Change of variables in differential equation?

Question

I have the following formula:

$$f(x) = \frac{d^2w(x)}{dx^2}$$

Now I would like to normalize $x$ by dividing it by $L$? This would be the substitution: $$\hat{x}=x/L$$

How would my formula change? (step by step please)

Answer 1

Using $x = L \hat{x}$ we have $dx = L d\hat{x}$ and $dx^2 = L^2 d\hat{x}^2$. Therefore, $$ \frac{d}{dx^2} = \frac{1}{L^2} \frac{d}{d\hat{x}^2}. $$ This simple "substitution" is not mathematically rigorous, but you could use the chain rule twice to obtain the same thing; since the change of variable consists only in the multiplication by a constant, this "heuristic" works.

The final expression is $$ f(L \hat{x}) = \frac{1}{L^2} \frac{d}{d\hat{x}^2} w(L\hat{x}). $$

If, for example, $w(x) = A \exp(Bx)$, we have $f(x) = AB^2 \exp(Bx)$. With the new variable, $$ A (BL)^2 \exp(BL \hat{x}) = \frac{d}{d\hat{x}^2} A \exp(BL \hat{x}). $$ If you define $\hat{B}=BL$, $\hat{f} = \hat{B}^2 \exp(\hat{B} \hat{x})$ and $\hat{w}=\exp(\hat{B} \hat{x})$, the equation is $$ \hat{f} = \frac{d \hat{w}}{d\hat{x}^2} . $$ Therefore, after you normalize $x$ with $\hat{x}$, you should also normalize the functions. If $w$ has units of Kelvin, for example, $f$ has units of $K/m^2$. After the normalization, both $\hat{w}$ and $\hat{f}$ are nondimensional.

Appendix: rigorous change of variable using chain rule

Let $x=L \hat{x}$. Therefore, $\hat{x}=x/L$. From the chain rule, $$ \frac{dw}{dx} = \frac{d\hat{x}}{dx} \frac{dw}{d\hat{x}} = \frac{1}{L} \frac{dw}{d\hat{x}} $$ and $$ \frac{d^2w}{dx^2}=\frac{d}{dx} \left(\frac{dw}{dx}\right) = \frac{d\hat{x}}{dx} \frac{d}{d\hat{x}}\left(\frac{1}{L} \frac{dw}{d\hat{x}} \right) = \frac{1}{L^2} \frac{d^2 w}{d \hat{x}^2}. $$