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Letwhere $\mathcal{H}$ is the Heaviside step function. Thenso the Eliassen equation for residual circulation isTake no normal residual flow at $y=0,L$ and $z=0$, decay as $z\to \infty$, and choose the streamfunction constant on the connected rigid boundary to be zero. Thus

The required Fourier series in sine modes isDefineFor each mode, the vertical equation isThe Dirac delta function requiresThe solution satisfying the boundary conditions is thereforewhere

The momentum equation givesThe jump of ${\overline{\chi }}_a^{\ast }$ supplies an equal positive delta function in $f_0{\overline{\chi }}_{a,z}^{\ast }$, so the singular terms cancel. The regular acceleration iswhereFinally, the transformed density equation givesThese exponentially decaying modes are the balanced mean response to wave-activity deposition at $z=H$.

The boundary data are $U\mathrm{sgn}(\sin \theta )$, whose Fourier series isRegular harmonic modes in a disk are $(r/R{)}^n\sin (n\theta )$ and $(r/R{)}^n\cos (n\theta )$. Matching the odd boundary data givesEvery term is regular at $r=0$, and the series approaches the prescribed values at every boundary point away from the two jump discontinuities.

The Fourier cosine series on $[0,L]$ isIt is the Fourier series of the even extension to $[-L,L]$, repeated with period $2L$. The Fourier sine series isIt comes from the odd extension and its $2L$-periodic extension. These Fourier coefficient formulas follow by Fourier orthogonality, since every nonconstant sine or cosine has squared integral $L/2$ on $[0,L]$, while the constant function has squared integral $L$.

For separation of variables, put $U=X(x)H(t)$ and set $X^{\prime \prime }/X=H^{\prime }/(\kappa H)=-\lambda$. The boundary conditions are $X(0)=0$, $X^{\prime }(L)=0$. Multiplying $-X^{\prime \prime }=\lambda X$ by $X$ and using integration by parts givesA nonzero solution therefore has $\lambda >0$: the zero case forces $X$ constant, and then $X(0)=0$ forces $X=0$. Writing $\lambda ={\mu }^2$, the eigenfunctions and their temporal factors areThese sine functions are orthogonal on $[0,L]$, with squared norm $L/2$. Expanding the initial condition and multiplying by $\sin ({\mu }_{m}x)$ gives, using the permitted interchange of the sum and integral,Here $\cos ({\mu }_{n}L)=0$ and $\sin ({\mu }_{n}L)=(-1{)}^n$. Thus the Fourier series solution isThe exponential decay permits differentiation of the series for $t>0$, and each term satisfies both boundary conditions. At $t=0$ it is the mixed-end Fourier expansion of $-T_{0}x/L$.

The gained heat is the increase in internal energy. Since the paper specifies no cross-sectional area, first give the answer per unit cross-sectional area; multiply by $A$ for a rod of cross-sectional area $A$. The initial temperature has integral $T_{0}L/2$, whereas ${\int }_0^L\sin ({\mu }_{n}x)dx=1/{\mu }_n$. At $t_{\ast }=L^2/\kappa$ this givesNumerically $\Delta Q/A\simeq 0.45623855\rho c{T}_{0}L$. This is positive and less than the eventual gain $\rho c{T}_{0}L/2$, as expected from relaxation towards the uniform steady state.

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.

The real Fourier series of $f$ is writtenwhere the Fourier coefficients areBy part (a)(i), any other interval of length $2\pi$ gives the same Fourier coefficients, because each integrand is a periodic function with period $2\pi$. The notation $\sim$ associates $f$ with its Fourier series; existence of the integrals alone does not assert pointwise convergence.

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.

By Dirichlet's theorem on Fourier series, a piecewise continuously differentiable periodic function has a Fourier series converging at each point to the average of its one-sided limits. In particular it converges to the function at every point of continuity. Our periodic extension satisfies these hypotheses, and $x=\pi /2$ is a point of continuity with $f(\pi /2)={\pi }^2/4$.

At this point $\sin (n\pi /2)=0$ for even $n$. For $n=2j-1$,Substituting in the Fourier series from part (c) provesThe sum is taken in its displayed order. Discarding its zero even-indexed terms is valid and does not rearrange the remaining terms.

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.

The function $x\sin (px)$ is an even function, so every Fourier sine coefficient is zero. Introduce the auxiliary integralFor a nonzero integer $k$, integration by parts givesConsequentlyFor $n\ge 1$, a product-to-sum formula yieldsIf $n\ne p$, this becomesThe exceptional Fourier coefficient at $n=p$ must be evaluated before dividing by $p-n$:Thus the full Fourier series isThis equality holds on $[-\pi ,\pi ]$ and by periodicity everywhere. Indeed the periodic extension is continuous at the joined endpoints, since both endpoint values are zero, and is piecewise smooth. The theorem on pointwise convergence of a piecewise smooth Fourier series therefore applies, including at the endpoints. The Fourier coefficients also decay as $n^{-2}$, making the resulting series absolutely and uniformly convergent.

