Inspired by a question I asked here, I am rethinking about a question:
Why heat equation is not time-reversible?
I don't know too much about PDE and physics but I guess there should be some "time arrow" in mathematics.
Consider the following initial value problem:
$$ \begin{cases}
\Omega: (x,t) \in \mathbb{R} \times (0,+\infty) \\
u(x,0) = \delta(x) \\
u_t – u_{xx} = 0
\end{cases} $$
The solution is given by
$$
u(x,t) = \frac{1}{\sqrt{4 \pi t}} \exp \Big( \frac{-x^2}{4t} \Big)
$$
I remember from my undergraduate PDE course, it is different from elliptic
equation which is time-reversible.
If I substitute $t \mapsto -t$ and change the domain $\Omega$ to $\mathbb{R} \times (-\infty,0)$, the above solution will not satisfy the PDE $u_t – u_{xx}$.
I know that we may recall second law of thermodynamics, but it is a physical law, not a mathematical theorem (or axioms). For a mathematical reason, there should be logical deduction from axioms to a "structure" that make heat equation different.
What is the reason behind that?
I also recall a second-order PDE on a domain $\Omega$:
$$
A(x,y) u_{xx} + 2B(x,y) u_{xy} + C(x,y) u_{yy} = W(u,u_x,u_y,x,y)
$$
We say it is
(i) parabolic if for all $x,y \in \Omega$, $B^2 – AC = 0$
(ii) hyperbolic if for all $x,y \in \Omega$, $B^2 – AC > 0$
(iii) elliptic if for all $x,y \in \Omega$, $B^2 – AC < 0$
Is it a pure analogy to conic section or there should be some structure behind that?
Best Answer
The heat equation $u_t-u_{xx}=0$ is not time-reversible because it involves an odd-order derivative of $t$. Under time reversal $t\mapsto-t$, we get $u_t\mapsto-u_t$. So if $u(x,t)$ is a solution to the heat equation, then $u(x,-t)$ is a solution to a different equation, namely $-u_t-u_{xx}=0$. The only way for $u$ to solve both equations is if $u_{xx}=0$ everywhere, which is not the case for most initial conditions of interest, such as your $\delta(x)$.
By contrast, the wave equation $u_{tt}-u_{xx}$ is time-reversible because it involves only even-order derivatives of $t$. Under time reversal $t\mapsto-t$, we get $u_{tt}\mapsto u_{tt}$. So if $u(x,t)$ is a solution to the wave equation, then $u(x,-t)$ is also a solution to the wave equation.