3
$\begingroup$

Consider a univariate continuous and differentiable function $f(x)$, such that $f(0)=0$. Additionally, it holds that

$$\frac{d f(x)}{d x}= a \left(\frac{f(x)}{x}\right)^b $$

where $a$ and $b$ are two real, positive constants.


An example of such function, and the one that motivated this question, is

$$ f(L) = A\left( B L^{\rho} + (1-B)K^{\rho} \right)^{\frac{1}{\rho}} $$

for a positive constant $K$, $L \geq 0$, and $\rho<0$ (the latter ensures that $f(0)=0$, which is computed using the limit). The notation correspondence between my function and the general one is $a=A^\rho B$ and $b=1-\rho$.


Say you want to find

$$\lim_{x \rightarrow 0} \dfrac{d f(x)}{d x} = \lim_{x \rightarrow 0} a \left(\frac{f(x)}{x}\right)^b$$

Because of the assumption $f(0)=0$, this derivative has an undefined expression inside the parenthesis, $\frac{0}{0}$. Thus, you want to use L'Hopital for this. But if you apply L'Hopital to the parenthesis, you get:

$$ \frac{\dfrac{d f(x)}{d x}}{1} $$

Thus, we end up where we started. In other words, we are in an infinite L'Hopital loop. How can we get around it?

Apparently the answer to this limit is $0$, but I fail to see who this is achieved.

PS: a fairly similar question asks for the solution for a particular $f(x)$. I want to know the solution for the general case.

$\endgroup$
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Jyrki Lahtonen Jun 3 '17 at 5:16
3
$\begingroup$

Let $f(x)$ satisfy the first order ordinary differential equation

$$f'(x)-a\left(\frac{f(x)}{x}\right)^b=0\tag 1$$

with initial condition $f(0)=0$, where $a>0$ and $b>0$.


Note that in general $\left(\frac{f(x)}{x}\right)^b\notin \mathbb{R}$ unless $x f(x)\ge 0$. We proceed, therefore, under the assumption that $x\ge 0$ and $f(x)\ge 0$.


CASE $(1)$: $\displaystyle b=1$

If $b=1$, then the general solution to $(1)$ with initial condition $f(0^+)=0$ is $f(x)=Cx^a$ for any value of $x\ge 0$.

Note that the solution is not unique since $C$ is arbitrary. Furthermore, we maintain the resriciton $x\ge 0$ since in general $f(x)$ is complex for $x<0$ (e.g. for $a=1/2$ and $x<0$, $f(x) =Cx^{1/2}=\pm iC\sqrt{|x|}$

Finally, we find that for $b=1$

$$f'(0^+)=\begin{cases}C&,a=1\\\\0&,a>1\\\\\text{sgn}(C)\,\infty&,0<a<1\end{cases}$$


If $b\ne 1$, then upon dividing both sides of $(1)$ by $(f(x))^b$ and integrating, we find the general solution of $(1)$ is given by

$$f(x)=(ax^{1-b}+C)^{1/{1-b}} \tag 2$$


CASE $(2)$: $\displaystyle 0<b<1$

If $0<b<1$, then the condition $f(0^+)=0$ requires that $C=0$ and we find

$$f(x)=a^{1/(1-b)}x \tag 3$$

Note that the solution given by $(3)$ is also valid for $x<0$.

Finally, taking the derivative yields $f'(x)=a^{1/(1-b)}$ for all $x$ and hence for $0<b<1$

$$f'(0)=a^{1/(1-b)}$$


CASE $(3)$: $\displaystyle 1<b$

If $1<b$, then unless $C=0$, $f(x)$ is not defined, in general, on the reals since $x^{1/(1-b)}$ is not real in general. Clearly for $x\ge 0$, $\lim_{x\to 0^+}f(x)=0$ when $b>1$. So, $(2)$ is the general solution to $(1)$ for $b>1$ and $x\ge 0$.

