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Both the canonical kinetic term and $V({\varphi }^{\mathsf{T}}\varphi )$ are invariant under the orthogonal group $O(n)$. The stated nonzero vacuum expectation value has length $|v|$. Its stabilizer is the orthogonal group $O(n-1)$ acting on the perpendicular subspace. HenceThe two nonzero coordinates of this particular vacuum do not break more generators than an axis-aligned vacuum: the two choices are related by a common orthogonal transformation.

Explicitly, with $s={\varphi }^{\mathsf{T}}\varphi$,stationary point at nonzero $v$ requires $V^{\prime }(v^2)=0$. The $n-1$ transverse eigenvalues vanish, while the radial squared mass is $4v^2{V}^{\prime \prime }(v^2)$. If this curvature also vanishes, there is an additional accidental massless radial mode; the number of symmetry-protected Goldstone bosons remains $n-1$. If $v=0$, the symmetry is unbroken and none is required.

Fix $c>0$. Stationary points satisfy $m(a+b{m}^2+c{m}^4)=0$. With $x=m^2$, the nonzero candidates areKeep only real positive roots. At such a point,so $x_{+}$ gives local minima and $x_{-}$ gives local maxima when both exist. The solution $m=0$ is a local minimum for $a>0$. To choose the phase, compare free energies: using the stationary point relation $a=-bx-c{x}^2$ givesFor $b>0$, this expression is negative whenever a nonzero stationary minimum exists. The paramagnetism phase is stable for $a>0$, and the ferromagnetism phase with $m=\pm \sqrt{x_{+}}$ is stable for $a<0$. The phase transition at $a=0$, $b>0$, is continuous and second order.

For $b<0$, equality with $F(0)$ requires $x=-3b/(4c)$. Substitution into the stationary equation gives the first-order coexistence curveThe disordered phase is the global minimum above this curve, and the ordered phase is the global minimum below it. At coexistence the factorizationexhibits three equally low minima, $m=0$ and $m=\pm \sqrt{-3b/(4c)}$, proving the discontinuous jump.

The ordered spinodal is $a=b^2/(4c)$, $b<0$, where its local minimum meets the intervening local maximum. The disordered spinodal is $a=0$, $b<0$. Thus the ordered phase is metastable for $a_{\mathrm{t}}<a<b^2/(4c)$; the disordered phase is metastable for $0<a<a_{\mathrm{t}}$. These spinodals are limits of metastability, not equilibrium transition curves.

The first-order curve meets the continuous transition line at the tricritical point $(a,b)=(0,0)$. On the specially tuned path $b=0$, the equilibrium ordered solution is $m=\pm (-a/c{)}^{1/4}$ for $a<0$ and tends continuously to zero at the origin.

The fixed common temperature, rigid external container and fixed particle numbers on either side make the total Helmholtz free energy the appropriate thermodynamic potential. Let the barrier have cross-sectional area $A$, and let increasing $x$ expand the left compartment. Then $d{V}_1=Adx$ and $d{V}_2=-Adx$. At fixed temperature and particle number, $d{F}_j=-p_{j}d{V}_j$, soThus mechanical equilibrium, $p_1=p_2$, makes $F$ stationary. The physical force on the barrier towards increasing $x$ is $f=A(p_1-p_2)=-F^{\prime }(x)$. If the barrier is displaced slightly from $x_{\ast }$, strict mechanical stability requires a restoring force:ConsequentlyThe Taylor expansion $F(x_{\ast }+\delta x)=F(x_{\ast })+\frac{1}{2}{F}^{\prime \prime }(x_{\ast })\delta x^2+O(\delta x^3)$ proves that thermodynamic equilibrium is a strict local minimum of the Helmholtz free energy, rather than merely a stationary point.

