The conformal map $\zeta =z^3$ sends the wedge $0<\arg z<\pi /3$ to the upper half-plane. The physical vortex maps to $\zeta =c^3$; an opposite image vortex at ${\overline{c}}^{,3}$ makes the real axis a streamline. Pulling this potential back givesSince $z^3-c^3=(z-c)(z^2+cz+c^2)$, splitting the first logarithm, up to an additive constant, gives exactly the required form. The image construction enforces zero normal velocity on both rays.

To find the vortex motion, hold $c$ fixed while differentiating with respect to $z$, remove the self-induced singular term $-i\Gamma /(2\pi (z-c))$, and then evaluate at $z=c$. The remaining complex velocity isA complex velocity $u-iv$ equals $d\overline{c}/dt$, rather than $dc/dt$. ThusThe overbar on the source derivative is essential to the trajectory calculation.

Write $c=r{e}^{i\theta }$. Since $c^3-{\overline{c}}^{,3}=2i{r}^3\sin (3\theta )$, multiply the equation by $e^{i\theta }$ to obtainTaking real and imaginary parts givesThereforeproving the trajectory invariantThe calculation also covers $\Gamma =0$, when the vortex position is stationary.

This is the sixty-degree case of a point vortex in a wedge.

The Milne-Thomson circle theorem says that if a complex potential $f(z)$ has no singularities on or inside $|z|=a$, then a circular impermeable boundary can be introduced by takingThe new singularities are the reflected ones inside the circle. On $|z|=a$, $a^2/z=\overline{z}$, so $W(z)=f(z)+\overline{f(z)}$ is real; the stream function is constant and there is no normal velocity. At infinity the added term tends to $f^{\ast }(0)$, a harmless potential constant, so it preserves the far-field flow. This proves the theorem with zero added net circulation about the cylinder. A logarithmic term may then specify a different cylinder circulation.

For $f(z)=-i\Gamma \log (z-b)/(2\pi )$, put $c=a^2/b$. The reflected term iswith locally consistent logarithm branches. The circle-theorem solution has zero cylinder circulation: its image vortex at $c$ and its central vortex cancel. To impose circulation $-\Gamma$, add $i\Gamma \log z/(2\pi )$. The result isIt consists of the physical vortex and an opposite image vortex inside the obstacle. On the circle, $|z-b|=(b/a)|z-c|$, so its stream function is indeed constant.

The complex velocity isInside the positively oriented obstacle contour, $q^2$ has only the double pole at $c$. Its residue isThe residue theorem and the supplied Blasius force theorem therefore giveConsequentlyThe force points toward the held vortex, independent of the sign of its circulation.

The residue calculation is the force on a circular cylinder from a held exterior vortex.

The stationary Schrödinger equation isA nonzero bound-state wave function has ${\int }_{\mathbb{R}}|\psi {|}^{2}dx<\infty$, so it can be rescaled to have integral one. Odd parity means $\psi (-x)=-\psi (x)$.

For $-V_0<E<0$, set $k=\sqrt{2m(V_0+E)}/\hslash >0$ and $\kappa =\sqrt{-2mE}/\hslash >0$. Oddness removes the interior cosine solution, and normalizability removes the growing exponentials outside. A continuous odd candidate can therefore be writtenContinuity of the derivative at $a$ gives $k\cos (ka)=-\kappa \sin (ka)$; oddness gives the other boundary condition. A nonzero solution cannot have $\sin (ka)=0$, since the derivative condition would then force its interior coefficient to vanish. Thus, with $\xi =ka$ and $\eta =\kappa a$,Physical bound states require $\xi ,\eta >0$. On $0<\xi <\pi /2$, the cotangent condition gives negative $\eta$, so no physical intersection occurs. On $(\pi /2,\pi )$, the curve $-\xi \cot \xi$ rises from zero to positive infinity, while the positive quadrant of the circle decreases. Their first intersection exists exactly when the circle radius $R=\sqrt{2m{V}_0{a}^2/{\hslash }^2}$ exceeds $\pi /2$.

In particular,There is also no normalizable odd state at equality: the putative intersection has $\eta =0$, hence $E=0$, and its exterior tail does not decay. Higher positive branches begin beyond this first interval and require still larger $R$.

At a regular constrained extremal, the first variations satisfy $\delta I=\lambda \delta J$ for a constant Lagrange multiplier $\lambda$. One way to justify this is to choose a variation for which $\delta J\ne 0$, use it to correct any other variation to preserve the constraint by the implicit function theorem, and differentiate $I$ along that corrected family. This gives the same ratio $\lambda$ for all variations. The exceptional case $\delta J=0$ for every variation requires the usual separate abnormal-extremal analysis.

Set $L=F-\lambda G$. Integration by parts in its first variation givesTaking compactly supported variations and applying the fundamental lemma of the calculus of variations yields the Euler-Lagrange equation. For the quadratic integrands, it isor $\mathcal{L}y=\lambda ry$ with $\mathcal{L}y=-(p{y}^{\prime }{)}^{\prime }-qy$. If endpoint values are free, independent endpoint variations give the natural conditionsAt a prescribed endpoint the variation instead vanishes and no natural condition is added there.

Under these natural conditions, integration by parts givesHence $\boxed{I=\lambda J}$. For two solutions satisfying the same natural conditions, subtract their cross-multiplied equations and integrate:The final equality is the vanishing endpoint boundary form. Thus distinct eigenvalues giveThe same argument works for other common homogeneous boundary conditions making that boundary form zero.

In the Legendre case, $((1-x^2)y_n^{\prime }{)}^{\prime }+{\lambda }_n{y}_n=0$. Comparing coefficients of $x^n$ in a degree-$n$ solution gives ${\lambda }_n=n(n+1)$, so different degrees have different eigenvalues. A degree-zero solution is $y_0=C_0$. A linear solution $Ax+B$ must be orthogonal to the nonzero constant solution, so $B=0$ and $y_1=C_{1}x$. Write $y_2=A{x}^2+Bx+C$. Orthogonality to $x$ gives $B=0$, and orthogonality to $1$ gives $2A/3+2C=0$. ThereforeThe constants are free normalization factors; for solutions of the stated exact degrees they are nonzero. The natural endpoint conditions hold automatically for these polynomials because $p(\pm 1)=0$.