A modification of He’s frequency-amplitude formulation
He’s original frequency formulation was derived from an ancient Chinese mathematical algorithm [10, 13], recently there was a hot discussion on He’s original frequency formulation [14-22]. Here we check Ren-Hu’s modification [14].
Rewrite the problem (1) in this form
\(\ddot{x}\left(1+x^{2}\right)+x^{3}=0\),\(\ x\left(0\right)=A\), \(\dot{x}\left(0\right)=0\). (10)
In view of He’s frequency-amplitude formulation [7-9], we use a trial solution
\(x_{1}=Acos\omega_{1}t\). (11)
Taking equation (11) into the first equation of problem (10) results in the residual
\(R_{1}={-\omega}_{1}^{2}A\cos{\omega_{1}t}\left(1+A^{2}\cos^{2}\omega_{1}t\right)+A^{3}\operatorname{}{\omega_{1}t}\)
=\(\left[\frac{3A^{3}}{4}\left(1-\omega_{1}^{2}\right)-\omega_{1}^{2}A\right]\cos{\omega_{1}t+\frac{A^{3}}{4}\left(1-\omega_{1}^{2}\right)}\operatorname{cos3}{\omega_{1}t}\). (12)
Introducing defined in [14]
\({\tilde{R}}_{1}=\frac{4}{T}\int_{0}^{\frac{T}{4}}{R_{1}\left(t\right)\text{dt}}\), (13)
with T the period of the oscillator.
Submitting equation (12) into equation (13), we obtain
\({\tilde{R}}_{1}=\frac{4}{T}\int_{0}^{\frac{T}{4}}{\left\{\left[\frac{3A^{3}}{4}\left(1-\omega_{1}^{2}\right)-\omega_{1}^{2}A\right]\cos{\omega_{1}t+\frac{A^{3}}{4}\left(1-\omega_{1}^{2}\right)}\operatorname{cos3}{\omega_{1}t}\right\}\text{dt}}\)
\(=\frac{2}{\pi}\left[\frac{3A^{3}}{4}\left(1-\omega_{1}^{2}\right)-\omega_{1}^{2}A\right]-\frac{\left(1-\omega_{1}^{2}\right)A^{3}}{6\pi}\)(14)
Let\(\text{\ \ x}_{2}=Acos\omega_{2}t\) with\(\omega_{1}\neq\omega_{2}\), by the similar operation as the above, we have
\({\tilde{R}}_{2}=\frac{2}{\pi}\left[\frac{3A^{3}}{4}\left(1-\omega_{2}^{2}\right)-\omega_{2}^{2}A\right]-\frac{\left(1-\omega_{2}^{2}\right)A^{3}}{6\pi}.\)(15)
Using He’s frequency-amplitude formulation [10, 13], we have
\(\omega^{2}=\frac{{{\tilde{R}}_{2}\omega}_{1}^{2}-{{\tilde{R}}_{1}\omega}_{2}^{2}}{{\tilde{R}}_{2}-{\tilde{R}}_{1}}\)\(=\frac{{\left\{\frac{2}{\pi}\left[\frac{3A^{3}}{4}\left(1-\omega_{2}^{2}\right)-\omega_{2}^{2}A\right]-\frac{\left(1-\omega_{2}^{2}\right)A^{3}}{6\pi}\right\}\omega}_{1}^{2}-{\left\{\frac{2}{\pi}\left[\frac{3A^{3}}{4}\left(1-\omega_{1}^{2}\right)-\omega_{1}^{2}A\right]-\frac{\left(1-\omega_{1}^{2}\right)A^{3}}{6\pi}\right\}\omega}_{2}^{2}}{\frac{2}{\pi}\left[\frac{3A^{3}}{4}\left(1-\omega_{2}^{2}\right)-\omega_{2}^{2}A\right]-\frac{\left(1-\omega_{2}^{2}\right)A^{3}}{6\pi}-\left\{\frac{2}{\pi}\left[\frac{3A^{3}}{4}\left(1-\omega_{1}^{2}\right)-\omega_{1}^{2}A\right]-\frac{\left(1-\omega_{1}^{2}\right)A^{3}}{6\pi}\right\}}\)\(=\frac{18A^{2}}{\left(9\pi-2\right)A^{2}+24}\) (16)
When , Eq. (16) becomes
\(\omega^{2}=\frac{18}{\left(9\pi-2\right)}\) (17)
while the exact one is \(\omega=1\). When A<<1, we have
\(\omega^{2}=\frac{18A^{2}}{24}=\frac{3A^{2}}{4}\) (18)
This agrees with that by the homotopy perturbation method [10-12]. So Ren-Hu’s modification is valid for A<<1.