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When $|k|h\gg 1$, the two shear layers are separated by many disturbance decay lengths. Since $\coth (|k|h)\to 1$, the roots reduce toThis is exactly the Kelvin-Helmholtz instability of one isolated vortex sheet separating equal-density streams of speeds $U$ and $0$: the pattern travels at their mean speed $U/2$, while its unstable amplitude grows at rate $|k|U/2$. The exponentially weak interaction between the two layers disappears in this short-wavelength limit.

Put $q=|k|$. Decay at infinity and the symmetry of the disturbance suggest the mode potentialsThe middle potential is an even function of $y$ because its vertical derivative must take opposite values at the two interfaces. The upper kinematic conditions giveThe lower conditions give the same two relations. The upper dynamic condition isEliminating $A$, $B$ and ${\eta }_0$ yields the dispersion relationWriting $C=\coth (qh)$, its two roots areFor every real $k\ne 0$, $C>0$ and one root has the positive growth rateThus every nonzero wavenumber is linearly unstable in this inviscid vortex sheet model; $k=0$ is neutral.

The linearized equations are homogeneous and have constant coefficients in $x$ and $t$. Consequently, spatial Fourier modes evolve independently, and the superposition principle reconstructs arbitrary sufficiently regular initial disturbances from them. It is therefore enough to seek the normal modesHere the real $k$ is the wavenumber, $\sigma$ is the possibly complex temporal exponent, and taking the real part produces a physical displacement.

The displacement must have small relative amplitude and small slope,Write the potentials as perturbations of the base flow,with $|\nabla {\phi }_{\pm }|/U$ and $|\nabla {\phi }_0|/U$ small. A linearization evaluates the conditions at the unperturbed lines $y=\pm h$ and discards products of perturbations. At $y=h$ this giveswhereas at $y=-h$ it givesThe perturbation potentials remain harmonic functions, and their gradients decay at the corresponding far fields.

Write the velocity potentials in the upper, middle and lower domains as ${\varphi }_{+}$, ${\varphi }_0$ and ${\varphi }_{-}$. Irrotational flow and incompressible flow imply the Laplace equationThe far-field boundary conditions are

Every shear layer is a material interface. The exact kinematic boundary condition at $y=h+\eta$ isand at $y=-h-\eta$ it isHere each equation is evaluated on the indicated moving interface.

Finally, pressure continuity is the dynamic boundary condition for an inviscid interface. The Unsteady Bernoulli equation givesafter absorbing spatially constant functions of time into the potentials.

The unperturbed fluid wake is the horizontal strip $-h<y<h$: the fluid is stationary there and moves to the right with velocity $U{\mathbf{e}}_x$ above and below it. Its two shear layers are the straight lines $y=\pm h$. After the symmetric disturbance, draw the upper layer at $y=h+\eta (x,t)$ and the lower one as its reflection at $y=-h-\eta (x,t)$, so the local wake width is $2(h+\eta )$.

Suppose ${\epsilon }_{ij}{e}_j={\epsilon }_e{e}_i$. Taking the scalar product of the equation in part (d) with $\mathbf{k}$ gives $\mathbf{k}\cdot \mathbf{e}=0$. The vector triple product then reduces the equation toso $k^2={\omega }^2\mu {\epsilon }_e$. From part (d),ConsequentlyThus the instantaneous flux, except at its zeros, and its time average both point in the wave-propagation direction $\mathbf{k}$.

Substitution of the sinusoidal plane waves into Faraday's law givesThe source-free Ampère-Maxwell equation, with $\mathbf{H}=\mathbf{B}/\mu$ and $D_i={\epsilon }_{ij}{E}_j$, givesEliminating $\mathbf{b}=(\mathbf{k}\times \mathbf{e})/\omega$ yields the wave equation for a plane wave in an anisotropic dielectric:

Because ${\epsilon }_{ij}$ and ${\mu }_{ij}$ are symmetric and time-independent, the linear anisotropic dielectric relations giveand similarlyIntegrate the identity from part (b) over $V$ and apply the divergence theorem to obtainWhen $\mathbf{J}=0$, this is conservation of electromagnetic energy. The Poynting theorem in a linear anisotropic medium identifiesas the energy density andas the energy flux, or Poynting vector.

Dot the Ampère-Maxwell equation with $\mathbf{E}$ and Faraday's law with $\mathbf{H}$:Adding and using the vector identitygivesThis is the local Poynting theorem before specializing the constitutive relations.

The macroscopic Maxwell equations in matter areHere $\rho$ and $\mathbf{J}$ are the free charge and current densities.

An adiabatic process has no heat transfer, $\delta Q=0$. With work done by the gas taken as positive, the first law of thermodynamics givesUsing $dE=C_{V}dT$ and $p=N{k}_{B}T/V$,Integration, followed by the Mayer relation $N{k}_B/C_V=\gamma -1$, yieldsEliminating $T$ with the equation of state gives the reversible ideal-gas adiabat

For $N$ particles, the ideal gas equation of state isThe internal energy of an ideal gas depends only on temperature and obeysFor constant heat capacity and a convenient choice of energy zero,

The Clausius statement of the second law says that no cyclic device can have as its sole effect the transfer of heat from a colder reservoir to a hotter one. The Kelvin-Planck statement of the second law says that no cyclic device can have as its sole effect the extraction of heat from one reservoir and its complete conversion into work.

Suppose first that the Clausius statement fails. Use the forbidden device to return to a hot reservoir the heat $Q_C$ rejected by an ordinary heat engine. The combined device extracts $Q_H-Q_C$ from the hot reservoir and converts all of it into work, violating the Kelvin--Planck statement.

Conversely, suppose the Kelvin--Planck statement fails. Use the work produced by the forbidden one-reservoir engine to drive an ordinary refrigerator. The engine's heat withdrawal and the refrigerator's heat discharge cancel by the work-energy balance, leaving only a transfer of heat from the cold reservoir to the hot reservoir with no external work. This violates the Clausius statement. Each statement therefore implies the other, so the two forms of the Second law of thermodynamics are equivalent.