= Solution
For zero <damping> the conserved <mechanical energy> is
$$
\boxed{H=\tfrac12\dot x^2+V(x)=\tfrac12v^2,\qquad V(x)=\tfrac12x^2-\tfrac\gamma4x^4.}
$$
For $\gamma<0$, the potential grows to infinity in both directions, so every nonstationary finite-energy trajectory is trapped between two <turning points> and is periodic; the zero-energy solution is stationary. The case $\gamma=0$ is the <harmonic oscillator>. The negative-$\gamma$ comparison extends the parameter range initially stated in the source.
For $\gamma>0$, the origin lies in a local potential well, with barriers at $x=\pm\gamma^{-1/2}$ of height $1/(4\gamma)$. A trajectory launched from the origin oscillates precisely when its initial energy is below that barrier:
$$
\boxed{|v|<\frac1{\sqrt{2\gamma}}.}
$$
At equality the trajectory approaches a barrier <equilibrium> asymptotically, rather than being periodic; for example the positive branch is $x=\gamma^{-1/2}\tanh(t/\sqrt2)$. Above the barrier the trajectory escapes the well. The source's inequality without the absolute value tacitly takes $v\geq0$; for signed initial velocity the energy criterion requires $|v|$.
Next set $\beta=\gamma=0$ and assume $\alpha>0$. The energy $H=(\dot x^2+x^2)/2$ satisfies $\dot H=-\alpha x^2\dot x^2\leq0$. This bounds the complete trajectory. The zero-dissipation set is the union of $x=0$ and $\dot x=0$, and its only invariant trajectory is the origin: a nonzero point on either axis immediately leaves it under the differential equation. The <LaSalle invariance principle> therefore gives $\boxed{x(t)\to0,\quad\dot x(t)\to0}$. If $\alpha=0$ as well, <damping> is absent and that conclusion fails for nonzero $v$; the positive order-one <damping> assumption is essential.
For the weakly nonlinear periodic motion, write $\gamma=\alpha\kappa$, set $\tau=\omega t$, $\omega=1+\alpha\omega_1+\cdots$, and expand $x=A\cos\tau+\alpha x_1+\cdots$. The first correction obeys
$$
x_1''+x_1=A(A^2/4-\beta)\sin\tau+
(2\omega_1A+3\kappa A^3/4)\cos\tau
+\frac{A^3}4\sin3\tau+\frac{\kappa A^3}4\cos3\tau.
$$
Resonant forcing at the <fundamental frequency> would generate <secular terms>. Eliminating both coefficients yields
$$
\boxed{A=2\sqrt\beta,\qquad\omega_1=-\frac{3\kappa A^2}{8},\qquad
\omega=1-\frac32\gamma\beta+O(\alpha^2).}
$$
Consequently the maximum displacement is $\boxed{2\sqrt\beta+O(\alpha)}$ for fixed positive $\beta$. A possible third-harmonic correction is $x_1=-A^3\sin3\tau/32-\kappa A^3\cos3\tau/32$, up to homogeneous <amplitude> and <phase> choices. The leading <amplitude> also balances averaged <damping>: its slow evolution is $\dot A=\alpha A(\beta/2-A^2/8)$, whose nonzero fixed point is attracting. The case $\beta=0$ is a degenerate zero-<amplitude> limit, rather than a nonzero <limit cycle>.
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