ODE initial conditions and integrating backwards in time

Question

This is a very basic question on initial value problems and ordinary differential equations. When solving an IVP, with $y$ as the dependent variable and $t$ as the independent variable (given some set of initial conditions $y(0)$, $y'(0)$,...), why do we "integrate forward" in time? That is, in an engineering/physics context the author would typically say something along the lines of $y(t)$ can be found for $t > 0$ (or generally $t > t_0$ where $t_0$ is the point at which the initial condition is specified). Doesn't this assume that the output was zero or undefined before the initial condition? Mathematically, can't we just as well find $y$ before $t = 0$ if we are given a set of initial conditions?

Answer 1

We often do integrate backwards as well as forwards. However, in applications it's more common to know where you're starting rather than where you're finishing.

Answer 2

If one carefully examines the usual statement and proof of the Picard-Lindelof Theorem, which is the standard result on existence and uniqueness of solutions to ordinary differential equations, one sees that a solution is typically guaranteed on some interval $I = (t_0 - a, t_0 + a)$, where the initial condition to

$\dot x = f(x, t) \tag 1$

is specified at $t_0$:

$x(t_0) = x_0; \tag 2$

thus, there is nothing essential about the "forward" solution $x(t)$, $t \ge t_0$; it is really an integral curve $x(t)$ through $(t_0, x_0)$ which is addressed by the fundamental theory, which says nothing intrinsic about the "direction of time". So the fact that, in practical applications, we like to think of things as starting out at some $t = t_0$ and then "evolving" in a future direction, is really based on our choice of point of view. Indeed, in more advanced physical theories, such as quantum electrodynamics, the notion of moving "backward in time" plays a fundamental and useful role.