Love wave

Remarks

The relation comes from $f(z) = A \cos(k s_1 z)$ in the layer and $f(z) = B e^{-k s_2 (z - H)}$ in the half-space, with $s_1 = \sqrt{c^2/\beta_1^2 - 1}$ and $s_2 = \sqrt{1 - c^2/\beta_2^2}$. The cosine satisfies the free-surface condition, and matching $f$ and $\mu f'$ at $z = H$ gives $\mu_1 s_1 \sin(k s_1 H) = \mu_2 s_2 \cos(k s_1 H)$.

Example: fix $k > 0$ and let $c$ run from $\beta_1$ to $\beta_2$. The left side starts at $0$ and grows, up to the first pole of the tangent if there is one, while the right side decreases from $+\infty$ to $0$. So the two sides are equal for at least one $c(k)$ in $(\beta_1, \beta_2)$, which gives the fundamental Love mode.

Non-example: a homogeneous half-space, $H = 0$. Then decay with depth needs $c < \beta_2$, and a traction-free surface forces $f' (0) = 0$, which together with $f = B e^{-k s_2 z}$ gives $B = 0$. So there is no Love wave without a slower layer.