1
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.