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Function

Ordinary Differerential Equations
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Papapetros Vaggelis
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Function

#1

Post by Papapetros Vaggelis » Fri Jan 06, 2017 12:43 pm

Find the differentiable function \(\displaystyle{f:\mathbb{R}\to \mathbb{R}}\) which satisfies

\(\displaystyle{f(0)=1}\) and \(\displaystyle{f^\prime(x)=f^2(x)\,f(-x)\,,\forall\,x\in\mathbb{R}}\).
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Re: Function

#2

Post by Riemann » Tue Feb 07, 2017 9:26 am

Let us prove first that the function $g(x)=f(x)f(-x)$ is constant. Indeed taking the derivative we have that

\begin{align*}
g'(x) &= f(x) f(-x) \\
&= f'(x) f(-x) - f(x) f'(-x)\\
&= \cancel{f^2(x) f^2(-x) - f^2(x) f^2(-x)}\\
&= 0
\end{align*}

Thus $g(x)=c$ forall $x \in \mathbb{R}$. For $x=0$ we have that $g(0)=f^2(0)=1$. Thus $g(x)=1$ forall $x \in \mathbb{R}$. Hence $f(-x) f(x)=1$. This relation tells us that $f$ cannot have any roots within its domain. Since $f(0)=1>0$ we also conclude that $f$ is positive. The initial condition gives us

$$f'(x)=f(x)$$

and this implies $f(x)=e^x , \; x \in \mathbb{R}$ a function that satisfies all conditions given.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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Re: Function

#3

Post by Papapetros Vaggelis » Tue Feb 07, 2017 11:42 am

Hi Riemann. Thank you for your answer.
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