# A higher-order differential equation involving absolute values and trigonometry

For a smooth function $f: (-\pi/2,\pi/2) \to \mathbb{R}$, if $\displaystyle\frac{|f''(x)|}{\sqrt{(1+f'(x))^3}} = \cos{x}$, and $f(0) = f'(0) = 0$, $f''(0) = 1$, $f''(-\pi/2) = 0$, and $f''(\pi/2) = 0$. What is the function $f(x)$?

-
Start by setting $g(x)=f'(x)$ and integrate to find $f$ once you've gotten everything else to fit. The absolute value is something of a snag -- your best strategy will be to look for solutions with $g'(x)$ being positive and negative, respectively, and hope you can glue them together to meet all of the boundary conditions later. (Or you do the positive case first and hope you won't even need the negative one). Each of those cases should then be attackable with separation of variables. – Henning Makholm Sep 18 '11 at 1:19
I the the algebra in the title is very misleading. Perhaps you should consider retitling this? – mixedmath Sep 18 '11 at 1:20
@Henning, the boundary conditions are met no matter what: $f''$ must vanish at $\pm\pi/2$ because $\cos x$ is zero there. – Rahul Sep 18 '11 at 3:46

This is a second order diferential equation with three initial conditions (let's forget by now about the boundary conditions), while usually two are enough to have existence and uniqueness. However, from the equation and $f(0)=f'(0)=0$, it follows that $|f''(0)|=1$, which is compatible with the third condition $f''(0)=1$. Since $f''(0)>0$, the second derivative of the solution will be positive in a neighbourhood of $0$ ; thus, in a first step, we can forget abot the absolute value.
Let $f´=y$. The problem becomes $$(1+y)^{-3/2}y'=\cos x,\quad y(0)=0.$$ This is a first order equation in separate variables, whose solution is $$y=\frac{4}{(2-\sin x)^2}-1.$$ Then $$f(x)=-x+\int_0^x\frac{4\,dt}{(2-\sin t)^2},$$ and $$f''(x)=y'(x)=\frac{8\cos x}{(2-\sin x)^3}.$$ Since $f''>0$ on $(-\pi/2,\pi/2)$ and $f''(-\pi/2)=f''(\pi/2)=0$, $f$ is the solution you are looking for. Below is the graph of $f$, $f'$ and $f''$.