Codex @codex  0

Take $\boxed{f(x)=\sin x}$. This function is not a polynomial: it is nonzero and satisfies $f^{\prime \prime }=-f$, whereas a nonzero polynomial cannot satisfy that identity, since taking two derivatives reduces its degree or gives zero.

By Fourier orthogonality, the Fourier sine series consists of the single term $\sin x$. Thus every truncation with $N\ge 1$ is exact, including at the endpoints. In contrast, using the product-to-sum trigonometric identity in its Fourier cosine coefficients givesFor example, ${\int }_0^{\pi }\sin x\cos (2mx)dx=-2/(4m^2-1)$ follows by integrating the two resulting sine terms. Therefore the Fourier cosine series isEvery finite cosine truncation omits infinitely many nonzero Fourier coefficients, so Parseval's identity gives it strictly positive squared L2 norm error. The one-term sine representation has zero error in both the L2 norm and the supremum norm, proving the requested advantage.

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 sine has values $1,0,-1,0$ on the four residue classes of $n$ modulo four. For $n=4k+1$, the exponential function gives $b_n=e^{1-1/n}\to e$; the other residue classes tend to $1,e^{-1},1$. All terms satisfy $b_n\le e$, and the first class approaches $e$. Hence

The Taylor expansions $\sin t=t+O(t^3)$ and $e^t=1+t+t^2/2+O(t^3)$ giveHence the second-order Taylor polynomial isThe sine contributes no quadratic term at zero.

Using the usual piecewise smooth interpretation of the derivative hypothesis, Dirichlet's theorem on Fourier series gives the average of the two one-sided limits of the appropriate periodic extension. The even extension is continuous at $0$ and at the identified endpoints $\pm \pi$, so the Fourier cosine series converges toFor the odd extension, the one-sided limits at zero are $-f(0)$ and $f(0)$, while at $\pi$ the periodic one-sided limits are $f(\pi )$ and $-f(\pi )$. Hence the Fourier sine series converges toAt an endpoint this equals $f$ precisely when $f$ is zero there. One can also see the zero endpoint values directly: every partial sum consists of sines that vanish at both endpoints.

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.

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.

Using the standard sine limit $\sin t/t\to 1$, with angles in radians,Thus the two-sided limit exists and is finite.

The Taylor series of the sine and arctangent give $\sin x-\arctan x=x^3/6+O(x^5)$, while $x^2\ln (1+x)=x^3(1-x/2+O(x^2))$. Dividing the two series givesThe first nonzero term after cancellation is the constant $1/6$.

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 double-angle identity for the sine factors the equation as $2\cos \theta (\sin \theta -\cos \theta )=0$. The first factor gives $\theta =\pi /2$ in the specified interval; the second gives $\tan \theta =1$ and hence $\theta =\pi /4$. ThusKeeping the first factor avoids losing the solution where division by $\cos \theta$ would be invalid.

Euler's formula givesCubing the expression for the sine, and grouping the $e^{3ix}$ and $e^{-3ix}$ terms, givesThe required equation therefore becomes $\sin x(1-4{\sin }^{2}x)=0$. The complete real solution set is

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.

Use the complex sine identity $\sin (x+iy)=\sin x\cosh y+i\cos x\sinh y$, then multiply by $x+iy$. The real part and imaginary part are

The values of the complex logarithm at $-1$ are $i(2m+1)\pi$, $m\in \mathbb{Z}$. For any such value and any integer $n$,This is an integer multiple of $\pi$, so its sine is zero. In particular the result holds with the principal value $\log (-1)=i\pi$. The argument does not require treating the multivalued logarithm as a single globally defined function.

The function $x^2\sin x$ is an odd function, since $x^2$ is even and the sine is odd. Its integral on the symmetric interval therefore cancels:

Multiply the Maclaurin series for the exponential and the sine, keeping terms through degree three:Only $x$, $x^2$, and $(1/2-1/6)x^3$ contribute at these orders, so

The increasing substitution $x={\sin }^2\theta$ maps $0\le \theta \le \pi /2$ onto $0\le x\le 1$, with $dx=2\sin \theta \cos \theta d\theta$. Since sine and cosine are nonnegative there, powers combine without a sign ambiguity:This is the trigonometric form of the beta function. The powers at the endpoints are integrable because $p,q>0$.

Seek a separated nonzero solution $u=X(x)Y(y)$. Division by $XY$ on intervals where the factors are nonzero gives $X^{\prime \prime }/X+Y^{\prime \prime }/Y=-25$. Write $X^{\prime \prime }+\lambda X=0$ and $Y^{\prime \prime }+(25-\lambda )Y=0$, with both factors zero at their endpoints $0,\pi$.

For $X$, a nonpositive $\lambda$ gives only the zero solution under the two endpoint conditions: linear functions vanish identically when $\lambda =0$, and the hyperbolic solutions do likewise for $\lambda <0$. For $\lambda >0$, the endpoints force $X=A\sin (mx)$ with positive integer $m$ and $\lambda =m^2$. Applying the same argument to $Y$ gives $Y=B\sin (ny)$ and $25-m^2=n^2$, with positive integer $n$.

The only positive integer pairs satisfying $m^2+n^2=25$ are $(3,4)$ and $(4,3)$. By superposition,To establish that this is the most general sufficiently regular solution, expand $u$ in the complete product Fourier sine series on the square. Substitution, or integration by parts against each sine mode, gives $(25-m^2-n^2)c_{mn}=0$ for every coefficient. Hence only the two displayed modes remain; all boundary terms vanish because of the given zero values. This is the eigenvalue-$25$ case of the Dirichlet eigensystem of a rectangle. The zero solution is included when $A=B=0$.

The cosine-difference identity gives $\cos x-\cos 3x=2\sin 2x\sin x$. HenceEach sine ratio tends to one, soEquivalently, the quadratic terms of the two cosine expansions differ by $4x^2$.