I would like to know if there are other methods to solve equations such as this one below. I don't really understand the theory behind the multiple scale analysis and why it works I understand some of the reasoning behind it just not the proof and theory. I kind of chose a differential equation for which I think it works and tried it out following a example from a similar equation. I think the answer seems strange but I was very thorough in my calculations. I welcome any comments or advice to problems like these thanks!

\begin{equation}u^{''} +\omega_0^2u=-2\epsilon \mu u^{'}-\epsilon u^2 u^{'} + \epsilon k \cos \Omega t\end{equation} First we introduce two time variables (fast and slow) and define them as \begin{equation}T_0=t\;\;\;T_1=\epsilon t\end{equation} where $T_0$ is the fast time scale and $T_1$ is the slow time scale.

The first order expansion of the problem is \begin{equation}u(t)=u_0(T_0,T_1)+\epsilon u_1(T_0,T_1)\end{equation} Differentiating the time derivative becomes \begin{align} \frac{du}{dt}=\frac{\partial u_0}{\partial T_0} \frac{dT_0}{dt}(1)+(\epsilon) \frac{\partial u_1}{\partial T_1} \frac{dT_1}{dt}=\frac{\partial u_0}{\partial T_0}+\epsilon \frac{\partial u_1}{\partial T_1} \end{align} Define the linear operators, and functions \begin{equation} D_0=\dfrac{\partial}{\partial T_0},\hspace{10pt} D_1=\dfrac{\partial}{\partial T_1}\end{equation}

The time derivative becomes $\dfrac{d}{dt}=D_0+\epsilon D_1$. The second order time derivative is then expressed as \begin{equation}\frac{d^2}{dt^2}=(D_0+\epsilon D_1)^2=D_0^2+2D_0D_1 \epsilon +\epsilon ^2 D_1^2\end{equation}

The $\epsilon ^2$ term is neglected. We are now ready to substitute our results in the original equation to obtain, \pagebreak

$$(D_0^2+2\epsilon D_0 D_1)(u_0+\epsilon u_1)+\omega^2_0(u_0+\epsilon u_1)=-2\epsilon \mu (D_0+ \epsilon D_1)(u_0+\epsilon u_1)-$$$$\epsilon (u_0+\epsilon u_1)^2(D_0+\epsilon D_1)(u_0+\epsilon u_1) + \epsilon k \cos \Omega t$$ $$\implies (D_0^2u_0+\epsilon D^2_0 u_1+2\epsilon D_1D_0 u_0 +2\epsilon ^2D_0 D_1 u_1^2)+\omega_0^2u_0+\epsilon \omega_0^2 u_1=-2 \epsilon \mu D_0u_0-2\epsilon^2\mu (...)$$ $$-\epsilon u^2_0D_0u_0+\epsilon^2(...)+\epsilon k \cos \Omega t $$ \begin{equation}\implies D_0 u_0+ \epsilon D_0 u_1 +2 \epsilon D_1 D_0 u_0+ \omega_0^2 u_0+\epsilon \omega_0^2 u_1=-2\epsilon \mu D_0 u_0-\epsilon u_0^2D_0u_0+\epsilon k \cos \Omega t\end{equation} Now we construct a system of differential equations in powers of $\epsilon$ called zero-order and first order problems and given by, \begin{align} D_0^2u_0+\omega_0^2u_0=0\end{align} \begin{align} D^2u_1 +\omega_0^2u_1=-2D_1D_0 u_0-2\mu D_0u_0-u_0^2D_0u_0+k \cos \Omega t \end{align}

The first equation is a second order linear, homogeneous equation with constant coefficients and has general solution \begin{equation}u_0(T_0,T_1)=A(T_1)e^{i\omega_0T_0}+\bar{A}(T_1)e^{-i\omega_0T_0}\end{equation} where $A(T_1)$ is the complex amplitude afterwards. Our next step is to investigate primary resonance of the system, which occurs when the actuation frequency $\Omega$ is near the natural frequency, $\omega_0$. This can be written as \begin{align}\Omega=\omega_0+\epsilon \sigma\\ \Omega T_0=\omega T_0 +\epsilon T_0 \sigma\\\Omega t= \omega T_0+T_1\sigma \end{align}

We substitute $u_0$ into the equation (2) to get, \begin{equation*}D_0^2u_1+\omega_0^2u_1= -2D_0D_1[A(T_1)e^{i \omega_0 T_0}+\bar{A}(T_1)e^{-i \omega T_0}]-2\mu D_0[A(T_1)e^{i\omega_0T_0}+\bar{A}(T_1)e^{-i \omega T_0}]- \end{equation*}\begin{equation} [A(T_1)e^{i\omega_0T_0}+\bar{A}(T_1)e^{-i\omega T_0}]^2D_0u_0+k\frac{e^{i \omega_0 T_0}e^{i\sigma T_1}+e^{-i\omega_0T_1}e^{-i\sigma T_1}}{2}\end{equation}

