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Let $Ay=(p{y}^{\prime }{)}^{\prime }+qy$ have real eigenfunctions $u_n$ obeying $A{u}_n={\lambda }_{n}w{u}_n$ and homogeneous Dirichlet boundary conditions, with $N_n={\int }_a^{b}w{u}_n^2>0$. Suppose $Ay+\mu wy=f$, with possibly nonzero endpoint values. The Lagrange identity for a Sturm-Liouville operator givesDivision by $({\lambda }_n+\mu )N_n$ gives the displayed coefficient whenever the denominator is nonzero. At resonance in a differential equation, equality of the forcing projection and the boundary term is the Fredholm solvability condition. The Sturm-Liouville eigenfunction expansion represents the interior solution in the weighted inner product sense; its zero endpoint terms do not reproduce nonzero boundary traces by pointwise substitution. A smooth lift of the endpoint data reduces the calculation to homogeneous Dirichlet boundary conditions.

For $\ell >0$, the logarithmic change of variable in a Sturm-Liouville problem sends this equation to $-Y^{\prime \prime }-\alpha Y=1$ on $(0,\ell )$. The normalized eigenfunctions are $\sqrt{2/\ell }\sin (n\pi t/\ell )$, with eigenvalues ${\lambda }_n=(n\pi /\ell {)}^2-\alpha$. If every ${\lambda }_n$ is nonzero, the Sturm-Liouville eigenfunction expansion givesIndeed the constant function has sine coefficients $2[1-(-1{)}^n]/(n\pi )$; division by each eigenvalue solves the equation in the operator domain. The series is uniformly convergent, with coefficients $O(n^{-3})$, while applying the differential operator is justified in the weighted square-integrable sense. At a zero eigenvalue the solvability condition at a Sturm-Liouville eigenvalue must instead be checked: an odd resonant index makes the constant forcing incompatible, whereas an even resonant index permits solutions with an arbitrary multiple of that zero mode.

A Neumann eigenfunction is an eigenfunction of a spatial differential operator whose normal derivative vanishes on the boundary. A Sturm-Liouville eigenfunction expansion in these functions gives a heat kernel for homogeneous Neumann boundary conditions.

For $L>0$ and $t>0$, the Sturm-Liouville eigenfunction expansion for $u_t=u_{xx}$ on $(0,L)$ with homogeneous Neumann boundary conditions givesThe constant eigenfunction mode preserves the spatial integral. This is the $\beta =0$ limit of the weighted Neumann heat kernel with constant drift, and its boundary inputs have the signs in the Neumann boundary-forcing formula.

Use the normalization from part (ii), and writeThe Sturm-Liouville eigenfunction expansion coefficient is $c_k=N_k^{-1}{\int }_1^{e}ry{\hat{y}}_{k}dx$. Although ${\hat{y}}_k$ has zero endpoint values, $y$ generally does not. Retain the endpoint term in the Lagrange identity for a Sturm-Liouville operator:The positive sign comes from subtracting the contribution at $x=1$. The differential equation also gives $\hat{L}y=r{e}^x-(\beta -1)ry$. Substitution into the preceding identity yieldsHence, away from resonance in a differential equation,This is the effect of boundary data in a Sturm-Liouville eigenfunction expansion: omitting the endpoint contribution would discard the dependence on $\alpha$.

If $\beta =1+n^2{\pi }^2$ for a positive integer $n$, the Fredholm solvability condition isWhen it fails, there is no solution. When it holds, $c_n$ is arbitrary and the other coefficients retain the displayed formula, giving infinitely many solutions differing by a multiple of ${\hat{y}}_n$. Sufficiency follows by subtracting the smooth lift $h(x)=\alpha x(1-\log x)$, which has the required endpoint values and satisfies $Lh=0$, and applying the Fredholm alternative to the resulting homogeneous Dirichlet boundary conditions.

Finally, for $\alpha \ne 0$, the displayed Sturm-Liouville eigenfunction expansion represents the solution on $1<x<e$ and in the weighted inner product sense. Every term is zero at $x=1$, so that series cannot give the nonzero prescribed value by substitution at the endpoint. The boundary trace must be imposed separately. Equivalently, expand $y-h$ with homogeneous Dirichlet boundary conditions and add $h$ back afterwards.

