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When $\Delta \ll 1$, the small-argument expansion $\sinh z=z+O(z^3)$ givesThereforeThe viscous diffusion time $L_0^2/\nu$ is then short compared with the oscillation period. Viscosity communicates the wall motion across the entire layer almost instantaneously, producing the linear profile of quasi-steady Couette flow.

Usingin the quotient from part (a), multiplying numerator and denominator by the complex conjugate of the denominator, and taking the real part after multiplication by $e^{i\omega t}$ giveswhereThe denominator follows from

For the parallel flow $\mathbf{u}=(u(y,t),0)$ with no pressure gradient, the Navier-Stokes equations reduce to the diffusion equationSubstitution ofgivesThe boundary velocities require $f(L_0)=1$ and $f(-L_0)=-1$. The problem is antisymmetric under $y\mapsto -y$, so $f$ is odd. Withthe unique odd solution is

The inverse Lorentz transformation of electromagnetic fields, from the primed rest frame to the unprimed frame, givesandEquivalently, $\mathbf{B}=\gamma \mathbf{v}\times {\mathbf{E}}^{\prime }/c^2$ for $\mathbf{v}=v{\mathbf{e}}_x$.

Using $1/{\mu }_0={ϵ}_0{c}^2$ and ${\gamma }^2=(1-v^2/c^2{)}^{-1}$,

The primed field is static, so $w$ depends on $t,x$ throughand has no explicit $t^{\prime }$ dependence. The chain rule givesConsequentlyThus the field-energy profile is transported rigidly with velocity $v{\mathbf{e}}_x$.

In a region with no current, the Ampère-Maxwell equation and Faraday's law areThereforewhere the last line uses the divergence and curl of a cross product. Thus Poynting's theorem takes the formThe vector $\mathbf{S}$ is the Poynting vector.

Part (a) makes ${\lambda }_k<0$ for all sufficiently large $k$, while part (b) would make ${\lambda }_k\ge 0$ whenever $P^{(k)}$ were nonzero. Hence some derivative of $P$ vanishes identically, so $P$ is a polynomial.

Let its degree be $n$. Then $P^{(n)}$ is a nonzero constant. Substituting $k=n$ into the differentiated equation annihilates both derivative terms and givesThus ${\lambda }_n=0$, andThe nonzero regular solutions are therefore constant multiples of the Legendre polynomial of degree $n$.

Multiplication by $(1-x^2{)}^k$ puts the equation in self-adjoint form:Multiply by $Q_k$ and integrate over $[-1,1]$. Regularity and the vanishing factor at both endpoints remove the boundary term in integration by parts, leavingIf $Q_k$ is not identically zero, the integral on the right without ${\lambda }_k$ is positive, whereas the left side is nonnegative. Therefore

Let $Q_k=P^{(k)}$. Differentiating Legendre's differential equation $k$ times and using the Leibniz rule givesThe induction step follows by differentiating this equation once: the derivative of the coefficient $-2(k+1)x$ contributes the additional term $-2(k+1)Q_k^{\prime }$. ThusFor fixed $\lambda$, this is negative for every sufficiently large $k$.

By the prime ideal quotient criterion,For a principal ideal, this is precisely the assertion that $p$ is a prime element. This closes the cycle and proves that (i)--(v) are equivalent.

Now let $\phi :R\to S$ be a surjective ring homomorphism, where $R$ is a principal ideal domain and $S$ is an integral domain. WriteIf $a=0$, the first isomorphism theorem for rings makes $\phi$ an isomorphism. Otherwiseis an integral domain, so the equivalences just proved show that $(a)$ is maximal and $S$ is a field.

Next suppose that the polynomial ring $R[X]$ is a principal ideal domain. It follows first that $R$ is an integral domain. For nonzero $a\in R$, the idealis principal. Since $f$ divides the nonzero constant $a$, it is constant; since it also divides $X$, that constant divides the coefficient $1$, so $f$ is a unit. Therefore $(a,X)=R[X]$, and there are polynomials $g,h$ such thatSetting $X=0$ gives $1=ag(0)$, so every nonzero $a\in R$ is a unit. Hence

Finally, let $R$ be an integral domain in which every two nonzero elements have a greatest common divisor. If an irreducible $p$ divides $ab$, then $\gcd (p,a)$ is either a unit or an associate of $p$. In the second case $p\mid a$. In the first case, the Euclid lemma in a greatest-common-divisor domain gives $p\mid b$. Thus every irreducible element of $R$ is prime.

Every field is an integral domain, so if $R/(p)$ is a field then it is an integral domain.

For any commutative ring with identity, the maximal ideal quotient criterion states that an ideal $I$ is maximal exactly when $R/I$ is a field. Applying it to $I=(p)$ proves

Suppose $p$ is irreducible andThen $p=ab$ for some $b\in R$. Irreducibility says that either $a$ is a unit, in which case $(a)=R$, or $b$ is a unit, in which case $(a)=(p)$. Thus no proper ideal lies strictly between $(p)$ and $R$, so

Assume that $p$ is a prime element. If $p=ab$, then $p$ divides $ab$, so $p$ divides $a$ or $b$. If $p\mid a$, write $a=pc$; cancellation in the integral domain gives $1=bc$, so $b$ is a unit. The other case is symmetric. Hence every prime element is irreducible.

A direct determinant calculation givesMoreover,has rank one, so its kernel has dimension two, while $(A-I{)}^2=0$ and $A\ne I$. Hence the minimal polynomial isThere are two Jordan blocks, because the eigenspace has dimension two, and the largest has size two, because the minimal polynomial has exponent two. Thus

Since $z(t)=h_I-\alpha t$, the free-surface area isuntil the emptying time $t_e=h_I/\alpha$. Equivalently,

If $z(t)$ is the water depth and $A(z)=\pi r(z{)}^2$ is the horizontal cross-sectional area, Torricelli's law gives the volume flux through the hole asConservation of volume givesFor a prescribed constant fall rate $\dot{z}=-\alpha$, $\alpha >0$, this requiresConsequentlyso the container radius must be proportional to the fourth root of height.

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