## Appendix A
For \(T_{e}\) = 2 eV we obtain \(\langle n_{e}\rangle_{\rm max}\) = 3 \(\times\) 10\({}^{19}\) cm\({}^{-3}\), and the total maximum value  
\[dN_{e}/dt\ =\ N_{e}/\theta(I), \tag{10}\]5  
It should be noted that the value \(N_{e}\) relates only to the partial ionization of the monomolecular surface layer of the bubble liquid wall. For the total ionization of the layer, we have the estimate \(4\pi{R_{0}}^{2}\times n^{2/3}\approx 1.2\times 10^{6}\), where \(n_{l}\) is the density of liquid molecules, \(n_{l}\) = 3.3 \(\times\) 10\({}^{22}\) cm\({}^{-3}\). Let us find now the distribution of \(n_{e}(r)\) for \(N_{e\rm max}\) = 2 \(\times\) 10\({}^{5}\). In accordance with Equation (A9), \(x\) = \(x_{\rm max}\)\(\approx\) 8.4, and therefore \(n_{e}(0)\)\(\approx\) 5 \(\times\) 10\({}^{7}\) cm\({}^{-3}\), \(n_{e}(R_{0})\)\(\approx\) 1.33 \(\times\) 10\({}^{20}\) cm\({}^{-3}\), \(a_{e}\) = 6 \(\times\) 10\({}^{-7}\) cm. Below, we will use the results (A6)-(A11) when examining the expansion of the bubstons during the SOC process. Let us now estimate the value of light wave intensity \(I\), for which the mean electron density \(\langle n_{e}\rangle\) = \(\langle n_{e}\rangle_{\rm max}\), resulting from the electron avalanche, accumulates inside the bubston prior to its expanding; we will term this stage as the process of inertial retention. As will be shown later, in this case the light intensity coincides with the threshold for the SOC effect, and therefore we denote it by \(I_{thr}\). We denote the inertial retention duration as \(t_{0}\). Assuming that, while expanding, the nanobubble keeps its spherical form, dependent on the initial radius \(R_{0}\), let us introduce the quantitative parameter \(\zeta\) of the retention extent as  
\[dN_{e}/dt\ =\ N_{e}/\theta(I), \tag{10}\]6 (A12)  
where \(u(t)\) = \(dR/dt\), and \(\zeta\) < 1; that is, the radius of the bubston within the time \(t_{0}\) grows less than twice. It means that the retention time \(t_{0}\) is a function of \(\zeta\). The specific value of \(\zeta\) is, of course, dependent on the required accuracy of the calculations. This question, however, is facilitated by a weak dependence of \(t_{0}\) on \(\zeta\). In order to evaluate \(t_{0}\) = \(t_{0}(\zeta)\) it is necessary to know the rate \(u(t)\). The bubston expansion develops due to quickly increasing electron pressure \(p_{e}\) = \(p_{e}(t)\), which is caused by the exponential growth of electron number \(N_{e}\) (according to Equation (A2)). Hence, under the inertial retention conditions, the mean electron density  
\[dN_{e}/dt\ =\ N_{e}/\theta(I), \tag{10}\]7 (A13)  
where \(L_{\rm max}\) = ln[\(N_{e\rm max}/N_{e}(0)\)] = ln(1.4 \(\times\) 10\({}^{5}\)) = 11.8 (it is assumed here that \(N_{e}(0)\) = 1). Providing that \(p\) >> \(p_{0}\), where \(p_{0}\) is the hydrostatic pressure, the velocity \(u(t)\) obeys the equation (see [52])  
\[dN_{e}/dt\ =\ N_{e}/\theta(I), \tag{10}\]8 (A14)  
Here, \(\rho\) is the density of liquid. At \(t\) = 0 the radius \(R\) = \(R_{0}\), and the velocity \(u(t)\) can be deduced from the linearized equation (A14), i.e., when the second term is neglected, and \(R\) = \(R_{0}\). Thus, for the velocity \(u(t)\) we obtain  
\[dN_{e}/dt\ =\ N_{e}/\theta(I), \tag{10}\]9 (A15)  
The pressure \(p_{e}\) is the sum of kinetic electron gas pressure \(p_{gas}\) = \(ne(R_{0})T_{e}\) and the pressure \(p_{Coul}\) = \(E^{2}(R_{0})/8\pi\), caused by the Coulomb repulsion of electrons. Here, \(E(r)\) = \(-d\rho/dr\) is the electric field strength inside a bubston. According to Equation (A6), we have  
\[\theta^{-1}(I)\ =\ \frac{\langle v_{e}\rangle w_{a,e}}{R_{0}}\bigg{(}\frac{I}{I_ {0}}-1\bigg{)}. \tag{11}\]0 (A16)and, using Equation (A9) at \(x>1\), we find for \(p_{Coul}\)  
\[\theta^{-1}(I)\ =\ \frac{\langle v_{e}\rangle w_{a,e}}{R_{0}}\bigg{(}\frac{I}{I_ {0}}-1\bigg{)}. \tag{11}\]1  
where \(\sigma=eN_{e}/4\pi R_{0}{}^{2}\) is the surface charge density.