Use the weighted inner product $⟨u,v{⟩}_r={\int }_a^{b}r(x)u(x)v(x)dx$. For real eigenfunctions, integration by parts with the zero endpoint values givesThus $({\lambda }_n-{\lambda }_m)⟨y_m,y_n{⟩}_r=0$. The regular separated Sturm-Liouville problem has simple eigenvalues: for a fixed eigenvalue, prescribing $y(a)=0$ leaves only the initial flux $p(a)y^{\prime }(a)$ free, so uniqueness of the associated first-order system makes any two eigenfunctions proportional. Distinct eigenfunctions are therefore orthogonal, and choosing their signs and scales so that $\int r{y}_n^2=1$ givesThe positive energy form $\int (p{y}^{\prime 2}+q{y}^2)$ realizes $r^{-1}\mathcal{L}$ as a positive self-adjoint operator in this weighted space, with compact inverse on the bounded interval. The spectral theorem for compact self-adjoint operators supplies a complete eigenbasis. This is the standard regular Sturm-Liouville eigenfunction expansion, also valid with continuous $p$ by working with the continuous flux $p{y}^{\prime }$.

For a source $F$, expand a zero-endpoint solution of $(\mathcal{L}-\mu r)y=F$ in that basis. Multiplication by $y_n$ and unweighted integration giveSince $\mu$ is not an eigenvalue, every coefficient is determined. The spectral Green function of a regular Sturm-Liouville problem is consequentlyWeighted completeness means $r(x){\sum }_n{y}_n(x)y_n(\xi )=\delta (x-\xi )$ as a distribution, so applying $\mathcal{L}-\mu r$ to this series produces the required delta source. The source integration is with $d\xi$, explaining the absence of an additional factor $r(\xi )$ in the stated kernel.

For the subsequent operator $L=d^2/d{x}^2$ the sign is explicitly reversed: $y_n=\sqrt{2/\pi }\sin (nx)$ and ${\lambda }_n=-n^2$. To cover every real non-eigenvalue $\mu$, define the homogeneous solutionIt satisfies $S_{\mu }(0)=0$, $S_{\mu }^{\prime }(0)=1$ and $S_{\mu }^{\prime \prime }=\mu S_{\mu }$. Matching the two endpoint solutions continuously at $x=\xi$ with derivative jump $G_x({\xi }^{+},\xi )-G_x({\xi }^{-},\xi )=1$ yieldsThe denominator is nonzero precisely off the Dirichlet eigenvalues $\mu =-n^2$. In particular $G_0=-x_{<}(\pi -x_{>})/\pi$. At $x=\xi =\pi /2$, its spectral expansion givesThis odd reciprocal-square sum from a Dirichlet Green function proves $\boxed{\sum _{j=0}^{\infty }(2j+1{)}^{-2}={\pi }^2/8}$.

Set $y=e^{-x/2}v$, the Dirichlet gauge transform for constant drift. Direct differentiation givesThe homogeneous endpoint conditions on $y$ become $v(0)=v(1)=0$. Thus the eigenvalue equation is $v^{\prime \prime }=(\lambda +1/4)v$. To exclude other branches without assuming the eigenvalue real, multiply by $\bar{v}$ and integrate by parts:for a nonzero eigenfunction. Therefore $\lambda +1/4=-k^2$ with $k>0$. The boundary conditions in $v=A\sin (kx)+B\cos (kx)$ give $B=0$ and $k=n\pi$, $n\ge 1$. HenceEquivalently, $L[y]=\lambda y$ is the Sturm-Liouville form $(e^x{y}^{\prime }{)}^{\prime }=\lambda e^{x}y$, with positive weight $e^x$. Its orthogonality relation isIt is the weighted inner product, not the unweighted one, that makes these eigenfunctions orthogonal.

Expand $x{e}^{-x/2}={\sum }_n{b}_n{y}_n$ with the coefficients from part a. Completeness and the nonzero eigenvalues give the Sturm-Liouville eigenfunction expansionThe coefficients are $O(n^{-3})$, so the function series converges uniformly and vanishes at both endpoints. Its first derivative also converges uniformly. Applying the transformed operator in the square-integrable sense gives the part-a expansion of $x$; equivalently $v^{\prime \prime }-v/4=x$ on the interior. Integrating this continuous right-hand side gives the classical twice-differentiable solution there. No nonzero homogeneous solution satisfies both endpoints, so this solution is unique. The mismatch of the forcing's sine series at a single endpoint does not affect the interior equation or these boundary conditions.

