Assume first that $f$ and $g$ are integrable functions, so that Fubini's theorem justifies changing the order of the two integrals. In the defining Fourier transform of the convolution, put $y=x-x^{\prime }$:This proves the convolution theorem with the stated normalization:

For $n\ne 0$, split the Green's function integral at $s=x$. Writing $\ell (x)=x^{-n}-2^n$ and $r(x)=x^{-n}-1$ givesFor $n\ne -1$, the two integrals areSubstitution and simplification yieldAt $x=1$ both contributions vanish. At $x=1/2$ the bracket is $-1/2+(2^n-1)/[2(2^n-1)]=0$. The differential operator kills $x^{-n}-1$ and sends $x/(n+1)$ to $1/x$, so this also verifies the inhomogeneous equation.

The exceptional integer is $\boxed{n=-1}$, because then the integrals contain $s^{-1}$. In this caseand the Green's function integral becomesHere $L=d^2/d{x}^2$, and $u^{\prime \prime }=1/x$. Explicitly $u(1)=0$ and $u(1/2)=-(\log 2)/2+(\log 2)/2=0$. Thus both exceptional and nonexceptional formulas satisfy the required boundary conditions.

Choose the convention $L_{x}G(x,s)=\delta (x-s)$, where $\delta$ is the Dirac delta distribution, together with $G(1/2,s)=G(1,s)=0$. The Green's function is continuous at $x=s$ and obeysThese conditions follow by integrating its differential equation across $s$; continuity prevents a derivative of the Dirac delta distribution. If $\ell$ and $r$ are homogeneous solutions satisfying the left and right boundary conditions, respectively, and $W=\ell r^{\prime }-{\ell }^{\prime }r$ is their Wronskian, thenFor $n\ne 0$ take $\ell (x)=x^{-n}-2^n$ and $r(x)=x^{-n}-1$. Their Wronskian is $W(s)=n(2^n-1)s^{-n-1}$, and henceFor $n=0$ take $\ell (x)=\log (2x)$ and $r(x)=\log x$, giving $W(s)=\log 2/s$ andBoth formulas satisfy the endpoint conditions, continuity, and the unit jump directly. The unique solution of the boundary value problem isThe factor $s^{n+1}$ in the first formula matters: this Green's function is for the original monic differential operator, with ordinary measure $ds$, rather than for the operator multiplied by $x^{n+1}$.

The homogeneous linear differential equation can be written in divergence form:One integration gives $u^{\prime }=C{x}^{-n-1}$. Therefore the two cases areIn each case the displayed functions have linear independence throughout the interval. Imposing both homogeneous Dirichlet boundary conditions forces $A=B=0$: for $n\ne 0$, $2^n\ne 1$, and for $n=0$, $\log 2\ne 0$. This also establishes uniqueness for the inhomogeneous boundary value problem.

The separation of variables solution with Dirichlet boundary conditions on the cuboid isThe constants are fixed by the initial density through Fourier sine series. The fastest term again has $p=q=r=1$. When $L=2L_{\mathrm{crit}}$, its growth rate becomesThus the precise condition for no exponentially growing term isFor example, $\lambda =1/4$ gives ${\lambda }^{-2}+2\lambda =33/2>12$, while $\lambda =6$ gives $12+1/36>12$: either a sufficiently flat or a sufficiently elongated cuboid suppresses growth despite its unchanged volume. The function ${\lambda }^{-2}+2\lambda$ decreases to its minimum $3$ at $\lambda =1$, then increases without bound. Hence there are two positive threshold values, with growth only between them; equality leaves one stationary term and all other terms decaying.

The fastest temporal growth occurs for the lowest spatial eigenvalue, with $p=q=r=1$. Its growth rate isFor a generic initial condition its coefficient is nonzero, so this term determines the long-time behavior. In fact, every nonzero, nonnegative initial density has $a_{111}>0$, since the product of the three first sine eigenfunctions is strictly positive in the interior. ConsequentlyFor $L<L_{\mathrm{crit}}$ every term decays. At equality the first term is stationary and all higher terms decay: equality does not produce exponential growth. This is a reaction-diffusion equation threshold arising from competition between production and loss through the Dirichlet boundary condition.

