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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.

Separation of variables givesHere $x>1$ makes the argument of the second logarithm positive. Substituting the initial condition yields $3y^2+1=4(x^2-1{)}^3$. The positive branch is thereforeThis is the maximal connected interval containing $\sqrt{2}$ on which this branch solves the original differential equation. At its left endpoint $y=0$, the original right-hand side has a zero denominator, so one cannot extend the solution through that point merely by continuing the squared equation.

For $y=x$, the first and second derivatives are one and zero, giving $-2x+2x=0$. To find an independent solution, use reduction of order and write $y=xv(x)$. Cancellation of the terms proportional to $v$ leavesWriting $w=v^{\prime }$ and using separation of variables givesSince $1/[x^2(1-x^2)]=x^{-2}+(1-x^2{)}^{-1}$, integration yields the general solutionThe second factor divided by $x$ has nonzero derivative $1/[x^2(1-x^2)]$, so the two solutions are linearly independent.

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.

Use separation of variables to seek a normal mode $y(x,t)=X(x)\cos (\omega t+\delta )$, with angular frequency $\omega >0$ and wavenumber $k=\omega /c$. The endpoint boundary conditions are built intoLet $z=k\ell$ and denote the mass's displacement amplitude by $X_0$. Continuity and the mass equation becomeThe two possibilities must be separated before dividing by $\sin z$.

Antisymmetric modes. If $\sin z=0$, then $z=n\pi$, $n=1,2,\dots$, and $X_0=0$. The force equation gives $A+B=0$, since $\cos z\ne 0$. Thus $X(-x)=-X(x)$ and the mass stays at rest. These allowed angular frequencies satisfyThe restoring forces on the mass cancel even though the string vibrates.

Symmetric modes. If $\sin z\ne 0$, continuity forces $A=B$. Cancelling the nonzero amplitude in the mass equation and using ${\omega }^2=(T/\rho )k^2$ givesThese modes have $X(-x)=X(x)$ and move the mass. In dimensionless form their equation is $\cot z=\alpha z$, where $\alpha =m/(2\rho \ell )>0$. This is the spectral splitting for a string with a central point mass.

For the antisymmetric family, the graph of $\sin z$ crosses zero at every positive multiple of $\pi$. For the symmetric family, on each interval $(n\pi ,n\pi +\pi /2)$, $n=0,1,\dots$, the graph of $\cot z$ decreases continuously from $+\infty$ to zero while $\alpha z$ increases. By the intermediate value theorem and strict monotonicity, there is exactly one intersection in each interval. On the remaining half-intervals $\cot z<0<\alpha z$, so there are no further positive roots. This proves graphically and analytically that both families contain infinitely many normal modes. The figure uses $\alpha =1$ for illustration.

The algebraic case $k=0$ is not a mode: a zero-frequency profile would be linear on each half, continuous and zero at the endpoints; the zero-acceleration mass equation also makes its slopes agree, forcing the entire profile to be zero. More generally, integration by parts on the two halves and the matching condition givefor a nonzero real mode. Thus no nonzero static or exponentially growing mode has been omitted.

Use the change of variables $u=x+y$. Then $u^{\prime }=1+y^{\prime }=1+\cos u=2{\cos }^2(u/2)$, a separable differential equation. Where $u$ is not an odd multiple of $\pi$, separation of variables givesIndeed the derivative of $\tan (u/2)$ with respect to $u$ is $1/(1+\cos u)$. The nonconstant branches can also be written $y=2\arctan (x+C)+2k\pi -x$, $k\in \mathbb{Z}$. Dividing by $1+\cos u$ excluded its zeros: the missing equilibrium solutions for $u$ give the additional solutions of a differential equationThey satisfy $y^{\prime }=-1=\cos (x+y)$ directly. The transformed right side is continuously differentiable, so the uniqueness theorem for ordinary differential equations prevents a nonconstant branch from crossing an equilibrium solution at finite $x$.

