## 1 Introduction
### Nonlinear Effects Under Ion-Relaxation Polarization Conditions
Furthermore, based on (8), we have  
\[\rho(\xi,\tau)=\sum_{\tau=1}^{\infty}\sum_{k=\tau}^{\infty}\
u^{k}\wp_{k}^{( \tau\omega)}(\xi,\tau). \tag{84}\]  
Expression (84) can be written in the form  
\[\rho(\xi,\tau)=\sum_{\tau=1}^{\infty}\rho^{(\tau\omega)}(\xi,\tau), \tag{85}\]  
where  
\[\rho^{(\tau\omega)}(\xi,\tau)=\sum_{k=\tau}^{\infty}\gamma^{k}\wp_{k}^{(\tau \omega)}(\xi,\tau). \tag{86}\]  
According to (86), we derive the following:  
\[\wp_{k}^{(\tau\omega)}(\xi,\tau)=\sum_{n=1}^{\infty}\Re_{k}^{(\tau\omega)}(n,\tau)\cdot\cos\Big{(}\frac{\gamma ma}{d}\xi\Big{)}. \tag{87}\]  
From (86) and (87), we have  
\[\rho^{(\tau\omega)}(\xi,\tau)=\sum_{n=1}^{\infty}\sum_{k=\tau}^{\infty}\
u^{ k}\Re_{k}^{(\tau\omega)}(n,\tau)\cdot\cos\Big{(}\frac{\gamma ma}{d}\xi \Big{)}. \tag{88}\]  
At \(r\) = 1,2,3, Expression (88) yields  
\[\rho^{(\omega)}(\xi,\tau)=\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\gamma^{k} \Re_{k}^{(\omega)}(n,\tau)\cdot\cos\Big{(}\frac{\gamma ma}{d}\xi\Big{)}, \tag{88a}\]  
\[\rho^{(2\omega)}(\xi,\tau)=\sum_{n=1}^{\infty}\sum_{k=2}^{\infty}\gamma^{k} \Re_{k}^{(2\omega)}(n,\tau)\cdot\cos\Big{(}\frac{\gamma ma}{d}\xi\Big{)}, \tag{88b}\]  
\[\rho^{(3\omega)}(\xi,\tau)=\sum_{n=1}^{\infty}\sum_{k=3}^{\infty}\gamma^{k} \Re_{k}^{(3\omega)}(n,\tau)\cdot\cos\Big{(}\frac{\gamma ma}{d}\xi\Big{)}. \tag{88c}\]  
Substituting (73) into (88a) and (75) into (88c) yields expressions coinciding with (A12a) and (A12b), respectively. When constructing function (83), by taking into account (88c), we obtain  
\[ \begin{array}{c}\rho^{(3\omega)}(\xi,\tau)=-\Big{(}\frac{4a}{d}\Big{)}^{3}n _{0}\gamma^{3}\times\sum_{n=1}^{\infty}\bigg{\{}n^{2}\wp_{3}(n)\times\sum_{k= 3}^{\infty}\frac{(k-1)(k-2)}{2}\cdot\Big{(}\frac{8n_{0}\phi_{1}\Lambda_{0} \gamma}{\pi^{2}}\Big{)}^{k-3}+\\ +\Lambda_{1}\sin^{2}(\frac{\gamma m}{2})\times\sum_{k=3}^{\infty}\Big{(}\frac{8 n_{0}\phi_{1}\Lambda_{0}\gamma}{\pi^{2}}\Big{)}^{k-3}\cdot\sum_{g-3}^{k-1}\frac{(g-1)(g-2)}{2} \Lambda_{0}^{g-3}\cdot\Lambda_{2}^{k-g-1}\bigg{\}}\times\\ \times\frac{\exp(3i\frac{\omega}{\pi^{2}(0)}\tau)}{\frac{1}{\pi_{2}}+i\Big{(} \frac{3}{\omega^{0}(0)}\tau)}\times\cos(\frac{\gamma ma}{d}\xi),\end{array}  \tag{89}\]  
whence, after calculating the sums of series, we obtain  
\[ \begin{array}{c}\rho^{(3\omega)}(\xi,\tau)=-\frac{64a^{3}\eta_{0}\gamma^{3}}{ \frac{3}{3}\Big{(}1-\frac{8n_{0}\phi_{1}\Lambda_{0}\gamma}{\pi^{2}}\Big{)}^{3} }\times\sum\limits_{n=1}^{\infty}\Bigg{\{}n^{2}\wp_{3}(n)+\frac{8n_{0}\phi_{1} \Lambda_{1}\gamma}{\pi^{2}}\cdot\frac{\sin^{2}(\frac{\gamma m}{2})}{1-\frac {8n_{0}\phi_{1}\Lambda_{2}\gamma}{\pi^{2}}}\Bigg{\}}\times\\ \times\frac{\exp(3i\frac{\omega}{\pi^{2}(0)}\tau)}{\frac{1}{\pi_{2}}+i\Big{(} \frac{3}{\omega^{0}(0)}\tau)}\times\cos(\frac{\gamma ma}{d}\xi).\end{array}  \tag{90}\]In continuation of (A13),  
\[\mathrm{P}^{(3\omega)}(\tau)=\frac{\mathrm{q}}{\mathrm{d}}\int_{0}^{\mathrm{d}} \mathrm{x}\rho^{(3\omega)}(\xi,\tau)\mathrm{d}\mathrm{x}, \tag{91}\]  
and by taking into account (90), we calculate the total polarization of the dielectric at the second odd frequency \((3\omega)\):  
\[\mathrm{P}^{(3\omega)}(\tau)=\frac{128\mathrm{a}^{3}\mathrm{a} \mathrm{r}\mathrm{o}\mathrm{r}^{3}}{\mathrm{d}^{2}\pi^{2}\left(1-\frac{8 \mathrm{a}\mathrm{r}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{r }\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{r}\mathrm{o}} \right)^{3}}\times\sum_{n=1}^{\infty}\left\{\varphi_{3}(\mathrm{n})+\frac{8 \mathrm{a}\mathrm{r}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{r }\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{r }\mathrm{o}}\right\}\times\sin^{2}\left(\frac{\pi n}{2}\right)\times \tag{92}\]  
Expression (92), in addition to expressions (A14a) and (A14b), confirms that odd relaxation modes \(\rho_{\mathrm{k}}^{(3\omega)}(\xi,\tau)\), as well as \(\rho_{\mathrm{k}}^{(\omega)}(\xi,\tau)\), make a non-zero contribution to polarization.