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At a fixed radius, Parseval's identity applied to the preceding Fourier series givesIn polar coordinates, $dA=rdrd\theta$. Integrating over radius and interchanging the nonnegative sum and integral givesThe interchange follows from the monotone convergence theorem. This is the special case of the area norm of the Poisson extension in the unit disk for the odd step boundary data.

Take $r>0$ and select a complex argument $\theta$ for $z$. To specify the complex exponentiation, use $z^{1+i}=\exp ((1+i)(\ln r+i\theta ))$. Expanding its exponent givesIts complex logarithm therefore has valuesThe continuous logarithm obtained by retaining the displayed exponent uses $k=0$. If the outer logarithm is the principal logarithm, choose $k$ to put the imaginary part in $(-\pi ,\pi ]$ instead; the real part is unchanged. In particular one cannot generally assert $\mathrm{Log}(e^w)=w$ without checking its imaginary part.

The condition on the real part gives $\ln r=\theta$, so the required logarithmic spiral has polar coordinatesAs $\theta$ increases, it runs from the negative real axis near radius $e^{-\pi }$ through $-i{e}^{-\pi /2}$, $1$ and $i{e}^{\pi /2}$ to the negative real axis at radius $e^{\pi }$. The angle increases by one complete turn. Using the conventional principal argument $-\pi <\theta \le \pi$, the inner negative-axis endpoint is excluded and the outer endpoint is included. The convention $-\pi \le \theta <\pi$ reverses those endpoint choices. If both limiting arguments are admitted by the paper's wording, both limiting endpoints are drawn; the interior curve is the same. The origin is never included because the complex logarithm is undefined there.

The most useful guide to the contour plot is the common saddle value. Direct polynomial factorization givesThus the level set $f=1/6$ is exactly the three complete linesTheir intersections are the three saddle points, and they enclose the origin in an equilateral triangle. Near the local minimum, the level sets are nearly circles because the leading term is $(x^2+y^2)/2$. For $0<c<1/6$, a closed contour surrounds the origin, while three separate open branches lie outside the triangular barrier. At the saddle level these components meet at the saddle points; above it the central closed contour has opened into the three outer channels.

The identity in polar coordinatesshows the rotation symmetry through $2\pi /3$. It identifies the function as the Hénon-Heiles potential and helps organize all three channels in the original sketch below. The dashed straight lines are the saddle level; labels on the other contours give $f$.

The cardioid has a nonnegative radial boundary, so each ray from the origin contributes $0\le r\le a(1+\cos \theta )$. Integrating the polar coordinates area element yieldsThe three angular integrals are $2\pi$, zero and $\pi$, respectively, using the double-angle formula for the last. Thus

Let $I$ be the positive Gaussian integral. Its improper integral converges because $e^{-x^2}\le e^{-|x|}$ for $|x|\ge 1$. Multiplying two copies and using the polar coordinates area element givesNonnegativity justifies the product double integral and the change of integration order. Since $I>0$,

For polar coordinates, $x=r\cos \theta$ and $y=r\sin \theta$. The Jacobian determinant is $r$, so the area element isThis accounts for the increasing length of an angular strip as its distance from the origin increases.

For $k=1$, the vector field is $(x^{2}y,x{y}^2,z{)}^T$. On the cone,The first term has zero integral by symmetry. Using polar coordinates for the remaining double integral givesOn the paraboloid $z={\rho }^2-1$,Again the $xy$ term cancels by symmetry, givingTherefore direct surface integral yieldsNo separate disk at $z=0$ is included: that rim joins the two surfaces and is not an additional face of the closed boundary.

In the translated coordinates $u=x-1$, $v=y+1$, the zero level set is the pair of lines $u=0$ and $v=0$. The function is positive in the quadrants $uv>0$ and negative in those with $uv<0$, and tends to zero as $u^2+v^2\to \infty$. Thus its nonzero contours form closed curves around the four off-centre stationary points.