Set $p=3$ and $x=0$ in the Fourier series from b. Since the function is continuous there, pointwise convergence of a piecewise smooth Fourier series givesIt follows thatThe $n=1$ and $n=2$ terms are $1/8$ and $-1/5$, respectively. Subtracting them gives the required absolutely convergent series:

For period two, write $f\sim a_0/2+{\sum }_{n\ge 1}[a_n\cos (n\pi x)+b_n\sin (n\pi x)]$. Direct integration gives $a_0=1/2$. For $0<x<1$, $f(x)+f(-x)=2-3x$ and $f(x)-f(-x)=x$, soThus the Fourier series isIt converges to the values specified in part a. Changing the assigned values at the jumps would not change these integral coefficients.

At $x=0$, the Fourier sum is $f(0)=1$, and only the odd cosine coefficients contribute. Therefore $1=1/4+(6/{\pi }^2){\sum }_{k\ge 0}(2k+1{)}^{-2}$, givingAt $x=1/2$, the sum is $f(1/2)=1/2$. All nonzero cosine terms vanish, and the sine terms have alternating signs over odd indices. Thus $1/2=1/4+(1/\pi ){\sum }_{k\ge 0}(-1{)}^k/(2k+1)$, soThis is the Leibniz formula for π obtained from the same Fourier series.

The Fourier cosine series and Fourier sine series use the respective formsTheir Fourier coefficients, obtained from Fourier orthogonality, areIn particular the constant term is $a_0/2={\pi }^{-1}{\int }_0^{\pi }f(t)dt$. The Fourier cosine series is the Fourier series of the even extension $f(|x|)$, made $2\pi$-periodic. The Fourier sine series is the Fourier series of the odd extension, equal to $f(x)$ for $0<x<\pi$ and $-f(-x)$ for $-\pi <x<0$, again made $2\pi$-periodic.

For the Fourier cosine series, integration by parts givesand, for $n\ge 1$,Only odd Fourier harmonics remain, so

For the Fourier sine series, another integration by parts givesConsequentlyAt $x=0$ and $x=\pi$, the Fourier sine series instead has value zero.

To describe the requested graphs precisely, take $t\in [-\pi ,\pi )$ congruent to $x$ modulo $2\pi$. The Fourier cosine series converges to the continuous triangular periodic extensionIts minima occur at the even multiples of $\pi$, with value $1$, and its maxima at the odd multiples, with value $1+\pi$. The Fourier sine series converges toThe jump discontinuities at even multiples of $\pi$ go from $-1$ to $1$; those at odd multiples go from $1+\pi$ to $-(1+\pi )$. Each jump is assigned its midpoint, zero, by the Fourier series.

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.

Write $T=T_s+T_u$. The transient satisfies the heat equation and homogeneous mixed boundary conditionsThe left condition is a Neumann boundary condition and the right one a Dirichlet boundary condition. Separation of variables, $T_u=X(x)G(t)$, givesThe condition $X^{\prime }(0)=0$ removes the sine term. The condition $X(2\ell )=0$ then selectsThere is no nonzero mode with $\mu =0$, because a linear $X$ satisfying both conditions is zero. There is no growing mode either: integration by parts gives ${\int }_0^{2\ell }(X^{\prime }{)}^{2}dx={\mu }^2{\int }_0^{2\ell }{X}^{2}dx$, so every nonzero mode has ${\mu }^2>0$.

These Sturm-Liouville eigenfunctions have orthogonality and squared normExpand the initial transient in this half-integer Fourier cosine series. Its coefficients areThe integration formula follows by integration by parts; the cosine at the endpoint is zero. Thus the half-integer cosine expansion of a linear profile givesAt $t=0$ the Fourier series reproduces $-T_s$ everywhere, including the endpoints; its $n^{-2}$ coefficient decay gives uniform convergence. For every $t>0$ the exponential factors allow termwise differentiation and verify the heat equation and both boundary conditions. The sudden switch-on of nonzero heat flux is incompatible with the spatial derivative of the constant initial temperature at the corner $(0,0)$. Accordingly, the positive-time boundary derivative must not be inferred by differentiating the initial Fourier series term by term at that corner.

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 $\pm \sin \pi a$.

The square wave alternates between a low interval of length five and a high interval of length five; its integral average is three. In the real Fourier series conventionintegration over one period givesandThus only odd harmonics occur:The Dirichlet-Jordan convergence theorem makes the sum equal to the function at its continuity points and to the mean of its one-sided limits at a jump. Consequently the sum is 3 at each of $x=-5,0,5$, regardless of any separately assigned values of the original function at those endpoints.