For individually stable gases, the positive isothermal compressibilities ${\kappa }_{T,j}=-(1/V_j)(\partial V_j/\partial p_j{)}_{T,N_j}$ make the sign especially transparent:If these inequalities hold throughout the accessible interval, $F$ is a strictly convex function and an interior stationary point is the unique global minimum of the Helmholtz free energy. Stability at one point alone proves only the local statement. A vanishing restoring derivative describes marginal stability and requires examining higher-order terms.

Write $P(N)=f(N,0)$. Its three positive zeros of a function are $1/2,1,2$, and expansion and differentiation giveThus its local minimum and maximum occur respectively atwith valuesThe function graph approaches the value $1$ as $N\downarrow 0$, decreases through its first zero to the minimum, increases through its second zero to the maximum, and then decreases through $2$ to $-\infty$. This fixes the shape and all axis crossings of the cubic polynomial.

Changing $a$ translates the function graph vertically without moving either stationary point. For $a=1/2$ both extrema are above the horizontal axis: their heights are approximately $0.421109$ and $0.764076$. For $a=-1/2$ both are below it, at approximately $-0.578891$ and $-0.235924$. The function graph for $a=1/2$ therefore has only its final positive-axis crossing, whereas the function graph for $a=-1/2$ has only its first positive-axis crossing. The original plots mark the extrema and also show the equilibrium point branches used below.

Yes: there is a quasistatic hysteresis loop. Equilibrium points lie on the mathematical model parameter curve $a=-P(N)$. Its two saddle-node bifurcations occur at the stationary points computed in part (a), at mathematical model parameter valuesBetween these values there are two attracting branches separated by a repelling branch. Starting below $a_{\mathrm{down}}$ and increasing $a$ slowly, a solution follows the lower attracting branch. This branch persists through the bistability interval and ends only at $a_{\mathrm{up}}$, where it meets the repelling branch in a saddle-node bifurcation. The solution then moves to the upper attracting branch.

If the direction of mathematical model parameter change is reversed, the upper branch persists until $a_{\mathrm{down}}$, where a different saddle-node bifurcation forces the downward transition. The two transition values differ, and both lie inside the allowed range $[-1/2,1/2]$. Thus the biological population selected at an intermediate mathematical model parameter value depends on its prior branch. This is the equilibrium point hysteresis shown in the branch diagram; finite-rate passage can additionally delay the jumps near the saddle-node bifurcations.

For positive edge lengths the volume is $V=abc$ and the surface area constraint is $2(ab+ac+bc)=S$. Introduce a Lagrange multiplier $\mu$:The stationary point equations areMultiplying these by $a,b,c$, respectively, and equating their left sides gives $\mu c(a-b)=0$ and $\mu b(a-c)=0$. Because all edge lengths are positive, the equations imply $\mu >0$; hence $a=b=c$. The constraint gives $6a^2=S$.

To establish a global maximum, rather than only a stationary point, apply the arithmetic-geometric mean inequality to $ab,ac,bc$:Equality holds exactly when $ab=ac=bc$, or $a=b=c$. Therefore

Write $f(x,y)=g(x)+g(y)$ with $g(s)=s^3/12-s$. On $[0,4]$, its derivative is $g^{\prime }(s)=s^2/4-1$, negative below two and positive above two. Compare the critical and endpoint values:Each coordinate can be chosen independently. Therefore the global minimum is $-8/3$ at $(2,2)$, and the global maximum is $8/3$ at $(4,4)$. This comparison includes the boundary of the rectangle, rather than classifying only interior stationary points.

The continuous function attains a global maximum and minimum on the closed disc by the extreme value theorem. All four unconstrained stationary points have squared distance eight, so none lies in its interior. The extrema must therefore lie on $x^2+y^2=6$.