Impermeability requires $\hat{w}(-1)=\hat{w}(1)=0$. Let $c=\omega /\alpha$ with real $\alpha \ne 0$, and set $a=|\alpha |$. If $c$ is nonreal, or real outside the channel, the Rayleigh equation for inviscid shear flow gives ${\hat{w}}^{\prime \prime }-a^2\hat{w}=0$ everywhere. The two wall conditions then force $\hat{w}=0$. In fact there is no nonzero globally smooth eigenfunction for any $c$: the complete family is the inviscid Couette continuous spectrum.

For each $\xi \in (-1,1)$, permit a localized vorticity sheet. Let $G_a$ be the Dirichlet Green function satisfying $(D^2-a^2)G_a(z,\xi )=\delta (z-\xi )$. An explicit generalized eigenfunction iswhere $z_{<}=\min (z,\xi )$ and $z_{>}=\max (z,\xi )$. It vanishes at both walls, is continuous at $z=\xi$, and has derivative jump $[{\partial }_z{G}_a{]}_{{\xi }^{-}}^{{\xi }^{+}}=1$. Consequentlyusing the Dirac delta multiplication identity. These are vorticity-sheet eigenfunctions of inviscid Couette flow, understood as generalized eigenfunctions, not as a discrete smooth Sturm-Liouville eigenfunction expansion.

To see completeness, define the vorticity variable $q=(D^2-a^2)w$. Its evolution is $({\partial }_t+i\alpha z)q=0$, so for any admissible initial vorticity $q_0$,The homogeneous Dirichlet problem for $D^2-a^2$ has only the zero solution, so this inversion reconstructs every initial vertical-velocity field in its usual function space. The generalized frequencies fill the interval with endpoints $\pm \alpha$; the endpoint values are understood as the closure of the continuous spectrum.

The drift makes the natural measure weighted. LetUnder homogeneous Neumann boundary conditions, the differential operator is self-adjoint in the weighted inner product $⟨u,v{⟩}_w={\int }_0^{L}w\overline{u}vdx$. Indeed integration by parts giveswhen the derivative boundary terms vanish. In particular its eigenvalues are nonpositive and it has the constant stationary eigenfunction. We first construct the weighted Neumann heat kernel with constant drift, then incorporate the two prescribed coordinate derivatives.

Put $b=\beta /2$. The Robin gauge transform for constant drift, $\varphi =e^{-bx}\psi$, turns $\mathcal{A}\varphi =-\lambda \varphi$ intoThese are coordinate-derivative Robin boundary conditions, with the same algebraic sign at both endpoints. The left condition for an oscillatory solution of wavenumber $k$ gives $\psi =\cos (kx)+(b/k)\sin (kx)$. The right condition then reduces to $(k+b^2/k)\sin (kL)=0$, giving $k_n=n\pi /L$. There is also a separate mode $\lambda =0$, $\psi =e^{bx}$, which must not be discarded by restricting to oscillatory $k$. For $0\le \lambda <b^2$, write $k=i\eta$: the endpoint determinant reduces to $\lambda \sinh (\eta L)=0$, leaving only that zero mode. At $\lambda =b^2$, the linear solution $\psi =A+Bx$ satisfies both endpoint conditions only if it vanishes, because $b>0$. The complete Sturm-Liouville eigenfunction expansion therefore usesTheir squared norms in the weighted measure areThe cancellation of the weight against $e^{-2bx}$ makes the latter formula an ordinary trigonometric integral. DefineFor $t>0$ this kernel is smooth in the interior. It integrates initial values against $w(y)dy$, not against unweighted Lebesgue measure. It is symmetric in $x,y$ as a weighted kernel, even though the drift operator does not look symmetric in an unweighted inner product.

