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The multivariable chain rule first givesTaking a second partial derivative in each variable, with the partial derivatives of $f$ evaluated at $(x(u,v),y(u,v))$, givesThe mixed partial derivatives cancel even without using their equality. Hence the Laplacians satisfyAway from the origin the factor $u^2+v^2$ is positive, so $g$ is a harmonic function if and only if $f$ is a harmonic function on the corresponding local coordinate patch. This is the Laplacian in planar parabolic coordinates.

Using the first partial derivatives from part (b),Also $uvg=yf$. Therefore the first partial differential equation becomesFor the second partial differential equation, the same chain rule givesAway from the origin, cancellation yields $f_x+2xf=0$. Treating $y$ as fixed, an integrating factor gives ${\partial }_x(e^{x^2}f)=0$. Thuswhere $F$ is an arbitrary sufficiently differentiable function. This is the local general solution, and the global form when $f$ is defined on the whole plane. If $f$ and its first partial derivatives are continuous at the origin, the reduced partial differential equation also holds there by continuity.

Here $\varphi$ is a scalar field: the source's description as a vector field is a typographical slip, since the displayed formula is for its gradient. For cylindrical coordinates, write $x=r\cos \theta$, $y=r\sin \theta$. Away from $r=0$, a local choice of $\theta$ gives the partial derivativesThe chain rule therefore givesUse the Cartesian coordinates definition $\nabla \varphi ={\varphi }_x{\mathbf{e}}_x+{\varphi }_y{\mathbf{e}}_y+{\varphi }_z{\mathbf{e}}_z$ and the unit vectorsCollecting their coefficients provesThis is the gradient in cylindrical coordinates. The factor $1/r$ converts a change in angle into a change per unit arc length.

Treat the explanatory variables as fixed and put $S_{xx}={\sum }_i{x}_i^2$, $S_{zz}={\sum }_i{z}_i^2$. To identify both slopes, assume $S_{xx}>0$ and $S_{zz}>0$. The given centring and orthogonality conditions make the three columns $1,x,z$ pairwise orthogonal.

The log-likelihood function in the normal linear model, up to a constant independent of the coefficients, isThus maximum likelihood estimation minimizes the residual sum of squares. Setting its partial derivatives to zero givesUsing $\sum x_i=\sum z_i=\sum x_i{z}_i=0$ decouples these equations:The Hessian matrix of the residual sum of squares is $2\mathrm{diag}(n,S_{xx},S_{zz})$, which is positive definite. These are therefore the unique minimizers. If either squared sum is zero, the associated explanatory variable vanishes identically and its coefficient cannot be identified.

The partial derivatives of the parametrization areTheir cross product isTaking its euclidean norm gives the surface area elementThe two poles and duplicated seam have zero surface area, so integrating over the stated parameter domain counts the surface area once. With the change of variables $u=\cos \theta$ and the supplied definite integral,Here $-\ln (\sqrt{2}-1)=\ln (\sqrt{2}+1)$ because $(\sqrt{2}-1)(\sqrt{2}+1)=1$.

At a regular point of the parametrized surface, the partial derivatives ${\mathbf{r}}_s$ and ${\mathbf{r}}_t$ are linearly independent tangent vectors. Their cross product gives a nonzero normal vector:Dividing by its Euclidean norm gives a unit normal; changing the sign chooses the opposite orientation. If the cross product vanishes, this parametrization is singular there and cannot determine a normal vector by this formula.

The partial derivatives of this parametrized surface areTheir cross product isThe mixed terms cancel in its squared Euclidean norm:Since $s>0$, the surface area element is $s(t^2+2)dsdt$. There is no repeated coverage: $z=2s$ recovers $s$, and $\sqrt{x^2+y^2}=st$ then recovers $t$. Therefore the surface area is

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 $r>0$, the partial derivatives $r_x=x/r$ and $r_y=y/r$ giveAdding these partial derivatives yields the divergence:The original vector field is undefined at the origin. Since $\Vert \mathbf{f}\Vert =|x|\le r$, it has the continuous extension $\mathbf{f}(0,0)=0$. This extension is not differentiable: on $y=0$ its first component is $|x|$, whose derivative at zero does not exist. Thus the displayed divergence has no classical value at the origin.

Write $u(x,t)=U(\xi ,\eta )$, where $\xi =x-t$ and $\eta =x+t$. The chain rule givesFor a twice continuously differentiable classical solution, mixed partial derivatives commute, so the wave equation becomesHence $U_{\xi }$ depends only on $\xi$. Integrating first in $\xi$ yields the countertravelling decomposition of the one-dimensional wave equation:On a rectangular domain in $(\xi ,\eta )$ this follows immediately; on the full line for $t\ge 0$, the connected characteristic slices give the same representation. The two terms travel at wave speed one in opposite directions.

