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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 $\sqrt{5}V/6$.

For $g=(x-1{)}^2+y^2-29$, the gradient $\nabla g=2(x-1,y)$ never vanishes on the circle. The Lagrange multiplier equations areIf $y=0$, the two possibilities are $(1\pm \sqrt{29},0)$. If $y\ne 0$, then $\lambda =1+x$, and the first equation gives $y^2=2x^2-2x-3$. Combining this with $y^2=28+2x-x^2$ yieldsPutBoth $h_{-}$ and $h_{+}$ are positive, giving the other four points $(r_{-},\pm h_{-})$ and $(r_{+},\pm h_{+})$. To classify all six, restrict $f$ to the circle by eliminating $y^2$:Its derivatives are $H^{\prime }=-3x^2+4x+31$ and $H^{\prime \prime }=4-6x$. In the interior, $H^{\prime \prime }(r_{-})=2s>0$ and $H^{\prime \prime }(r_{+})=-2s<0$. At either endpoint $H^{\prime }<0$; along the circle, $x$ has a strict local minimum at the left endpoint and a strict local maximum at the right endpoint. Consequently the six constrained extrema areThe circle is a compact set, so the extreme value theorem guarantees that its global minimum and global maximum occur among these six points. Their values give the corrected sharp boundThe printed lower bound $-20$ is false. A particularly short counterexample isThe sole local minimum in part (a), $(-1/2,0)$, has value $-1/4$, larger than values on this circle; it is therefore not a global minimum. In fact $f(-2,y)=2-y^2\to -\infty$, so $f$ has no global minimum on the plane.

For an observed random sample with all $x_i>0$, which happens almost surely, independence gives the likelihood functionIgnoring a term independent of $\theta$, the log-likelihood and its derivative areThis derivative is positive for $0<\theta <S/n$ and negative for $\theta >S/n$. Thus there is a unique global maximum, and the maximum-likelihood estimator is

Let $F=x+y+z^2$ and $g=x^2+y^2+z^2$. The unit sphere is compact, so the continuous function $F$ has a global maximum and global minimum. Since $\nabla g\ne 0$ on the unit sphere, all such points satisfy the Lagrange multiplier equationsThus $x=y$ and either $z=0$ or $\lambda =1$.

If $z=0$, then $x=y=\pm 1/\sqrt{2}$, giving $F=\pm \sqrt{2}$. If $\lambda =1$, then $x=y=1/2$ and $z=\pm 1/\sqrt{2}$, giving $F=3/2$. These exhaust the candidates, so

On the constraint $x^2-z^2=1$, substitute $x^2=1+z^2$ to obtainTherefore the two global minima areEvery pair $(y,z)$ is allowed on either smooth branch $x=\pm \sqrt{1+z^2}$, so these are also the only local extrema. There is no local maximum or global maximum: the displayed positive quadratic in $(y,z)$ has no other stationary point and is unbounded above. At the minima $F_z=-2z=1$, consistent with the hypothesis in part i.

Write $A=a^2>0$ and $E=e^{-(x^2+y^2)/A}>0$. The partial derivatives factor asA critical point has both partial derivatives zero. If $x,y\ne 0$, this would require $x^2-2y^2=A$ and $x^2-2y^2=-2A$, which is impossible. On the axes the remaining conditions giveAt either critical point with $x\ne 0$, the Hessian matrix isIt is negative definite, so the second-derivative test shows that both $(\pm |a|,0)$ are strict local maxima, of value $a^2/e$. They also attain the global maximum: $f\le x^2{e}^{-x^2/A}\le A/e$. For completeness, the Hessian matrix at $(0,\pm |a|)$ is $\mathrm{diag}(6/e,8/e)$, giving strict local minima, while at the origin it is $\mathrm{diag}(2,-4)$, giving a saddle point.

For $x>0$, logarithmic differentiation givesSince $f(x)>0$, the derivative vanishes precisely when $\ln x=1$. HenceThe derivative is positive before $e$ and negative after $e$, so this is a global maximum on the given positive domain.

The Hessian matrix has entriesAt each stationary point the mixed entry vanishes, so the signs of the diagonal entries directly classify the local quadratic change:In particular the two local minima are also global minima: the squared term is nonnegative and $\ln (3+y^2)\ge \ln 3$, with equality in both bounds exactly at $(\pm 1,0)$. The local maximum at the origin is not a global maximum, since $f$ grows without bound as $x^2+y^2\to \infty$. The opposite diagonal signs at the other two points establish their saddle point character.

