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The $-\mu \psi$ term means that the generator is the grand potential, rather than just the physical energy. With boundary variations vanishing, the Hamiltonian functional isIts functional derivative isso the Gross–Pitaevskii equation is $i\hslash {\psi }_t=\delta H/\delta {\psi }^{\ast }$. Treating $\psi$ and ${\psi }^{\ast }$ as independent variables in this variation gives the factor $g$, not $2g$.

For repulsive interactions $g>0$ and $\mu >0$, complete the square in the local energy:The gradient contribution is nonnegative, so the homogeneous ground state has arbitrary constant complex argument andThe infinite-volume constant may be subtracted to define a relative grand potential; it does not affect the functional derivative. At fixed particle number one instead minimizes the physical energy $H+\mu N$, with $\mu$ as its Lagrange multiplier. For $g>0$, $\mu \le 0$ the grand-potential minimum over $n\ge 0$ is the vacuum $n=0$. For $g<0$ the displayed unconstrained functional is unbounded below; the positive-density stable uniform state assumed in the following parts requires repulsion.

Suppress tildes in the normalized equation and let $X=x-vt$. A Gross–Pitaevskii solitary wave stationary in these translating coordinates obeysThe boundary fixes the bulk complex argument and describes a localized disturbance with zero net vortex winding number; a vortex pair is possible. No extra plane-wave factor is needed when only the coordinates are changed. A full Galilean transformation of the field would additionally change the bulk flow and its complex argument convention.

Subtract the uniform background from the grand potential. A convenient dimensionless energy and the renormalized momentum of a condensate areThe subtraction in $\mathbf{p}$ removes the troublesome background complex argument gradient. In particular, for the usual subsonic localized far field $\psi -1=O(r^{-1})$, $\nabla \psi =O(r^{-2})$, and $1-|\psi {|}^2=O(r^{-2})$, the energy density is $O(r^{-4})$ and the momentum integrand is $O(r^{-3})$, both integrable in two dimensions. Merely writing $\psi \to 1$ without sufficient decay is not by itself a convergence proof. The common solitary-wave branch is subsonic: linearizing this normalization gives ${\omega }^2=k^2/2+k^4/4$ and speed of sound $1/\sqrt{2}$.

Integration by parts, with decaying variations and their boundary terms vanishing, givesThus the translating equation is precisely $\delta (E-v{p}_x)=0$. Along a differentiable family ${\psi }_{\lambda }$ with fixed bulk normalization, evaluate this stationarity on ${\partial }_{\lambda }\psi$ to obtainwhere $d{p}_x/d\lambda \ne 0$. This energy–momentum slope of a solitary wave does not require differentiating the Lagrange multiplier $v$ inside the field variation. At a turning point the parametrized identity remains true, while $E$ need not be a single-valued function of $p_x$ locally. The chosen factors make $i{\psi }_t=\delta E/\delta {\psi }^{\ast }$, consistent with the normalized equation; changing the normalization of $E$ alone would change the claimed slope.

Assume the holonomic constraints have linearly independent gradients, so their common level set is locally a smooth constraint manifold parametrized by $q_1,\dots ,q_n$. The multiplier form of the Euler-Lagrange equations isAt fixed time ${\sum }_A{\partial }_A{f}_{\alpha }{\partial }_i{x}_A=0$. The chain rule for the reduced Lagrangian givesConversely, if these reduced equations hold, the Cartesian Euler-Lagrange residual annihilates every tangent direction. Linear independence of the constraint gradients implies that it is a linear combination of those gradients, giving the required Lagrange multipliers. This proves the local equivalence, including time-dependent constraints; singular constraints would need separate analysis.

For the bead, $\dot{y}=f^{\prime }(x)\dot{x}$, soThe autonomous conservation of energy gives $E=\frac{1}{2}(1+f^{\prime 2}){\dot{x}}^2$. If $E>0$, the motion has a fixed sign $s=\mathrm{sgn}\dot{x}$ andThe integral is the signed arc length divided by the constant speed, and is locally invertible. The special solution $E=0$ is a stationary bead, for which writing time as a function of its unchanging position is not meaningful.

For $f=x^2+y^2+z^2$ and $g=x^2+2y^2-z^2=1$, the Lagrange multiplier equations areThe four constrained stationary points areOn the constraint, $f=2x^2+3y^2-1$ with $x^2+2y^2\ge 1$. The two $y$-axis points are global minima, while the $x$-axis points are saddles: motion along the waist decreases $f$, whereas motion away from it increases $f$. There is no maximum.

