Suppose I want to solve the below equation numerically.

$$ \frac{dy}{dx}=y $$

I'd like to normalize the space discretization by choosing $$ a\bar{x}=x $$ where I assume $\bar{x}$ is unity. Then the equation is

$$ \frac{1}{a}\frac{dy}{d\bar{x}}=y \qquad\rightarrow\qquad \frac{dy}{d\bar{x}}=ay $$

Thus, my discretization becomes,


$$ \frac{y_{i+1}-y_{i}}{\Delta \bar{x}}=ay $$

which I don't think it makes sense, since $\Delta \bar{x}$ must be unity, so is


$$ y_{i+1}-y_{i}=ay $$


  • $\begingroup$ You cannot set the variable $\bar x$ to unity. This is not how scaling works. If $A \leq x \leq B$, then by employing the scaling $a \bar x = x$, you have $A/a \leq \bar x \leq B/a$. The value of $\Delta \bar x$ is independent of the scaling you choose. $\endgroup$
    – namu
    Apr 16, 2016 at 22:58
  • $\begingroup$ I see... I cannot really set $\bar{x}$ to unity. Then how does $\frac{dy}{d\bar{x}}$ transform into? $\endgroup$
    – user65452
    Apr 17, 2016 at 2:53
  • $\begingroup$ $\frac{dy}{d\bar x} = \frac{dy}{d(x/a)} = a \frac{dy}{dx}$ $\endgroup$
    – namu
    Apr 18, 2016 at 0:52
  • 1
    $\begingroup$ I agree, you cannot set $\bar{x}$ to unity. The purpose of scaling is to collapse solutions together onto one assuming the non-dimensional groups are equal. You can think of $\bar{x}$ as a "scaled" version of the coordinate $x$. $\endgroup$
    – Charles
    Apr 18, 2016 at 2:32


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