The conformal map $\zeta =z^3$ sends the wedge to the upper half-plane. A fluid vortex at $c^3$ there has an opposite image at ${\bar{c}}^3$, so the pulled-back potential isThe two boundary rays map to the real axis, where the imaginary part is constant, giving the required impermeability. Factoring $z^3-c^3=(z-c)(z^2+cz+c^2)$ gives the printed form up to branch constants.

The fluid vortex moves with the regular part of the complex velocity, excluding its own singular term. Since $w^{\prime }=u-iv$, its motion is the conjugated derivative. Differentiating the remaining terms and setting $z=c$ givesThe overbar on the time derivative is essential.

Write $c=r{e}^{i\theta }$. Using $c^3-{\bar{c}}^3=2i{r}^3\sin 3\theta$ and multiplying the velocity equation by $e^{i\theta }$ givesThus $\dot{r}=3\Gamma \cot 3\theta /(4\pi r)$ and $\dot{\theta }=-\Gamma /(4\pi r^2)$. Direct differentiation now yieldsHence $\boxed{r\sin (3\theta )=\text{constant}}$ along the trajectory while it stays inside the wedge.

The circle theorem says that if a complex potential $f$ has no singularities on or inside $|z|=a$ (and has the required analytic continuation there), thenadds image singularities inside the circle and makes it a streamline. On the circle $a^2/\bar{z}=z$, so $w=f(z)+\overline{f(z)}$ is real and its imaginary part is constant. This gives zero normal velocity. The reflected term is holomorphic where defined, as follows from its power series in $1/z$, and its singularities are precisely the reflected ones; the external original singularities are retained. This proves the construction, with an additive constant and any independently prescribed circulation handled separately.

For the fluid vortex use $f(z)=-i\Gamma \log (z-b)/(2\pi )$. Reflecting gives an opposite image at $d=a^2/b$ and also a term $-i\Gamma \log z/(2\pi )$, up to constants. The zero-circulation circle-theorem construction has both these terms. To impose the specified circulation $-\Gamma$ about the obstacle, add $+i\Gamma \log z/(2\pi )$, canceling the central term. ThusIt has the stated circulation and the same impermeable boundary, since the added central-fluid vortex streamfunction is constant on the circle.

For the given Blasius integral take the obstacle contour counterclockwise. Inside it the sole singularity ofis at $d$. Its residue is ${\Gamma }^2/[2{\pi }^2(d-b)]$, from the cross term. The residue theorem then givesTherefore $\boxed{F_x=\rho {\Gamma }^{2}b/[2\pi (b^2-a^2)],F_y=0}$. The force on the obstacle points toward the external fluid vortex.

The stationary Schrödinger equation is $-{\hslash }^2{\psi }^{\prime \prime }/(2m)+V\psi =E\psi$. A nonzero bound-state function is square-integrable on the real line and can be scaled to ${\int }_{\mathbb{R}}|\psi {|}^{2}dx=1$. Odd parity means $\psi (-x)=-\psi (x)$.

Set $k=\sqrt{2m(V_0+E)}/\hslash$ and $\kappa =\sqrt{-2mE}/\hslash$, both positive. Oddness and decay giveContinuity at $a$ gives $B=A\sin (ka)$, and derivative continuity gives $-\kappa B=Ak\cos (ka)$. A nonzero solution cannot have $\sin (ka)=0$, since that would force both the value and derivative there to vanish. Dividing therefore yields $\kappa =-k\cot (ka)$. With $\eta =\kappa a$ and $\xi =ka$,The matching at $-a$ follows automatically by odd parity.

In the required sketch, the second equation is a first-quadrant circle arc of radius $R=\sqrt{2m{V}_0{a}^2}/\hslash$. The curve $-\xi \cot \xi$ starts with limiting value $-1$ at zero, rises to zero at $\pi /2$, and then rises to $+\infty$ as $\xi \uparrow \pi$. Thus a positive-$\eta$ intersection in this range must have $\xi >\pi /2$. The accompanying plot shows the circle below threshold and one above it.

