## 3 Exact Solution for Free Undamped Vibrations and Conclusions for Forced Vibrations  
As part of the descriptive explanation of the observations announced in the previous section, we consider the exact solution in the case of free undamped vibrations:  
\[x^{\prime\prime}(t)+\omega_{0}{}^{2}\;x(t)+\gamma x^{3}(t)\ =\ 0 \tag{6}\]  
This solution can be found in numerous textbooks and papers, often generalized for arbitrary conservative nonlinear restoring or the pendulum equation, e.g., [5]. It is, in fact, based thereon that Equation (6) describes a conservative system, i.e., conservation of energy gives the following:  
\[\frac{1}{2}x^{\prime 2}+\frac{1}{2}\omega_{0}x^{2}+\frac{1}{4}\gamma x^{4}\ =\ E_{0}\ =\ \text{const}. \tag{7}\]  
with total energy \(E_{0}\) depending on the chosen amplitude due to initial conditions. The total energy for a free undamped oscillation with amplitude A is given according to (7) for a state with zero velocity and therefore only potential energy by  
\[E_{0}\ =\ \omega_{0}{}^{2}\;A^{2}+\gamma A^{4} \tag{8}\]  
From Equation (7) follows  
\[\frac{\text{d}x}{\text{d}t}\ =\ x^{\prime}\ =\ \pm\sqrt{2\bigg{(}E_{0}-\frac{1}{ 2}\omega_{0}{}^{2}x^{2}-\frac{1}{4}\gamma x^{4}\bigg{)}} \tag{9}\]  
By separation of variables, the time \(t\) can be computed as a function of the displacement \(x\). Integrating from \(x(t\ =\ 0)\ =\ 0\) to \(x(t)\), we find  
\[t\ =\ \int_{0}^{t}\text{d}\hat{t}\ =\ \int_{0}^{x}\frac{\text{d}\hat{x}}{\pm \sqrt{2(E_{0}-\frac{1}{2}\omega_{0}{}^{2}\hat{x}^{2}-\frac{1}{4}\gamma\hat{x}^ {4}}}, \tag{10}\]  
which is the incomplete elliptic integral of the first kind. Evaluation of Equation (10) with limits 0 and \(A\) gives a quarter of the period \(T\), from which the fundamental circular frequency \(\omega\) of the free vibrations can be calculated by  
\[\omega\ =\ \frac{\pi}{2}\frac{1}{\int_{0}^{A}\frac{\text{d}\hat{x}}{\pm \sqrt{2(E_{0}-\frac{1}{2}\omega_{0}{}^{2}\hat{x}^{2}-\frac{1}{4}\gamma\hat{x}^ {4}}}} \tag{11}\]  
In the following, the inverse function \(x(t)\) to (10) shall be calculated, which requires the usage of Jacobi elliptic functions. Using the prefactor:  
\[c\ =\ \sqrt{\frac{2}{A^{2}\gamma+2\omega_{0}^{2}}} \tag{12}\]  
as well as the parameters:  
\[\varphi\ =\ \text{arcsin}\bigg{(}\frac{x}{A}\bigg{)} \tag{13}\]  
\[m\ =\ -\frac{A^{2}\gamma}{A^{2}\gamma+2\omega_{0}^{2}} \tag{14}\]  
Equation (10) can be written in compact form as  
\[t(x;A,\omega_{0})\ =\ cF(\varphi,m) \tag{15}\]  
In the results in Table 1, it can be seen that for the exact solutions, the lower the fundamental circular frequency \(\omega\) (and the larger the amplitude \(A\)) becomes, the greater the influence of the higher harmonics, and the representation of the solution by just the basic harmonics becomes more and more erroneous. Looking at Figure 3 (left), this means that the closer the amplitude comes to the unstable equilibrium position, the harder it is to approximate the restoring characteristics by a linear one. It should be mentioned again that the plotted solutions show results for \(x(t~{}=~{}0)~{}=~{}0\) and corresponding velocity \(x^{\prime}(t~{}=~{}0)\) in such a way that the given \(\omega\) and \(A\) are realized. Of course, the same solution with an arbitrary phase shift in time is also an exact solution of Equation (6), as there is no trigger by an external excitation. Comparing these observations with the properties of the nose solutions in case of forced vibrations described in the section before, there are direct relations. Question 1 from the previous section for a descriptive explanation of the nose solution can be answered in that way, that the nose solutions are a resonance phenomenon (i.e., large response compared to the excitation) in that sense, that we get, for a certain excitation frequency an amplitude which is close to the amplitude of free vibrations with same fundamental frequency.