An equation involving first order derivate

Question

How can one solve the following equation for $f(x)$:

$$\frac{f'(x)}{f(2x)-f(x)}= \ln x$$

where $f'(x)$ is the first derivate of $f(x)$?

I've tried but couldn't do anything.

Answer 1

Just some thoughts on the problem, not an answer.

Since this equation is "random" and no initial condition is provided, there's quite a bit of freedom in searching for possible solutions.

One interesting feature is the behaviour of this function around $x=0$. On the one hand, there's $\ln x$ which is not defined for $x=0$, on the other hand, $f(2x)=f(x)$ for $x=0$, so let's see how this could cancel out:

$$f'(x)=\left( f(2x)-f(x) \right) \ln x \tag{1}$$

If $f(x)$ goes to $0$ as $ a x$ for some constant $a$ then the limit on the right exists because $\lim_{x \to 0} x \ln x=0$ (we can also add an arbitrary constant or course, since it would cancel out on both sides of the equation). However, the derivative on the left will be approximately $a$ which means $a=0$ and we have a problem. So around $x=0$ the function has to go at least as $a x^{1+r}$ where $r>0$ if we want the solution to be defined at $x=0$.

This points out to some variation of Frobenius method to find a solution around $x=0$ in the form:

$$f(x)=\sum_{n=0}^\infty a_n x^{n+r}$$

However, because of $f(2x)$ we do not get a polynomial equation for $r$, but rather a transcendental one. Not to mention the $\ln x$ term can't be expanded this way.

We can try expanding around $x=1$, but it's too cumbersome in my opinion.

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Another way is to look at asymptotics for $x \gg 1$. Let us search for a function $f(x)$ which monotonically increases with $x$. The defining equation allows for such solutions because it's consistent with both sides of the equation being positive.

Let us introduce a new variable:

$$y=\ln x, \qquad x=e^y$$

Then we can introduce a new function:

$$g(y)=f(e^y) \\ g(y+\ln 2) = f(2e^y)$$

$$\frac{df(x)}{dx}=e^y\frac{df(e^y)}{dy}=e^y \frac{dg(y)}{dy}$$

Our equation becomes:

$$g'(y)=y e^{-y} \left(g(y+\ln 2)-g(y) \right) \tag{2}$$

For $y \gg \ln 2$ we can write:

$$g(y+\ln 2) \asymp g(y)+g'(y) \ln 2$$

Or, more correctly (where we used our assumption that $f(x)$ and thus $g(y)$ are monotonically increasing):

$$g(y+\ln 2) = g(y)+g'(y) \ln 2+o(g(y))$$

So our equation becomes (asymptotically):

$$g'(y)=(\ln 2 )y e^{-y} g'(y)$$

But that's certainly can't be solved unless $g'(y)=0$ which means $g(y)$ approaches a constant asymptotically as $y \to \infty$ (same for $f(x)$ by definition). This is one possible solution consistent with the original equation.

Let's upgrade our asymptotic expansion with another term:

$$g(y+\ln 2) \asymp g(y)+g'(y) \ln 2+g''(y) \frac{\ln^2 2}{2}$$

Introducing a new function:

$$h(y)=g'(y)$$

We obtain from (2):

$$h(y)=y e^{-y} \left((\ln 2) h(y)+\frac{\ln^2 2}{2} h'(y) \right)$$

$$\frac{dh}{dy}=\frac{2}{\ln^2 2} \left( \frac{e^y}{y}- \ln 2 \right) h$$

$$\frac{dh}{h}=\frac{2}{\ln^2 2} \left( \frac{e^y}{y}- \ln 2 \right) dy$$

$$\ln h = \frac{2}{\ln^2 2} \int \left( \frac{e^y}{y}- \ln 2 \right) dy + \text{const}$$

$$g(y)=\int h(y)dy +\text{const}$$

This is (in some way) an asymptotic solution for (2) if $ y \to \infty$. Nothing pretty, but some information can be gained from this.

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We can use (2) to search for a solution around $y=0$, or $x=1$ as well, expanding for $y \ll \ln 2$:

$$g(y+\ln 2) \approx g(\ln 2)+g'( \ln 2) y+g''(\ln 2) \frac{y^2}{2} + \dots$$

We don't know the values of any of the derivatives. However, we could also search for $g(y)$ in the form of a series:

$$g(y)=a_0+a_1 y+a_2 y^2 +\dots$$

Now we can expand every term in equation (2) as a series around $y=0$, and search for relation between $a_k$ which make the equation true for any $y$. The added complication is the derivatives $g^{(n)}( \ln 2)$ which should be explicitly written in terms of $a_k$ as well, using the above series representation.

This seems like the most promising way, but you would have to solve a large system of equations for $a_k$, instead of getting a nice recurrence relation as it usually happens with linear ODEs.