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Consider the equation \begin{cases} \displaystyle y'(x)=\cos\left(y(x)\right)\\ y(0)=0 \end{cases} I've found explicitely $y$ and I know this equation has a unique solution. However is there a way to prove unicity of such an $y$ without using Cauchy-Lipschitz Theorem ? I've tried to suppose two solutions $y_1$ and $y_2$ to show $y_1-y_2=0$. I only found $$ (y_1-y_2)'=-2\sin\left(\frac{y_1+y_2}{2}\right)\sin\left(\frac{y_1-y_2}{2}\right) $$ But it did not help me. Can anyone got an idea to show it this way ?

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    $\begingroup$ If you set $z:=y_1-y_2$ and integrate what you found, using that $z(0)=0$, you should be able to conclude via Gronwall's inequality (integral form). $\endgroup$ – Giuseppe Negro Jan 24 '19 at 19:38
  • $\begingroup$ @GiuseppeNegro I wonder how you would use it ? Because $\cos\left(y\right)$ does satisfy $\leq y$ ? $\endgroup$ – Atmos Jan 25 '19 at 17:32
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In order to prove the unicity in this particular case let $U$ be an open interval around $0\in{\mathbb R}$, where $|y(x)|<{\pi\over2}$, and consider the auxiliary function $$u(x):=\log{1+\sin\bigl(y(x)\bigr)\over1-\sin\bigl(y(x)\bigr)}\qquad(x\in U)\ .\tag{1}$$ One computes $$u'(x)=\left({1\over1+\sin\bigl(y(x)\bigr)}-{-1\over1-\sin\bigl(y(x)\bigr)}\right)\cos\bigl(y(x)\bigr)y'(x)\equiv2\qquad(x\in U)\ .$$ I think you will accept that this together with $u(0)=0$ implies $u(x)=2x$ for all $x\in U$. From $(1)$ we then get $${1+\sin\bigl(y(x)\bigr)\over1-\sin\bigl(y(x)\bigr)}=e^{2x}\qquad(x\in U)\ ,$$ and this leads to $$y(x)=\arcsin\bigl(\tanh x\bigr)\qquad(x\in U)\ .\tag{2}$$ As $|\tanh x|<1$ for all $x\in {\mathbb R}$ we see that the |RHS| of $(2)$ is $<{\pi\over2}$ for all $x\in{\mathbb R}$, so that we may choose $U={\mathbb R}$.

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  • $\begingroup$ I found that $u'(x)=-sin(y(x))$ $\endgroup$ – Atmos Jan 24 '19 at 20:51

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