# Ancient Heat equation and Liouville's theorem

I encounter a difficulty when proving the bounded solution of ancient heat equation implying constant function. Suppose $$u(t,x)$$ is the solution of ancient heat equation:

$$$$u_{t} = \Delta u \quad \mathrm{in} ~\ (- \infty, 0] \times \mathbb{R}^{n}$$$$

In order to prove it is constant function, we can mimic the proof of Liouville's theorem in complex analysis by showing that $$\nabla_{x} u(t,x)$$ goes to zero. In usual heat equation defined in $$(0, T] \times \mathbb{R}^{n}$$, we have the following gradient estimation:

$$$$\sup_{x \in \mathbb{R}^{n}}| \nabla_{x} u(T,x)| \leq \frac{C}{\sqrt{T}} \sup_{x \in \mathbb{R}^{n}} |u_{0}|$$$$ where C is a constant relates to its dimension $$n$$ and $$u_{0}$$ is the initial value of $$u(t,x)$$. Originally, I want to change the variable $$t \rightarrow -t$$ such that I can apply the above gradient estimation to show that the solution of ancient heat equation $$u(t,x)$$ is constant function. However, there is a problem that $$\sqrt{T}$$ would be complex under this change of variable. Therefore, how can I combined these two concepts to show bounded solution for ancient heat equation is constant function, similar to Liouville's theorem in Complex analysis?

• Yes, this inequality is false if $T<0$ ... when you look at the proof, you will see that if $t<0$, it doesn't work since the disspative term is now no more negative. – LL 3.14 Apr 13 '20 at 8:04
• To see that $u$ is constant you have to apply the estimates for the derivatives in the following form $$|D_t^n D_x^k u(t_0,x_0)| \le \frac{C_{n,k} \|u\|_{\infty, Q(t_0,x_0; R)}}{R^{k+2n}},$$ where $Q(t_0,x_0;R)=(t_0-R^2,t_0)\times B(x_0,R)$ adn let $R \to \infty$ for every fixed $(t_0,x_0)$ (you need only $n,k$ to be 0 or 1). – Giorgio Metafune Apr 13 '20 at 8:15

$$$$\sup_{x \in \mathbb{R}^{n}}| \nabla_{x} u(t,x)| \leq \frac{C}{\sqrt{T}} \sup_{x \in \mathbb{R}^{n}} |u(t-T,x)| ,\quad t\le 0,\ T>0.$$$$