For separation of variables, take $n=T(t)X(x)Y(y)Z(z)$ and divide the reaction-diffusion equation by this product wherever it is nonzero:Each spatial term depends on a different variable and must be constant. Write $X^{\prime \prime }=-k_x^{2}X$ and similarly for the other two factors. The Dirichlet boundary conditions give $X(0)=X(L)=0$. A zero separation constant permits only a linear function and hence the zero solution; a positive value of $X^{\prime \prime }/X$ likewise permits no nonzero function vanishing at both endpoints. The nontrivial choices are thereforeThe temporal ordinary differential equation isso, absorbing its constant multiplier into $a_{p,q,r}$,The superposition principle and completeness of the Fourier sine series give the stated threefold expansion with these factors. For an initial density $n_0$, orthogonality also determines the constants:

It is enough to choose an admissible parameter, without solving the minimizing equation. Take $a=V_0^2>0$ in the Gaussian variational bound for an attractive Gaussian well. The Rayleigh-Ritz variational principle giveswhere an explicit remainder isThe denominator tends to $2$ as $V_0\to 0$, proving $\epsilon =O(V_0^4)$ in Big O notation. More precisely the binomial series givesand thereforeThis is an upper bound on the exact ground state energy, rather than an assertion that the leading coefficient of its exact weak-potential expansion is $-1/2$.

Differentiating the Gaussian variational bound for an attractive Gaussian well givesThus an interior stationary point obeysThe left side of the first equation increases strictly from zero to infinity, so there is exactly one such point. More explicitly,so $E$ is a strictly convex function on $a>0$ and $a_{\ast }$ is its global minimum. At that minimum the stationarity equation givesThere is a continuous limiting zero at $a=0$, although $y_T=1$ at exactly $a=0$ is not a square-integrable function and hence is not an admissible trial. For a positive zero, division by $\sqrt{a}$ and squaring gives $a(1+a)=4V_0^2$, sois the unique positive zero. The other quadratic root is negative and inadmissible. Near zero,so the graph leaves the origin with slope $-\infty$, falls to its unique negative minimum, rises through $a_z$, and grows as $a/2-V_0+O(a^{-1})$ at infinity.

By the Rayleigh-Ritz variational principle,Thus the attractive well supports a negative-energy bound state for every $V_0>0$.

For the trial function, $y_T^{\prime }=-ax{y}_T$. The required Gaussian integrals areThe second formula follows by differentiating the first with respect to $a$. Thereforeand division by $G[y_T]$ gives the Gaussian variational bound for an attractive Gaussian well:This is a good trial family because it consists of smooth, positive, square-integrable functions decaying at both infinities. Each is an even function, matching the symmetry of the lowest bound state in this even potential, and the parameter varies its width. It also permits exact Gaussian integrals. Its Gaussian tail need not equal the exact exponential tail of a negative-energy bound state for the Rayleigh-Ritz variational principle to give a valid bound.

Take nonzero real trial functions $y$ with $y,y^{\prime }$ square-integrable and sufficient decay for integration by parts. For a compactly supported variation $y\mapsto y+\epsilon \eta$, the first variations areThe quotient rule for the functional then yieldsBy the fundamental lemma of the calculus of variations, stationarity for all such $\eta$ is equivalent toConversely, multiplying a decaying eigenfunction equation by $y$ and using integration by parts gives $F[y]=EG[y]$, so its Rayleigh quotient equals $E$ and its first variation vanishes. Thus the stationary values of the functional are the energy eigenvalues. The word "extrema" in the question includes stationary points here: higher eigenfunctions can be saddle points of the unrestricted Rayleigh quotient.

The Rayleigh-Ritz variational principle givesChoose an admissible trial family, calculate its Rayleigh quotient, and minimize over its parameters to obtain the best upper bound within that family. Equality holds for the exact ground state; a restricted family generally gives only an upper bound.

