, the effective tension is $\tau =2\gamma$. Choose a unit normal $N$ from the minus side to the plus side, and $\Delta p=p_{-}-p_{+}$. Translating the Young–Laplace equation to $S=-dN$ givesA single interface uses $\tau =\gamma$. An outward-oriented spherical soap bubble has $H=-1/R$, giving $\Delta p=4\gamma /R$. The positive-convex curvature sum in the Young–Laplace equation article is $-2H$. ^[ref]

An open film with the same resting gas on both sides has zero pressure jump. Neglecting film weight, soap film pressure balance gives $H=0$: a minimal surface. A pressurized bubble instead has nonzero mean curvature. These statements assume equilibrium, constant surface tension, and negligible normal viscous stress. Surfactant gradients or motion require the fuller normal stress balance. ^[ref]

The variational outline is short. With fixed boundary, a normal displacement $fN$ gives $\delta A=-2{\int }_{M}HfdA$ by the first variation of area formula. For a closed interface oriented outward, $\delta V={\int }_{M}fdA$, so stationarity of $\tau A-\Delta pV$ gives $-2\tau H-\Delta p=0$. This is a first variation calculation; testing for a stable minimal surface needs the second variation, and deciding a global minimum needs a comparison of competitors.

A point is stationary for a differentiable function or functional when its first derivative or first variation vanishes in every admissible direction.

At a freely moving regular junction of three smooth image edges with equal isotropic length weights, the endpoint first variation requires the sum of their outward unit tangent vectors to vanish. Three unit vectors with zero sum have pairwise angles of $120$ degrees. Different interface weights instead give a weighted force balance, and constrained or singular junctions require separate hypotheses.