To determine the boundary forcing, set $a_n(t)=⟨{\varphi }_n,q(\cdot ,t){⟩}_w/N_n$. Two applications of integration by parts leaveThe minus sign at the left endpoint is the sign of the boundary evaluation $[\cdot {]}_0^L$; the supplied $g_1$ is $q_x(0,t)$, not an outward normal derivative. Solving this linear differential equation by an integrating factor and recombining the modes gives the requested space-time integral representation:This is the Neumann boundary-forcing formula. The zero mode also gives the useful consistency checkOrdinary unweighted mass instead obeys $d{\int }_0^{L}qdx/dt=h_1-g_1+\beta [q(L,t)-q(0,t)]$. Replacing the weighted measure by $dy$ or deleting the constant mode would therefore give an incorrect solution.

There is a subtle point when checking nonzero derivative data: every homogeneous eigenfunction has zero endpoint derivative, but differentiating the boundary-forcing sum and its time integral term by term at an endpoint is not valid. The kernel is singular as $t-s\downarrow 0$, and the prescribed derivatives are interior limits of the full solution. An explicit Laplace transform form verifies those limits without this interchange and gives an alternative purely contour-integral version.

For $\mathrm{Re}p>0$, set $\kappa =\sqrt{p+b^2}$ using the principal square root, and defineBoth solve $(p-\mathcal{A})f=0$, with $u_p(0)=v_p(L)=1$, $u_p^{\prime }(0)=v_p^{\prime }(L)=0$. Their weighted Wronskian is the nonzero constantHence the resolvent kernel for Neumann advection-diffusion on an interval isThe first derivative in $x$ has jump $-e^{-\beta y}$, giving $(p-{\mathcal{A}}_x){\mathcal{R}}_p=e^{-\beta y}{\delta }_y$ in the sense of distributional derivatives. If $G(p)={\int }_0^T{e}^{-ps}{g}_1(s)ds$ and $H(p)={\int }_0^T{e}^{-ps}{h}_1(s)ds$, an equivalent transformed solution isIn fact $v_p^{\prime }(0)=-D_p$ and $u_p^{\prime }(L)=e^{-\beta L}{D}_p$, so $Q_x(0,p)=G(p)$ and $Q_x(L,p)=H(p)$ exactly. The inverse Bromwich contour integral of $e^{pt}Q(x,p)/(2\pi i)$ is an alternative final answer. Its poles at $p=0$ and $p=-{\lambda }_n$ reproduce the constant and decaying modes of $K_{\beta }$, respectively; apparent square-root singularities in the normalized hyperbolic factors are removable. This also independently checks the kernel normalization and boundary signs.

As in question 1, the printed dot on $q_0$ denotes its spatial derivative: the two corner conditions are $q_0^{\prime }(0)=g_1(0)$ and $q_0^{\prime }(L)=h_1(0)$. The initial eigenfunction expansion converges to $q_0$ in the weighted space and, for the stated smooth compatible data, has the appropriate classical initial and boundary limits. Differences of solutions with zero data satisfy $\frac{1}{2}d\Vert q{\Vert }_w^2/dt=-{\int }_0^{L}w|q_x{|}^{2}dx$, proving uniqueness in the usual regular solution class. In the limit $\beta \to 0$, $N_0\to L$, ${\varphi }_n\to \cos (n\pi x/L)$, and the ordinary Neumann heat kernel on an interval and its boundary signs are recovered.

The Dirichlet Laplacian eigenfunctions and positive Dirichlet Laplacian eigenvalues on the interval of length two areA Sturm-Liouville eigenfunction expansion gives $u={\sum }_j{q}_j(t){\varphi }_j$, with $q_j^{\prime \prime }+({\lambda }_j-\alpha )q_j=0$. Thus all normal modes oscillate precisely when $\alpha <{\lambda }_1$.

To justify a spatially uniform bound for arbitrary appropriate data, assume the usual wave equation energy class $u(0)\in H_0^1(-1,1)$ and $u_t(0)\in L^2(-1,1)$, or smoother compatible data. The energy method conservesThe sharp Poincaré inequality is $\Vert u{\Vert }_{L^2}^2\le {\lambda }_1^{-1}\Vert u_x{\Vert }_{L^2}^2$. For $0\le \alpha <{\lambda }_1$, this makes $E$ coercive with constant $1-\alpha /{\lambda }_1$; for $\alpha <0$, coercivity is immediate. The one-dimensional Sobolev representative satisfies $|u(x,t)|\le \sqrt{2}\Vert u_x(t){\Vert }_{L^2}$ by the fundamental theorem of calculus and the Cauchy-Schwarz inequality. Hence the conserved energy bounds $u$ uniformly in both variables.

