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In cylindrical coordinates, $(-y,x,0)=r{\mathbf{e}}_{\theta }$ and $x+r=r(1+\cos \theta )$. Hence the vector field simplifies toTake the orientation of $C$ to be increasing $\theta$, as in its given parametrization. On the circle, $d\mathbf{r}=a{\mathbf{e}}_{\theta }d\theta$, so its line integral isWe used $\sqrt{1+\cos \theta }=\sqrt{2}|\cos (\theta /2)|$; the cosine is nonnegative on this interval. In particular, the answer does not depend on $a$.

For $0<y<\pi$, define $J(y)={\int }_0^{\pi }{u}_y(x,y)dx$. Differentiate under the integral and use Laplace's equation:Thus $J$ is constant. The top Neumann boundary condition gives $J(\pi )=0$, soFor the included endpoint $y^{\ast }=0$, the integral is interpreted as an improper integral at the bottom corners if necessary. The Fourier series constructed in c gives, on $0<x<\pi$,The second series is absolutely and uniformly convergent; the first term has only integrable logarithmic singularities. Its integral is zero since ${\int }_0^{\pi }\log (2\sin x)dx=0$, while every cosine term also has zero integral. This proves the endpoint as well:For completeness, the logarithmic integral follows without any endpoint differentiation. Put $I={\int }_0^{\pi /2}\log (\sin x)dx$ and substitute $x\mapsto \pi /2-x$ to obtain $I={\int }_0^{\pi /2}\log (\cos x)dx$. Adding and using $\sin x\cos x=\sin (2x)/2$ gives $2I=I-(\pi /2)\log 2$, hence $I=-(\pi /2)\log 2$ and the required full-interval result.

On $(-\pi ,\pi )$ the function is odd, so its mean and all its cosine Fourier coefficients vanish:The assigned endpoint value does not affect these integrals. The product of the two odd functions $x|x|$ and $\sin nx$ is even, hence the sine Fourier coefficients areTwo applications of integration by parts giveThereforeThe periodic extension has a jump discontinuity at every odd multiple of $\pi$: its left and right limits are ${\pi }^2$ and $-{\pi }^2$. Its assigned value at each such point is $-{\pi }^2$. Thus it is odd away from these points, but it is not a globally odd function pointwise. The Fourier series will take the midpoint value zero at these jumps.

For a nonzero integer $k$, direct integration giveswhereas ${\int }_{-\pi }^{\pi }1dx=2\pi$. The product-to-sum formulastherefore establish Fourier orthogonality: for positive integers $m,n$,Here ${\delta }_{mn}$ is the Kronecker delta. In particular the constant function is orthogonal to every nonconstant sine and cosine in this system. Integrating the given Fourier series, then multiplying it by $\cos (mx)$ or $\sin (mx)$ and integrating, isolates its Fourier coefficients. Using the permitted interchange of the sum and integral givesIf $f$ is an even function, then $f(x)\sin (nx)$ is an odd function, while $f(x)\cos (nx)$ is an even function. ThusThe Fourier series consequently reduces to a Fourier cosine series.

For a nonzero separable solution of the Laplace equation in polar coordinates, write $T(r,\theta )=R(r)\Theta (\theta )$ with $r>0$. Multiplying the partial differential equation by $r^2$ and dividing where $R\Theta \ne 0$ giveswhere $\lambda$ is a real separation constant. The resulting ordinary differential equations areThe equations continue across zeros of the factors by continuity. The radial equation is a Cauchy-Euler equation; putting $s=\log r$ changes it to $d^{2}R/d{s}^2-\lambda R=0$.

For positive separation constant, write $\lambda =k^2$ with $k>0$. The radial factors are powers, while the angular factors are sine and cosine:For zero separation constant, each transformed equation has zero second derivative, givingFor negative separation constant, write $\lambda =-k^2$ with $k>0$. The radial factors oscillate in $\log r$, and the angular factors are exponential functions:These exhaust the nonzero real separated solutions before imposing angular periodicity or regularity at the origin. The zero solution is also allowed. Since Laplace's equation is linear, sums of these separated solutions are solutions wherever the sums and required derivatives converge appropriately.

We verify that the augmented partition of an interval in part i has vanishing mesh of a partition. Define ${\varphi }_n(t)=t^2/(t^2+n)$ for $t\ge 0$. Its derivative isSince $2u\le 1+u^2$ for $u\ge 0$, we have $0\le {\varphi }_n^{\prime }(t)\le 1/\sqrt{n}$. The mean value theorem therefore givesThe appended interval has length $1/(n+1)$, so $\Vert {\mathcal{P}}_n\Vert \le \max \{n^{-1/2},(n+1{)}^{-1}\}\to 0$ for the augmented partition of an interval.

