Use the fast time and slow time , and write
There is no resonant forcing. Solving
gives a convenient particular solution
At , the coefficient of in the forcing must vanish by the solvability condition in the method of multiple scales. This gives
or
The two slow exponents satisfy
The second instability tongue of a weak Mathieu oscillator has real when . At either endpoint the repeated zero exponent permits a linearly growing slow solution, so boundedness for every initial condition requires the strict stable ranges

Articles by others on the same topic (0)

There are currently no matching articles.