Laplace transforms and IVP solutions

Question

When deriving the Laplace transform, we note that the Laplace transform of a function is only defined for $x\ge 0$ (unless in the bilateral case you set $f(x)=0$ for $x \lt 0$). However, we often solve initial value problems over all reals with Laplace transforms. How can we be sure that the Laplace transform gives a solution for all x in this case when we originally assumed our solution was equal to $0$ when $x \lt0$?

Answer 1

Basically, the Laplace transform is unilateral but the location of $t=0$ in the graph of the function $f$ can be anywhere you want. So for example if I want to solve an equation on $[t_0,\infty)$ I can apply the Laplace transform in the form $\int_{t_0}^\infty e^{-s(t-t_0)} f(t) dt = \int_0^\infty e^{-s u} f(u+t_0) du$, and then I get a solution for $u \in [0,\infty)$ which is to say $t \in [t_0,\infty)$. That $t_0$ can be as negative as you like.

Answer 2

I don't understand the question. It is an INITIAL value problem, so time starts at some point, which we can shift to $0$ if we like.