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Take a horizontally polarized shear displacement $u(y)e^{i(kx-\omega t)}$, with $k>0$ and $c=\omega /k$. In the layer put $q=\sqrt{c^2/{\bar{c}}_s^2-1}$ and in the half-space put $\gamma =\sqrt{1-c^2/c_s^2}$. The required inequality makes both positive. The shear-wave equations givewhere the first form satisfies zero traction at $y=h$ and the second decays as $y\to -\infty$. Displacement and traction continuity at zero give $A\cos (kqh)=B$ and $\bar{\mu }kqA\sin (kqh)=\mu k\gamma B$. Hence the Love wave dispersion relation isThe left-hand square root uses the layer speed ${\bar{c}}_s$, as in the original PDF; the TeX aid incorrectly omits its bar there.

Let $K=kh>0$ and $q_{max}=\sqrt{c_s^2/{\bar{c}}_s^2-1}$. The allowed interval is $0<q<q_{max}$, and the right side becomesa positive strictly decreasing function from infinity to zero. On each positive branch of $\tan (Kq)$ beginning at $Kq=m\pi$, the left side is strictly increasing from zero to infinity, or to a positive endpoint value if $q_{max}$ comes first. There is exactly one intersection on that branch whenever $m\pi <K{q}_{max}$. The fundamental branch $m=0$ thus supplies a real solution for every positive $kh$.

The second branch exists precisely when $K{q}_{max}>\pi$. Thus the strict cutoff of the second Love-wave mode isAt equality the second solution would have $c=c_s$, zero half-space decay exponent, and fails the required disturbance decay. Consequently the cutoff is the infimum, rather than an attained smallest value, under the printed strict speed inequality. Also $kh=0$ itself is a limiting case, not a trapped propagating mode; the “any value” assertion is valid for positive $kh$.