# Solving $\forall x \in \mathbb{R}_+^*, f'(x) = f\left(\frac1{x}\right)$

I recently came across this equation : $$\forall x \in \mathbb{R}_+^*, f'(x) = f\left(\frac1{x}\right)$$where $f \in \mathcal{C}^1(\mathbb{R}, \mathbb{R})$.

I've done the following, but I'm stuck at the end. Could you give me pointers? Thanks!

Differentiating yields $$\forall x, f''(x) = -\frac1{x^2}f(x) \tag{S_0}$$Solutions in the form $$x \mapsto \frac1{x^\phi}$$ work iff $\phi(\phi+1) = -1$, ie. $\phi = \frac{-1 \pm i \sqrt{3}}{2} =e^{\pm 2i\pi/3} = j, \overline{j}$. Elements of the vector space generated by the free pair $(x^j, x^\overline{j})$ are therefore solutions of ($S_0$).

I then feed $\lambda x^j + \mu x^\overline{j}$ in the original equation, which yields $-\lambda j\frac1{x^{j+1}}-\mu\overline{j}\frac1{x^{\overline{j} + 1}} = \frac{x^{j + \overline{j}}}{\lambda x^\overline{j} + \mu x^j}$, then $(-\lambda j x^{\overline{j}+1} - \mu \overline{j} x^{j+1})(\lambda x^{\overline{j}} + \mu x^j) = x^{1+j+\overline{j}} = x^0 = 1$, and $-\lambda^2 j x^{2\overline{j} + 1} - \mu^2 \overline{j} x^{2j+1} - \lambda\mu(j + \overline{j}) = 0$. Thus, $$\lambda^2 j x^{-2i\sin(2\pi/3)} + \mu^2 \overline{j} x^{2i\sin(2\pi/3)} = \lambda\mu$$

Does that mean that no solutions can be found to the original equation, except the trivial $x \mapsto 0$ one? Or that I didn't take the right approach? I can't figure out how to handle the last equality.

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Sorry for the size of the equations; it seems that \displaystyle doesn't work. Can someone help me make this look better, or tell me how to do so? Thanks! –  Clément Apr 7 '11 at 8:21
@Clément: Is this what you wanted? All I did was change some of your single-dollar signs to double-dollar signs. –  TonyK Apr 7 '11 at 8:30
I think $(S_0)$ is wrong. The argument on the right-hand side should still be $1/x$. –  joriki Apr 7 '11 at 8:44
@TonyK: Thanks! –  Clément Apr 7 '11 at 8:48
Suggestion for a different solution: I think that if you put $g(x)=f(x)+f(1/x)$ and $h(x)=f(x)-f(1/x)$ then you get $g'(x)=g(x)$ and $h'(x)=-h(x)$. You can solve these two equations and obtain $f(x)$ from it. –  Martin Sleziak Apr 7 '11 at 8:51

There is a mistake in Clément's calculation: The Eulerian differential equation $y''+y/x^2=0$ has solutions of the form $y(x)=x^\lambda$ (resp. $=\exp(\lambda\log x)$ ) where $\lambda$ satisfies the "index equation" $\lambda(\lambda-1)+1=0$, so $\lambda={1\over2}\pm i{\sqrt3\over2}$. The general solution is $$f(x)=c_1\exp(\lambda_1\log x)+c_2\exp(\lambda_2\log x)\>.$$ If we confront this with the original functional equation $f'(x)=f(1/x)$ then we see that the latter even has real solutions, namely $$f(x)=C\>\sqrt{\mathstrut x}\>\cos\Bigl({\sqrt3\over2}\log x-{\pi\over6}\Bigr)\>,\qquad C\in{\mathbb R}.$$ Of course it is easy to check a posteriori that these are indeed solutions.

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You're right, actually forgot a minus sign in my calculation. I calculated $\phi$ using the $1/x^\phi$, and plugged $x^\phi$ in the equation. Thanks for your answer! –  Clément Apr 7 '11 at 22:24

I don't understand how you got the equation after "in the original equation, which yields"; it seems there might be something wrong there but I'm not sure exactly what you did. I think this all gets a bit easier if you transform to $y=\ln x$ and $g(y)=f(x)$; then the condition reads

$$g'(y)=\mathrm{e}^yg(-y)\;,$$

and differentiating as you did yields

$$g''(y)=g'(y)-g(y)\;.$$

The solutions of the characteristic equation are the same $j,\overline{j}$ that you got, so the original equation becomes

$$\left(c_1\mathrm{e}^{jy}+c_2\mathrm{e}^{\overline{j}y}\right)'=\mathrm{e}^y\left(c_1\mathrm{e}^{-jy}+c_2\mathrm{e}^{-\overline{j}y}\right)\;.$$

Since $1-j=\overline{j}$ and $1-\overline{j}=j$, this is satisfied if $jc_1=c_2$ and $\overline{j}c_2=c_1$, and these conditions are actually equivalent, since $j\overline{j}=1$. So the solution is

$$c \left(\mathrm{e}^{jy}+j\mathrm{e}^{\overline{j}y}\right)=c\left(x^j+jx^{\overline{j}}\right)\;.$$

For this to be real, we must have $c=b/\sqrt{j}$ with $b\in\mathbb{R}$, and thus

$$f(x)=a\Re\left(x^j/\sqrt{j}\right)$$

with $a\in\mathbb{R}$. Plugging this back into the original equation shows that this is indeed a solution.

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By the way, the problem is a lot more interesting than it looks at first sight :-) –  joriki Apr 7 '11 at 10:00
Thanks for your thorough explanation and solution; it is a pretty approach indeed, one which I didn't think of at all. Nice work! As you may have seen in my comment to Christian's answer, my mistake is that I plugged $x^j$ instead of $x^{-j}$, which gave me the eventual impression that there were no solutions. Thanks again for your help! –  Clément Apr 7 '11 at 22:26
i don't know if i can comment to clement's question so i will write my answer here. what you have shown is that any solution to $\frac{df(x)}{dx} = f(\frac{1}{x})$ satsifies the cauchy-euler equation $\frac{d^2 f(x)}{dx^2}=-\frac{1}{x^2} f(x)$ whose solutions are $\sqrt x \cos(\sqrt 3 \ln x /2)$ and $\sqrt x \sin(\sqrt 3 \ln x /2)$. the trouble, i think, is the converse statment that the solutions of cauchy-euler equation satisfies the $\frac{df(x)}{dx} = f(\frac{1}{x})$ is false.
Where did you read that $f''(x)=-f(x)$? –  Did Apr 7 '11 at 15:19
i meant to write $\frac{d^2 f}{dx^2} = -\frac{1}{x^2}f(x)$ making it a cuachy-euler equation. –  abel Apr 7 '11 at 16:56
Hi abel, thanks for your help! In fact, the reasoning only showed that the condition obtained was necessary, but not a priori sufficient. Hence my question, since it seemed to me that no combination of the Euler equation was a solution to the initial problem, and I wanted to make sure that there was no mistake; but the actual mistake was pretty trivial: I swapped a $x^j$ and $x^-j$ :/ –  Clément Apr 7 '11 at 22:32