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A Riemann integrable function on a closed interval is bounded and has equal upper and lower Darboux integrals. More explicitly, for a partition of an interval $P:a=x_0<x_1<\cdots <x_m=b$, define the Darboux sumsThen Riemann integrability meansand this common value is the Riemann integral. Equivalently, the Riemann integrability criterion requires that for every $\epsilon >0$ some partition of an interval satisfies $U(f,P)-L(f,P)<\epsilon$. Indeed, the common partition refinement of two partitions increases each lower Darboux sum and decreases each upper Darboux sum, so equality of the two extremal values is equivalent to arbitrarily small gaps.

Set ${\omega }_1=u$, ${\omega }_2=\Omega +v$, ${\omega }_3=w$ and retain terms linear in the perturbations in Euler equations for a torque-free rigid body. Using $I_3=I_1+I_2$ givesConsequently $\ddot{u}={\beta }^2{\Omega }^{2}u$, where $\beta =\sqrt{(a^2-b^2)/(a^2+b^2)}>0$. For $\Omega >0$ the growing solution isThere is also a decaying mode with the opposite exponent. For negative $\Omega$, select the exponent $+\beta |\Omega |$ as the growing one. This is the intermediate axis theorem: steady rotation about the intermediate principal inertia axis is unstable whenever $\Omega \ne 0$.

Take the sheet to lie in the $x_1{x}_2$ plane, with lengths $a,b$ in those directions. Its surface density is $M/(ab)$. Reflection symmetry about the coordinate axes makes the off-diagonal entries of the inertia tensor zero. Its principal moments of inertia areand similarly $I_2=M{a}^2/12$. Since $I_3$ integrates $x_1^2+x_2^2$, the perpendicular axis theorem gives

The given vectors satisfyand $y_k,y_{k+1}$ are independent, so their span is invariant under $A$. On that plane the characteristic polynomial is ${\lambda }^2-2\lambda +2$. ThereforeIndeed $A{v}_{\pm }=y_{k+2}-(1\mp i)y_{k+1}=(1\pm i)v_{\pm }$. No third eigenvalue is determined by these data.

The displayed $x_k=(1,1,1)$ has norm $\sqrt{3}$, although a large-$k$ normalized iterate should have norm one. This harmless source normalization is resolved by dividing all three displayed vectors by $\sqrt{3}$; the recurrence, eigenvalues and eigenvector directions are unchanged.

Since $A$ is diagonalizable over the complex numbers, write $x^{(0)}={\sum }_i{a}_i{v}_i$. Provided iteration does not encounter zero, its direction is that ofWith one strictly dominant eigenvalue and $a_n\ne 0$, all other contributions are smaller by factors $(|{\lambda }_i|/|{\lambda }_n|{)}^k$. The dominant eigenvalue is real, since nonreal eigenvalues of a real matrix occur in conjugate pairs of equal modulus. The normalized power method therefore converges to a signed unit eigenvector if ${\lambda }_n>0$, and alternates between opposite unit eigenvectors if ${\lambda }_n<0$. A quotient such as $(x^{(k)}{)}^{T}A{x}^{(k)}$ approaches ${\lambda }_n$ in either case. If $a_n=0$, replace it by the largest active modulus; if $A{x}^{(k)}=0$, the prescribed iteration is undefined.

With two dominant eigenvalues of equal modulus and both components active, the iterates approach their real invariant plane rather than one direction. Distinct real equal-modulus eigenvalues are $\rho ,-\rho$ and produce a limiting two-cycle. A conjugate pair $\rho e^{\pm i\theta }$ produces a rotation in suitable coordinates on that plane: the normalized directions are periodic when the angle is rational relative to $2\pi$ and otherwise generally keep rotating without convergence. If the two eigenvalues are equal and real, the initial projection is itself one eigenvector in their repeated eigenspace, and converges or alternates as in the single-value case; one orbit cannot recover every independent eigenvector of that eigenspace.

For distinct dominant values, retain unnormalized consecutive vectors $y,Ay,A^{2}y$ in the nearly invariant plane and solveThe roots of ${\lambda }^2-s\lambda +p$ estimate the two eigenvalues. Corresponding eigenvectors are $Ay-{\lambda }_{2}y$ and $Ay-{\lambda }_{1}y$, since applying $A$ and using the recurrence multiplies them by ${\lambda }_1$ and ${\lambda }_2$. This is dominant eigenpair extraction from a two-step Krylov recurrence. Equivalent extraction uses the $2\times 2$ matrix of $A$ on the computed plane. It needs two independent iterates and nonzero projections onto both distinct eigenvectors.