For an explicit description of the contour plot, use polar coordinates $u=r\cos \theta$, $v=r\sin \theta$:At fixed direction within a quadrant, the magnitude increases until $r=\sqrt{2}$ and then decreases towards zero. Also $|u|e^{-u^2/2}\le e^{-1/2}$, with equality exactly when $|u|=1$, and similarly for $v$. Consequently $|f|\le e^{-1}$; each positive level below $e^{-1}$ encloses a positive peak in each positive quadrant, with the analogous negative contours in the negative quadrants.

To classify the stationary points, the Hessian matrix isAt $u=v=0$ it is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, with eigenvalues $1$ and $-1$, so $(1,-1)$ is a saddle point. At $u,v\in \{-1,1\}$ it is $-2fI$. Therefore $(2,0)$ and $(0,-2)$ are strict local maxima, and $(2,-2)$ and $(0,0)$ are strict local minima. These are also global maxima and minima by the bound just proved.

For the nonconstant boundary data, use separation of variables in polar coordinates and put $\varphi =R(r)\cos 2\theta$. Laplace's equation reduces toThis Cauchy-Euler equation has independent solutions $r^2$ and $r^{-2}$. Unlike a disk containing the origin, the annulus permits both. The zero inner boundary value requires $R=A(r^2-a^4{r}^{-2})$, and the outer value requires $A=V{b}^2/(b^4-a^4)$. HenceSubstituting $r=a$ gives zero, while $r=b$ gives $V\cos 2\theta$. This is the Dirichlet problem on an annulus for one Fourier mode; no other angular mode is needed.

For the changed inner boundary value, use superposition to add a radial harmonic function that is $-V$ at $a$ and zero at $b$. The radial zero-mode equation is $(r{R}^{\prime }{)}^{\prime }=0$, giving $R=C+D\log r$. Imposing the two constant boundary values givesTherefore the second answer isThe logarithmic term is independent of $\theta$, so it changes only the angular average of the electric potential. Both answers are unique for these Dirichlet boundary conditions, by uniqueness for Laplace's equation.

Apply the Laplacian in polar coordinates twice to the proposed separated function. For any real exponent $m$,so the biharmonic equation becomesIts four distinct roots are $m=-2,0,2,4$. The separated polar solutions of the biharmonic equation therefore give the complete radial family in this angular sector,Equivalently, its radial equation is a fourth-order Cauchy-Euler equation; four distinct powers supply four linearly independent solutions, so no logarithmic terms are required.

Interior. Continuity with value zero at the origin removes both $r^{-2}\cos 2\theta$ and $\cos 2\theta$. The latter depends on the approach direction even though it is bounded. The two boundary conditions at unit radius then give $C+D=0$ and $2C+4D=1$. ConsequentlyThis extends smoothly through the origin: $r^2\cos 2\theta =x^2-y^2$ and $r^4\cos 2\theta =(x^2+y^2)(x^2-y^2)$ are polynomials. Its radial derivative at $r=1$ is $(2r^3-r)\cos 2\theta =\cos 2\theta$.

Exterior. The condition on the radial derivative at infinity excludes the $r^2$ and $r^4$ terms. A nonzero $B\cos 2\theta$ is permitted, since the required decay concerns the radial derivative. Now $A+B=0$ and $-2A=1$, givingIndeed ${\partial }_r{\psi }_{\mathrm{out}}=r^{-3}\cos 2\theta \to 0$, and its value at unit radius is zero. These are particular interior and exterior solutions for the biharmonic streamfunction; no condition requiring $\psi$ itself to vanish at infinity was imposed.

For a separated harmonic function $R(r)\Theta (\theta )$ in polar coordinates, single-valuedness forces integer angular frequencies. The radial factors for frequency $m\ge 1$ are $r^m,r^{-m}$, while frequency zero has a constant and $\ln r$. Finiteness at the origin excludes $r^{-m}$ and $\ln r$.

