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November 27th, 2017, 01:13 PM  #1 
Member Joined: Jan 2016 From: Blackpool Posts: 92 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  
Math Team Joined: Jul 2011 From: Texas Posts: 2,723 Thanks: 1376  Quote:
integrate both sides ... $\lnf(x) = \alpha \ln{x} + C_1$ $\lnf(x) = \ln(x^{\alpha}) + \ln(e^{C_1})$ $\lnf(x) = \ln(x^{\alpha} \cdot e^{C_1})$ $f(x) = C_2 \cdot x^{\alpha}$  
November 28th, 2017, 06:00 PM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 312 Thanks: 161 Math Focus: Dynamical systems, analytic function theory, numerics  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 
Math Team Joined: Jul 2011 From: Texas Posts: 2,723 Thanks: 1376  
November 29th, 2017, 06:45 PM  #5 
Senior Member Joined: Sep 2016 From: USA Posts: 312 Thanks: 161 Math Focus: Dynamical systems, analytic function theory, numerics  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: 2,723 Thanks: 1376 
To be honest, that explanation goes far beyond what I learned in my undergraduate analysis class in 1975. Appreciate you sharing it, SDK.

December 1st, 2017, 01:59 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 18,704 Thanks: 1529 
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. 

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