If instead $g=x^2+2y^2-z^2$ is optimized on $x^2+y^2+z^2=1$, the stationary points are all six coordinate-axis points. The $y$-axis points are maxima with value $2$, the $z$-axis points are minima with value $-1$, and the $x$-axis points are saddles with value $1$.

Unbiasedness requires ${\sum }_i{\alpha }_i=1$. Let $S={\sum }_i{\sigma }_i^{-2}$ and minimize $V(\alpha )={\sum }_i{\sigma }_i^2{\alpha }_i^2$. The Lagrange multiplier equations for $V-\lambda ({\sum }_i{\alpha }_i-1)$ are $2{\sigma }_i^2{\alpha }_i=\lambda$, so normalization gives $\lambda =2/S$. HenceThe gradient at this point is $(2/S)1$, whose inner product with every feasible displacement $h$ satisfying $1^{T}h=0$ is zero. The Hessian matrix is $2\mathrm{diag}({\sigma }_1^2,\dots ,{\sigma }_N^2)$, and $h^T{\nabla }^{2}Vh=2{\sum }_i{\sigma }_i^2{h}_i^2>0$ for every nonzero feasible $h$. These are the first- and second-order constrained minimum conditions. Strict convexity also proves global uniqueness of this inverse-variance weighted mean; normal errors are not needed.

Use the Minkowski metric ${\eta }_{mn}=\mathrm{diag}(-1,1,\dots ,1)$ and units with speed of light one. The contractions are $P^2={\eta }^{mn}{P}_m{P}_n$ and $(T{X}^{\prime }{)}^2=T^2{\eta }_{mn}(X^m{)}^{\prime }(X^n{)}^{\prime }$; $(X^m{)}^{\prime }{P}_m$ and ${\dot{X}}^m{P}_m$ pair a vector with a covector without another metric. The Lagrange multipliers impose the two first-class constraintsThey generate worldsheet diffeomorphisms, so the phase space contains both constrained directions and gauge redundancy. Two first-class constraints remove two canonical pairs, leaving $D-2$ physical degrees of freedom per point.

In Monge gauge, $X^0=t$ and $X^1=\sigma$. Write the transverse canonical variables as $\mathbf{X}=(X^2,\dots ,X^{D-1})$ and $\mathbf{P}=(P_2,\dots ,P_{D-1})$. Solving the first-class constraints givesThe negative root selects positive energy. Substitution into the phase-space action gives the Hamiltonian reductionFor a static segment, $\mathbf{P}=0$, its proper length element is $d\ell =\sqrt{1+|{\mathbf{X}}^{\prime }{|}^2}d\sigma$, and $\mathcal{E}d\sigma =Td\ell$. Thus the string tension is the rest energy per unit proper length. In particular a straight resting segment has $\mathcal{E}=T$. Monge gauge is a local choice on a string embedding map for which $X^1$ is a valid coordinate; it need not cover folded strings or all endpoint configurations.

The real canonical variables $x^m,p_m$ describe the center-of-mass position and total momentum of the open string. The nonzero string oscillators are complex Fourier modes with ${\alpha }_{-k}^m=({\alpha }_k^m{)}^{\ast }$; the Lagrange multipliers obey ${\lambda }_{-j}={\lambda }_j^{\ast }$. The Virasoro constraints satisfy $L_{-j}=L_j^{\ast }$, so the multiplier term is real.

For each $k>0$, the oscillator kinetic term differs from the manifestly real expressionby $\frac{i}{2k}\frac{d}{dt}({\alpha }_k\cdot {\alpha }_{-k})$. Thus the written action is real up to a boundary term, which does not change its symplectic form or bulk dynamics. Adding the corresponding endpoint term makes reality exact.

The independent nonzero Poisson brackets areAll brackets between these independent center-of-mass and oscillator variables vanish. The dependent zero mode is ${\alpha }_0^m=p^m/\sqrt{\pi T}$, so, if it is used, $\{x^m,{\alpha }_0^n\}={\eta }^{mn}/\sqrt{\pi T}$.

