Insert the inverse Fourier transforms of $f$ and its complex conjugate, initially under enough decay to justify interchanging integrals:The middle step uses the stated Dirac delta distribution identity, and the last uses its sampling property. This is the Parseval identity for the chosen normalization. Approximation extends the result to square-integrable functions, including the function used below.

With the paper's Fourier transform convention, the Fourier inversion theorem givesThis holds at continuity points under the usual inversion hypotheses; at a jump with one-sided limits it gives their mean. The factor $1/(2\pi )$ and the sign change in the exponential are part of this normalization.

The original PDF has the absolute-value hypothesis $|g(\xi )|<C{e}^{-s\xi }$, which is missing from the local OCR. From the Green's function formula and $|\sin |\le 1$,Because $s>a>0$, the upper bound tends to zero. The squeeze theorem proves $\boxed{y(x)\to 0\text{ as }x\to \infty }$.

For the zero-data initial value problem, choose a causal Green function satisfying $L_{x}G(x,\xi )=\delta (x-\xi )$ and $G=0$ for $x<\xi$. Continuity at $x=\xi$ prevents a derivative of the Dirac delta distribution, while integration across the source givesFor $x>\xi$, part (i) gives the shifted homogeneous solutions. The zero value at the join removes the cosine term, and the unit derivative jump fixes the sine coefficient. Therefore the causal Green function of a damped oscillator isEquivalently, multiply the first expression by the Heaviside step function $H(x-\xi )$. The required solution isIts initial value and first derivative vanish: the interval has zero length at $x=0$ and the kernel vanishes at its upper limit. Differentiating twice shows that the upper-limit contribution to $y^{\prime \prime }$ is precisely $g(x)$, while the remaining homogeneous kernel terms cancel in $Ly$.

For an exponential trial $y=e^{rx}$, the characteristic equation isIts distinct complex roots are $r=-a\pm ib$. Thus the general real solution of the homogeneous second-order linear differential equation isThe two displayed solutions have nonzero Wronskian $b{e}^{-2ax}$ and hence linear independence. The same formula permits complex solutions when $C_1,C_2$ are complex.

Let $f(x)=\rho (x,0)$. The first normalized Fourier sine series coefficient isStrict positivity follows because $f\ge 0$, has positive total mass, and $\sin (\pi x/L)>0$ everywhere inside the interval. In particular, there is no cancellation that could remove the slowest term. This is the positivity of the first Dirichlet reaction-diffusion coefficient.

For finite initial mass $M={\int }_0^{L}f(x)dx$, all coefficients satisfy $|a_n|\le \sqrt{2/L}M$. The eigenfunction expansion therefore gives the uniform boundThe positive first profile thus controls the long-time behavior. We recover exactly the threshold $\boxed{{\gamma }_c=D{\pi }^2/L^2}$, with decay below it and unbounded positive growth at every interior point above it. The finite-mass condition is the usual meaning of an initial density profile here; standard positive-time smoothing gives the same conclusion in the usual broader admissible data classes.

The smallest decay-rate eigenvalue is ${\lambda }_1=D{\pi }^2/L^2-\gamma$, so the threshold isBecause $0<x_0<L$, $a_1>0$. For every fixed interior point $0<x<L$, the first sine eigenfunction is positive and the remaining terms have strictly faster decay relative to it. HenceOne can justify the relative-error limit directly: after factoring out the first exponential, the absolute remainder is bounded by $(2{\rho }_0/L){\sum }_{n\ge 2}{e}^{-D{\pi }^2(n^2-1)t/L^2}$, which tends to zero. Thus $\boxed{\rho \to 0\text{ if }\gamma <{\gamma }_c}$, and $\boxed{\rho (x,t)\to +\infty \text{ for each interior }x\text{ if }\gamma >{\gamma }_c}$. The boundary values stay zero. At equality the first term is stationary and the density approaches its positive sine profile.

By orthogonality and the sampling property of the Dirac delta distribution,Consequently the reaction-diffusion equation solution isFor every $t>0$ this eigenfunction expansion is smoothed by the Gaussian decay in $n$. As $t\downarrow 0$, it recovers the specified initial condition in the sense of distributions.

