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For incompressible inviscid flow without body force,Taking curl givesequivalently ${\partial }_t\omega =\nabla \times (u\times \omega )$.

Direct differentiation of the stated field gives $\omega =u$. Hence $u\times \omega =0$, so the vorticity equation gives ${\partial }_t\omega =0$ and the field is steady. FinallyEuler's equation therefore impliesso $H$ is spatially uniform.

Maxwell equations areTaking the divergence of the Ampère-Maxwell equation and using Gauss's law givesThereforebecause $J$ vanishes on the boundary. Componentwise integration by parts also giveswith the boundary term again zero.

Let $r_1,r_2$ be the roots of $\alpha r^2+\beta r+\gamma =0$. The causal Green function is $G(t,\tau )=H(t-\tau )g(t-\tau )$, wherefor distinct roots, and $g(s)=s{e}^{rs}/\alpha$ for a repeated root. These formulas enforce $g(0)=0$ and $\alpha g^{\prime }(0)=1$, hence the required delta jump.

For the oscillator,If $\lambda \ne \omega$, this isAt resonance $\lambda =\omega$, the limiting expression is

Taking curls givesBoth have zero divergence and curl, so they satisfy the static vacuum equations. Their upward fluxes through the loop areFor clockwise current $I$, the net force is zero in $B_1$, while in $B_2$

Put $K=2\pi b_2{r}^2$. Motion in $B_1$ changes no flux, so no current is induced and $z=z_0-g{t}^2/2$. In $B_2$, using counterclockwise current as positive,ThusWith $\Gamma =K^2/R$, the velocity tends to $-mg/\Gamma$ andAt terminal speed, $I^{2}R=mg|\dot{z}|$, so Joule loss equals gravitational-energy loss. For $R=0$, Ohm's quasistatic formula is singular: flux is conserved, persistent current stores magnetic energy, and ideal motion has no resistive terminal state (inductance must be retained).

The commutator follows directly on a test function:Expansion givesTo make $V_{-}=2m$, takeThenwhich tends to $2m$ at both ends.

Applying $p+if$ to $e^{ikx}$ gives a partner scattering state with asymptotic amplitudes $\hslash k-2mi$ at $-\infty$ and $\hslash k+2mi$ at $+\infty$, and no reflected wave. Their moduli agree. Subtracting the common constant $2m$ therefore shows that the displayed $-4m{\mathrm{sech}}^2$ potential is reflectionless:

Transverse force balance on a short element givesWith fixed ends,Here $A_n=0$. If $b_n$ are the sine coefficients of initial velocity, thenThe total deposited energy is $E=\rho v^2/2$, whileThe seventh mode is eliminated by striking at a node $l=jL/7$, $j=1,\dots ,6$. For a narrow hammer, each fixed-mode fraction is $O(\epsilon )$; beyond the inverse-width cutoff the envelope falls as $n^{-2}$, modulated by the two sine factors.