Consider the Duffing oscillator

\[\ddot{x} + x + \epsilon x^3 = 0\]

with boundary conditions $x(0) = 0, \ \dot{x}(0) = 1$.

To find an approximate analytical solution, we will use the method of multiple scales. Here $\epsilon$ is a small number $(«1)$ and the oscillator is normalized appropriately to get the above form. We start by defining two time steps $T_0 = t$ (the original time scale) and $T_1 = \epsilon t$ and treat them as independent variables. So, the solution of the oscillator $x$ will be of the form $x(T_0, T_1)$. Therefore,

\[\dot{x} = {\partial{x}\over \partial{T_0}} + \epsilon {\partial{x}\over \partial{T_1}}\]

and

\[\ddot{x} = {\partial\over\partial{T_0}} \left({\partial{x}\over \partial{T_0}} + \epsilon {\partial{x}\over \partial{T_1}}\right) + \epsilon{\partial\over\partial{T_1}} \left({\partial{x}\over \partial{T_0}} + \epsilon {\partial{x}\over \partial{T_1}}\right)\]

we take only the terms till $O(\epsilon)$, hence

\[\ddot{x} = {\partial^2 x\over\partial{T_0}^2 } + 2 \epsilon {\partial^2{x}\over \partial{T_0 T_1}}\]

Also, we expand $x$ as

\[x = x_0\left(T_0, T_1\right) + \epsilon x_1\left( T_0, T_1\right)\]

substituting all of these in the original equation, we obtain

\[\epsilon x_0^3+\epsilon x_1+2 \epsilon {\partial^2{x_0}\over \partial{T_0 T_1}}+\epsilon {\partial^2 x_1\over\partial{T_0}^2 }+x_0+{\partial^2 x_0\over\partial{T_0}^2 } + O(\epsilon^2) = 0\]

Now we make the leading order term $0$, giving us

\[x_0+{\partial^2 x_0\over\partial{T_0}^2 } = 0\]

This gives

\[x_0\left( T_0, T_1\right) = P\left(T_1\right) \sin\left(T_0 + Q(T_1)\right)\]

Now, we take $\epsilon$ order term and substitute the solution for $x_0$ and simplify to obtain,

\[x_1+{\partial^2 x_1\over\partial{T_0}^2 }+\frac{1}{4}\left(8 P' \cos (Q+{T_0})-8 P Q' \sin (Q+{T_0})+3 P^3 \sin (Q+{T_0})-P^3 \sin (3 Q+3 {T_0})\right) = 0\]

Since the solution should be bounded, we make the coefficient of $\cos(Q+T_0)$ and $\sin(Q+T_0)$ (called secular terms) $0$, otherwise the solution will blow up. Therefore, we obtain,

\[P^\prime = 0\]

and

\[3P^3 - 8PQ^\prime = 0\]

this gives

\[P = C_1\]

and

\[Q = {3\over 8} T_1 C_1^2 + C_2.\]

Now $T_1 = \epsilon T_0$ and $T_0 = t$, therefore,

\[x_0 = P \sin\left(t + Q\right) = C_1\sin\left(t+ {3\over 8}\epsilon t C_1^2 + C_2 \right)\]

applying boundary conditions,

\[x(0) = 0 \Rightarrow C_1 \sin(C_2) = 0,\] \[\dot{x}(0) = 1 \Rightarrow C_1 \left(1 + {3\over 8}\epsilon C_1^2\right)\cos\left(t+ {3\over 8}\epsilon t C_1^2 + C_2 \right) =1\]

we compare the leading order terms with leading order and $\epsilon$ order terms with itself, and obtain,

\[C_2 = 0, \ C_1 = 1\]

Therefore the leading order solution will give,

\[x(t) = \sin\left({1+{3\over 8}\epsilon}\right)t\]