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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

Put $x=Vt$ with fixed $V>0$. The relevant stationary point will have $k=O(V^2/S^2)$, so the assumption $ϵV^2/S^2\ll 1$ allows the long-wave approximationUsing $\sin A\cos B=[\sin (A+B)+\sin (A-B)]/2$, the phase $S{k}^{3/2}+Vk$ has no stationary point for $k>0$. The other phase iswithandThe stationary part of the integral is thereforeBy the one-dimensional stationary-phase formula, this is asymptotic toSincewe obtainUsing $V=x/t$, this has the requested formwhere

The equilibrium magnetization is a global minimum of the Landau free energy. Its candidates are the stationary pointsThus $m=0$, or $x=m^2\ge 0$ is a nonnegative root ofOne retains candidates with nonnegative curvature $f^{\prime \prime }(m)$ and compares their free energies; the candidate with the smallest value is the equilibrium state. Equal global minima describe phase coexistence.

With Euclidean source convention $+\int d^{d}x{J}_a{\varphi }_a$, the generating functional isHere $J_a(x)$ is a nondynamical source. Its functional derivatives insert fields:This is why $Z$ “generates” correlation functions.

With the notation of the question, $W[J]=-\log Z[J]$ is the Euclidean connected generating functional; it is distinct from a Wilsonian effective action. Define the classical fieldThe quantum effective action is the Legendre transformwhere $J$ is eliminated in favor of $\Phi$. Its first derivative isso at zero source the quantum expectation value is a stationary point of $\Gamma$. Moreover,where $G_{ab}=⟨{\varphi }_a{\varphi }_b{⟩}_{J,\mathrm{conn}}$ is the exact connected correlation function. Thus the second derivative of $\Gamma$ is the inverse exact propagator.

Perturbatively, $Z[J]$ sums all Feynman diagrams, including disconnected ones. The exponential formula for combinatorial structures says that its logarithm selects connected Feynman diagrams, so $W$ sums connected diagrams with the sign dictated by the convention above. The Legendre transform removes diagrams that disconnect when one internal line is cut; consequently $\Gamma$ sums one-particle-irreducible Feynman diagrams. Equivalently, every connected diagram is a tree assembled from one-particle-irreducible vertices and full propagators, and the Legendre transform inverts that tree construction.

Now let $J_a^{\prime }=J_b{U}_{ba}$. ThenChanging variables to ${\varphi }_b^{\prime }=U_{ba}{\varphi }_a$ and using invariance of both the action and functional measure gives $Z[J^{\prime }]=Z[J]$, hence $W[J^{\prime }]=W[J]$. In the Legendre transform, the pairing obeysChanging the source variable and using the invariance of $W$ therefore givesThe symmetry of the classical action and measure is inherited by the full quantum effective action when it has no quantum anomaly.

The multivariable chain rule gives, at $(x,y)=(a+t\cos \gamma ,b+t\sin \gamma )$,Sufficient conditions for a strict local minimum at $t=0$ are $g^{\prime }(0)=0$ and $g^{\prime \prime }(0)>0$.

At a stationary point, $g^{\prime }(0)=0$ in every direction. Its second derivative is the quadratic form of the Hessian matrix. Ifthen completing the square givesfor every nonzero direction $(u,v)$. The Hessian is therefore positive definite, and the stationary point is a strict local minimum.

Write the base-state gradients asThe base heat flux and Stefan condition relations are

Under the quasi-stationary approximation, and with advective heat transport neglected, each temperature perturbation satisfies the Laplace equation. The decaying normal-mode solutions areKeeping the displaced ice interface at $T_m$ omits the curvature correction described by the Gibbs--Thomson relation.
The fixed melting temperature at the displaced ice interface givesExpanding temperature and conductive heat flux continuity at the displaced outer interface gives

Put $r=k_a/k_w$. Since $G_a=G_w/r$, the temperature condition in the limits $r\ll 1$ and $\alpha h\ll 1$ givesAlthough $r$ is small, the product $rC$ is therefore order one and cannot be discarded. The flux condition then givesThis is why the stated asymptotic warning matters.

