Chapter 7.6, Problem 18E

Interpretation Introduction

Interpretation:

Consider the Mathieu equation $\ddot{\text{x}}\text{+}\left(\text{a+ εcost}\right)\text{x = 0}$ with $\text{a}\approx 1$. Using two-timing with a slow time ${\text{T = ε}}^{\text{2}}\text{t}$, show that the solution becomes unbounded as $\text{t}\to \infty$ if $\text{1-}\frac{\text{1}}{\text{12}}{\text{ε}}^{\text{2}}\text{+O}\left({\text{ε}}^{\text{4}}\right)\le \text{a}\le \text{1+}\frac{\text{5}}{\text{12}}{\text{ε}}^{\text{2}}\text{+O}\left({\text{ε}}^{\text{4}}\right)$.

Concept Introduction:

The expression of the Mathieu’s Equation is $\frac{{\text{d}}^{\text{2}}\text{x}}{{\text{dt}}^{\text{2}}}\text{+}\left(\text{δ+εcost}\right)\text{x = 0}$. Here, $\text{x}$ is the position, $\text{δ}$ and $\text{ε}$ is the constant parameter.

The power series expansion of x is $\text{x}\left(\text{τ,T}\right){\text{ = x}}_{\text{0}}\left(\text{τ,T}\right){\text{+εx}}_{\text{1}}\left(\text{τ,T}\right)\text{+}\mathrm{....}$.

The partial fraction of the function $\frac{{\text{d}}^{\text{2}}\text{x}}{{\text{dt}}^{\text{2}}}$ is $\frac{{\text{d}}^{\text{2}}\text{x}}{{\text{dt}}^{\text{2}}}\text{ = }{\partial }_{\text{τ}}^{\text{2}}{\text{x}}_{\text{0}}\text{+ε}{\partial }_{\text{τ}}^{\text{2}}{\text{x}}_{\text{1}}{\text{+ε}}^{\text{2}}\left({\partial }_{\text{τ}}^{\text{2}}{\text{x}}_{\text{2}}\text{+2}{\partial }_{\text{τ}}{\partial }_{\text{T}}{\text{x}}_{\text{0}}\right)\text{+O}\left({\text{ε}}^{\text{3}}\right)$.