The second approximation

Let us consider the equation to second order in $\varepsilon$:

$$\frac{\partial^2\varphi_2^{\vphantom{+}}(x,\tilde t)}{\parti... ...^{\vphantom{+}}(x,\tilde t) \varphi_0^2(x,\tilde t). \eqno(10)$$

If all diagonal coefficients of $\varphi_1^{\vphantom{+}}(x,\tilde t)$ are zeros: $\forall n$ : $b_{nn}^{\vphantom{+}}=0$, then $\forall j,n$ : $b_{jn}^{\vphantom{+}}=-b_{nj}^{\vphantom{+}}$, and the function $\varphi_1^{\vphantom{+}}(x,\tilde t)\varphi_0^2(x,\tilde t)$ has no diagonal harmonics. Hence, selecting $\omega_2^{\vphantom{+}}=-\frac{1}{2}\omega_1^2$ , we obtain a periodic solution to equation (10):

$$\varphi_2^{\vphantom{+}}(x,\tilde t)\equiv\sum_{n=1}^\infty ... ...}^{\vphantom{+}}\sin(nx)\sin(n\tilde t) \mbox{, \ \ \ where }$$

$$H(x,\tilde t)\equiv 2\omega_1^{\vphantom{+}}\frac{\partial^2 ... ...1}^\infty\:\sum_{j=1}^\infty H_{nj} \sin(nx)\sin(j\tilde t).$$

It should be noted that all diagonal coefficients $h_{nn}^{\vphantom{+}}$are arbitrary numbers.