I have asked this questions: Change of variables in differential equation?
…but after thinking about it, I am still a little confused of how to rigorously use the chain rule to calculate the derivative(s) of a function for a change of variable.
I have the following derivative:
$f(x) = \frac{dw(x)}{dx}$
Now I introduce the change of variable: $\hat{x}=\frac{x}{L}$
and I apply the chain rule:
- I write: $g(\hat{x}) = L \hat{x} = x$
- I substitute: $f(g(\hat{x})) = \frac{dw(g(\hat{x}))}{d(g(\hat{x}))}$
…but this does not help me… I am confusing something.
I would be glad, if someone could show me in detail and step by step how to do this rigorously.
Thanks a lot.
Best Answer
Let's say you want to change $x$ for $y$, where $y$ is a function of $x$, i.e., $y=g(x)$. I will use $y$ for the nondimensional variable because it's way easier to type in my phone. The chain rule tells us that if we want to calculate the derivative $$\frac{df}{dx}$$ in terms of $y$, we need to use the formula $$\frac{df}{dx}=\frac{dy}{dx}\frac{df}{dy}.$$ The $dy/dx$ part is equal to $dg(x)/dx$, therefore, $$\frac{df}{dx}=\frac{dg(x)}{dx}\frac{df}{dy}.$$ See that it works for any possible change of variables. In your particular case, $g(x)=x/L$ and $dg/dx=1/L$. Therefore, we have $$\frac{df}{dx}=\frac{1}{L}\frac{df}{dy}.$$ It leads to $$\frac{df}{dy}=Lw(g^{-1}(y)),$$ in which $g^{-1}(y)=Ly$. Finally: $$\frac{df}{dy}=Lw(Ly).$$ It's up to you to show that this equation is dimensionally consistent. Remember: $L$ has the same dimension of $x$ and $y$ is nondimensional. Let me know if you have any question.