The Fourier coefficients of the Virasoro constraints areTheir Poisson brackets areThey form the classical Witt algebra, with no Virasoro central extension. In particular they are first-class constraints, closing on the constraint surface, and generate the remaining worldsheet diffeomorphisms rather than independent physical excitations.

The action contains neither $\dot{N}$ nor ${\dot{N}}^i$, so their conjugate momentum vanishes. They impose constraints rather than independent evolution equations: after the Legendre transform in mechanics, the lapse function and shift vector are Lagrange multipliers in the Hamiltonian formulation of general relativity.

Use precisely the paper's sign convention for extrinsic curvature, $K_{ij}=({\dot{h}}_{ij}-D_i{N}_j-D_j{N}_i)/(2N)$. At fixed $h_{ij},N,N^i$,Consequently the canonical momentum of the spatial metric isThis is a weight-one tensor density, with indices raised and lowered using $h_{ij}$. Its trace and inverse relation areThe sign of $K_{ij}$ used here is opposite to the convention in extrinsic curvature of a spatial hypersurface; mixing the two conventions would reverse the momentum formula.

Associate nonnegative Lagrange multipliers $\alpha$ with $-Pz-1\le 0$ and $\beta$ with $Pz-1\le 0$. The Lagrangian dual problem is obtained fromIts infimum over unrestricted $z$ is finite exactly when $P^T(\alpha -\beta )=x$. Thus the dual isThe precise finiteness condition is $x\in \mathrm{range}{P}^T$. If $x=P^{T}q$, then $x^{T}z=q^{T}Pz\ge -\Vert q{\Vert }_1$ on the feasible set. The primal is feasible and bounded below, and splitting $q$ into positive and negative parts makes the dual feasible. Linear programming duality therefore gives attained equal finite optima. Equivalently, the strictly feasible point $z=0$ satisfies the Slater condition, with the required finiteness hypothesis. This is strong duality in the usual finite sense.

If $x\notin \mathrm{range}{P}^T=(\ker P{)}^{\perp }$, there is $d\in \ker P$ with $x^{T}d<0$. Every $z=td$, $t\ge 0$, is feasible and the objective tends to $-\infty$. The dual is then infeasible. Equality of extended values still holds if its supremum over the empty feasible set is defined as $-\infty$, but there are no finite optimizers. The paper does not assume full column rank of $P$, so this qualification is necessary. The same calculation gives the support function of an inverse image of an infinity-norm ball, ${\inf }_r{r}^{T}x=r_0^{T}x-{\min }_{P^{T}q=x}\Vert q{\Vert }_1$ for finite decisions.

In Cartesian coordinates the Lagrangian isIntroduce a Lagrange multiplier $\lambda$ for $x^2+y^2+z^2=c^{-2}$. The Euler-Lagrange equations becomeorDifferentiating the constraint twice givesTaking the scalar product of the vector equation with $\mathbf{x}$ therefore yieldsSubstitution gives

For $c=0$, the equations have the first integralsWriting $X=x+B$ and $Y=y-A$ gives $\dot{X}=-Y$ and $\dot{Y}=X$. Hence every trajectory isThese are helices of unit angular frequency about a line parallel to the $z$ axis, including circles when $C=0$ and a straight line when $R=0$.

Introduce the orthogonal coordinatesThen the constraint and objective becomeThe Lagrange multiplier equations for a stationary point areIf $\lambda =0$, then $u=y=0$ and $v=\pm 1$, givingIf $\lambda \ne 0$, then $v=0$. The first two equations and the constraint implyConverting back to $(x,y,z)$ gives the other four points. Therefore all stationary points arewith matching signs, together withThe middle pair has $\varphi =1/2$, the last pair has $\varphi =-1/2$, and the first pair has $\varphi =0$.

The Lagrange sufficiency theorem for maximizing $f$ subject to $g_i\ge 0$ and $h_j=0$ states that if a feasible point $x^{\ast }$ and multipliers ${\lambda }_i\ge 0,{\mu }_j$ satisfy complementary slackness and $x^{\ast }$ globally maximizes the Lagrangian function in constrained optimizationthen $x^{\ast }$ globally maximizes $f$ on the feasible set.