The constraint gradient is nonzero on that circle, so Lagrange multipliers apply:Subtracting gives $(x-y)((x+y)/4-2\lambda )=0$. If $x=y$, the constraint gives $x=y=\pm \sqrt{3}$. Otherwise $2\lambda =(x+y)/4$; substitution into either equation gives $xy=-4$. This is impossible, since $|xy|\le (x^2+y^2)/2=3$ on the circle. Thus the diagonal points are the only candidates. Comparing their values gives

The extreme value theorem guarantees extrema on the closed interval. Compare its endpoints and the stationary points $0,\pm \sqrt{10/3}$:ThusThe two-point answer in each case also follows from the even function symmetry.

Let $h(x,y)=2y^3+x{y}^2-y^2-4x$, so $g=e^h>0$. Its gradient is $\nabla g=g\nabla h$, andThus a stationary point must have $y=\pm 2$ and $x=1-3y$, giving $(-5,2)$ and $(7,-2)$. To classify them, useAt either stationary point, $\nabla h=0$ and $12y+2x-2=6y$. Hence $\det H_g=g^2(-4y^2)=-16g^2<0$. The Hessian has one positive and one negative eigenvalue, so both are saddle points. There are no other stationary points.

Let $q=x^2+4y^2$. The objective has the exact sum of squares decompositionFor every feasible point, $q\le 8$, so this is at least $3-8/2=-1$. Equality requires simultaneously $q=8$ and $x+2y=0$. Substitution gives $y^2=1$ and $x=-2y$. Both resulting points are feasible. Therefore the global minimum is $-1$, attained exactly at $(2,-1)$ and $(-2,1)$. The lower bound and its equality conditions prove global attainment of the minimum, rather than merely identifying stationary points on the boundary.

Use the optimization LagrangianSymmetry gives the gradient equations $2Ax=\lambda 1$ and $2By=\lambda 1$. The positive-definite matrices are invertible, so defineThe affine constraint gives $\lambda =2/D$, henceFor a global certificate, write any other feasible point as $(x_{\ast }+u,y_{\ast }+v)$. Feasibility implies $1^{T}u+1^{T}v=0$. Expansion cancels the cross terms and leavesBy positive definiteness this is positive unless $u=v=0$. Thus the solution is the unique global minimum, not merely a stationary point. This is a two-block instance of positive quadratic minimization under one affine constraint.

Let $T(x)$ be the weighted average on the right, and define $F$ and $\Phi =\log F$ as in part (i). Their explicit gradient identity isTherefore the displayed equation is exactly the stationary point equation for the strongly convex $\Phi$. It has one solution, which is the unique minimizer from part (i). Reuse the BFGS iteration, withThe gradient is explicit, the weights can again be scaled by their largest exponent, and the usual full-memory BFGS assumptions give local superlinear convergence.

Simply setting $x_{n+1}=T(x_n)$ is not automatically a convergent fixed-point iteration. Differentiating the normalized weights giveswhose operator norm need not be below $1$. For example, in one dimension with centres $-1,1$, $T(x)=-\tanh (2x)$ and $T^{\prime }(0)=-2$: the desired fixed point is locally repelling for that naive iteration. If exact second derivatives are affordable, multivariable Newton iteration on $x-T(x)=0$ is another valid choice, with Jacobian matrix $I-J_T$ as given above and local quadratic convergence.

The stationary point equations give $y=x$ and $x^3=x$, hence $(0,0),(1,1),(-1,-1)$. At $(0,0)$ the Hessian matrix has negative determinant $-16$, so $(0,0)$ is a saddle point. At each of $(1,1)$ and $(-1,-1)$, the Hessian matrix has positive leading minors $12$ and $32$, proving a strict local minimum. More strongly,Thus $(1,1)$ and $(-1,-1)$ are exactly the global minima, with value $3$. There are no local maxima, since any differentiable interior local maximum would be a stationary point, and all three have been classified.

