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November 27th, 2017, 01:13 PM   #1
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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.
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November 27th, 2017, 01:31 PM   #2
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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 ...

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

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

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

$f(x) = C_2 \cdot x^{\alpha}$
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November 28th, 2017, 06:00 PM   #3
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Quote:
Originally Posted by skeeter View Post
$\dfrac{f'(x)}{f(x)} = \dfrac{\alpha}{x}$

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.
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November 29th, 2017, 06:09 AM   #4
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This doesn't quite do the job though. You have shown that $cx^{\alpha}$ is one solution, but not that no other solutions exist.
please continue ...
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November 29th, 2017, 06:45 PM   #5
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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.
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Last edited by SDK; November 29th, 2017 at 07:00 PM.
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November 30th, 2017, 12:03 PM   #6
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To be honest, that explanation goes far beyond what I learned in my undergraduate analysis class in 1975. Appreciate you sharing it, SDK.
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December 1st, 2017, 01:59 AM   #7
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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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