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Will be glad for a little hint for this one; f is a real valued differentiable function defined on $[1,\infty)$ with $f(1)=1.$ Suppose moreover f satisfies $$f'(x)=\frac{1}{x^2+f^2(x)}$$ Show that $f(x)\le 1+\frac{\pi}{4}$

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Note that, $f'(x) \leq \frac{1}{x^2+1},$ since $f$ is increasing. – Ehsan M. Kermani Apr 16 '12 at 18:36
up vote 4 down vote accepted

From the given condition $f'(x)={1\over x^2+f^2(x)}$, it follows that $f'$ is positive on $[1,\infty)$. So $f$ is increasing on $[1,\infty)$, and since $f(1)=1$, we have $f(x)\ge1$ on $[1,\infty)$. The aforementioned condition also implies that $f'$ is continuous on $[1,\infty)$. We thus have $$ f(t)-f(1)=\int_1^t {1\over x^2+f^2(x)}\,dx\le \int_1^\infty{1\over 1+x^2}\,dx=\tan^{-1} x\,\bigl|_1^\infty={\pi\over 4}; $$ whence the result follows.

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By the way, this initial value problem has a closed-form implicit solution: $$ \frac{x Y_{1/4}(f(x)^2/2) - f(x) Y_{-3/4}(f(x)^2/2)}{x J_{1/4}(f(x)^2/2) + f(x) J_{-3/4}(f(x)^2/2)} = \frac{Y_{1/4}(1/2) + Y_{-3/4}(1/2)}{J_{1/4}(1/2) + J_{-3/4}(1/2)}$$ where $J$ and $Y$ are Bessel functions of the first and second kinds. The actual limit of $f(x)$ as $x \to \infty$ is approximately $1.729444022$.

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