The source's angular factor is $\cos \theta$ in every term, as printed, rather than $\cos (n\theta )$. The power-law angular mode for the planar Poisson equation givesHence, assuming the series permits termwise differentiation, the general finite, single-valued solution isThe arbitrary harmonic series must itself converge to a finite function on the prescribed disk. If ${\alpha }_0\ne 0$, the source has a direction-dependent limit at the origin; the displayed potential still extends finitely there, but the equation at that point is understood weakly rather than as a smooth classical source.

For the special source, a particular potential is $r^3\cos \theta /8$. Its boundary cosine term is canceled by $-a^{2}r\cos \theta /8$, and the boundary sine term is supplied by $(r/a{)}^2\sin 2\theta$. The resulting Dirichlet boundary conditions has solutionIt is finite at zero, has the required boundary values, and its Laplacian is $r\cos \theta$.

For Laplace's equation in polar coordinates, put $\Phi =R(r)\Theta (\theta )$. Separation of variables gives ${\Theta }^{\prime \prime }+n^2\Theta =0$ and $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$. Periodicity selects integers $n\ge 0$. For $n>0$, the radial solutions are $r^n,r^{-n}$; for $n=0$, they are $1,\ln r$, while a term linear in $\theta$ is excluded by periodicity. Their superposition isInside the disk, boundedness at the origin removes the logarithm and all negative powers. The separately biased semicircular boundary values have mean $(V_1+V_2)/2$, no nonconstant cosine coefficients, and sine coefficients $2(V_1-V_2)/(\pi n)$ for odd $n$, zero for even $n$. The disk potential with two semicircular boundary values is thereforePut $w=(r/a)e^{i\theta }$. The odd logarithmic series sums to ${\sum }_{m\ge 0}{w}^{2m+1}/(2m+1)=\frac{1}{2}\log [(1+w)/(1-w)]$. The quotient has positive real part for $|w|<1$, so its argument is the principal arctangent of $2(r/a)\sin \theta /[1-(r/a{)}^2]$. Taking imaginary parts givesThe boundary limits away from the two voltage jumps are $V_1,V_2$. Along the diameter through a jump the value is the mean; incompatible adjacent boundary limits prevent a continuous extension at the junctions.

The time dependence gives $u_{tt}=-{\omega }^{2}u$, and the angular derivative vanishes for these radial normal modes. The Laplacian in polar coordinates therefore yieldsMultiplication by $r$ puts the Sturm-Liouville problem in the formThe relevant self-adjoint operator is $\mathcal{A}R=-r^{-1}(r{R}^{\prime }{)}^{\prime }$ in the weighted inner product with weight $r$. A physical displacement is regular at the origin, excluding the logarithmic singular solution. In particular $R$ is bounded and $r{R}^{\prime }\to 0$ there; regular eigenfunctions also have $R^{\prime }(0)=0$. This origin condition is needed in addition to the fixed outer boundary.

Direct differentiation givesOn the unit disk $D$, polar coordinates give ${\int }_D{y}^{2}dA=\pi /4$, ${\int }_{D}xdA=0$ and ${\int }_{D}dA=\pi$. Consequently the volume integral isFor an independent surface integral calculation, the relevant cross product isThe bottom face contributes ${\int }_{D}xydA=0$ and the top contributes ${\int }_D(-h{y}^2-xy)dA=-\pi h/4$. On the cylindrical side, put $x=\cos \theta$, $y=\sin \theta$, $\hat{\mathbf{n}}=(x,y,0)$ and $dS=d\theta dz$. The flux density is $x^3+2xy{z}^2+2y^{2}z-x{y}^2$. All terms except $2y^{2}z$ integrate to zero because they are odd functions of $x$ or $y$, givingThus minus the total outward flux is $-(\pi h^2-\pi h/4)$, in agreement with the volume integral.