The Hessian matrix is diagonal:Its eigenvalues are $c$ and $d$. Hence the second-derivative test gives a strict local minimum if $c,d>0$, a strict local maximum if $c,d<0$, and a saddle point of a scalar function if $cd<0$. No degenerate case occurs because both are nonzero.

The local minimum is the unique global minimum, since both separated summands are strictly convex functions and tend to $+\infty$ at either end of the real line. Similarly, when both signs are negative there is a unique global maximum.

A central cross-section of the sphere gives the constraint $r^2+h^2/4=R^2$, and the cylinder's volume is $V=\pi r^{2}h$. At an interior optimum with $r,h>0$, the Lagrange multiplier equations for this constrained optimization problem areThe first gives $\lambda =\pi h$; substitution into the second gives $r^2=h^2/2$. The constraint then givesThe feasible nonnegative $(r,h)$ form a compact arc, whose endpoints have zero volume. Since this is the only positive stationary candidate, it is the global maximum. Its volume and fraction of the sphere's volume are

Use the Lagrange multiplier condition $\nabla f=\lambda \nabla g$ for $g=4x^2+y^2/16$:Neither coordinate can vanish at a constrained stationary point: either would force both to vanish, contradicting the constraint. Substitution therefore gives ${\lambda }^2=1$, so $y=8x$ or $y=-8x$. In either case $8x^2=1$, givingEqual signs give the two points attaining the global maximum $f=1$; opposite signs give the two global minima with $f=-1$. These exhaust the extrema because the ellipse is compact, its constraint gradient never vanishes on it, and $f$ is smooth. The bound can also be checked by setting $X=2x$, $Y=y/4$: then $X^2+Y^2=1$ and $f=2XY$, so $|f|\le 1$.

An antiderivative of the initial velocity is $-2c{e}^{-s^2/2}$. Thus D'Alembert's formula givesThis is a positive Gaussian function moving right and a negative Gaussian function moving left. Their cancellation at $t=0$ gives the zero initial displacement, while direct differentiation gives ${\psi }_t(x,0)=2cx{e}^{-x^2/2}$.

For every $t>0$, the profile is an odd function of $x$, vanishes at $x=0$, is positive for $x>0$ and negative for $x<0$, and tends to zero at both spatial infinities. The signs follow from $(x+ct{)}^2-(x-ct{)}^2=4xct$. The positive and negative lobes move farther apart as $t$ increases; for large separation they approach the individual unit-height Gaussian functions centred at $x=\pm ct$. At finite separation the actual extrema lie slightly beyond these centres: writing $a=ct$, the positive global maximum occurs at the unique $x>a$ satisfying $(x-a)/(x+a)=e^{-2ax}$.

For the upper function graph, the vertical cross-section of the enclosed region has length $y(x)$, so its area is $J[y]={\int }_0^{2a}y(x)dx$. A maximizing nonnegative arch therefore uses $F=y$ and $G=\sqrt{1+y^{\prime 2}}$ in the normal Euler-Lagrange equation. It becomesIntegrating once gives $y^{\prime }/\sqrt{1+y^{\prime 2}}=(c-x)/\lambda$. The concave upper branch has $\lambda >0$ and integrates to $y=\sqrt{{\lambda }^2-(x-c{)}^2}+d$. The equal zero endpoint heights force $c=a$ and $d=-\sqrt{{\lambda }^2-a^2}$. HenceIt is a circular arc. To find its parameter, substitute into the length integral: $\sqrt{1+y^{\prime 2}}=\lambda /\sqrt{{\lambda }^2-(x-a{)}^2}$, givingThis scalar equation determines $\lambda$ numerically when needed; no explicit solution is required.

The arch is a global maximum in its smooth function graph regime. For $\lambda >0$, the integrand $y-\lambda \sqrt{1+y^{\prime 2}}$ is concave and strictly concave in its slope. Its supporting-tangent inequality, integrated against any competing function graph with the same endpoint values, has first-order term zero by the Euler-Lagrange equation. Thus $J[y]-\lambda I[y]\le J[y_{\ast }]-\lambda I[y_{\ast }]$. On the fixed-length class this is $J[y]\le J[y_{\ast }]$; strict slope concavity also gives uniqueness.