For the equality constraint, takeThe Lagrange multiplier equations giveso positivity and the constraint implyFor this multiplier, $L$ is a strictly concave function on the positive octant and its stationary point is its unique global maximum. The sufficiency theorem therefore proves that the constrained maximum is

For $x^2+y^2+z^2\le 1$, use $g=1-x^2-y^2-z^2$. The same point and multiplier satisfy feasibility, stationarity, $\lambda \ge 0$, and complementary slackness, and they again globally maximize the Lagrangian. Hence the answer is unchanged. Equivalently, any point with strict inequality can be radially scaled outward, increasing all three positive coordinates and therefore increasing $\log (xyz)$.

The first term is the local viscous force per unit length from resistive-force theory. With unit tangent vector $\mathbf{t}={\mathbf{r}}_s$, the drag tensor gives $-c_{\parallel }{\mathbf{v}}_{\parallel }-c_{\perp }{\mathbf{v}}_{\perp }$, where $\mathbf{v}={\mathbf{r}}_t$ is velocity relative to the fluid.

The second term is the bending force of an inextensible filament. Varying its energy $E_b=(B/2)\int |{\mathbf{r}}_{ss}{|}^{2}ds$ gives bulk force density $-B{\mathbf{r}}_{ssss}$. The final term, ${\partial }_s(T{\mathbf{r}}_s)$, is the force density from filament tension; $T$ is the Lagrange multiplier enforcing $|{\mathbf{r}}_s|=1$. At low Reynolds number, these forces balance without a filament acceleration term.

With mass, length, and time dimensions $M,\mathsf{L},\mathsf{T}$, the coefficients haveThus the drag coefficients have the dimensions of dynamic viscosity, the filament bending modulus has dimensions force times length squared, and tension has dimensions force.

Keep $y\ge 0$ as the domain restriction, introduce nonnegative Lagrange multipliers $s_i$ for $y_i\le 1$, and introduce a free multiplier $t\in \mathbb{R}$ for ${\sum }_i{y}_i=k$. The maximizing Lagrangian isIts supremum over $y\ge 0$ is finite exactly when $s+t1\ge x$, and then equals $kt+{\sum }_i{s}_i$. By linear programming duality, since the primal is feasible and bounded,Minimizing over $s$ for fixed $t$ gives $s_i=(x_i-t{)}_{+}$, the positive part of a real-valued function. Thus the threshold formula for the sum of the largest components isFor a direct attainment check, order the coordinates $x_{[1]}\ge \cdots \ge x_{[n]}$. Any $t\in [x_{[k+1]},x_{[k]}]$ is optimal for $k<n$, while any $t\le x_{[n]}$ is optimal for $k=n$.

Use the stable parameters $b,\kappa >0$. The bulk stationary points satisfy $f^{\prime }(\varphi )=a\varphi +b{\varphi }^3=0$. For $a<0$, $\varphi =0$ has negative curvature, while the two minima areBoth have $f(\pm {\varphi }_B)=-a^2/(4b)$ and $f^{\prime \prime }(\pm {\varphi }_B)=-2a>0$. To establish minimization at fixed total composition, writeFor mean composition $\bar{\varphi }$ with $|\bar{\varphi }|<{\varphi }_B$, a mixture of the two minima with volume fraction $\lambda =(\bar{\varphi }+{\varphi }_B)/(2{\varphi }_B)$ in the positive phase attains this lower bound in the bulk. Macroscopic slabs have only a subextensive interfacial excess free energy, so the bound is attained per unit volume in the thermodynamic limit. Thus constrained minimization gives phase coexistence in the mean-field approximation; at the endpoints only one phase is needed.

The two equilibrium conditions also follow directly by constrained optimization. Introduce a Lagrange multiplier $\mu$ for $\lambda {\varphi }_{+}+(1-\lambda ){\varphi }_{-}=\bar{\varphi }$ in the mixture's bulk free-energy density. Varying ${\varphi }_{+}$ and ${\varphi }_{-}$ gives $f^{\prime }({\varphi }_{+})=f^{\prime }({\varphi }_{-})=\mu$; varying $\lambda$ gives $f({\varphi }_{+})-\mu {\varphi }_{+}=f({\varphi }_{-})-\mu {\varphi }_{-}$. These equate the chemical potentials and pressures, respectively.

Equivalently, their bulk chemical potentials are both zero and their pressures are equal:These are precisely the conditions of the common-tangent construction for phase coexistence: the horizontal supporting tangent touches both minima. Equal chemical potentials prevent net composition exchange, and equal pressure gives mechanical equilibrium at a flat interface.