Close the hyperboloid segment with the two disks at $z=\pm 1$, and let $V$ be its interior. The flux through either disk is zero because $A_z=0$. Thus the desired surface integral is the total outward flux through the closed boundary. Sincethe divergence theorem and cylindrical coordinates giveThe extra factor $r$ is the volume integral measure in cylindrical coordinates; omitting it would give the wrong flux.

The divergence theorem states that, for a continuously differentiable vector field $\mathbf{H}$ on a bounded region $V$ with piecewise smooth boundary $S$ and outward unit normal $\hat{\mathbf{n}}$,To apply it, first derive the divergence of a cross product using the Levi-Civita symbol and the product rule:In the second term the minus sign comes from interchanging the first two indices of the Levi-Civita symbol. The divergence theorem with $\mathbf{H}=\mathbf{F}\times \mathbf{G}$ now yields

Use Einstein summation convention, writing ${\partial }_j=\partial /\partial x_j$. The contraction of two Levi-Civita symbols givesThe last equality is the product rule. Recognizing the divergences and directional derivatives proves the curl of a cross product identity:

The center of a group contains $e$. If $z,w$ lie in the center, then for every $g\in G$,It is therefore closed under the group operation and inverse elements, and inherits associativity, proving it is a subgroup. Each of its elements commutes with every element of $G$, in particular with every other central element. HenceThus the center is an abelian subgroup.

In additive notation every element $(a,b)\in {\mathbb{Z}}_2\times {\mathbb{Z}}_4$ obeys $4(a,b)=(0,0)$. Its order of a group element therefore divides four, whereas $1\in {\mathbb{Z}}_8$ has order eight. A group isomorphism preserves element orders, soFor the second product, define the group homomorphismIt is well-defined because multiples of six are divisible by both two and three. Its kernel contains only zero: simultaneous divisibility by two and three implies divisibility by six. Thus $\Psi$ is injective, and equal finite group orders make it surjective. Explicitly the six images areThereforeFor general positive integers $n,m$, the order of a direct-product element is the least common multiple of its component orders. Hence every element order divides $\mathrm{lcm}(n,m)=nm/\gcd (n,m)$, while $(1,1)$ attains this order. It generates the whole product precisely when $n,m$ are coprime:This is the cyclicity of a product of two finite cyclic groups, with the explicit isomorphism furnished by the Chinese remainder theorem.

Both subgroups contain the identity element, so $H\cap K$ is nonempty. If $x,y\in H\cap K$, then $xy\in H$ and $xy\in K$, so $xy\in H\cap K$. Similarly $x^{-1}$ belongs to both subgroups and hence to their intersection. Associativity is inherited from $G$. ThusThis verifies the group axioms; finiteness is unnecessary.

Closure of the componentwise operation follows from closure in each factor. For three pairs, associativity holds coordinatewise:The identity element is $(e_1,e_2)$ and the inverse element of $(g_1,g_2)$ is $(g_1^{-1},g_2^{-1})$. These verify all group axioms for the direct product of groups. Counting independent choices givesThe subset $N=G_1\times \{e_2\}$ contains the identity and is closed under products and inverses, so it is a subgroup. For any $x=(a,b)$ in the product and $(g,e_2)\in N$,Conjugation by $x^{-1}$ gives the reverse inclusion, and henceThis is the definition of a normal subgroup.

By (d), every group of group order three is a cyclic group. Writing its elements as $e,a,a^2$, every product is determined by $a^3=e$; the multiplication table isThe map $a^j\mapsto j(\mathrm{mod}3)$ is a bijective group homomorphism to addition modulo three. Thereforeso there is exactly one such group up to isomorphism.

Choose $g\ne e$ in a group of prime group order $p$. Its order of a group element divides $p$ by (c) and is not $1$, hence equals $p$. Consequently its cyclic subgroup already contains all elements of $G$:Thus every nonidentity element is a group generator.