Faraday's law is $\mathcal{E}={\oint }_C(E+v\times B)\cdot dl=-d{\Phi }_B/dt$. Here $I=-L(B\dot{x}+x\dot{B})/R$ (sign relative to the chosen circuit orientation), and the bar force is $ILB$. For $B=B_0/(1+t)$ this gives the displayed equation with $\alpha =B_0^2{L}^2/(mR)$. At $\alpha =1$, the independent solutions are $u$ and $u{e}^{1/u}$, $u=1+t$; the data give $x=x_{0}u{e}^{1/u-1}$. Thus $\dot{x}\to x_0/e$ and total mechanical work is $m{x}_0^2/(2e^2)$. The current circulates to reinforce the decaying $+z$ flux, and its force drives increasing area, both opposing the flux decrease as Lenz's law requires.

With the proton fixed, $[-{\hslash }^2{\nabla }^2/(2m_{\mu })-e^2/(4\pi {\epsilon }_{0}r)]\psi =E\psi$: kinetic energy plus Coulomb attraction, assuming nonrelativistic point particles and neglecting spin, radiation, and proton motion (or replacing $m_{\mu }$ by the reduced mass). Substitution of the $2s$ form gives $E=-{\hslash }^2{\alpha }^2/(8m_{\mu })$ and $\alpha =m_{\mu }{e}^2/(4\pi {\epsilon }_0{\hslash }^2)$. Direct differentiation verifies it. The curve starts positive, crosses at $r=2/\alpha$, then approaches zero from below; ordinary hydrogen has the same shape but the smaller electron mass gives a larger radial scale.

Separation gives angular modes $e^{im\theta }$ and $-[rR\prime ]\prime +m^{2}R/r=\lambda rR$. Frobenius gives $p=m$ and $a_k=-a_{k-2}/[k(k+2m)]$, producing $J_m$; termwise differentiation gives the two stated recurrences. A radial solution is $u={\sum }_n[A_n\cos (c{j}_{0n}t)+B_n\sin (c{j}_{0n}t)]J_0(j_{0n}r)$. Here $B_n=0$ andequivalently $A_n=2[j_{0n}{J}_0\prime (j_{0n}{)}^2{]}^{-1}{\int }_0^{1}r{J}_1(j_{0n}r)dr$ after integration by parts.

The inverse function theorem says that a $C^1$ map with invertible derivative at $x$ restricts to a $C^1$ diffeomorphism between neighbourhoods of $x$ and $f(x)$. It makes every zero isolated. Compactness of $\bar{\Omega }$, together with separation from zero on the boundary, then makes the zero set finite and interior. Continuity of $Df$ gives the displayed radii. Choose disjoint balls around the zeros and a positive lower bound for $|f|$ on their complement and boundary. A sufficiently small $C^1$ perturbation has no zeros outside, while the quoted local lemma gives exactly one zero in each ball with the same determinant sign. Summing proves $N(g)=N(f)$.

Replace the quadratic constraint by the two linear inequalities $-2\le x_1-x_2+x_3\le 2$. Simplex pivots lead to the active constraints $x_1=0$, $-x_2+x_3=2$, and $3x_2+x_3=4$. Thus $x^{\ast }=(0,1/2,5/2)$ and the value is $3$. With right-hand side $4+\epsilon$, the same basis remains optimal for small nonzero $\epsilon$, giving $x^{\ast }=(0,1/2+\epsilon /4,5/2+\epsilon /4)$.