Inverse problem and heat equation In several books dealing with the subject of evolution equations we always find this expression: 'the equation of heat is irreversible in time' or 'the equation of the waves is reversible in time'.
This expression means what? My first impression when I read this expression is: can we find the initial state of the system from the final state (inverse problem).
is it true ?
Thank you.
 A: I'm not sure if reversible is a technical term, but I'd interpret it the way you did. 
Wave equation. It's clear that if $u$ is a solution of $u_{tt} = \Delta u$, then $v(t,x) = u(-t,x)$ is also a solution. Thus we can reverse the time. Given any initial conditions $f \in C^2(\mathbb{R}^n)$, $g \in C^1(\mathbb{R}^n)$, the problem 
$$
\begin{cases}
u_{tt} = \Delta u \\
u(0,\cdot) = f \\
u_t(0,\cdot) = g
\end{cases}
$$
has a solution defined not only on $[0,\infty) \times \mathbb{R}^n$, but even on $\mathbb{R} \times \mathbb{R}^n$ (i.e., for all times $t \in \mathbb{R}$); the usual formula works. Then one can take $u(-1,\cdot)$ and $u_t(-1,0)$ as initial conditions and obtain $f,g$ at time $1$. 
Heat equation. The same trick $t \mapsto -t$ doesn't work here. Also, one cannot solve the inverse problem in general. The formula for the solution implies that it's a smooth function for all $t>0$. Therefore you can take some function $u_0 \in C_c^2(\mathbb{R}^n) \setminus C^\infty(\mathbb{R}^n)$ to be an initial condition (which is perfectly valid even in the classical sense), but finding a solution $u \colon [0,\infty) \times \mathbb{R}^n \to \mathbb{R}$ such that $u(t,\cdot) = u_0$ for some $t>0$ is impossible. 
As a side comment, I've never really understood how this irreversibility pops up when one derives the heat equation from the kinetic equations (i.e. as a limit of the model with finitely many colliding particles). 