At $\alpha ={\lambda }_1$, the smooth solution $u=t{\varphi }_1$ is unbounded. At $\alpha >{\lambda }_1$, $u=e^{\sqrt{\alpha -{\lambda }_1}t}{\varphi }_1$ is unbounded. ThereforeThe bound depends on the initial data; it is not a single bound for all arbitrarily rescaled solutions. The energy-class hypothesis supplies pointwise meaning and excludes undefined rough-data interpretations.

Use the weighted Neumann heat kernel with constant drift. The spatial operator has the self-adjoint formso its homogeneous Neumann boundary conditions give a regular Sturm-Liouville problem in the weighted inner productThe constant eigenfunction is ${\varphi }_0=1$, with eigenvalue ${\lambda }_0=0$ and squared norm $N_0=(e^{\beta L}-1)/\beta$. For $k\ge 1$, put $q_k=k\pi /L$. Substitution of $\varphi =e^{-\beta x/2}v$ in $\mathcal{L}\varphi =-\lambda \varphi$ gives $v^{\prime \prime }+(\lambda -{\beta }^2/4)v=0$ and the Robin boundary conditions $v^{\prime }=\beta v/2$ at both endpoints. HenceIndeed,The Sturm-Liouville eigenfunction expansion is complete and orthogonal. Its heat kernel, relative to the measure $e^{\beta y}dy$, isFor a coefficient $a_k(t)=⟨{\varphi }_k,u(\cdot ,t){⟩}_{\beta }/N_k$, two applications of integration by parts yieldSolving these scalar linear differential equations and summing gives the Neumann boundary-forcing formula:This depends only on the given data. In particular, the zero eigenvalue retains the changing weighted mean:The endpoint derivatives of the boundary integral are understood as interior limits. Differentiating each homogeneous Neumann eigenfunction at the endpoint before summing would incorrectly discard the prescribed boundary forcing: the short-time heat kernel makes that interchange invalid. The stated compatibility conditions give the classical boundary traces.

Set $y=e^{-x}u$. Thenso the eigenvalue equation becomesThe Dirichlet eigenvalues and corresponding eigenfunctions are thereforeThey form an infinite discrete increasing set. Their weighted inner products satisfyThus the Sturm-Liouville eigenfunction expansion coefficient is

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 $x=L$.

Use separation of variables with $u=W(x)T(t)$ and choose $W^{\prime \prime }/W=T^{\prime }/T=k$. Then $T=e^{kt}$ and $W^{\prime }(0)=-W(0)$, $W^{\prime }(1)=-W(1)$.

If $k=0$, write $W=A+Bx$. The first condition gives $B=-A$, and the second gives $A+2B=0$. Together they force $A=B=0$, so this case contributes only the zero solution.

If $k={\alpha }^2>0$, write $W=A\cosh (\alpha x)+B\sinh (\alpha x)$. The first condition gives $B=-A/\alpha$. The second becomesA nonzero solution therefore requires $\alpha =1$, giving $W\propto e^{-x}$ and time factor $e^t$.

If $k=-{\beta }^2<0$, write $W=A\cos (\beta x)+B\sin (\beta x)$. Now $B=-A/\beta$, and the remaining condition is $-A(\beta +{\beta }^{-1})\sin \beta =0$. Thus $\beta =n\pi$, $n\ge 1$. Choosing a convenient amplitude gives $W_n=n\pi \cos (n\pi x)-\sin (n\pi x)$. The complete Sturm-Liouville eigenfunction expansion is consequentlyThese are self-adjoint separated Robin boundary conditions: the boundary form $[f^{\ast }{g}^{\prime }-f^{\prime \ast }g{]}_0^1$ vanishes because both functions satisfy $y^{\prime }=-y$ at each end. The regular interval problem has a complete orthogonal eigenbasis; the calculation includes its one negative eigenvalue for $-D^2$, not only its positive ones.

To project the initial data, calculate the squared norms:The second identity uses the half-unit integrals of ${\cos }^2(n\pi x)$ and ${\sin }^2(n\pi x)$ and the zero integral of their product. By orthogonal projection,For square-integrable $f$, the initial expansion is understood in that norm. Classical agreement of boundary derivatives at $t=0$ requires compatible smoother data.