For $k>0$, the two ordinary differential equations have solutionsTheir products $h(x)g(y)$ and $g(x)h(y)$ solve Laplace's equation, since the two second derivatives contribute opposite $k^2$ terms. For $k=0$, both factors are linear functions. This is separation of variables with positive, negative and zero separation constants.

On an interval avoiding $x=\pm 1$, separation of variables gives, for $y\ne -1$,Absorb the sign into the integration constant and include the equilibrium solution $y=-1$ by allowing $C=0$. The complete real family isEach solution is considered on a connected interval in $\mathbb{R}\setminus \{-1,1\}$, where the original equation is defined.

For a nonzero solution, separation of variables gives $dy/y=kdx/(1+x^2)$, so $\ln |y|=k\arctan x+C$. Restoring the zero solution by allowing an arbitrary real multiplicative constant givesIt is valid for all real numbers $x$, including the constant solutions when $k=0$.

For a nonzero separated solution $U=X(x)T(t)$, separation of variables gives $X^{\prime \prime }/X=T^{\prime }/T=-\lambda$. The zero Dirichlet boundary conditions force $X=\sin nx$, $\lambda =n^2$, with integers $n\ge 1$; the time factors are $e^{-n^{2}t}$. Expanding the initial profile in this Fourier sine series requires only the two already displayed spatial modes. ThusBoth terms individually solve the heat equation, vanish at the endpoints and give the required initial profile at $t=0$. The initial function is also $4{\sin }^{3}x\ge 0$ on the bar; the later solution stays nonnegative and tends to zero as heat leaves through the fixed-temperature ends.

First retain the equilibrium solution $y=0$. For $y\ne 0$ and $x\ne \pm 1$, separation of variables gives $\int y^{-2}dy=\int x/(1-x^2)dx$, hence $-1/y=-\frac{1}{2}\ln |1-x^2|+C$. Renaming the constant yieldsA nonzero solution is considered on a regular interval avoiding $x=\pm 1$ and zeros of its denominator. The zero solution itself extends through both singular points of the divided equation.

For the negative-sign equation, take a nonzero separated solution $\varphi =X(x)Y(y)$. Separation of variables gives $X^{\prime \prime }=-\mu X$ and $Y^{\prime \prime }=-(25-\mu )Y$. The zero Dirichlet boundary conditions on $X$ force $\mu =m^2$ with a positive integer $m$ and $X=\sin mx$, up to a constant. Zero boundary values on $Y$ similarly require $25-m^2=n^2$ with a positive integer $n$. A zero or negative separation eigenvalue would give linear or hyperbolic factors, whose two zero endpoint values force the trivial solution.

The only positive integer pairs with $m^2+n^2=25$ are $(3,4)$ and $(4,3)$. By the complete Dirichlet eigensystem of a rectangle, these span the whole eigenspace, soFor the positive-sign equation there is only the zero solution. One can use the same strictly positive eigenvalues of $-\Delta$, or directly apply the zero Dirichlet solution of the modified Helmholtz equation. Multiplying by $\varphi$ and using Green's first identity giveswhere $\Omega =(0,\pi {)}^2$. Both nonnegative integrals must vanish, proving $\boxed{\varphi =0}$.

Use separation of variables on intervals avoiding zeros of $\cos x$:Taking the tangent gives the real general solutionEach solution is restricted to an interval where both the original equation and this expression are finite. There is no real constant solution because $1+y^2$ never vanishes for real $y$.

Rearranging gives $y^{\prime }=2x(y-1{)}^2$. First retain the equilibrium $y=1$. For every other solution, separation of variables gives $\int (y-1{)}^{-2}dy=\int 2xdx$, so $-1/(y-1)=x^2+C$. HenceA nonconstant solution is defined on any interval where its denominator is nonzero. Dividing by $(y-1{)}^2$ without retaining the equilibrium would lose one solution.

