Inspired by a question I asked [here](https://math.stackexchange.com/questions/626927/how-a-group-represents-the-passage-of-time), 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?
['https://math.stackexchange.com/questions/628720', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/62765/']
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.
This is going to be more of a "Why I think it's not reversible" than a rigorous argument:

At $t = 0$, let the temperature for a 1-dimensional homogeneous medium be piece-wise constant:

$T(0,x) = 1$ for $x =< 0$ ;

$T(0,x) = 3$ for $x > 0$ .

For any value of x, as $t \to \infty$, $T(t,x) \to 2$ .

Now, take any value of $t > 0$, and $T(t,x)$ will not, as a function of $x$, be identically equals 2. Numerically or analytically, you can run the heat diffusion equation backwards, to smaller values of t. But once you reach $t = 0$, you will regain the original discontinuity at $x = 0$. What does $T(0 - 
\epsilon,x)$ look like? I don't believe there is a reasonable answer to that.