The thin-layer condition isIntegrate the differential equation once: $(f^{\prime \prime }+f{f}^{\prime }{)}^{\prime }=0$. Since $f(0)=f^{\prime \prime }(0)=0$, its integration constant is zero. A second integration gives $f^{\prime }+f^2/2=b$. The outward-jet solution with $b>0$ and $f(0)=0$ isUsing the given fourth-power integral in the normalization gives $1=4k^4[4/(3k)]=16k^3/3$, so $k=(3/16{)}^{1/3}$. Thus the similarity profile of an axisymmetric radial viscous free jet isIt is symmetric, outward, decays in $z$, and has exactly the conserved flux $F$. It describes the region beyond the small injection slit where the stated boundary-layer approximation applies; its formal $r=0$ singularity is outside that approximation.

Let $U(r)$ be the radial velocity scale. Advection-viscosity balance gives $U^2/r\sim \nu U/{\delta }^2$, while the conserved momentum flux gives $F\sim U^{2}r\delta$. Eliminating $U$ yields $\delta /r\sim ({\nu }^2/F{)}^{1/3}$, a constant, so $\delta =Cr$ and $U\propto r^{-1}$.

The streamfunction automatically satisfies incompressibility. With the proposed form it giveswhere $u_r,u_z$ are the two velocity components. Differentiation in the momentum equation cancels the terms proportional to $\eta f^{\prime }{f}^{\prime \prime }$ and leavesChoose $A{C}^2=\nu$. The momentum integral is $F=A^{2}C\int f^{\prime 2}d\eta$, so choosing $A^{2}C=F$ makes its normalization one. ThereforeReflection symmetry gives $u_z=0$ and ${\partial }_z{u}_r=0$ at zero. Taking the streamfunction's odd representative makes $f(0)=0$, and symmetry gives $f^{\prime \prime }(0)=0$; far-field decay gives $f^{\prime }\to 0$. The factor $f$ multiplying $f^{\prime \prime }$ is present in the original PDF but missing from the TeX aid; the direct substitution confirms the PDF equation.

Axisymmetric incompressibility is ${\partial }_r(r{u}_r)+r{\partial }_z{u}_z=0$. Multiply the radial momentum equation by $r$ and use this to put advection in conservative form:Integrating through the layer givesThe boundary terms vanish because radial velocity and its shear decay into the surrounding fluid; an entrainment velocity need not itself vanish as long as its product with $u_r$ does. Hence

For the difference of two solutions the equation is $w_{tt}+w_t-\Delta w=0$ with zero initial data. The same shrinking-ball energy now obeysAgain $E(0)=0$ forces the difference to vanish at every chosen spacetime point. Therefore the two $C^2$ solutions agree for every $t>0$. Damping contributes an additional negative energy term and does not weaken the local uniqueness proof.

Let $w$ be the difference between a $C^2$ solution and the formula in part (ii). It satisfies $w_{tt}-\Delta w=0$ and zero initial data. Fix $(T,x_0)$ and $\epsilon >0$, and put $B_t=B(x_0,T+\epsilon -t)$. Define the local energyDifferentiating on the shrinking ball and integrating by parts givessince $2w_t{\partial }_{n}w\le w_t^2+({\partial }_{n}w{)}^2\le w_t^2+|\nabla w{|}^2$. Initially $E=0$, hence $w_t=\nabla w=0$ inside each ball. In particular $w(t,x_0)$ is constant from its initial value zero, so $w(T,x_0)=0$. Since the point was arbitrary, every $C^2$ solution equals the Kirchhoff solution. This shrinking cone energy argument uses only energy on compact sets, so it does not assume global decay of the competing solution.

Rotate the polar axis to $\xi$. Since $W_t$ is a finite sphere measure, its Fourier transform iswith value $t$ at $\xi =0$. Fourier transforming the wave equation gives ${\hat{u}}_{tt}+|\xi {|}^2\hat{u}=0$ with initial displacement zero and velocity $\hat{g}$. Its solution is this multiplier times $\hat{g}$, so inversion gives $u=W_t\ast g$. Thus the Kirchhoff formula isThis extends to arbitrary smooth $g$, without any growth restriction: for each finite spacetime neighborhood, multiply $g$ by a compactly supported smooth cutoff equal to one on every sphere occurring in that neighborhood. The formula is unchanged there, so its verified differential equation remains valid locally. Its initial values follow from $u(t,x)=tg(x)+o(t)$ and differentiation under the sphere integral. Since the neighborhood was arbitrary, it is a smooth solution everywhere for $t\ge 0$.

