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 November 27th, 2017, 01:13 PM #1 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 104 Thanks: 2 Differentiation question: Let f(x) be a differentiable function such that $xf'(x)=\alpha f(x)$ for all x>0, show that f(x)=$cx^{\alpha}$for some constant c. November 27th, 2017, 01:31 PM   #2
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 Originally Posted by Jaket1 Let f(x) be a differentiable function such that $xf'(x)=\alpha f(x)$ for all x>0, show that f(x)=$cx^{\alpha}$for some constant c.
$\dfrac{f'(x)}{f(x)} = \dfrac{\alpha}{x}$

integrate both sides ...

$\ln|f(x)| = \alpha \ln{x} + C_1$

$\ln|f(x)| = \ln(x^{\alpha}) + \ln(e^{C_1})$

$\ln|f(x)| = \ln(x^{\alpha} \cdot e^{C_1})$

$f(x) = C_2 \cdot x^{\alpha}$ November 28th, 2017, 06:00 PM   #3
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 Originally Posted by skeeter $\dfrac{f'(x)}{f(x)} = \dfrac{\alpha}{x}$ integrate both sides ... $\ln|f(x)| = \alpha \ln{x} + C_1$ $\ln|f(x)| = \ln(x^{\alpha}) + \ln(e^{C_1})$ $\ln|f(x)| = \ln(x^{\alpha} \cdot e^{C_1})$ $f(x) = C_2 \cdot x^{\alpha}$
This doesn't quite do the job though. You have shown that $cx^{\alpha}$ is one solution, but not that no other solutions exist. November 29th, 2017, 06:09 AM   #4
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 Originally Posted by SDK This doesn't quite do the job though. You have shown that $cx^{\alpha}$ is one solution, but not that no other solutions exist. November 29th, 2017, 06:45 PM   #5
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Quote:
 Originally Posted by skeeter please continue ...
There might be a more elegant method but here is a sketch which gets the job done using Picard iteration.

1. As you have already proved, $f(x) = cx^{\alpha}$ is a solution.

2. Define the operator $h: C^0(0,\infty) \to C^1(0,\infty)$ by
$h_y(x) = y(1) + \int_1^x \frac{\alpha y(s)}{s} \ ds$
and notice that if $y$ is any solution then it is a fixed point of $h_y$ since we can take a derivative to obtain
$\frac{d}{dx} h_y = \frac{d}{dx} \int_1^x \frac{\alpha y(s)}{s} \ ds = \frac{\alpha f(x)}{x} = y'$ implying $h_y$ and $y$ have the same derivative and noting that $h_y(1) = y(1)$.

3. It suffices to prove that $h$ has only a single fixed point which is equivalent to saying the DE has only one solution. To do this fix $y(1)$ to be the initial data for the known solution and define the sequence of functions $f_0(x) = y(1)$ and
$f_{n+1}(x) = h_{f_{n}}(x) = y(1) + \int_1^x \frac{\alpha f_n(s)}{s} \ ds$

4. Compute a few of these functions:
$f_1(x) = y(1) + \alpha \int_1^x \frac{y(1)}{s} \ ds = y(1) + \alpha y(1) \log x$
$f_2(x) = y(1) + \alpha \int_1^x \frac{y(1) + \alpha y(1) \log s}{s} \ ds = y(1) + \alpha y(1) \log x + \frac{1}{2} \alpha^2 y(1) \log ^2 x$
and you can easily prove by induction that in general
$f_n(x) = y(1) + y(1)\sum_{k=1}^n \frac{\alpha^k \log ^k(x)}{k!}$.

5. Show that $f_n$ is uniformly convergent on compact subsets of $\mathbb{R}$. This follows by noting that
$||f_{n+1} - f_n||_{\infty} = y_0 ||\frac{\alpha^{n+1} \log^{n+1}(x)}{(n+1)!}||_{\infty}$. Since $C^0$ is a Banach space it follows from the Banach fixed point theorem local uniqueness restricted to compact subsets is proved which immediately implies global uniqueness.

Edit: In fact, you can recognize the expression for $f_n$ above as the convergent partial sums for the power series
$y(1)(1 + \sum_{k=1}^{\infty} \frac{\alpha^k \log^k(x)}{k!}) = y(1) \exp(\alpha \log(x)) = y(1)x^{\alpha}$
as we expected.

Last edited by SDK; November 29th, 2017 at 07:00 PM. November 30th, 2017, 12:03 PM #6 Math Team   Joined: Jul 2011 From: Texas Posts: 3,016 Thanks: 1600 To be honest, that explanation goes far beyond what I learned in my undergraduate analysis class in 1975. Appreciate you sharing it, SDK. Thanks from Joppy December 1st, 2017, 01:59 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,972 Thanks: 2222 For $x$ > 0, the equation is equivalent to $x^{-\alpha}f^\prime\!(x) - \alpha x^{-\alpha - 1}f(x) = 0$, integrating which gives $x^{-\alpha}f(x) = c$, where $c$ is a constant. Hence $f(x)= cx^\alpha$ for $x$ > 0. There are other solutions that differ from the above by having a different domain that includes $x$ > 0. Tags differentiation, question Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post rath Calculus 5 May 14th, 2016 07:25 AM Tooperoo Calculus 9 May 7th, 2012 12:17 AM ads98765 Calculus 7 February 15th, 2012 03:10 PM zoopie2 Calculus 1 August 26th, 2010 04:53 PM Kiranpreet Algebra 1 December 13th, 2008 05:52 AM

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