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Write $q_i=a^{ij}{D}_{j}u$. If $u$ is a classical solution, multiply $-{\partial }_i{q}_i=f$ by a smooth function $v$ and apply the divergence theorem. The Neumann boundary condition $q_i{N}_i=0$ removes the boundary term and givesThe density of smooth functions in a Sobolev space and boundedness of the coefficients extend this identity to every $v\in H^1(U)$, so $u$ is a weak solution.

Conversely, take $v$ to be a test function compactly supported in $U$. The weak formulation saysThe fundamental lemma of the calculus of variations gives the equation in $U$. Under the regularity implicit in the stated notion of a classical solution, it holds pointwise. Applying integration by parts again with arbitrary $v\in H^1(U)$ leaveswhere $T$ is the trace operator. Traces of smooth functions can be chosen arbitrarily on the boundary, so the boundary fundamental lemma of the calculus of variations gives $(a^{ij}{\partial }_{j}u)N_i=0$. Thus $u$ is a classical solution.

Varying the Polyakov action with respect to the worldsheet metric givesIn conformal gauge, variation of $X$ gives the wave equationwhile the Virasoro constraints become $\dot{X}\cdot X^{\prime }=0$ and ${\dot{X}}^2+X^{\prime 2}=0$. The boundary variation is $-T\int d\tau ,X^{\prime }\cdot \delta X$ at each endpoint. It vanishes through a Neumann boundary condition $X^{\prime \mu }=0$, a Dirichlet boundary condition $\delta X^{\mu }=0$, or a direction-by-direction mixture of the two.

Multiply the Poisson equation by $v\in V$ and apply Green's first identity. The Dirichlet boundary condition makes the trace of $v$ vanish on ${\Gamma }_1$, while the Neumann boundary condition makes the boundary flux vanish on ${\Gamma }_2$. Thus the weak formulation is: find $u\in V$ such thatwhere

False. The constant function $u=1$ is a harmonic function and has zero normal derivative on every smooth boundary, but it is nonzero in $V$.

In fact, the first of Green's identities givesHence $\nabla u=0$; path connectedness makes $u$ constant throughout $V$. Thus this homogeneous Neumann boundary condition determines a harmonic function only up to an additive constant.

Let $w$ be the difference of two solutions. It is a harmonic function in the open rectangle, continuous on its closure, withInterpret these Neumann boundary conditions classically on the open edges, with the corresponding first derivatives extending there. We use the strong maximum principle for harmonic functions together with even reflection across a homogeneous Neumann boundary. This avoids imposing differentiability at the two bottom corners, where the original data fail a corner compatibility condition for mixed boundary data.

By continuity on the compact rectangle, $w$ attains a maximum $M$. Suppose $M>0$. It cannot occur on the bottom edge, where $w=0$. If it occurs in the interior, the strong maximum principle for harmonic functions makes $w$ constant, contradicting its bottom value. If it occurs on an open vertical or top edge, the zero normal derivative allows an even reflection across that edge. The reflected harmonic function then has an interior maximum, giving the same contradiction.

The only remaining possibilities are the two top corners. Reflect evenly across both meeting edges. This gives a bounded harmonic function on a punctured neighbourhood of the corner. The removable singularity for a bounded harmonic function fills the centre with its continuous value. The corner is now an interior maximum, again impossible. Therefore $w\le 0$. Applying the identical argument to $-w$ gives $w\ge 0$.

Thus $\boxed{w=0}$, and the solution is unique among the continuous classical solutions with the stated edge conditions.

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.

For separation of variables, set $u=X(x)Y(y)$, with $X^{\prime \prime }/X=-Y^{\prime \prime }/Y=-\lambda$. The side Neumann boundary conditions give $X^{\prime }(0)=X^{\prime }(\pi )=0$. As in the standard Sturm-Liouville problem, integration by parts shows $\lambda \ge 0$. The zero eigenvalue gives constant $X$, and $Y^{\prime \prime }=0$ together with $Y^{\prime }(\pi )=0$ makes $Y$ constant too.

For positive eigenvalues, the side conditions give $\lambda =n^2$, $X_n=\cos (nx)$, $n\ge 1$. The top condition chooses the hyperbolic cosine factor $\cosh (n(\pi -y))$. Normalize it to one at $y=0$, obtainingThe orthogonality of integer Fourier modes and the permitted interchange of summation and integration giveFor the remaining Fourier coefficients, integrate by parts twice. The sine boundary term vanishes, and with $f=x(\pi -x)$,The last integral vanishes because $f^{\prime \prime }=-2$ and $n$ is a positive integer. Thus $a_n=-2(1+(-1{)}^n)/n^2$: only even Fourier modes survive, as expected from reflection symmetry about $x=\pi /2$. Writing $n=2m$ givesThe factor involving hyperbolic cosine lies between zero and one on the closed rectangle. Comparison with $\sum m^{-2}$ proves uniform convergence, so the solution is continuous on the entire boundary and takes the prescribed bottom values. On every strip $y\ge \epsilon >0$, the Fourier coefficients have additional exponential decay, allowing differentiation and verifying Laplace's equation and all three Neumann boundary conditions.

