What is $\ddot y(t) = a(t), $y(t) defined for $t \ge 0$, with boundary condition? I have the following question which I do not know how to tackle. Any hints are welcome.
Consider the second-order differential equation
$\ddot y(t) = a(t), $y(t) defined for $t \ge 0$ with the boundary condition $\lim_{t \to \infty} \dfrac{y(t)}{t} = 1$ at $\infty$.
Show that if $a(t) \ge 0$ on $[0, 1]$ and vanishes for $t\ge 1$ is a smooth function, then the equation admits a nonnegative solution, i.e., $y(t) \ge 0$.
 A: Start with a solution $z$ with $z(0)=z'(0)=1$. The point $(z,z')$ in phase space starts in the first quadrant and thus will remain there, moving away from the origin. So $z'(1)$ is positive.
Now set $y=z/z'(1)$ to get $y'(1)=1$. As $y''(t)=0$ for $t>1$, this slope remains constant, $y(t)=y(1)+(t-1)$ is linear and $y(t)/t$ converges to the slope value $1$.
A: Require $y'(1)=1$ for the BC at infinity, then write
$$y(t)=y(1) + y'(1)(t-1) + \int_1^t \int_1^x y''(s) ds dx = y(1) - 1 + t + \int_1^t \int_1^x a(s) y(s) ds dx.$$
This is now in the form of a fixed point equation. Note that the double integral is nonnegative (no matter whether $t>1$ or $t<1$) provided that $y$ is, and $y(1) - 1 + t$ is nonnegative if $y(1) \geq 1$. So by induction, the fixed point iterates starting from, say, $y=y(1)-1+t$ are all nonnegative if $y(1) \geq 1$.
These iterates converge on some neighborhood of $1$ to a nonnegative function which solves the ODE IVP.
Now you need to propagate this observation to all of $[0,\infty)$. This is doable, but there is no new insight to be gained, you just do the same procedure over again starting from a point to the left of $1$ and continue until you actually get to $0$. (To the right of $1$ there is no difficulty because the double integral will just be $0$ for $t>1$.)