The source leaves the length range implicit. A twice continuously differentiable function graph of this form needs $\lambda >a$, giving $2a<l<\pi a$. At $l=2a$ the straight function graph is the degenerate solution. The semicircle limit $l=\pi a$ has vertical endpoint tangents and is outside that smooth function graph class. For $l>\pi a$, no arc in the claimed smooth function graph family exists, so the stated formula cannot describe a twice continuously differentiable maximizing function graph under those unrestricted data. Allowing overhanging curves changes the admissible class.

The objective is continuous on the compact disk, so a global maximum exists. Its gradient is $(e^x,2e^{2y})$, which is never zero, so a maximum cannot lie in the interior. It must have $x^2+y^2=1$.

Both coordinates of a maximizing point are positive. Reflecting a negative coordinate to its positive counterpart strictly increases the objective without changing the constraint; at an endpoint with one coordinate zero, a small move along the circle into the positive quadrant also increases the objective to first order. Thus it suffices to consider $x>0,y>0$.

The boundary Lagrange multiplier equations are $e^x=2\lambda x$ and $2e^{2y}=2\lambda y$. Eliminate $\lambda$ to obtain the two-dimensional nonlinear system of equationsKeeping this quadrant restriction excludes the boundary minimum that an unconstrained elimination could also produce. Under the stated uniqueness assumption for the admissible solution, the global maximum must be that solution. Approximate it with multivariable Newton iteration, solving $J_F(z_n){\Delta }_n=-F(z_n)$ and setting $z_{n+1}=z_n+{\Delta }_n$, with a suitable safeguard if needed. For example its explicit Jacobian matrix isNo numerical solution of the system is required.

Here a dense $100000$-by-$100000$ matrix has $10^{10}$ entries, about $80$ GB even with eight-byte storage per entry. Use limited-memory BFGS on the negative of the objective, or equivalently its positive-definite approximate-metric form of gradient ascent. It stores only a fixed small number $\ell$ of vector pairs, costing $O(\ell d)$ memory rather than $O(d^2)$.

Let $G(x)={\sum }_{i=1}^{100}{e}^{e^{-\Vert x-a_i{\Vert }^2}}$ and put $c_i(x)=e^{e^{-\Vert x-a_i{\Vert }^2}}{e}^{-\Vert x-a_i{\Vert }^2}$. Direct use of the chain rule givesIf $H_n$ denotes the positive-definite inverse-Hessian matrix approximation for $-G$, the iteration iswith a line search increasing $G$ and the usual safeguarded limited-memory updates. No exact Hessian matrix is required. With fixed memory and the usual bounded-metric and line search hypotheses, expect local linear convergence near a nondegenerate strict maximum; fixed-memory limited-memory BFGS does not generally inherit the superlinear theorem for full-memory BFGS.

Unlike the preceding objective, $G$ is not generally a concave function. Thus this is a method for finding a local maximum; comparison of different converged candidates or a global search is needed to certify a global maximum for arbitrary centres. The symbolic problem supplies no assumption that would justify an unconditional global-convergence claim.^[ref]

The gradient vanishes at $0$. Decompose any vector as $x=tb+w$, with $w\perp b$. Symmetry of $A$ and $Ab=\frac{1}{2}b$ imply $b^{T}Aw=(Ab{)}^{T}w=0$. Since $\Vert b\Vert =1$,If $d\ge 2$, choose a nonzero $w\perp b$. Arbitrarily close to zero, the direction $b$ gives negative values and the direction $w$ gives positive values. Thus $0$ is a saddle point for $d\ge 2$. Equivalently, the Hessian matrix has one negative eigenvalue and all its other eigenvalues positive.

The dimension matters: if $d=1$, the orthogonal complement is zero-dimensional and $g(tb)=-t^2/2$. In that case $0$ is the unique global maximum. This edge case is not excluded by the stated hypotheses. The decomposition is a rank-one downdate along an eigenvector.

If a twice differentiable real function on $\mathbb{R}$ satisfies $f^{\prime \prime }(x)\le -c$ for some fixed $c>0$, then it has a unique global maximum. The mean value theorem shows that $f^{\prime }$ is strictly decreasing, and that $f^{\prime }(x)+cx$ is nonincreasing. Hence $f^{\prime }(x)\le f^{\prime }(0)-cx$ for $x>0$ and $f^{\prime }(x)\ge f^{\prime }(0)-cx$ for $x<0$. These inequalities force opposite derivative signs at the two ends. The intermediate value theorem supplies a unique zero of $f^{\prime }$, at which $f$ changes from increasing to decreasing. Merely having $f^{\prime \prime }<0$ does not guarantee existence: $f(x)=-e^{-x}$ is increasing and approaches its unattained supremum zero.