Thus applying the linear operations to the above to get $$-2[A^{'}e^{i\omega_0T_0}(i\omega_0)+\bar{A}^{'}e^{-i\omega_0T_0}(-i\omega_0)]-2\mu[Ae^{i\omega_0 T_0}(i\omega_0)+\bar{A}e^{-i\omega_0T_0}(-i\omega_0)]+$$$$\bigg(-A^2e^{2i\omega_0T_0}-2A\bar{A}-\bar{A}^2e^{-2i\omega T_0}\bigg)\bigg(Ae^{i\omega_0T_0}(i\omega_0)-\bar{A}e^{-i\omega_0T_0}(i\omega_0)\bigg)+\frac{k}{2}\bigg(e^{i(\omega T_0+\sigma T_1)}+e^{-i(\omega_0 T_0+\sigma T_1)}\bigg)$$ $$\implies -2A^{'}e^{i\omega_0T_0}(i\omega_0)+2\bar{A}^{'}e^{-i\omega_0T_0}(i\omega_0)-2\mu Ae^{i\omega_0 T_0}(i\omega_0)+2\bar{A}e^{-i\omega_0T_0}(i \omega_0)$$$$-A^3e^{3i\omega_0T_0}(i\omega_0)+A^2\bar{A}e^{i\omega_0T_0}(i\omega_0)-2A^2\bar{A}e^{i\omega_0T_0}(i\omega_0)+2A\bar{A}^2e^{-i\omega T_0}(i\omega_0)-A\bar{A}^2e^{-3i\omega_0T_0}(i\omega_0)+\bar{A}^3e^{-3i\omega_0T_0}(i\omega_0)$$\begin{equation}+\frac{k}{2}\bigg(e^{i\omega T_0}e^{i\sigma T_1}+e^{-i\omega_0} e^{-i\sigma T_1}\bigg)\end{equation}

Consider $A(T_1)=\dfrac{1}{2}ae^{i\beta}$ where $a$ and $\beta$ are the real amplitude and phase from the secular\ terms equation. Secular terms are nonhomogenous terms that make the function unbounded they are inconsistent with the behavior of the physical system, thus they must be eliminated. A term is secular if it is a solution of the homogenous equation. Therefore, the sum of the coefficient of $e^{i \omega T_0}$ must be $0$.

$$-2A^{'}(i\omega_0)-2\mu A(i\omega_0)-A^2\bar{A}(i\omega_0)+\dfrac{k}{2}e^{i\sigma T_1}=0$$ $$e^{-i\beta}\bigg(-a^{'}(i\omega_0)e^{i\beta}+a\beta^{'}\omega_0e^{i\beta}-\mu ae^{i\beta}(i\omega_0)-\frac{1}{8}a^3e^{i\beta}(i\omega_0)+\frac{k}{2}e^{-i\sigma T_1})=0\bigg) $$ \begin{equation}-a^{'}(i\omega_0)+a\beta^{'}\omega_0-\mu a(i \omega_0)-\frac{1}{8}a^3(i \omega_0) +\frac{k}{2}\bigg(\cos(\sigma T_1-\beta)+i\sin(\sigma T-\beta)\bigg)=0\end{equation} For the entire function to be zero Real and Imaginary parts must be 0. \begin{equation}a\beta^{'}\omega_0+\frac{k}{2}\cos(\sigma T_1-\beta)=0 \end{equation} \begin{equation}-a\omega_0-\frac{1}{8}a^3\omega_0 + \sin(\sigma T_1-\beta)-\mu a \omega_0=0\end{equation} We make a change of variables and set $\gamma=\sigma T_1-\beta$ so that $\beta^{'}=\sigma T_1-\gamma^{'}$. Therefore we have that, \begin{equation}a\gamma^{'}=a\sigma +\frac{k}{2\omega_0}\cos \gamma\end{equation} \begin{equation}a^{'}=-\mu a-\frac{1}{8}a^3+ \frac{\sin \gamma}{\omega_0}\end{equation} Steady states occur when $a\gamma^{'}=0$ and $a^{'}=0$ therefore we have that \begin{equation}-\mu a-\frac{1}{8}a^3 +\frac{\sin \gamma}{\omega_0}=0\end{equation} \begin{equation}a\sigma+\frac{k}{2\omega_0} \cos \gamma=0\end{equation} There fore the solution is \begin{equation}\begin{cases} a+\dfrac{1}{8}a^3=\dfrac{\sin \gamma}{\mu \omega_0} \\ a\sigma=-\dfrac{k}{2\omega_0}\cos \gamma \end{cases}\end{equation}


1 Answer 1


In the case of this particular equation, there is another 'multiple scale' approach, known as averaging, to obtain an (asymptotic) approximation for periodic solutions to the equation, without having to consider things like secular terms.

For more information, you could read this paper, and/or search for analysis of periodic solutions of the 'Van der Pol - Mathieu equation'. For more information on the averaging technique, see e.g. F. Verhulst, Nonlinear Differential Equations and Dynamical Systems, Springer, 2006.


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