To justify the endpoint calculation in b, differentiate at $0<x<\pi$, $y=0$ in the interior-edge sense:Subtracting $1$ from the hyperbolic tangent leaves an absolutely convergent series. For the remaining part, the power series $-\log (1-z)={\sum }_{m\ge 1}{z}^m/m$ gives, by taking real parts at $z=r{e}^{2ix}$ and then $r\uparrow 1$,The same formula follows as the limit of the differentiated interior solution, since its leading multiplier is $e^{-2my}$. This establishes the boundary normal derivative used in b. It also explains why the failure of the corner compatibility condition for mixed boundary data affects first derivatives at the bottom corners without spoiling continuity of $u$ itself.

Subtract the constant steady state profile: set $v=T-T_0$. ThenThe last equation follows from Fourier's law and zero heat flux at the insulated end. For a nonzero separated solution $v=X(x)H(t)$, separation of variables givesTo see that $\mu >0$ for a nonzero eigenfunction, multiply the spatial equation by $X$ and use integration by parts:Both boundary terms vanish. If the right side were zero, $X$ would be constant and $X(0)=0$ would make it zero. Thus write $\mu ={\lambda }^2$ with $\lambda >0$. Solving gives $X=A\sin (\lambda x)$, and $X^{\prime }(L)=0$ requires $\cos (\lambda L)=0$. The mixed Dirichlet-Neumann modes on an interval therefore haveThe eigenvalues of $-d^2/d{x}^2$ are ${\mu }_n={\lambda }_n^2$; ${\lambda }_n$ are their positive square roots. By superposition, with coefficients for which the resulting series and its required derivatives converge at positive times,The eigenfunctions are orthogonal on $[0,L]$ and have squared norm $L/2$. Expanding the initial excess temperature $v(x,0)=T_{0}x/L$ givesThus the requested temperature isAt the insulated endpoint $\sin ({\lambda }_{n}L)=(-1{)}^{n-1}$ cancels the alternating coefficient sign, leavingEvery summand is positive. The endpoint temperature never drops below $T_0$; it decreases towards $T_0$ as $t\to \infty$ and starts at $T(L,0)=2T_0$.

For completeness, the Sturm-Liouville eigenfunction expansion gives the initial profile in the $L^2$ sense. Here the coefficients are absolutely summable, so the Weierstrass M-test also gives uniform convergence on $[0,L]\times [0,\infty )$, and the initial values are recovered everywhere. For positive times the exponential factors justify all required termwise derivatives, giving a classical solution. There is, however, an initial incompatibility with a Neumann boundary condition at $(L,0)$: the initial slope is $T_0/L$, while the insulated Neumann boundary condition requires zero slope for $t>0$. The temperature is continuous to the initial edge, but $T_x$ cannot extend continuously to that corner. One must not differentiate the initial series term by term at $x=L$.

The trigonometric identityshows that only angular Fourier modes $1$ and $3$ are required. Differentiating the harmonic expansion on a disk at $r=1$ multiplies the coefficient of each nonconstant Fourier mode by $n$. Thus $B_1=3/4$ and $3B_3=-1/4$, givingEach term is a regular harmonic function, and the radial derivative isThe Neumann compatibility condition for Laplace's equation holds becauseas is necessary by the divergence theorem applied to the gradient of a harmonic function.

The additive constant is undetermined. This Neumann boundary condition is unchanged by $T\mapsto T+C$, so the paper's blanket uniqueness assertion requires a normalization in this case. The uniqueness of the Neumann problem gives uniqueness up to a constant; prescribing $T(0)=0$, or zero mean temperature, selects the displayed answer with $C=0$.

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.

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 a short string element of length $\Delta x$, the small-slope approximation makes the upward components of tension at its two ends $T{y}_x(x+\Delta x,t)$ and $-T{y}_x(x,t)$. Its mass is $\rho \Delta x$. Newton's second law givesDivide by $\Delta x$ and take the limit to obtain the wave equation on a string,Here $\rho$ is linear mass density, and the derivation assumes small transverse slopes, uniform tension, and a flexible string.

At the right endpoint the string pulls the ring toward the interior. Its vertical force is therefore $-T{y}_x(L,t)$ to first order; the frictionless wire supplies no vertical force. The ring obeysThis is the dynamic endpoint law for a string with an endpoint mass.

