Find functions f(x) continuous on the real line satisfying given conditions

by RoRi

January 31, 2016

Find functions $f(x)$ continuous everywhere on $\mathbb{R}$ such that

$(f(x))^2 + (f'(x))^2 = 1.$

(Note that $f(x) = -1$ is one solution.)

First, we note that $f(x) = \pm 1$ is a solution. To find the other solutions assume that $f(x) \neq \pm 1$ . Then letting $y = f(x)$ we have

$\begin{align*} (y')^2 + y^2 = 1 && \implies && y' &= \sqrt{1-y^2} \\ && \implies && \frac{y'}{\sqrt{1-y^2}} &= 1 \\ && \implies && \int \frac{1}{\sqrt{1-y^2}} \, dy &= \int \, dx \\ && \implies && \arcsin y &= x + C \\ && \implies && f(x) &= \sin (x+C). \end{align*}$

Stumbling Robot

Find functions f(x) continuous on the real line satisfying given conditions

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