The perturbed Stefan condition at $x=0$ isUsing $B\sim {\eta }_2{G}_w$ and $-k_w{G}_w={\rho }_i{L}_{f}V$ yieldsSubstitution of the interface-amplitude ratio from part i produces the thin-film icicle-ripple instability dispersion relation

Set $b=\Gamma h{\alpha }^3$. Rationalizing this complex number givesso its growth rate and imaginary part areDifferentiating the growth rate with respect to the wavenumber shows that its positive stationary point satisfiesIt is the unique global maximum, and henceAt this wavenumber $b=1/\sqrt{5}$. With the convention $e^{i\alpha z+\sigma t}$, a constant phase travels with phase velocityThereforeBecause $z$ increases downward, the negative sign means that the icicle ripples migrate upward with speed $\sqrt{5}V/6$.

In the high-tension, narrow-gap regime, the almost spherical membrane has area $4\pi (R+\delta {)}^2$. Its spherical bending energy $8\pi k_c$ is independent of radius, while Helfrich repulsion over area $4\pi R^2$ costsUp to terms independent of $\delta$, a suitable free energy is consequentlyIts stationary point satisfiesUsing $\delta /R\ll 1$ gives the fluctuation-supported membrane--particle gapThe inverse-square entropic repulsion prevents contact, while membrane tension limits the area gained by opening the gap.

Stationary points satisfy $x/t={\omega }^{\prime }(k_j)$. For nondegenerate points,

Expanding and taking partial derivatives gives the gradient and Hessian matrixAt $p_0=(-1,0{)}^T$, these are $\nabla f(p_0)=(-6,2{)}^T$ andThus the Newton step solvesThe second equation gives $s_y=1-2s_x$; substituting into the first gives $22s_x=10$. HenceAt $p_{\ast }=(-1/2,-1/2{)}^T$, however,The eigenvalues of this Hessian matrix are $-1$ in direction $(1,1)$ and $7$ in direction $(1,-1)$. It is invertible but indefinite. Thus the Newton step is zero, andThis illustrates that stationary points are fixed points of Newton's method. Here the fixed point is a saddle point of a scalar function, not a local minimum. Indeed, direct substitution along the two eigenvector directions givesFor sufficiently small nonzero $t$, one direction decreases $f$ and the other increases it. The global minimum is instead zero: the sum of squares vanishes at $(0,0)$ and $(-1,-1)$. These are the only real zeros, since $x=-y^2$ and $y=-x^2$ imply $y(y^3+1)=0$. Thus a zero gradient is an insufficient stopping certificate for minimization problem without classifying the stationary point.

For an unconstrained differentiable function of two real variables, an interior local minimum must solve the simultaneous stationary point equationsApply Newton's method in optimization to this gradient equation. The Jacobian matrix of the gradient is the Hessian matrixAt $p_n=(x_n,y_n{)}^T$, solve the linear system $H(p_n)s_n=-\nabla f(p_n)$, then set $p_{n+1}=p_n+s_n$. When the Hessian matrix is invertible, this isThis follows by setting the first order Taylor expansion $\nabla f(p_n+s_n)\simeq \nabla f(p_n)+H(p_n)s_n$ equal to zero. Solving the linear system avoids explicitly computing a matrix inverse. A stationary point need not be a local minimum: for example, a positive-definite matrix $H(p)$ suffices to classify a stationary point as a strict local minimum, whereas an indefinite Hessian matrix identifies a saddle point of a scalar function.

A maximum or minimum can occur at an endpoint. The rule that a derivative vanishes at an extremum requires an interior extremum and differentiability there. The hypotheses give no two-sided derivative at the endpoints, and even an existing one-sided derivative need not vanish there.