Apply separation of variables with $\rho (x,t)=X(x)T(t)$ and temporal factor $T=e^{-\lambda t}$. The spatial eigenvalue equation isThe nontrivial solutions of these Dirichlet boundary conditions are sine eigenfunctions. Normalizing in the usual inner product on $(0,L)$ giveswith ${\int }_0^L{\psi }_n{\psi }_{m}dx={\delta }_{nm}$. The superposition principle and the Fourier sine series then giveThe constants are the initial-data coefficients $a_n={\int }_0^L\rho (x,0){\psi }_n(x)dx$. Here physical diffusion means $D>0$.

Put $s_n(x)=\sin (q_{n}x)$ and $s_m(x)=\sin (q_{m}x)$. Their differential equations are $-s_n^{\prime \prime }=q_n^2{s}_n$ and $-s_m^{\prime \prime }=q_m^2{s}_m$. Subtract the two cross-multiplied equations and use integration by parts:The lower boundary term is zero because both functions vanish there. At the upper endpoint, the Robin boundary condition gives $s_j^{\prime }(1)=s_j(1)/\alpha$, soSince distinct positive roots have $q_n^2\ne q_m^2$, we concludeThis orthogonality holds for every $\alpha >0$, even when the sine family requires the additional nonpositive mode for completeness as explained in part (c).

Write the endpoint condition in its linear Robin boundary condition form $y^{\prime }(1)=y(1)/\alpha$. For a positive eigenvalue $\lambda =q^2$, the left Dirichlet boundary condition gives $y=C\sin (qx)$, and the right condition givesFor each positive root $q_n$, the squared norm isConsequently the sine coefficients in any complete expansion areWhen $\alpha >1$, these are all modes and the Sturm-Liouville eigenfunction expansion gives exactly the stated sine series, with the usual convergence interpretation. Positivity of all eigenvalues follows from integration by parts and the Cauchy-Schwarz inequality:for a nonzero admissible $y$, since $y(1)={\int }_0^1{y}^{\prime }dx$.

However, the original PDF states only $\alpha >0$, and with that range its sine-only completeness claim is false. The missing modes are as follows. At $\alpha =1$, $\lambda =0$ admits $y=x$, so the correct expansion isFor $0<\alpha <1$, a negative eigenvalue $\lambda =-{\kappa }^2$ admits $y=\sinh (\kappa x)$, withThere is exactly one positive root: $\tanh \kappa /\kappa$ decreases strictly from $1$ to zero. For example, its derivative has the sign of $\kappa {\mathrm{sech}}^2\kappa -\tanh \kappa <0$, because $\sinh \kappa \cosh \kappa >\kappa$. Thus the corrected expansion isThese are the nonpositive modes in a Dirichlet-Robin interval expansion. As an explicit counterexample to the printed claim, at $\alpha =1$ take $f(x)=x$: it satisfies both boundary conditions, and each positive root obeysbecause $\tan q_n=q_n$. Thus all $a_n$ vanish despite $f\ne 0$. Likewise $f=\sinh (\kappa x)$ is a counterexample for $\alpha <1$, by orthogonality of Sturm-Liouville eigenfunctions with distinct eigenvalues. The formulas above give the full solution without silently assuming away the omitted modes.

The Sturm-Liouville eigenfunction expansion of a suitable real function $f$ isComparing this identity with the sampling property of the Dirac delta distribution gives the completeness kernelThis is an identity of distributions, rather than a pointwise-convergent series of ordinary functions.

Assuming zero is not an eigenvalue, define the spectral Green function of a regular Sturm-Liouville problem byApplying ${\mathcal{L}}_x$ gives $w(x){\sum }_n{y}_n(x)y_n(\xi )=\delta (x-\xi )$: multiplying the preceding completeness kernel by $w(x)/w(\xi )$ leaves the Dirac delta distribution unchanged. Each term also obeys the same homogeneous boundary conditions. There is no extra weight factor in $G$ for the ordinary source measure $d\xi$.

If zero is an eigenvalue, the inverse Green's function does not exist on all forcing functions: the forcing must obey the solvability condition at a Sturm-Liouville eigenvalue. Omitting the zero mode instead constructs a reduced inverse, with its projection removed from the right-hand side. This exception occurs in part (c) at $\alpha =1$.

Use the standard regular Sturm-Liouville problem hypotheses: positive weight $w$, real coefficients, and real self-adjoint boundary conditions. The positive weight and boundary-domain assumptions are implicit in the requested real orthonormal basis; real differential coefficients alone would not guarantee them for arbitrary complex boundary conditions.

