L'Hospital's rule for Laplacian

[TheCanadian]

In the above expression for the Laplacian, how exactly does the author apply l'Hospital's rule? And is this transformation only valid for $\rho = 0$?

 

[fresh_42]

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 $ρ=1$?

 

[TheCanadian]

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 $ρ=1$?

Got it, thank you. Although why exactly did you specify $\rho = 1$?

 

[fresh_42]

Got it, thank you. Although why exactly did you specify $\rho = 1$?

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.

 

[TheCanadian]

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.

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.

 

[TheCanadian]

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.

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 $\rho$?

 

[fresh_42]

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?

 

[TheCanadian]

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?

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 $\rho \rightarrow 0$.

 

[fresh_42]

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.

 

[TheCanadian]

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.

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?

 

[fresh_42]

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

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

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

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

since L'Hôpital's rule requires both parts to yield an indefinite value?

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 . But this mistake shows, that we have to use L'Hôpital and not just the definition of $\partial$.

 

[TheCanadian]

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 . But this mistake shows, that we have to use L'Hôpital and not just the definition of $\partial$.

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?

 

[fresh_42]

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?)

 

[vanhees71]

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!