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I L'Hospital's rule for Laplacian

  1. Jul 25, 2016 #1
    Screen Shot 2016-07-25 at 11.19.35 AM.png

    In the above expression for the Laplacian, how exactly does the author apply l'Hospital's rule? And is this transformation only valid for ## \rho = 0##?
     
  2. jcsd
  3. Jul 25, 2016 #2

    fresh_42

    Staff: Mentor

    To start with: I'm not sure.
    But ##ρ=0## looks like the point where the expression isn't defined. How about differentiating according to the product rule, application of L'Hôpital on the first summand ##\frac{1}{ρ} \frac{\partial e}{\partial ρ}## to get the second derivative at ##ρ=1##?
     
  4. Jul 25, 2016 #3
    Got it, thank you. Although why exactly did you specify ##\rho = 1##?
     
  5. Jul 25, 2016 #4

    fresh_42

    Staff: Mentor

    I had no better idea to get rid of ##\frac{1}{\rho}## and get the ##2## of the solution. Otherwise it would have been ##(1+\frac{1}{\rho})## instead.
    But as I said, I'm not sure and it was a bit of a quick and dirty calculation. You've possibly done it better.
     
  6. Jul 25, 2016 #5
    It's as you said: use product rule, but then only apply L'Hôpital's rule to ##\frac {1}{\rho} \frac {\partial e}{\partial \rho}##. This gives the extra term needed to get a coefficient of 2.
     
  7. Jul 26, 2016 #6
    Just to add to my previous post: since L'Hôpital's rule is only applicable at ## \rho = 0## to derive their expression, wouldn't it only be valid for ## \rho = 0## and no other values of ## \rho ##?
     
  8. Jul 26, 2016 #7

    fresh_42

    Staff: Mentor

    Without going through the list of conditions I used
    $$\lim_{\rho \rightarrow a} \frac{f(\rho)}{g(\rho)} = \lim_{\rho \rightarrow a} \frac{f'(\rho)}{g'(\rho)}$$
    and simply substituted ##\frac{\partial e}{\partial \rho}## by ##\frac{\partial^2 e}{\partial \rho^2}## which left me with a factor ##\frac{1}{\rho}## that I adjusted by letting ##\rho \rightarrow a = 1##. (I told you it was quick and dirty.) How did you get rid of this factor?
     
  9. Jul 26, 2016 #8
    I got the expression:

    $$\frac {1}{\rho} ( \frac {\partial e}{\partial \rho} + \rho \frac {\partial ^2 e}{\partial \rho ^ 2})$$

    This gives me:

    $$\frac {1}{\rho} \frac {\partial e}{\partial \rho} + \frac {\partial ^2 e}{\partial \rho ^ 2} $$

    Now this is where I could be wrong, but I applied L'Hôpital's rule to only the first term here for when ## \rho = 0## which gives me

    $$ \lim_{\rho \rightarrow 0} \frac {\partial^2 e}{\partial \rho ^2} + \frac {\partial ^2 e}{\partial \rho ^ 2} $$

    Thus it only seems to match their result when ##\rho \rightarrow 0##.
     
  10. Jul 26, 2016 #9

    fresh_42

    Staff: Mentor

    But this isn't the rule de L'Hôpital. It looks like the definition of differentials. If we take ##f(\rho)= \frac{\partial e}{\partial \rho}## you took ##\lim_{\rho \rightarrow 0} \frac{f(\rho)}{\rho} = f'(\rho)## but wouldn't it be ##\lim_{\rho \rightarrow 0} \frac{f(\rho_0 + \rho)}{\rho} = f'(\rho_0)##?
    Thus you are also left with a choice, at which point ##\rho_0## you want to evaluate the differentials. Omitting the index ##0## only disguises this fact by implicitly assuming ##f(\rho_0 + \rho) = f(\rho + \rho) = f(2\rho) = f(\rho)## or as you said ##\rho_0 = 0##
    L'Hôpital gives a general case with factor ##(1+\frac{1}{\rho})##. What makes more sense in the context? ##\rho_0 = 0## or ##\rho_0 = 1##?
    I think it is in any case, how ever we may turn it, a bit of a sloppy calculation.
     
  11. Jul 26, 2016 #10
    I agree with most of what you've said, but wouldn't:

    $$\lim_{\rho \rightarrow 0} \frac{f(\rho_0 + \rho)}{\rho} = f'(\rho_0)$$

    only be a valid expression if in this case

    $$ \lim_{\rho \rightarrow 0} f(\rho_0 + \rho) = 0 $$

    since L'Hôpital's rule requires both parts to yield an indefinite value?
     
  12. Jul 26, 2016 #11

    fresh_42

    Staff: Mentor

    No. I simply changed the variable names we all are used to:

    ##\lim_{\rho \rightarrow 0} \frac{f(\rho_0 + \rho)}{\rho} = \lim_{h \rightarrow 0} \frac{f(x_0+ h)}{h} = f'(x_0)##

    No. L'Hôpital's rule can be formulated as I did above in #7 (plus a bunch of conditions that have to hold). It doesn't require any infinity (beside those implied by taking differentials of course). It simply holds for ##a \in \{± \infty \}## as well.

    Edit: You are right, I forgot the ##-f(x_0)## part in ##\frac{df}{dx}##. Shame on me :sorry:. But this mistake shows, that we have to use L'Hôpital and not just the definition of ##\partial##.
     
  13. Jul 26, 2016 #12
    Haha that makes more sense. Also, just to clarify, although there is no general constraint on ## a ##, when evaluating

    $$\lim_{\rho \rightarrow a} \frac{f(\rho)}{g(\rho)} = \lim_{\rho \rightarrow a} \frac{f'(\rho)}{g'(\rho)}$$

    these are only equivalent when

    $$ \lim_{\rho \rightarrow a} \frac{f(\rho)}{g(\rho)} = \frac {\pm \infty}{\pm \infty}$$ or

    $$ \lim_{\rho \rightarrow a} \frac{f(\rho)}{g(\rho)} = \frac {0}{0}$$

    correct?
     
  14. Jul 26, 2016 #13

    fresh_42

    Staff: Mentor

    Essentially. (##g'(x) \neq 0## for ##x \neq a## in a neighborhood of ##a## must also hold.)
    It probably is sufficient to require the latter, for one may switch nominator and denominator.
    (It's been quite a while since I ... what was I about to say?)
     
  15. Jul 27, 2016 #14

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Just a remark. The above Laplacian is the radial part in cylinder coordinates, and one must be utmost careful with points on the axis ##\rho=0##, because there the coordinates are singular. If in doubt with an expression, it's better to translate it into Cartesian coordinates and analyse the situation there!
     
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