Impose the initial conditions. At $t=0$,Differentiate the first identity and combine:If $H(x)={\int }_{x_{\ast }}^{x}g(s)ds$, then $F=(f-H)/2+C$ and $G=(f+H)/2-C$. Substitution cancels $C$ and the arbitrary base point $x_{\ast }$, giving D'Alembert's formula:For example, $f\in C^2$ and $g\in C^1$ make this a classical solution of the wave equation satisfying both initial conditions. The same formula defines a weak solution for less regular data, including the continuous piecewise smooth data of part (b).

Treat the boundary data as $2\pi$-periodic in the angular variable. Denote the outer data by $g(\theta )$, and set $h(\theta )=f(a,\theta )$ for the inner data. Define their mean and nonconstant Fourier coefficients byThe outer data have reflection symmetry, so all their sine Fourier coefficients vanish. Their mean is $1/2$, and their cosine Fourier coefficients are

The angular Fourier harmonics for a single-valued harmonic function in an annulus are $\cos (n\theta )$ and $\sin (n\theta )$, with integer $n\ge 1$. Their radial factors are independent combinations of $r^n$ and $r^{-n}$; the zero angular mode has radial factor $A_0+B_0\log r$. For arbitrary inner data, the sine and cosine modes need independent radial coefficients: forcing them to share one radial factor, as in the product form supplied in the paper, would impose an extra restriction. For either sine or cosine mode, if its outer and inner coefficients are $v$ and $w$, the radial coefficients $\alpha ,\beta$ obeySolving this linear system givesIt is convenient to collect the result into the annular harmonic interpolation factorsThey satisfy $H_n(1)=1$, $H_n(a)=0$, $K_n(1)=0$ and $K_n(a)=1$. The zero mode satisfies $A_0=1/2$ and $B_0=(p_0-1/2)/\log a$. Therefore the required separation of variables solution isAll dependence on $f(a,\theta )$ is given by the displayed integrals. Substituting $r=a$ leaves its Fourier series, and substituting $r=1$ leavesthe Fourier series of the specified outer step. The radial factors satisfy $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$, so each nonconstant term satisfies Laplace's equation; $1$ and $\log r$ do also. For integrable boundary data the radial factors give exponential decay on every closed subannulus strictly inside $a<r<1$, justifying the necessary termwise partial derivatives there.

There is a qualification at the two outer jump discontinuities. By pointwise convergence of a piecewise smooth Fourier series, the outer series gives $1/2$ at $\theta =\pm \pi /2$, whereas the printed data assign $1$ at those two points. Thus the displayed solution attains the outer data at their continuity points, and has radial limit $1/2$ at the jumps. One may separately assign the boundary value $1$ at the two exceptional points without changing the interior harmonic function, but no continuous extension can match the discontinuous data there. For piecewise smooth inner data, the analogous boundary trace agrees with $h$ at continuity points and with its midpoint at jumps. Some regularity of $h$ is necessary for a classical interpretation; the problem imposes none explicitly.

Put $u(r,\theta )=R(r)\Theta (\theta )$ in Laplace's equation and multiply by $r^2/(R\Theta )$ on intervals where the factors are nonzero. Separation of variables givesand henceThe angular Dirichlet boundary conditions are $\Theta (0)=\Theta (\pi /2)=0$. Multiplication of the angular equation by $\Theta$, followed by integration by parts, givesFor a nonzero angular eigenfunction this forces $\lambda >0$: equality would make $\Theta$ constant and then zero. Write $\lambda ={\mu }^2$. The first angular condition removes the cosine term, and the second requires $\sin (\mu \pi /2)=0$. ThusThe radial Cauchy-Euler equation has the two solutions $r^{2n}$ and $r^{-2n}$. By superposition, the general separated series is thereforeThe coefficients are arbitrary subject to convergence with enough partial derivatives on compact subsets to give a harmonic function and to the prescribed zero values on the two open radial edges. Completeness of the sine eigenfunctions gives the usual series representation for solutions smooth up to those edges on closed radial subintervals. There is no constant or logarithmic zero mode because its angular factor would be linear in $\theta$ and would vanish at both ends only if it were zero.

The negative powers must be retained: the stated domain omits $r=0$, and no boundedness at that vertex is prescribed. For example, $r^{-2}\sin (2\theta )$ is a harmonic function satisfying both edge conditions. An additional boundedness condition as $r\to 0$ would remove all $B_n$; an additional value on the circular arc would then determine the remaining $A_n$. With only the conditions in the paper, neither restriction is available and the solution is not unique.

Set $u=X(x)Y(y)$ in Laplace's equation. Separation of variables givesThe Sturm-Liouville problem in $x$ has the constant mode $X_0=1$ with ${\lambda }_0=0$, and the eigenfunctions $X_n=\cos (nx)$ with ${\lambda }_n=n^2$, $n\ge 1$. Indeed, integration by parts makes $\lambda$ nonnegative; for $\lambda >0$ the first Neumann boundary condition removes the sine term and the second requires $\sqrt{\lambda }$ to be an integer. For $\lambda =0$, both derivative conditions make $X$ constant.

