Let $f$ be defined for $x \in \mathbb{R}$ by $$ f\left(x\right)=2\text{arctan}\left(e^x\right)-\frac{\pi}{2} $$ I've shown that $f$ is odd and satisfies for $x \in \mathbb{R}$ $$f'\left(x\right)=\cos\left(f\left(x\right)\right)$$ To prove it, i've used that $$ \cos\left(f\left(x\right)\right)=2\sin\left(\text{arctan}\left(e^x\right)\right)\cos\left(\text{arctan}\left(e^x\right)\right) $$ And then use that $$ \cos\left(\text{arctan}\left(e^x\right)\right)=\frac{1}{\sqrt{1+e^{2x}}} $$ I've two questions :

$\bullet$ Is that the unique solution for $f(0)=0$ ? I've tried to prove it by supposing to different solutions and trying to prove there are infact equals with trigonometric formula but it does not seem to work.

$\bullet$ Is there another way ( even wiser or faster ) to prove it ?

Thanks for those who take time to answer.

  • $\begingroup$ If I have understood your question correctly, I feel you are asking whether (1) the solution to the DE is correct and (2) solutions to f(x)=0? $\endgroup$ – Rohan Dec 28 '17 at 10:56
  • $\begingroup$ No, that's not what i meant. I want to know if $f$ is the unique solution for the problem $f'(x)=\cos\left(f\left(x\right)\right)$ with $f(0)=0$ and if there's other proof that $f$ is indeed a solution. $\endgroup$ – Atmos Dec 28 '17 at 11:00

Do you ask if $f\left(x\right)=2\text{arctan}\left(e^x\right)-\frac{\pi}{2}$ is the unique solution of $$ \begin{cases} f'(x)=\cos(f(x))\\ f(0)=0 \end{cases} $$ ?

Since $\cos$ is Lipschitz continuous, you can use Picard-Lindelöff theorem to see that the solution is unique.

For your second question: If you have to show that your function is a solution of the ODE above, then the only way is to check that the equations hold for your function. There is no other way.


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