For a separated oscillatory solution, write $y=X(x)q(t)$ with $q^{\prime \prime }=-{\omega }^{2}q$ and $X^{\prime \prime }+(\omega /c{)}^{2}X=0$. The fixed left end removes the spatial cosine, so the nonzero normal modes haveThe endpoint law gives $M{\omega }_n^2\sin ({\omega }_{n}L/c)=T({\omega }_n/c)\cos ({\omega }_{n}L/c)$. Since $\sin ({\omega }_{n}L/c)=0$ would also force the cosine to vanish, it can be divided out for a nonzero mode. Using $T=\rho c^2$ givesFor physical $M\ge 0$, these give all positive modal frequencies. There is no nonzero static linear mode: the left value and right slope conditions would force it to vanish. Positivity also follows by integration by parts from

For $M=0$, the right endpoint is free, with the Neumann boundary condition $y_x(L,t)=0$. At $L=\pi$, the allowed wave numbers are $k_n=n+1/2$, $n\ge 0$, and ${\omega }_n=c{k}_n$. Zero initial displacement removes every cosine-in-time coefficient. The initial velocity contains only the modes $k=1/2$ and $k=5/2$, each with coefficient one. Dividing those velocity coefficients by their angular frequencies givesAt $x=0$ each term vanishes, at $x=\pi$ each spatial derivative vanishes, and at $t=0$ the displacement and its time derivative reproduce the two initial conditions.

For the heat equation ${\Theta }_t=\kappa {\Theta }_{xx}$, separation of variables with homogeneous Neumann boundary condition at both ends gives spatial modes $\cos (n\pi x/L)$ and decay factors $e^{-\kappa (n\pi /L{)}^{2}t}$. The $n=0$ mode is constant and does not decay.

The initial profile already contains only two such modes:Evolving each mode therefore yieldsThe endpoint derivatives vanish, and substitution checks both the equation and the initial profile. The spatial integral remains $L{T}_0/2$: no heat escapes through insulated ends, so the limiting uniform temperature is $T_0/2$.

The wave equation is a homogeneous linear partial differential equation. Consequently, if ${\psi }_1$ and ${\psi }_2$ solve it, every constant linear combination $A{\psi }_1+B{\psi }_2$ also solves it. This is the superposition principle, because partial differentiation distributes over the linear combination.

The same statement holds for the homogeneous Neumann boundary conditions. Initial conditions combine in the same way; an arbitrary combination need not retain a fixed nonzero initial displacement.

The equation as printed requires $v\ne 0$; its sign does not affect the result. Seek separated solutions $\psi =X(x)T(t)$. Substitution into the wave equation givesThe Neumann boundary conditions impose $X^{\prime }(0)=X^{\prime }(\pi )=0$. Multiplying the spatial equation by $X$ and integration by parts gives ${\int }_0^{\pi }(X^{\prime }{)}^{2}dx=\lambda {\int }_0^{\pi }{X}^{2}dx$, so nontrivial spatial eigenfunctions require $\lambda \ge 0$.

For $\lambda =0$, $X$ is constant and $T=A_0+B_{0}t$. For $\lambda =k^2>0$, $X=C\cos kx+D\sin kx$. The first boundary condition gives $D=0$, and the second gives $\sin k\pi =0$. Thus $k=n$ is a positive integer and the spatial eigenfunctions are $\cos nx$. Their time dependence is $A_n\cos (nvt)+B_n\sin (nvt)$. By superposition,The zero initial velocity sets $B_0=B_n=0$. The trigonometric identity gives the initial profile ${\cos }^2(x/2)=(1+\cos x)/2$, so only the constant mode and the first cosine mode occur. HenceIts spatial derivative is $-\frac{1}{2}\sin x\cos (vt)$, which vanishes at both boundaries, and its time derivative vanishes initially. Thus it satisfies all the required data.

For the two travelling profiles, the initial conditions implySince $v\ne 0$, $g-f$ is constant. One may therefore choosewhere the constants cancel in the sum. The D'Alembert formula then givesThis agrees exactly with the restricted-domain solution from part (b). The whole-line cosine profile already has zero spatial derivative at both $x=0$ and $x=\pi$ for every time, so its restriction satisfies the Neumann boundary conditions without further modification.

Boundary form. For real functions and $\mathcal{L}y=-(p{y}^{\prime }{)}^{\prime }+qy$, integration by parts givesThe integral terms containing $p{y}_1^{\prime }{y}_2^{\prime }$ cancel, as do the $q$ terms. This is the Lagrange identity for a Sturm-Liouville operator. The boundary conditions must make this boundary form zero for every admissible pair. Several suitable choices are homogeneous Dirichlet boundary conditions at both endpoints; homogeneous Neumann boundary conditions at both endpoints; one of each; or separated homogeneous Robin boundary conditionswhere each fixed real pair $(\alpha ,\beta )$ is nonzero. At each endpoint the two boundary data vectors are then proportional, making their antisymmetric product zero. Another choice is periodic data $y(a)=y(b)$ and $p(a)y^{\prime }(a)=p(b)y^{\prime }(b)$, which cancel the two endpoint contributions. These domains give the required self-adjoint differential operator in the regular setting.

