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For inviscid flow with a conservative body-force potential $\Phi$, the Euler momentum equation and the identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times (\nabla \times \mathbf{u})$ apply. In an irrotational region write $\mathbf{u}=\nabla \varphi$. For constant density they giveThis is the Unsteady Bernoulli equation; its constant is spatially uniform, with time-dependent gauge absorbed into $\varphi$ if desired.

For the intended linearly elastic balloon discharge model, take an initially stationary, inviscid plug of water in the constant-area tube. Put $B=\pi a^2$ and $p_{\mathrm{in}}-p_{\mathrm{atm}}=KV$ with $K>0$. At the two tube ends the speeds are equal, so their kinetic terms cancel in the Bernoulli difference. Since $\varphi (L)-\varphi (0)=Lu$, this givesHence $\ddot{V}+{\Omega }^{2}V=0$, where ${\Omega }^2=KB/(\rho L)$. With $V(0)=V_0$ and $u(0)=0$,ThusAcceleration stays positive until emptying, so the maximum speed is reached at that instant. The time is independent of initial volume, while the maximum speed is not.

The initial-rest condition is implicit in the claimed volume independence, rather than specified in the text. For an independently prescribed positive initial speed $u_0$, the first zero instead occurs at ${\Omega }^{-1}\arctan [\Omega V_0/(B{u}_0)]$. The calculation also takes the given balloon pressure as the pressure directly at the tube entrance; a reservoir-to-exit kinetic-head or entrance-loss model would add other terms and is not this equal-speed tube model.

The Drude model treats carriers as classical independent particles, accelerated between collisions by the electric force. Collisions occur at constant rate ${\tau }^{-1}$ and randomize the directed velocity to zero mean; the ions are an effectively stationary background. Over a short time $dt$, directed velocity gains $(e/m)\mathbf{E}dt$ while the fraction $dt/\tau$ that collides loses its mean directed motion. ThereforeHere $e$ is the signed carrier charge. For the stated complex harmonic convention,The conductivity is positive in its dissipative real part regardless of the sign of $e$.

For zero charge density, Gauss's law makes the electric amplitude transverse. Faraday's and Ampère-Maxwell's laws, with displacement current retained, giveSubstitution of $e^{i\mathbf{k}\cdot \mathbf{x}-i\omega t}$ yieldsFor propagation in a fixed direction, $k$ is the possibly complex scalar wave number; equivalently the spatial equation uses $\mathbf{k}\cdot \mathbf{k}$, without complex conjugation. The printed real-vector notation $|\mathbf{k}|$ must be analytically continued in an absorbing or evanescent medium.

In the high-frequency Drude plasma cutoff regime, $\omega \tau \gg 1$ gives ${ϵ}_r\simeq 1-{\omega }_p^2/{\omega }^2$. Below the plasma frequency, the physical decaying branch isFinite collisions add a small phase component while maintaining positive attenuation. Very near the threshold one should retain the full complex permittivity rather than the purely imaginary approximation.

Using the momentum operator $\hat{p}=-i\hslash d/dx$ and its commutator with multiplication by $A$, expansion of the factorized quantum Hamiltonian givesThe first-order zero-mode equation integrates toFor $A=x^n$, its exponent is $-s{x}^{n+1}/[\hslash (n+1)]$. It decays at both ends exactly when $n+1$ is even. If $n$ is even it grows at negative infinity. Thus a nonzero square-integrable zero mode exists exactly for odd $n$.

For $n=1$ the Gaussian zero mode is $\chi =(s/(\pi \hslash ){)}^{1/4}{e}^{-s{x}^2/(2\hslash )}$. More generally, on normalized states with finite second moments and the usual integration-by-parts boundary conditions, $\hat{H}=Q^{†}Q/(2m)$ is nonnegative. With zero means this implies, for every $s>0$,Minimizing the quadratic in $s$ at $s=\hslash /[2(\Delta x{)}^2]$ yieldsThe Gaussian mode attains equality. The argument concerns the same state for the whole positive family of factorized operators, which permits the minimization.

Write $p(z)=c{\prod }_{j=1}^{2m}(z-a_j)$, including root multiplicities. For $|z|>R$ the convergent logarithm series for every factor $1-a_j/z$ definesChoosing either square root of the nonzero leading coefficient gives the holomorphic square root outside all polynomial rootsIts square is $p$. The even degree makes $z^m$ single-valued, avoiding the winding obstruction that an odd degree would create on the exterior annulus.

For the branch behaving as $z^2$ at infinity, the Laurent series isIt converges uniformly on the circle of radius two. Only the coefficient of $z^{-1}$ contributes to its anticlockwise integral, sofor this branch, and $+\pi i$ for the other. These are the two allowed answers.

For uniform convergence to $k$ on $A$, for every $\eta >0$ there is $N$ such that $|k_n(x)-k(x)|<\eta$ for all $n\ge N$ and all $x\in A$. The bound $|\sin (nx)/\sqrt{n}|\le 1/\sqrt{n}$ proves uniform convergence to zero, whereas its derivative is $\sqrt{n}\cos (nx)$. On an interval containing zero these derivatives diverge at zero, so they cannot converge uniformly. This supplies a counterexample to differentiating a uniformly convergent sequence without additional hypotheses.

For uniform derivative convergence with an anchored value, put $g=\lim k_n^{\prime }$ and $a=\lim k_n(x_0)$. A uniform limit of continuous functions is continuous. The fundamental theorem of calculus givesThus $k_n(x)\to k(x)=a+{\int }_{x_0}^{x}g(t)dt$, with error bounded byThis is uniform on bounded subintervals, and on all of $A$ if it is bounded. The fundamental theorem of calculus now gives $k^{\prime }=g$, so $k$ is continuously differentiable. For an unbounded $A$, no global uniform convergence of $k_n$ follows merely from the displayed hypotheses, but local uniform convergence and the claimed differentiability do.

For the series, use partial sums and differentiate their individual terms:For $|x|\le 1$, their absolute values are at most $2n/(n^3+1)$, a summable bound. The Weierstrass M-test makes the derivative series uniformly convergent, while every partial sum is zero at $x=0$. The anchored-value result therefore proves continuous differentiability on $(-1,1)$, with derivative equal to the displayed termwise derivative series.