A special equation : f^[t](x) = f(t x)/t

[tommy1729]

hey tommy
just got home from uni
i took some time to solve for this:
By assumption, \[f^t(x)=\frac{f (t x)}{t},f(0)=0,f(0)=\sum_{n=1}^{\infty}\frac{f_n x^n}{n!},f_n=f^{(n)}(0)\]
Take derivative of x at 0, \[f'(0)^t=f'(0)\Longrightarrow f'(0)=1\]
We can write \[t  f^t\left(x\right)=\sum _{n=1}^{\infty } \frac{f_n t^n x^n}{n!}=f (t x)\]
And recall a formula about the Julia function:\[\frac{d f^t(x)}{d t}=\frac{d f^t(x)}{d x}[t^1]f^t(x)\text{ where }[t^1]f^t(x)=\lambda_f(x)\]
Thus by the series and this equation we have, \[[t^1]f^t(x)=f_2\frac{x^2}{2}=\lambda_f(x)=\frac{\frac{\mathrm{d}f^t(x)}{\mathrm{d}t}}{\frac{\mathrm{d}f^t(x)}{\mathrm{d}t}}=\frac{\frac{\mathrm{d}\frac{f(tx)}{t}}{\mathrm{d}t}}{\frac{\mathrm{d}\frac{f(tx)}{t}}{\mathrm{d}t}}=\frac{tx-\frac{f(tx)}{f'(tx)}}{t^2}\]
use tx=y to reduce this, we see that \[f_2\frac{y^2}{2}=y-\frac{f(y)}{f'(y)}\]
solve it as ODE, we can show the only possible solutions are (C1 and f2 are constants):
\[f(x)=\frac{C_1x}{2-f_2x}\]
As f(x) is a rational function, we can solve it very easily now:
if C_1 is not 2, we put t=2 then we suddenly get C1=0 is the only way to equalize the 2 sides, which gives f(x)=0
if C_1 = 2, we suddenly get f_2 can be any constant, so that we have a series of solutions
So we conclude that \[f(x)=0\text{  or  }f(x)=\frac{2x}{2-f_2x}\] are the solutions of this equation.

regards
leo

Thank you !

I will look at the details later ...

regards

tommy1729