= Solution
Away from $z=0,\pm1$, the base profile is linear or constant, so $U''=0$ and $\phi''-\alpha^2\phi=0$. Across a corner $z=z_0$, integration of the Rayleigh equation gives the jump condition
$$
(U(z_0)-c)[\phi']=[U']\phi.
$$
For the even mode, at $z=0$ this gives
$$
\phi'(0^+)=-\frac{\phi(0)}{1-c}.
$$
For $0<z<1$, therefore,
$$
\phi(z)=A\left[
\cosh(\alpha z)-
\frac{\sinh(\alpha z)}{\alpha(1-c)}
\right].
$$
For $z>1$, decay requires $\phi\propto e^{-\alpha(z-1)}$. The jump at $z=1$, where $[U']=1$, gives
$$
\phi'(1^-)=\left(\frac1c-\alpha\right)\phi(1).
$$
Substitution and elementary simplification yield
$$
\boxed{
2\alpha^2c^2+\alpha(1-2\alpha-e^{-2\alpha})c
-[1-\alpha-(1+\alpha)e^{-2\alpha}]=0}.
$$
The coefficients are real, so instability occurs when the quadratic discriminant is negative. At $\alpha=1$ it is
$$
(1+e^{-2})^2-16e^{-2}<0,
$$
whereas at $\alpha=2$, using $e^{-4}\simeq1/55$, it is positive. By continuity there is a first threshold $1<\alpha_s<2$ at which the discriminant vanishes. Hence one conjugate root has $c_i>0$ for
$$
\boxed{1\leq\alpha<\alpha_s<2}.
$$
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