Assume $f$ and $g$ are continuously differentiable near the point. At a regular feasible point, meaning $g(x)=0$ and $\nabla g(x)\ne 0$, a differentiable constrained local extremum satisfies the Lagrange multiplier equationsThey arise by differentiating the optimization Lagrangian $L(x,\lambda )=f(x)-\lambda g(x)$ with respect to all coordinates and the multiplier. Geometrically, the objective has zero derivative along the tangent space of the constraint surface, so its gradient must lie in the normal direction.

Solve these equations for candidates, examine feasible points where the regularity condition fails separately, and compare objective values or use a second-order test restricted to feasible directions. The equations locate stationary points; they do not by themselves establish which are minima or whether a global minimum is attained.

Use $F=x^6+\frac{3}{2}{y}^4$ and $G=x^4-2y^2-3$. Every feasible point has $x\ne 0$, so $\nabla G=(4x^3,-4y)$ cannot vanish. The Lagrange multiplier equations areThe first gives $\lambda =\frac{3}{2}{x}^2>0$. If $y\ne 0$, the second would give $\lambda =-\frac{3}{2}{y}^2<0$, a contradiction. Therefore $y=0$, and the constraint gives $x^4=3$. The only constrained stationary points are $(\pm 3^{1/4},0)$.

For a global certificate, feasibility implies $x^4\ge 3$, and henceEquality requires $y=0$ and $x^4=3$. ThusThe case split is essential: dividing by $y$ would discard precisely the minimizers.

Here the constraint surface $x^4+2y^2=3$ is compact and regular, so a global minimum exists and satisfies the Lagrange multiplier equationsKeep the zero-coordinate cases. If $y=0$, the two points are $(\pm 3^{1/4},0)$, with value $3\sqrt{3}$. If $x=0$, the two points are $(0,\pm \sqrt{3/2})$, with value $27/8$. If both coordinates are nonzero, the equations give $x^2=y^2=t>0$. The constraint becomes $t^2+2t=3$, whose only positive solution is $t=1$. These are the four points $(\pm 1,\pm 1)$, all with value $5/2$.

These cases exhaust the stated eight constrained stationary points. Comparing their values givesFor additional classification, substituting $u=x^2\in [0,\sqrt{3}]$ gives $F(u)=u^3+\frac{3}{8}(3-u^2{)}^2$, withIt decreases from $u=0$ to $u=1$ and increases from $u=1$ to $u=\sqrt{3}$. Thus the two $y$-axis points are constrained local maxima, the two $x$-axis points are global maxima, and the four mixed-coordinate points are global minima.

The matrix $C=A-b{b}^T$ is an invertible matrix. Indeed, on the line spanned by $b$ it has eigenvalue $-1/2$, and on $b^{\perp }$ it agrees with the positive-definite restriction of $A$. Therefore the full Newton step is well defined and givesThis is the unique stationary point because $C$ is an invertible matrix. The algebra applies to every starting point, but it does not make the answer a minimum: for $d\ge 2$ it is the saddle point just classified, and for $d=1$ it is a maximum. A safeguarded minimization algorithm that replaces an indefinite matrix used as its Hessian matrix or damps the step need not take this exact step; the assertion concerns the ordinary full Newton step formula.

Write $(x^{T}b{)}^2=x^{T}b{b}^{T}x$ and put $C=A-b{b}^T$, which is symmetric. Differentiating this quadratic form givesUnlike $A$, $C$ need not be a positive-definite matrix. Its behaviour must be checked before classifying the stationary point.

Substitution givesIn particular, $1$ is a stationary point. For $x$ sufficiently close to $1$, Taylor's theorem with fifth-order remainder giveswhere $\xi$ lies between $1$ and $x$. The fifth derivative is bounded near $1$, so the remainder is $O((x-1{)}^5)$ and is smaller in magnitude than half the positive quartic term for sufficiently small nonzero $x-1$. Therefore $x=1$ is a strict local minimum, despite the vanishing second derivative.

An exact algebraic check gives the stronger global conclusion:Equality holds only at $x=1$, so it is the unique global minimum on the actual domain.