The coordinate tangent vectors are the partial derivatives of the position vector:Their pairwise dot products vanish. Their lengths, the scale factors of orthogonal coordinates, are $1,r,1+r\sin \alpha$. Since $R<1$, the last is positive throughout the body. Thus these circular torus coordinates are orthogonal wherever $r>0$; at $r=0$ the cross-section angle degenerates, as in ordinary polar coordinates.

The ordered Jacobian determinant isIts sign is positive for $r>0$, so the volume element is $dV=r(1+r\sin \alpha )drd\alpha d\beta$. The coordinate degeneracy on the central circle has zero volume and does not obstruct volume integration.

Let the evaluation point be ${\mathbf{r}}_0=(x,y,z)$, $z>0$, and reflect it to ${\mathbf{r}}_0^{\ast }=(x,y,-z)$. The method of images constructsThe two Euclidean distances coincide on the plane, so $G=0$ there. The negative image source lies outside the domain; inside, only the real source contributes to $-{\nabla }_{\mathbf{r}}^{2}G={\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}_0)$. The difference also decays at infinity.

On the plane write $\mathbf{r}=(\xi ,\eta ,s)$ and set $s=0$. The outward unit normal of the upper half-space is $n=-e_z$. Direct differentiation givesSubstituting into the boundary representation from part (c) yields the Poisson kernel for the upper half-space:Its sign is positive, as expected from the maximum principle for nonnegative boundary data. Its normalization can be checked in polar coordinates:As $z\downarrow 0$ the kernel is an approximate identity, recovering $f$ at its continuity points. Boundary data must have the decay and regularity compatible with the stated condition at infinity.

First compute the curl directly:Reflection symmetry in $x$ makes the last component integrate to zero. The middle component gives the hemisphere's volume, soFor an independent surface integral, the closed boundary consists of the curved hemisphere and its flat base. The right-hand side in part (b) is ${\int }_S\mathbf{n}\times \mathbf{F}dS$.

On the curved part use spherical coordinates with $0\le \theta \le \pi /2$ and $0\le \varphi <2\pi$. There,The last component has zero angular integral. The first two components giveOn the base, $z=0$ and the outward normal vector is $-{\mathbf{e}}_z$. Thus $\mathbf{n}\times \mathbf{F}=(x^2+y^2,0,0)$. In polar coordinates its integral isThe first components cancel, and therefore both required sides equal

Put $\Psi =U(r)V(\varphi )$ in the Laplace equation in polar coordinates. After multiplication by $r^2/(UV)$, separation of variables givesThe angular function is $2\pi$-periodic and single-valued, so its nonconstant modes are $\cos (n\varphi ),\sin (n\varphi )$ with integer $n\ge 1$. Their radial Euler differential equation has solutions $r^n,r^{-n}$; finiteness at the origin excludes $r^{-n}$.

The zero angular mode has constant angular part and radial solutions $1,\log r$. The logarithmic term is again excluded at the origin. Superposition therefore yieldsThis is the regular harmonic expansion inside the disk.

For a nondegenerate surface take $R>0$, and abbreviate $C=\cosh R$, $H=\sinh R$. Close the upper oblate spheroid by its equatorial disk $D$, obtaining a half-ellipsoid $V$. The divergence isThe integral of $2y$ vanishes by reflection symmetry. At height $z$, the cross-section is a disk of radius $q=C\sqrt{1-z^2/H^2}$. Using planar polar coordinates,ConsequentlyThe outward unit normal vector on $D$ is $-{\mathbf{e}}_z$, so its flux is $-2\pi C^2$. Applying the divergence theorem to the closed surface and subtracting this disk contribution givesThis is a special case of polynomial flux through a half-ellipsoid. Including the closing disk is essential: the constant vertical component has zero divergence but nonzero flux through the open curved surface.

For a holomorphic function with $z=r{e}^{i\theta }$, the chain rule givesWriting $f=u+iv$ and equating real parts and imaginary parts,For $r>0$, division by $r$ gives the Cauchy-Riemann equations in polar coordinates:The restriction $r>0$ reflects the coordinate singularity at the origin; the Cartesian equations have no such restriction.