For example, $f(x)=x$ on $[0,1]$ has $m=0$ at $x_m=0$ and $M=1$ at $x_M=1$, but $f^{\prime }(x)=1$ at every interior point; its endpoint one-sided derivatives are also one. Its graph meets both horizontal bounding lines without being tangent to either. Thus the claimed conclusion about two zeros of the derivative fails.

There is a second logical issue: the argument never establishes two distinct extremizing points. For a constant function, the maximum and minimum coincide and may be represented by the same point. The constant function does have many stationary points, but that needs its own argument; it does not follow from counting the same selected point twice. In Rolle's theorem, equal endpoint values ensure an interior extremum when the function is nonconstant, which is the missing mechanism here.

For $f=x\ln x$, the derivatives are $f^{\prime }=1+\ln x$ and $f^{\prime \prime }=1/x>0$. Its unique stationary point is the global minimum $(e^{-1},-e^{-1})$. For $g=\ln x/x$, $g^{\prime }=(1-\ln x)/x^2$ changes from positive to negative at $x=e$, giving the global maximum $(e,e^{-1})$. Their endpoint limits areAn intersection requires $(x^2-1)\ln x=0$, and on $x>0$ both factors vanish only at $x=1$. There $f=g=0$ and $f^{\prime }=g^{\prime }=1$, proving a common tangent line $y=x-1$. In fact,because the two numerator factors always have the same sign. Thus the curves touch at $(1,0)$ without exchanging their order.

At an interior local extremum of a differentiable function, $f^{\prime }(x)=0$. The second-derivative test classifies a stationary point: a positive second derivative gives a strict local minimum, and a negative second derivative gives a strict local maximum. A zero second derivative is inconclusive: $x^4$, $-x^4$, and $x^3$ at zero illustrate a minimum, a maximum, and neither.

HereThus $f^{\prime \prime }(1/3)=-8$ and $f^{\prime \prime }(3)=8$. Substitution into $f$ gives

For two independent smooth constraints $g=h=0$, the tangent direction is orthogonal to both gradients. At a constrained stationary point, the gradient of the objective is also orthogonal to that direction, hence lies in their span: $\nabla f=\lambda \nabla g+\mu \nabla h$. This is the two-constraint Lagrange multiplier condition. Alternatively, distance extrema on a central plane section of an ellipsoid follow from a two-dimensional eigenvalue problem.

Minimizing distance is equivalent to minimizing $f=x^2+y^2+z^2$. Put $g=x-y+z$ and $h=x^2+(y^2+z^2)/4-1$. The Lagrange multiplier equations areAdding the latter two gives $(2-\mu /2)(y+z)=0$. If $\mu =4$, those equations give $\lambda =x=0$; the constraints then give $y=z=\pm \sqrt{2}$ and $f=4$. Otherwise $z=-y$ and $x=2y$. The ellipsoidal constraint gives $y=\pm \sqrt{2}/3$, and henceThe intersection is compact. The two constraint gradients are independent there: if $\nabla h=t\nabla g$, then $(x,y,z)=(t/2,-2t,2t)$, and $g=0$ forces $t=0$, contradicting $h=0$. Thus every global extremum is among these candidates. Taking square roots gives

Near a nondegenerate stationary point, the quadratic part of the Taylor expansion determines the local contour plot. For a strict local maximum, the model $f=f_0-X^2-Y^2$ gives nested closed level sets, with values decreasing as one moves outward. For a strict local minimum, the model $f=f_0+X^2+Y^2$ gives the same nested shape with values increasing outward. Unequal nonzero quadratic coefficients stretch the circles into ellipses after a rotation to the Hessian matrix's eigenvectors.

For a nondegenerate saddle point of a scalar function, the model $f=f_0+X^2-Y^2$ gives hyperbolas. Levels above $f_0$ open along the positive-curvature direction; levels below $f_0$ open along the negative-curvature direction. The critical level set is the two crossing lines $Y=\pm X$. Degenerate stationary points can have different local geometry, so these are the standard nondegenerate sketches.

