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Away from the first parametric resonance, one forcing interaction changes a harmonic frequency by one and produces no resonant coupling of the two conjugate oscillations. Two interactions can couple frequency $+1$ to frequency $-1$, so the next parametric resonance is near $\omega =1$ and its effect first appears at order ${ϵ}^2$. This gives $p=2$ and the appropriate slow time $T={ϵ}^{2}t$.

Write $\omega =1+\kappa {ϵ}^2$ and $y_0=a(T)\cos t+b(T)\sin t$. The first correction satisfies $({\partial }_t^2+1)y_1=-\cos t{y}_0$, whose bounded particular solution isAt second order,The resonant part of $-\cos t{y}_1$ is $(5a/12)\cos t-(b/12)\sin t$. Applying the solvability condition in the method of multiple scales givesThe squared slow eigenvalue isThe second instability tongue of a weak Mathieu oscillator therefore has the leading intervalHere the two little-$o$ terms describe the respective exact edge locations; the strict inequality in $\kappa$ is the leading interior criterion. In particular $\omega =1$ is unstable, with physical growth rate ${ϵ}^2\sqrt{5}/24+o({ϵ}^2)$. The general leading solution for bounded $T$ again has $y_0=a(T)\cos t+b(T)\sin t$, where the matrix exponential of the displayed two-by-two amplitude system evolves arbitrary $a(0),b(0)$. Adding $ϵy_1$ identifies the first nonresonant harmonic correction. The leading expression alone has error $O(ϵ)$ on $t=O({ϵ}^{-2})$.

For comparison, at fixed $\omega \ne 1/2,1$, write $y_0=A(T)e^{i\omega t}+\overline{A}(T)e^{-i\omega t}$. The first forced harmonics have coefficients $A/[2(2\omega +1)]$ and $A/[2(1-2\omega )]$. Their return to the original harmonic at the next order givesIt is a phase correction, not amplitude growth. Consequently, away from the first two resonance neighborhoods, a general leading solution valid for bounded ${ϵ}^{2}t$ isNarrower higher-order instability intervals are invisible on this time scale. The detuned resonant calculation above is needed near $\omega =1$ because both the phase shift and the conjugate-mode coupling then matter.

Apply the method of multiple scales with independent fast time $t$ and slow time $T={ϵ}^{2}t$. A preliminary slow time $ϵt$ produces no resonant forcing at first order, so its leading amplitudes are constant; the first nontrivial modulation is at $T$. Write $x=x_0+ϵx_1+\cdots$, withAt order $ϵ$, the inhomogeneous linear differential equation is $({\partial }_t^2+1)x_1=-\cos t{x}_0$. Absorbing its homogeneous part into the amplitudes givesAt order ${ϵ}^2$, the forcing is $-2{\partial }_t{\partial }_T{x}_0-k{x}_0-\cos t{x}_1$. Its resonant Fourier series coefficients areThe solvability condition in the method of multiple scales sets both coefficients to zero, giving the amplitude equations. DefineFor bounded initial conditions $x(0)=x_{\ast }$, $\dot{x}(0)=v_{\ast }$, the requested leading approximation isThis holds on bounded intervals of $T$, hence $t=O({ϵ}^{-2})$, with the constant in the error allowed to depend on the bounded $T$ interval. If desired, retain the displayed $ϵx_1$ and adjust $A(0),B(0)$ by $O(ϵ)$ to match initial conditions at that order. No particular initial conditions were supplied.

Since $M^2=abI$, the matrix exponential is $I\cosh (\sigma T)+(M/\sigma )\sinh (\sigma T)$ when ${\sigma }^2=ab>0$. When $ab<0$, put $\omega =\sqrt{-ab}$ and replace this with $I\cos (\omega T)+(M/\omega )\sin (\omega T)$. At $ab=0$ it is $I+TM$. Thus the leading second instability tongue of a weak Mathieu oscillator has exponentially growing solutions for $-1/12<k<5/12$, and the leading modulation is bounded for every initial condition precisely whenAt either displayed endpoint, $M$ is a nonzero nilpotent operator; some initial conditions give a secular term proportional to $T$. This leading calculation alone must not be mistaken for an exact finite-$ϵ$ stability decision at the endpoints.

For an exact small-$ϵ$ interpretation, put $t=2s$. The standard Mathieu equation parameters are $a_{\mathrm{M}}=4+4k{ϵ}^2$ and $q=-2ϵ$. Its two relevant Mathieu characteristic values are obtained by projecting $-d^2/d{s}^2+2q\cos 2s$ onto even and odd Fourier series modes. Near the unperturbed value $4$, the even modes $1,\cos 2s,\cos 4s,\cos 6s$ have diagonal entries $0,4,16,36$ and adjacent couplings $\sqrt{2}q,q,q$ in an orthonormal basis; the odd modes $\sin 2s,\sin 4s,\sin 6s$ have diagonal entries $4,16,36$ and couplings $q,q$. Substituting $4+c_2{q}^2+c_4{q}^4$ into their characteristic polynomials givesHigher modes of the Fourier series first contribute at higher order. Therefore the fourth-order edges of the second Mathieu instability tongue areFloquet theory gives boundedness of every solution outside the closed gap, $k<k_{-}(ϵ)$ or $k>k_{+}(ϵ)$, in this fixed-$k=O(1)$ neighborhood. At the actual gap edges the second independent solution is generically unbounded. In particular, if $k$ is fixed exactly at $-1/12$ or $5/12$, it lies just outside the true gap and is stable for each sufficiently small positive $ϵ$. The strict boxed criterion is the robust leading-order, uniformly separated criterion; the endpoints require this higher-order refinement, and their amplification bounds need not remain uniform as $ϵ\to 0$.

Use the fast time $t$ and slow time $T={ϵ}^{2}t$, and writeThere is no $O(ϵ)$ resonant forcing. Solvinggives a convenient particular solutionAt $O({ϵ}^2)$, the coefficient of $e^{it}$ in the forcing must vanish by the solvability condition in the method of multiple scales. This givesorThe two slow exponents satisfyThe second instability tongue of a weak Mathieu oscillator has real $s$ when $-1/12<f<5/12$. At either endpoint the repeated zero exponent permits a linearly growing slow solution, so boundedness for every initial condition requires the strict stable ranges