For $x_0>0$, separation of variables gives $1/x=kt+1/x_0$, hence $x(t)=x_0/(1+k{x}_{0}t)$. Setting $x(t)=x_0/2$ gives the half-lifeThis is the same reciprocal decay law as a second-order reaction with the specified disappearance rate coefficient. A positive initial value is needed for a positive half-time; $x_0=0$ instead gives the identically zero solution.

On an interval avoiding zero put $v=y/x$. A real solution requires $|v|\le 1$, and the principal square root is $\sqrt{x^2-y^2}=|x|\sqrt{1-v^2}$. The homogeneous first-order differential equation reduces toWhere $|v|<1$, separation of variables gives $\arcsin v=\mathrm{sgn}(x)\log |x|+C$. Thus the nonconstant pieces arerestricted to intervals where the cosine of the phase is nonnegative. This sign condition is essential when substituting back into the original square root.

There are also the solutions $y=x$ and $y=-x$. They can join the nonconstant pieces with a continuous first derivative, because the transformed rate vanishes but is not Lipschitz continuous at $v=\pm 1$. For example, on the positive axis every $A>0$ givesThe sticking solutions of a bounded square-root differential equation describe the same joining mechanism on either half-axis after using the increasing variable $t=\mathrm{sgn}(x)\log |x|$.

For $c>0$ and $l>0$, the zero-endpoint Dirichlet boundary conditions select the Fourier sine basis functions $\sin (n\pi x/l)$. Separation of variables gives temporal factors with angular frequencies ${\omega }_n=n\pi c/l$. Zero initial displacement removes the temporal cosine terms, givingDifferentiating at $t=0$ and using Fourier orthogonality givesThe evaluation property of the given initial velocity means $v(x)=\delta (x-l/2)$ in the sense of a Dirac delta, an idealized impulsively struck fixed-end string. Thus $a_n=2\sin (n\pi /2)/(n\pi c)$: every even mode vanishes, while the lowest two excited modes have $n=1,3$. Their ordinary frequencies and amplitude ratio areThe signed coefficient ratio is $a_3/a_1=-1/3$. The impulse solution is interpreted as a distributional weak solution, rather than one with a smooth initial velocity.

Use separation of variables, $f=X(x)T(t)$, and setFor $\kappa =-{\lambda }^2$, with $\lambda >0$, the separated solutions areFor $\kappa =0$, they areFor $\kappa =+{\lambda }^2$, they areArbitrary admissible superpositions of separated solutions are again solutions. An individual product is only a separated family, not the entire general solution. Without prescribed boundaries, the general twice-differentiable solution is equivalently D'Alembert's formula,For example, changing to $\xi =x-ct$, $\eta =x+ct$ makes the differential equation $f_{\xi \eta }=0$, whose two integrations give this form.

For the solution through $x=0$, work on the connected domain $x>-1$. While $y>0$, separation of variables givesThe initial condition gives $C=2$, soThe restriction matters: the unsquared equation requires $1-\frac{1}{2}\ln (1+x)\ge 0$. The solution reaches zero at $x_{\ast }=e^2-1$. Continuing the squared formula beyond that point would give a positive derivative, contradicting the original negative square-root right-hand side.

The nonnegative forward continuation is insteadThe two pieces join with $y^{\prime }=0$, and both satisfy the original equation. Because the right-hand side is nonpositive for $x>-1$, a nonnegative solution cannot leave zero in the forward direction. This is a finite-time extinction phenomenon for a separable differential equation.

With the same separation of variables ansatz, $x{X}^{\prime }Y-2yX{Y}^{\prime }=0$ givesOn $x,y>0$, integration yields $X=A{x}^{\lambda }$ and $Y=B{y}^{\lambda /2}$. The choice $\lambda =2$, $AB=1$ gives a polynomial that extends smoothly to all real $x,y$:It is nonconstant and real everywhere, and $x{u}_x-2y{u}_y=2x^{2}y-2x^{2}y=0$, including on the axes.