= Solution
Eliminating the <phase velocity> from the first two relations yields
$$
\ell^2=k^2\left(1-\frac{c_1^2}{c_2^2}\right)-\frac{c_1^2}{c_2^2}m^2.
$$
With $\theta=mh$, put
$$
\boxed{a=h\sqrt{\frac{c_2^2}{c_1^2}-1}
=h\sqrt{\frac{\mu_2\rho_1}{\mu_1\rho_2}-1},\qquad
\epsilon=\frac{\mu_1c_2}{\mu_2c_1}
=\sqrt{\frac{\mu_1\rho_1}{\mu_2\rho_2}}.}
$$
Then $\ell=(c_1/(c_2h))\sqrt{a^2k^2-\theta^2}$. The <dispersion relation> becomes
$$
\boxed{\tan\theta=-\frac{\epsilon\theta}{\sqrt{A^2-\theta^2}},\qquad
0<\theta<A,\qquad A=ak.}
$$
Here $a>0$ follows from $c_1<c_2$. The parameter $a$ has dimensions of length; $A$ and $\epsilon$ are dimensionless.
For the <strict mode count for a rigid-backed elastic layer>, consider
$$
g(\theta)=-\frac{\epsilon\theta}{\sqrt{A^2-\theta^2}},\qquad
g'(\theta)=-\frac{\epsilon A^2}{(A^2-\theta^2)^{3/2}}<0.
$$
The right side is negative, so intersections occur only on the negative branches of the <tangent>,
$$
p_n<\theta<n\pi,\qquad p_n=(n-\tfrac12)\pi,\quad n=1,2,\ldots.
$$
On each such branch $\tan\theta-g(\theta)$ is strictly increasing. If $p_n<A$, its limit at $p_n^+$ is negative infinity. At the other end of the available interval, it is positive: either $\theta\to n\pi^-$ with $n\pi<A$, or $\theta\to A^-$ with $A\leq n\pi$. The <intermediate value theorem> and strict <monotonicity> give exactly one intersection. No such interval exists when $p_n\geq A$.
\Image[/past-exam-of-the-elasticity-and-plasticity-course-of-the-university-of-oxford/2023-root-count.png]
{title=Intersections of the negative tangent branches with the interface loading curve. A branch beginning exactly at the cutoff contributes no trapped mode.}
Thus the exact answer, including all cutoff values, is
$$
\boxed{N=\#\{n\geq1:(n-\tfrac12)\pi<A\}
=\left\lceil\frac{A}{\pi}-\frac12\right\rceil.}
$$
Away from $A=(n-1/2)\pi$, this equals the printed <floor function> expression $\lfloor A/\pi+1/2\rfloor$. At $A=(n-1/2)\pi$, the printed expression is one too large: the $n$th interval is empty and the limiting state has $\ell=0$, contradicting <exponential decay>. This exception is present in the native paper, rather than introduced by transcription.
Now hold $A$ fixed and let $\epsilon\to0^+$. Neglecting the right side gives $\theta_n\simeq n\pi$ and hence the candidate <phase velocities>
$$
\boxed{c_n\simeq c_1\sqrt{1+\frac{n^2\pi^2}{k^2h^2}}.}
$$
The <zero> $\theta=0$ is excluded: it would require $c=c_1$, already ruled out. For every $n\pi<A$, the candidate is a valid leading <asymptotic expansion>. A first correction, obtained by expanding the <tangent> about $n\pi$, is
$$
\theta_n=n\pi-\frac{\epsilon n\pi}{\sqrt{A^2-n^2\pi^2}}+O(\epsilon^2).
$$
The <big O notation> here assumes a fixed positive gap between $n\pi$ and $A$. In particular, when $A>N\pi$, all $N$ leading approximations are valid. The <phase velocity> increases strictly with $\theta$, so only the last branch can cause an endpoint difficulty.
If $(N-1/2)\pi<A<N\pi$, the formal last <equation root> $N\pi$ lies outside the allowed interval. The actual last <equation root> approaches $A$, where the small factor $\epsilon$ is balanced by the vanishing denominator. This is a <singular perturbation>. For fixed $A$ strictly inside that interval, write $\theta_N=A-\delta$. Since $\tan A<0$,
$$
\tan A\sim-\frac{\epsilon A}{\sqrt{2A\delta}},\qquad
\boxed{\delta\sim\frac{\epsilon^2A}{2\tan^2A},\quad
\theta_N=A-\frac{\epsilon^2A}{2\tan^2A}+O(\epsilon^4).}
$$
The valid <endpoint approximation for a rigid-backed shear mode> consequently gives
$$
\boxed{c_N=c_2\left[1-\frac{\epsilon^2}{2\tan^2A}
\left(1-\frac{c_1^2}{c_2^2}\right)+O(\epsilon^4)\right].}
$$
Here we used $c^2=c_1^2(1+\theta^2/(kh)^2)$ and $A^2/(kh)^2=c_2^2/c_1^2-1$. Thus this <normal mode> approaches the upper <wave speed> from below; it does not approach the inadmissible value obtained from $\theta=N\pi>A$.
There is a further exceptional fixed value $A=N\pi$. Put $\theta_N=A-\delta$ again. Now $\tan(A-\delta)\sim-\delta$, so the balance changes to
$$
\delta\sqrt{2A\delta}\sim\epsilon A,\qquad
\boxed{\delta\sim\left(\frac{\epsilon^2A}{2}\right)^{1/3},\qquad
c_N\sim c_2\left[1-\left(1-\frac{c_1^2}{c_2^2}\right)\frac{\delta}{A}\right].}
$$
In the last formula the bracket specifies the first nonzero correction. The zeroth-order candidate $c_N=c_2$ is correct as a limit but misses this correction and the positive exterior <exponential decay> rate. More generally, close to $A=N\pi$, let $\Delta=A-N\pi$ and $\theta_N=N\pi-\delta$. For $|\Delta|,\delta\ll1$, a transition approximation valid on both sides is
$$
\boxed{\delta^2(\Delta+\delta)\simeq\frac{\epsilon^2N\pi}{2},\qquad
\delta>0,\quad\Delta+\delta>0,\qquad
c_N\simeq c_1\sqrt{1+\frac{(N\pi-\delta)^2}{(kh)^2}}.}
$$
It recovers the $\epsilon^{2/3}$ balance at $\Delta=0$ and the ordinary expansion above the transition. The fixed-endpoint formula involving $\tan A$ should be used when $A$ is below, and bounded away from, $N\pi$.
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