Consequently $R<\pi /2$ rules out all odd states: the first positive branch already requires that radius, and later branches require still larger $\xi$. This is precisely $\boxed{2m{V}_0<(\pi \hslash /(2a){)}^2}$. At equality the intersection has $\eta =0$ and would not decay at infinity, so it too is not a normalizable bound state. An odd state first appears only for a strict increase beyond threshold.

By reflection symmetry, $G$ is the midpoint of the cylinder axis. With position $r$ measured from $G$, define $J_{ij}^G=\int (|r{|}^2{\delta }_{ij}-r_i{r}_j)dm$. Symmetry makes the off-diagonal entries zero. Uniform density gives $⟨x^2⟩=⟨y^2⟩=a^2/4$ and $⟨z^2⟩=b^2/3$, by integrating over a disk and the interval $[-b,b]$. HenceFor parallel axes based at $P$, the tensor form of the parallel-axis theorem is $J^P=J^G+M(|d{|}^{2}I-d{d}^T)$, where $d=\overrightarrow{PG}$. This follows by substituting $r+d$ in the defining integral and using $\int rdm=0$. Here $d$ is an axial vector of length $b$, soThe Euler angles $\psi ,\varphi$ are cyclic coordinates. Their conserved conjugate momenta are $p_{\psi }=C(\dot{\psi }+\dot{\varphi }\cos \theta )=Cn$ and $p_{\varphi }=A\dot{\varphi }{\sin }^2\theta +Cn\cos \theta =h$. The Lagrangian has no explicit time dependence, so the conserved energy is kinetic plus gravitational potential, namely the expression given with constant $E$.

Eliminate the angular velocities using $\dot{\varphi }=(h-Cn\cos \theta )/(A{\sin }^2\theta )$ and $\dot{\psi }+\dot{\varphi }\cos \theta =n$. The energy equation becomesThusThe elimination is valid away from the coordinate poles; the multiplied equation extends by continuity to any pole reached by a regular motion.

At a regular constrained extremum, choose an admissible variation $k$ with $\delta J[k]\ne 0$. The implicit function theorem applied to $J[y+\epsilon h+sk]$ permits a compensating $s(\epsilon )$ keeping $J$ fixed. Differentiation at zero gives $\delta I[h]=\lambda \delta J[h]$ for every $h$, where $\lambda =\delta I[k]/\delta J[k]$. Integrating the first variation of $I-\lambda J$ by parts and using arbitrary compactly supported $h$ givesSmoothness alone does not exclude an abnormal constraint: if $\delta J$ vanishes identically, one needs the general multiplier form with two multipliers rather than automatically setting the coefficient of $I$ to one. The displayed normal form uses the usual regular-constraint hypothesis.

For the given quadratic functionals, the equation is $\boxed{(p{y}^{\prime }{)}^{\prime }+qy+\lambda ry=0}$. If the endpoint values of $y$ are free, the boundary variation $[p{y}^{\prime }h{]}_a^b$ gives the natural conditions $\boxed{p(a)y^{\prime }(a)=p(b)y^{\prime }(b)=0}$, interpreted by limits at singular endpoints. Multiply the equation by $y$ and integrate. These boundary terms vanish, leavingFor two eigenfunctions, multiply their equations by the other function and subtract. Integration givesThe natural conditions remove the bracket, so distinct eigenvalues imply $\boxed{{\int }_a^{b}r{y}_m{y}_{n}dx=0}$.

In the Legendre case, comparison of leading polynomial coefficients gives ${\lambda }_n=n(n+1)$, so the degrees correspond to distinct eigenvalues. A degree-zero polynomial is $y_0=C_0$. Write $y_1=Ax+B$; orthogonality to $1$ on $[-1,1]$ gives $B=0$. Write $y_2=A{x}^2+Bx+C$. Orthogonality to $x$ gives $B=0$, and orthogonality to $1$ gives $2A/3+2C=0$. ThusThese polynomials satisfy the natural endpoint limits because $1-x^2$ vanishes at both endpoints.