The chemical potential is the change in internal energy when one particle is added at fixed entropy and volume:Equivalently, $\mu =(\partial F/\partial N{)}_{T,V}$ for the Helmholtz free energy.

Let a microstate $s$ have energy $E_s$ and particle number $N_s$. Maximizing the Gibbs entropy subject to normalization and fixed mean values of $E$ and $N$ is the maximum-entropy derivation of equilibrium ensembles. With Lagrange multipliers $\alpha$, $\beta$, and $-\beta \mu$, varyThe stationarity equation is $-\log p_s-1-\alpha -\beta (E_s-\mu N_s)=0$. Normalization therefore gives the grand canonical ensembleHere $\mathcal{Z}$ is the grand canonical partition function and $k_B$ is the Boltzmann constant.

For one fermionic quantum state of energy $\epsilon$, the Pauli exclusion principle permits occupation numbers only $n=0,1$. Its two grand-canonical weights are $1$ and $e^{-\beta (\epsilon -\mu )}$, so its mean occupation is the Fermi-Dirac distribution

For a free nonrelativistic particle, $\epsilon ={\hslash }^2{k}^2/(2m)$. In a region of area $A$, the number of wave-vector states in the annulus from $k$ to $k+dk$, including the two spin angular momentum states, isSince $kdk=(m/{\hslash }^2)d\epsilon$, the two-dimensional free-electron density of states is constant:

We now use units in which $k_B=1$, as in the question. At zero temperature, the Fermi-Dirac distribution is a step function, and henceAt positive temperature the same fixed particle number satisfies the exact relationThusThis is the low-temperature particle-number cancellation for constant density of states. Therefore, for $T\ll {\epsilon }_F$, the mean number of particles in the energy interval $[\epsilon ,\epsilon +d\epsilon ]$ is

Finally, compare the internal energy with its zero-temperature value. The thermally excited particles above ${\epsilon }_F$ and holes below ${\epsilon }_F$ have equal leading particle numbers, so their terms proportional to ${\epsilon }_F$ cancel. With $x=|\epsilon -{\epsilon }_F|/T$,where the integral follows from the fermion-to-boson thermal integral ratio. Differentiation gives the linear low-temperature heat capacity of a two-dimensional Fermi gas:The requested power law is therefore linear in $T$, with exponent one.

Because the constraint determinesthe constrained function isIts stationary points satisfyWhen $yz=0$, these equations giveWhen $yz\ne 0$, subtracting the two bracketed equations gives $z^2=a^2{y}^2$, and substitution then giveswith the two signs chosen independently. These are exactly the constrained stationary points; equivalently they follow from the Lagrange multiplier equations.

Under the additional restriction $x\ge 0$, the constraint gives the compact elliptical disc $a^2{y}^2+z^2\le b^2$. On its boundary $x=0$, the objective is zero. At the four nonzero interior stationary points,with the sign determined by $yz$. Therefore

The time-zero budget constraint isso ${\theta }^0=w_0-{\theta }^T{S}_0$. The terminal wealth is thereforePutThenand the target mean is the linear constraint ${\theta }^{T}b=c$.

Since the invertible covariance matrix $V$ is positive definite, the Lagrange multiplier equation for minimizing ${\theta }^{T}V\theta$ isThusThe constraint determinesThis is the one-period Gaussian minimum-variance portfolio; the riskless holding is then ${\theta }^0=w_0-{\theta }^T{S}_0$.

Introduce a Lagrange multiplier $\lambda$ for the normalization and varyFor a smooth variation $u+\epsilon \eta$ with $\eta =0$ on $\partial D$, integration by parts givesThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equation

Multiplying by $u$ and integrating, while usingthe divergence theorem and $u=0$ on the boundary giveThe normalization is one, so the multiplier equals the stationary value:

Replacing a minimizer by its absolute value does not increase its gradient norm, so choose a nonnegative minimizer $P$. The Euler-Lagrange equation isfor a Lagrange multiplier $\mu$. Multiplication by $P$ and integration show thatso $\mu >0$. Set $P_{\eta }=\sqrt{\mu }P$. Then $P_{\eta }$ is nonzero, belongs to $H^1({\mathbb{R}}^2)$, and satisfies the trapped focusing cubic ground-state equationStandard elliptic regularity and the strong minimum principle for elliptic operators make the nonnegative solution positive.