The sine is continuous, hence Riemann integrable on $[0,1]$. By the Riemann sum convergence theorem, $T_n\to {\int }_0^1\sin tdt=1-\cos 1$, using the fundamental theorem of calculus and the cosine antiderivative. As $T_n-S_n=\sin 1/(n+1)\to 0$, the answer is

Use polar coordinates with $0\le r\le {\sin }^2\theta$ and $0\le \theta \le \pi$. The Jacobian determinant gives $dA=rdrd\theta$, so the area isThe trigonometric identityreduces the integral to its constant term: both cosine terms have zero integral over this interval. Therefore

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.

For the cardioid, $r^{\prime }=-a\sin \theta$. The expression under the square root simplifies toSince $0\le \theta \le {\theta }_2\le \pi$, the cosine is nonnegative. ThereforeThe upper half of the cardioid starts at $(2a,0)$, passes through $(0,a)$, bends to negative $x$, and ends at the cusp $(0,0)$ when ${\theta }_2=\pi$. Its full upper-half arc length is $4a$; reflecting gives $8a$ for the full cardioid.

The fundamental theorem of calculus givesThis decreases from $2a$ to zero, so the arc length increases fastest at ${\theta }_2=0$, interpreted as the right-hand rate at the initial endpoint. If only positive endpoints are allowed, the rate has supremum $2a$ as ${\theta }_2\downarrow 0$, rather than an attained interior maximum.

For the Fourier cosine series, first take the even extension $F_e(x)=f(|x|)$ on $[-\pi ,\pi ]$, and then its $2\pi$-periodic extension. Evenness eliminates the sine terms. With the constant term written as $a_0/2$, the Fourier coefficients and series areFor the Fourier sine series, take the odd extension $F_o(x)=f(x)$ for $0<x<\pi$ and $F_o(x)=-f(-x)$ for $-\pi <x<0$, and extend with period $2\pi$. Values assigned at isolated endpoints do not change the Fourier coefficients. Oddness eliminates the constant and cosine terms, leavingBoth formulas follow by orthogonality on $[-\pi ,\pi ]$ and the parity of the extensions.

For every real $\alpha$, $|\cos (\alpha n)/n^2|\le 1/n^2$. The bounding P-series with exponent two converges. The comparison test for series therefore givesOscillations of cosine are unnecessary for convergence here.

Use the complex exponential representation of cosine. For real $\alpha$ and $\beta >0$, exponential decay makes the improper integral absolutely convergent, andFor the second integral, use differentiation under the integral sign with respect to $\beta$. The minus sign is essential: ${\partial }_{\beta }{e}^{-\beta x}=-x{e}^{-\beta x}$. HenceTo justify this step near any fixed ${\beta }_0>0$, restrict $\beta \ge {\beta }_0/2$. The absolute value of the differentiated integrand is at most $x{e}^{-{\beta }_{0}x/2}$, an integrable bound, so dominated convergence permits the differentiation. At $\alpha =0$ the answers reduce to $1/\beta$ and $1/{\beta }^2$, respectively.

For an odd integrable function, both $f(x)$ and $f(x)\cos nx$ are odd. Their integrals over $[-\pi ,\pi ]$ therefore vanish, so the constant and cosine Fourier coefficients are zero. For $f(x)=\sin ax$, the remaining Fourier sine coefficients areThe product-to-sum identity in the first step reduces the integral to two cosines. Here $a$ is real and noninteger, so no denominator vanishes. Consequently the Fourier series of a noninteger-frequency sine isAt $x=\pm \pi$ every term is zero. The Fourier series converges there to the average of the two limits of the periodic extension, namely zero, rather than to the nonzero endpoint values $\pm \sin \pi a$.

Factor the numerator as $(e^x-1{)}^2=x^2+O(x^3)$. The Taylor series of the cosine gives $\cos 3x-2\cos 2x+\cos x=(-9/2+4-1/2)x^2+O(x^4)=-x^2+O(x^4)$. HenceThe first nonzero term is the constant $-1$.

Use the complex exponential forms of the sine and cosine, and put $w=e^{2iz}$. Multiplying the equation by $2e^{iz}$ gives $(2-i)(w-1)/i+(w+1)=0$, which simplifies to $w=1-i$. Taking all branches of the complex logarithm,Thus every solution isThe exponential factor used to multiply the equation is nonzero, so no solutions were introduced or lost.

The chain rule for the cosine gives

The square is on the cosine: near zero, $\ln ({\cos }^{2}x)=2\ln (\cos x)$. Substituting $\cos x=1-x^2/2+x^4/24+O(x^6)$ into the logarithmic series givesTherefore

The product-to-sum identity for the sine gives $\sin 2x\sin 4x=\frac{1}{2}(\cos 2x-\cos 6x)$. Each cosine has zero integral on this interval, soThis is also an instance of Fourier orthogonality.

The double-angle identity $\cos (2x)=2{\cos }^{2}x-1$ reduces the equation to $\cos x(1+2\cos x)=0$. On the specified interval the cosine takes each of the values $0$ and $-1/2$ once, giving