Expand the top data using a Fourier cosine series:The constant coefficient isFor $n\ge 1$, the constant part has zero integral against $\cos (nx)$, and two integrations by parts give

Since the bottom value is zero, a convenient factor depending on $y$ for the nonconstant modes is $\sinh (n(y+1))$, which vanishes at $y=-1$. Its value at $y=1$ is $\sinh (2n)$. The zero mode is linear in $y$, so matching both constant boundary values givesTo obtain exactly the paper's form, use the hyperbolic-function identityThus

Each product of hyperbolic sine or hyperbolic cosine with $\cos (nx)$ has $u_{xx}+u_{yy}=0$. The zero mode is also a harmonic function. On $-1\le y\le 1$, the factor $\sinh (n(y+1))/\sinh (2n)$ lies between zero and one. Since $\sum n^{-2}$ converges, the Weierstrass M-test gives uniform convergence on the closed rectangle. The series therefore gives zero at the bottom and the continuous Fourier cosine series of ${\pi }^2-x^2$ at the top, including its endpoints. On compact subsets with $y<1$, the factors decay exponentially with $n$, justifying termwise partial derivatives. In particular $u_x=0$ on both open vertical edges because every differentiated cosine factor vanishes there. The solution has finite Dirichlet energy: Fourier orthogonality gives the $n$th nonconstant mode's energy as $(\pi /2)n{g}_n^2\coth (2n)=O(n^{-3})$, whose sum is finite. Here $\coth$ is the hyperbolic cotangent. Thus a proves uniqueness in that class.

There is a corner compatibility condition for mixed boundary data in the printed wording. The top data have $g^{\prime }(\pi )=-2\pi$, but the paper writes $u_x(\pi ,y)=0$ even at $y=1$. The series solves the interior equation and all the open-edge boundary conditions; the literal derivative condition at $(\pi ,1)$ is incompatible with the top data. A function differentiable in $x$ along the top edge would necessarily have $u_x(\pi ,1)=g^{\prime }(\pi )=-2\pi$. Hence it cannot simultaneously satisfy the printed zero derivative at that corner, and there is no solution with all those literal corner requirements. The standard interpretation of this mixed boundary condition imposes the Neumann boundary condition on the open sides and the Dirichlet boundary condition continuously on the horizontal edges. Differentiating the series term by term on the top edge would be invalid at this corner; the uniform convergence of the undifferentiated series does not license that operation.

The partial derivatives areThus the Jacobian determinant isThis vanishing reflects the functional dependence found next.

Minimize the residual sum of squaresSetting its three partial derivatives to zero gives the normal equationsThese equations characterize the linear least squares minimizers, since $Q$ is a convex quadratic. A unique coefficient vector additionally requires full column rank of the design matrix.

Let $D$ be a bounded planar region with a piecewise smooth, simple closed boundary $C$, oriented counterclockwise, and let $P,Q$ have continuous first partial derivatives on an open neighbourhood of its closure. Green's theorem states thatFor a region with holes, orient every boundary component with $D$ on its left: the outer component is counterclockwise and the inner components are clockwise.

Choose $P=-y/2$ and $Q=x/2$. Their indicated partial derivatives differ by one, so the area isThis is signed area as a boundary integral; reversing the direction of $C$ changes the sign of the integral.

Let $\Sigma$ be a piecewise smooth oriented surface, and let $\mathbf{F}$ be continuously differentiable on an open neighbourhood of it. Stokes theorem states thatThe boundary orientation is induced by the chosen unit normal vector through the right-hand rule. Smoothness on a neighbourhood makes the spatial partial derivatives in the curl well defined.

Choose the outward orientation of a surface on the lower half of the ellipsoid. At its lowest point the normal points down, so its boundary ellipse is traversed clockwise when viewed from above. UseThe line integral is

For an independent surface integral calculation, first compute $\nabla \times \mathbf{F}=(2y,-2x,1)$. Parametrize the lower half byThe outward vector area element isThusThe two sides agree. Choosing the inward unit normal vector and its corresponding counterclockwise boundary orientation changes both answers to $6\pi$.

Let $a$ be an arbitrary constant vector. By the cyclic identity for the scalar triple product,Apply Stokes theorem to $\mathbf{F}\times a$. Since $a$ is constant, the curl product identity givesConsequentlyFor precision, the operator in the last line has componentsThe partial derivatives act on $\mathbf{F}$, with $\mathbf{n}$ multiplying their values; they do not differentiate $\mathbf{n}$. Since the dot products agree for every constant $a$, the cross-product form of Stokes theorem follows:Here $\wedge$ in the paper denotes the cross product, and the printed surface integral has its area element $dS$ understood.

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.

The fundamental theorem of calculus differentiates the upper endpoint, while differentiation under the integral sign differentiates the parameter:These formulas hold, for example, when $g$ and its displayed parameter partial derivative are continuous on a neighborhood of the integration interval and the parameter value.