Multiply the eigenfunction equation by $\overline{y_n}$ and integrate. By self-adjointness, the number ${\int }_a^b\overline{y_n}\mathcal{L}{y}_{n}dx$ is real, whereasThe second integral is strictly positive, so $\boxed{{\lambda }_n\in \mathbb{R}}$. Because the coefficients and boundary conditions are real, the real and imaginary parts of any complex eigenfunction separately obey the same equation and conditions. At least one is nonzero, so a real eigenfunction can be chosen. More generally, each eigenspace has a real basis, which can be orthonormalized using the weighted inner product.

The orthogonality of Sturm-Liouville eigenfunctions, with normalization chosen to one, isFor the real choice, omit the complex conjugate.

Choose the symmetric tensor fieldThe product rule and equality of mixed partial derivatives giveThis is the Euclidean scalar-gradient stress identity. Substitute it into part (e), and use ${\Sigma }_{ki}{\hat{n}}_k=({\partial }_n\psi ){\partial }_i\psi -\frac{1}{2}|\nabla \psi {|}^2{\hat{n}}_i$, where ${\partial }_n\psi =\hat{\mathbf{n}}\cdot \nabla \psi$. Reassembling the components gives the required pure surface integral:Sufficient smoothness of $\psi$ is understood for the displayed derivatives and applications of the divergence theorem.

Apply the product rule to the tensor field, without assuming any symmetry of $\Sigma$:Integrating, the divergence theorem converts each of the two divergences into a boundary term:The repeated indices are summed according to Einstein summation convention, while $i$ labels the component of the resulting vector identity.

Use Einstein summation convention and write ${\partial }_j={\nabla }_j$. Since ${\partial }_j{r}_i={\delta }_{ij}$, where ${\delta }_{ij}$ is the Kronecker delta, the product rule givesThis proves $\boxed{P_i={\partial }_j(r_i{w}_j)-w_i}$. Now take $w_j={\partial }_j\psi$. Integrating this identity and applying the divergence theorem to the first term givesPart (b) changes the last volume integral of the gradient into a surface integral, yieldingHere $d{S}_i={\hat{n}}_{i}dS$; the Laplacian is ${\nabla }^2\psi ={\partial }_j{\partial }_j\psi$.

The divergence of a cross product and $\nabla \times \mathbf{a}=0$ give $\nabla \cdot (\mathbf{a}\times \mathbf{A})=-\mathbf{a}\cdot (\nabla \times \mathbf{A})$. On the boundary the scalar triple product gives $(\mathbf{a}\times \mathbf{A})\cdot \hat{\mathbf{n}}=-\mathbf{a}\cdot (\hat{\mathbf{n}}\times \mathbf{A})$. Thus the divergence theorem givesBecause $\mathbf{a}$ is arbitrary, the curl identity follows:

Since $\mathbf{a}$ is constant, the product rule gives $\nabla \cdot (\mathbf{a}\psi )=\mathbf{a}\cdot \nabla \psi$. Applying the divergence theorem therefore givesThis holds for every constant vector $\mathbf{a}$, so each component of the difference vanishes. Consequently the gradient has the boundary representation

For a continuously differentiable vector field $\mathbf{u}$ on a bounded region $V\subset {\mathbb{R}}^3$ with piecewise smooth closed boundary $S=\partial V$, the divergence theorem statesHere $\nabla \cdot \mathbf{u}$ is its divergence, $dV$ is the volume element, $dS$ is the surface area element, $\hat{\mathbf{n}}$ is the outward unit normal, and $d\mathbf{S}=\hat{\mathbf{n}}dS$. Thus a volume integral of the divergence equals the outward flux through the boundary.

The group is the direct product of groups ${\mathbb{Z}}_p\times {\mathbb{Z}}_q$. Repeatedly adding $(1,1)$ returns to $(0,0)$ for the first time when both $p$ and $q$ divide the step number. Its order of a group element is therefore $\mathrm{lcm}(p,q)$. If $p,q$ are coprime, this is $pq$, so it generates every element:Explicitly, $[t{]}_{pq}\mapsto ([t{]}_p,[t{]}_q)$ is the Chinese remainder theorem group isomorphism.

If $d=\gcd (p,q)>1$, every pair satisfiesThus every element order divides $pq/d<pq$, and no element can generate the whole group. HenceThis is the cyclicity of a product of two finite cyclic groups.