Set both first partial derivatives equal to zero:Thus $x=0,\pm 1/\sqrt{2}$ and $y=0,\pm \sqrt{2}$, independently, giving exactly nine stationary points. The Hessian matrix is diagonal everywhere:Its diagonal entries are its eigenvalues, so the second-derivative test immediately classifies the points:

In particular,All nine stationary points are nondegenerate, so no higher-order test is needed. Neither type of local extremum is global: $f(x,0)=x^4-x^2\to +\infty$ as $|x|\to \infty$, whereas $f(0,y)=-y^4+4y^2\to -\infty$ as $|y|\to \infty$.

To organize the contour plot, complete the squares:The level set through all four outer saddle points of a scalar function factors exactly:It is therefore the union of the circle $x^2+y^2=5/2$ and the hyperbola $y^2-x^2=3/2$, which cross at those four saddle points of a scalar function. Near the origin, $f=-x^2+4y^2+O((x^2+y^2{)}^2)$, so the zero contours have tangents $y=\pm x/2$. These two exact structural features, together with the nested contours around the four local extrema, determine the requested sketch. They are an example of a separable quartic landscape with nine stationary points.

For $f(x)=x^3{e}^{-3x^2/2}/3$, the product rule gives $f^{\prime }(x)=x^2(1-x^2)e^{-3x^2/2}$, so the stationary points occur at $x=-1,0,1$. Its sign is negative for $|x|>1$ and positive for $0<|x|<1$. ConsequentlyThey are also the global extrema, since the function tends to zero at both infinities and is a monotone function on the intervals just identified. At zero the derivative does not change sign. Since $f(x)=x^3/3+O(x^5)$, $(0,0)$ is a stationary inflection point, not an extremum.

The stationary point equations are $x^2+2y-21/4=0$ and $y^2+2x-21/4=0$. Subtracting them factors as $(x-y)(x+y-2)=0$, so all solutions lie either on $x=y$ or on $x+y=2$. The first case gives $x=y=3/2$ or $x=y=-7/2$; the second gives $(x,y)=(-1/2,5/2)$ or $(5/2,-1/2)$. This also verifies the two stated points and exhausts the possible solutions.

The Hessian matrix is $H=\left(\begin{array}{cc}2x& 2\\ 2& 2y\end{array}\right)$, with determinant $4(xy-1)$. The second-derivative test gives:

These extrema are local. Along $x=y=t$ the cubic term makes the function unbounded above and below, so it has no global maximum or minimum on the whole plane.

The fourth Chebyshev polynomial eliminates the parameter:Its derivatives are $y^{\prime }=16x(2x^2-1)$ and $y^{\prime \prime }=16(6x^2-1)$. The interior stationary points are a local maximum $(0,1)$ at $\theta =\pi /2$, and two minima $(\pm 1/\sqrt{2},-1)$ at $\theta =\pi /4,3\pi /4$. At the maximum $y^{\prime \prime }=-16<0$, while at either minimum $y^{\prime \prime }=32>0$.

The two inflection points areThe curvature changes sign there. The endpoint limits are $(\pm 1,1)$, at $\theta =0,\pi$; they are excluded from the open curve. The central maximum and the two minima are also the attained global extremal values on that open curve.

Differentiation and factorization give $f^{\prime }(x)=e^{-x^2}(1+x^2-2x^4)=e^{-x^2}(1-x^2)(1+2x^2)$. The exponential and $1+2x^2$ are positive, so the only stationary points are at $x=\pm 1$. The derivative is positive for $|x|<1$ and negative for $|x|>1$, givingThe function is odd, crosses the origin with slope one, and tends to zero from below as $x\to -\infty$ and from above as $x\to \infty$. These facts determine the sketch.