Exploring Multi-Scale Perturbation Techniques

[Dustinsfl]

This is in relation to multi-scale perturbation techniques.

$x''+\epsilon f(x,x') + x = 0$

$f(x,x') = (x^2-1)x'$

\begin{alignat*}{3}
f\left(x_0 + \epsilon x_1 + \cdots , \left(\frac{\partial}{\partial t} + \epsilon\frac{\partial}{\partial T}\right)(x_0 + \epsilon x_1 + \cdots)\right) & = & f(x_0 + \epsilon x_1 + \cdots , x_{0t} + \epsilon(x_{0T} + x_{1t}) + \cdots)\\
& = & f(x_0,x_{0t}) + \epsilon x_1f_x(x_0,x_{0T}) + \epsilon (x_{0T} + x_{1t})f_{x'}(x_0,x_{0T})\\
& & + \cdots
\end{alignat*}

In line two, why/how does $x_1$ move outside of $f(,)$? I can't remember why terms are what they are after the order 1 term. Could someone explain it to me?

 

[jvicens]

The reason why $x_1$ moves outside of $f(,)$ in the second line is because of the way the perturbation technique is applied. In this technique, we are expanding the solution of the original equation in powers of a small parameter $\epsilon$. This means that we are assuming the solution can be written as a series of terms, each multiplied by a different power of $\epsilon$.

In the first line, we have expanded $f(x,x')$ in powers of $\epsilon$ as well. This means that $f(x,x')$ can be written as $f(x_0,x_{0t}) + \epsilon f_1(x_0,x_{0t}) + \cdots$, where $f_1$ is the term multiplied by $\epsilon$. So, in the second line, we are simply replacing the terms $f(x,x')$ with their expanded forms.

Furthermore, the notation used in the second line, with the subscript $x$ and $x'$, indicates that we are taking partial derivatives with respect to these variables. So, when we expand $f(x,x')$, we are also expanding the derivatives of $f$ with respect to $x$ and $x'$.

Therefore, in the second line, when we write $f(x_0 + \epsilon x_1 + \cdots , x_{0t} + \epsilon(x_{0T} + x_{1t}) + \cdots)$, we are essentially expanding $f$ and its derivatives in terms of the powers of $\epsilon$. This is why $x_1$ moves outside of $f(,)$ in the second line.