# Does scaling factor affect discretization?

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,

1)

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

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

2)

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

right?

• 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.
– namu
Apr 16, 2016 at 22:58
• I see... I cannot really set $\bar{x}$ to unity. Then how does $\frac{dy}{d\bar{x}}$ transform into? Apr 17, 2016 at 2:53
• $\frac{dy}{d\bar x} = \frac{dy}{d(x/a)} = a \frac{dy}{dx}$
– namu
Apr 18, 2016 at 0:52
• 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 $x$. Apr 18, 2016 at 2:32