Eigenfunctions. Set $x=\cos \theta$, $0\le \theta \le \pi$, and $Y(\theta )=y(\cos \theta )$. The chain rule givesThe Chebyshev differential equation therefore becomes $Y^{\prime \prime }+n^{2}Y=0$, with $Y(0)=Y(\pi )=0$. For a positive integer $n$, the first condition excludes the cosine term and the second is satisfied by $\sin (n\theta )$. ThusThere is no nonzero $n=0$ eigenfunction: the transformed solution is linear in $\theta$ and vanishes at both endpoints only if identically zero. Negative $n$ merely repeats a positive mode with the opposite sign, so it is excluded from the indexing of an independent eigenbasis.

Weight and normalization. Multiplying the differential equation by $-1/\sqrt{1-x^2}$ puts it into Sturm-Liouville form:Hence $p=\sqrt{1-x^2}$, $q=0$ and $w=(1-x^2{)}^{-1/2}$. Under $x=\cos \theta$, the measure $w(x)dx$ becomes $-d\theta$, soThereforeThese endpoints are singular because $p$ vanishes and $w$ diverges there, but the weight is integrable. Moreover $p{y}_n^{\prime }=-N_{n}n\cos (n\theta )$ remains finite, so the vanishing values of $y_n$ make the boundary form zero. These are the Dirichlet eigenfunctions of the Chebyshev differential operator, not the first-kind Chebyshev polynomials $\cos (n\arccos x)$, which have nonzero endpoint values.

Adapted homogeneous solutions. First assume real $\kappa \ne 0$. The functions $u(x)=\cosh (\kappa x)$ and $v(x)=\cosh (\kappa (1-x))$ solve $z^{\prime \prime }-{\kappa }^{2}z=0$ and obey $u^{\prime }(0)=v^{\prime }(1)=0$. Their Wronskian isIt is independent of $x$ and nonzero.

Green function. For $0<\xi <1$, continuity at $x=\xi$ and the jump condition $G_x({\xi }^{+},\xi )-G_x({\xi }^{-},\xi )=1$ follow by matching the Dirac delta function source. In particular a jump in $G$ itself would produce a derivative of the Dirac delta function, which is absent. Write $G=A(\xi )u(x)$ for $x<\xi$ and $G=B(\xi )v(x)$ for $x>\xi$. Solving the continuity and derivative-jump equations gives $A=v(\xi )/W$, $B=u(\xi )/W$. Hence the Neumann Green function for the modified Helmholtz equation on an interval isThe two pieces coincide at the source. Their derivative difference is $W/W=1$ and their endpoint derivatives vanish. The same expression has continuous one-sided limits as the source approaches an endpoint; the interior-source construction is what is needed in the integral solution.

Solution for a general source. By superposition, $y(x)={\int }_0^{1}G(x,\xi )f(\xi )d\xi$. Splitting the integral at $\xi =x$ givesNo nonzero homogeneous solution satisfies both endpoint conditions when $\kappa \ne 0$: the condition at zero eliminates $\sinh (\kappa x)$, and the condition at one then eliminates $\cosh (\kappa x)$. Thus the Green function gives the unique solution.

The printed zero-parameter case. The statement permits real $\kappa$, but $\kappa =0$ is exceptional. Integrating a putative equation $G^{\prime \prime }=\delta (x-\xi )$ would give $G^{\prime }(1)-G^{\prime }(0)=1$, contrary to the two zero derivatives. Thus this Green function does not exist at zero. For $y^{\prime \prime }=f$ with homogeneous Neumann boundary conditions, solvability instead requires ${\int }_0^{1}f=0$, and a compatible solution has an arbitrary additive constant. This is the Fredholm solvability condition associated with the constant zero mode.

Explicit forcing. For $f(x)=x$ and $\kappa =1$, the particular solution is $-x$. Write $y=C\cosh x+D\sinh x-x$. Then $y^{\prime }(0)=0$ gives $D=1$ and $y^{\prime }(1)=0$ gives $C=(1-\cosh 1)/\sinh 1=-\tanh (1/2)$. ThereforeDirect differentiation gives $y^{\prime }=\cosh (x-1/2)/\cosh (1/2)-1$, which vanishes at both endpoints, and $y^{\prime \prime }-y=x$, verifying the solution independently of the Green function integration.