Going back in time in an initial value problem

Question

Consider an initial value problem (IVP) $y'=f(t,y)$ with the initial value given by $y(t=0) = 0$.

If I need to find $y(t^*)$, hence finding the path for $y$ in $t \in [0,t^*]$ and $t^*<0$; is the problem then still an initial value problem? In other words can you go back in time in IVP.

The reason I ask is that I have an algorithm where in each step I need to solve for different $t^*$. I intend to use solve_ivp in Python which is based on Runge-Kutta 45 method and I want to know if there are any theoretical contradictions when I apply RK45.

I know that if I will use Eulers method then there is no problem. But what about RK45. Will I get the desired result.

Answer 1

This is technically still an IVP if you do an appropriate change of variables. Given your time is between $t \in [t^*, 0]$, make a new time variable $\tau = -t$ so that $\tau \in [0, -t^*]$ and you can modify the time derivatives accordingly. This means that you should have the differential equation $\frac{dy}{d\tau} = -f(-\tau, y)$ with $y(\tau = 0) = 0$ as your new IVP. This implies that you should be able to use Runge-Kutta approaches just fine.

Answer 2

There is no direction prescribed in the solution of an initial-value problem. A function $y:(\alpha,\beta)\to\Bbb R^n$ solves an IVP if $α<t_0<β$ and $y(t_0)=y_0$, and of course $y'(t)=f(t,y(t))$ for all $t\in (α,β)$.

While numerical solvers have directionality built-in, publicized solvers usually treat negative step sizes as well as positive step sizes. Only switching the direction during the integration is usually not supported. So calling

res = solve_ivp(f,[0,ts],y0)

will work equally well for positive and negative ts$=t^*$.