Choose the Fourier transform conventionIt maps the Schwartz space to itself. For a tempered distribution $u$, define the transform by bilinear distributional duality:For an ordinary integrable function this agrees with the first formula by interchanging its integrals. The absence of complex conjugation is the distribution-pairing convention.

Use the Runge exhaustion of a slit disk, with the principal argument $-\pi <\arg z<\pi$:These are nested compact subsets of $\Omega$. Their complements are connected: the inner disk and exterior region join through the omitted sector around the negative real radius. Every compact subset of $\Omega$ has positive distance from zero, the unit circle and the slit, so it is contained in $K_m$ for all sufficiently large $m$.

Since $f$ is holomorphic on $\Omega$, it is holomorphic on a neighborhood of each $K_m$. The polynomial Runge theorem gives $P_m$ with ${\sup }_{K_m}|P_m-f|<1/m$. Given any compact $K\subset \Omega$, eventually $K\subseteq K_m$, so the same estimate holds on $K$. ThereforeThe nested exhaustion makes this one sequence work simultaneously on all compact subsets.

The polynomial Runge theorem states that, for a compact set $K\subset \mathbb{C}$ with connected complement and a function $f$ holomorphic on a neighborhood of $K$, for every $\epsilon >0$ there is a polynomial $P$ such thatThe connected-complement hypothesis permits polynomial approximation rather than approximation by rational functions with poles in bounded complementary components.

Assume the riskless gross return $q=1+r>0$, as required by the positive-wealth model, and a finite moment $m_b=\mathbb{E}{X}_0^b$. DefineThis maximum exists: the control interval is compact and dominated convergence gives continuity, using the integrable bound $(X_0+q{)}^b\le X_0^b+q^b$ for $0<b\le 1$.

The terminal value is $V_N(w)=U(w)$. If $V_{n+1}(w)=a_{n+1}U(w)$, the Bellman equation and independence of the next return giveBackward induction therefore provesAny maximizing proportion can be used each day; allowing adapted choices cannot exceed the Bellman bound. For $\alpha =1$, $C_{\ast }=m_b$, exactly agreeing with part (b). If $m_b=\infty$ and at least one day remains, the allowed choice $\theta =1$ already gives infinite expected utility, so the value is infinite rather than a finite coefficient. This is the necessary moment qualification to the finite optimization formula.

Put $b=1-R>0$ and $m_b=\mathbb{E}{X}_0^b$. Homogeneity of the constant relative risk aversion utility gives $U(z_N)=U(z_n){\prod }_{j=n}^{N-1}{X}_j^b$. ThusAs before this is an extended-valued conditional expectation if the required moment is infinite and days remain.

The product $z_n=z_0{\prod }_{j=0}^{n-1}{X}_j$ uses only past multipliers, so all $X_n,\dots ,X_{N-1}$ are independent of it. With $m=\mathbb{E}{X}_0$, independence givesThe formula is finite when $m<\infty$; since the variables are positive, it also has the extended value $+\infty$ for $m=\infty$ and $n<N$. At $n=N$, it simply reads $z_N$.

For equilibrium, let $t\to \infty$ in part (e). Monotone convergence of the tail integral and the finite mean giveand consequentlyThis limiting argument uses the equilibrium age and residual-service distribution produced by the queue, not just a fresh independent copy of $S$ for each mosquito present. Replacing their residual service by fresh service would generally be wrong unless the service law is memoryless.

After closing, the count can only decrease. Thus $V(t)>v$ is the event that at least one mosquito remains at $t+v$. The zero probability of a Poisson count yields the clearance time after closing an infinite-server queue:The clearance time has an atom at zero of mass $e^{-m_t}$, corresponding to an empty room when the window closes.

Only arrivals at times $u\le t$ remain after the window closes. Such an arrival survives until $t+v$ exactly when its service time exceeds $t+v-u$. The same independent Poisson thinning givesAt $v=0$ this agrees with part (b), while for large $v$ the parameter tends to zero.