Finally, we have for $b>1$

$$\begin{align} f'(0^+)&=\lim_{h\to 0^+}\frac{f(h)}{h}\\\\ &=\lim_{h\to 0^+}\frac{(ah^{1-b}+C)^{1/{1-b}}}{h}\\\\ &=\lim_{h\to 0^+} a^{1/(1-b)}\left(1+\frac{Ch^{b-1}}{a}\right)^{1/(1-b)}\\\\ &=a^{1/(1-b)} \end{align}$$


SUMMARY:

The general solution to $(1)$ with initial condition $f(0^+)=0$ is given by

$$f(x)=\begin{cases} a^{1/(1-b)}x&,0<b<1, x\in \mathbb{R}\\\\ Cx^a&,b=1,x\ge 0\\\\ (ax^{1-b}+C)^{1/(1-b)}&,1<b, x\ge 0 \end{cases}$$

and the derivative at $0^+$ of $f(x)$ is given by

$$f'(0^+)=\begin{cases} a^{1/(1-b)}&,b>0, b\ne 1\\\\ C&,b=1, a=1\\\\ 0&,b=1,a>1\\\\ \text{sgn}(C)\,\infty&,b=1,0<a<1 \end{cases}$$

$\endgroup$
4
$\begingroup$

Suppose that $f\in C^1$, i.e., $f$ is differentiable and the derivative is continuous (this is a small additional assumption). Moreover, we assume that $\frac{f(x)}{x}\geq0$ for all $x$ so we don't need complex numbers. Then,

$$ f'(0)=\lim_{x\rightarrow 0}\frac{df(x)}{dx}=\lim_{x\rightarrow 0}a\left(\frac{f(x)}{x}\right)^b=a\left(\lim_{x\rightarrow 0}\frac{f(x)}{x}\right)^b. $$ Since $f(0)=0$, this equals $$ a\left(\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x}\right)^b. $$ But, we observe that the limit is actually a derivative, so we get $$ f'(0)=a(f'(0))^b. $$ Therefore, $f'(0)$ is a root of the polynomial $at^b-t=t(at^{b-1}-1)$ if $b\not=1$ and $x(a-1)$ if $b=1$. Therefore, if $b\not=1$, then either $f'(0)=0$ or $f'(0)=a^{\frac{1}{1-b}}$. In the other case $(b=1)$, either $a=1$ as well or $f'(0)=0$.

  • For example, in the case $a=1$ and $b=1$, then $f(x)=Kx$ satisfies the conditions and the derivative at $0$ is $K\not=0$.

  • The constant zero function, $f(x)=0$ gives a case where $f'(0)=0$.

  • And @MarkViola gives the function $f(x)=a^{\frac{1}{1-b}}x$ whose derivative at $0$ is $a^{\frac{1}{1-b}}$.

$\endgroup$
  • 1
    $\begingroup$ It might be instructive to alert the OP and others to an issue here that $f(x)$ must be of the same sign as $x$, else $\left(\frac{f(x)}{x}\right)^b$ is not defined on the reals in general. $\endgroup$ – Mark Viola Jun 1 '17 at 15:19
  • $\begingroup$ @MarkViola Thanks for the observation, I've made the appropriate change. $\endgroup$ – Michael Burr Jun 1 '17 at 15:38
2
$\begingroup$

If(!) $f'(0)$ exists, then $$f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\frac{f(x)}{x} =\lim_{x\to 0}\left(\frac1af'(x)\right)^{1/b}=\left(\frac 1a\lim_{x\to 0}f'(x)\right)^{1/b}.$$ In particular, $\lim_{x\to 0}f'(x)$ must exist. But then this limit equals $f'(0)$ and we see that $$ f'(0)=\left(\frac1af'(0)\right)^{1/b}$$ and hence (if $b\ne 1$) $$ f'(0)= \frac 1{a^{1/(b-1)}}.$$

$\endgroup$
  • $\begingroup$ @MarkViola That is not true, as per my other comments. $\endgroup$ – luchonacho Jun 2 '17 at 8:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.