# Regularity of stochastic heat equation

For the heat equation:

$$\frac{\partial u}{\partial t} = \Delta u + \zeta(t,x)$$

where $$u:\mathbb{R}^{+}\times \mathbb{R}^{n} \rightarrow \mathbb{R}$$ and $$\zeta$$ is a space-time white noise.

I'm wondering how to understanding the following words:

''We can ‘trade’ space-regularity against time-regularity at a cost of one time derivative for two space derivatives.''

I want an example to understand these words. Could you please show me some examples?

The presence of noise allows to exchange ("trade") time regularity for space regularity. For a simple example, consider the map $$(t,x)\mapsto \int_0^t b(s,x+W_s)\,ds,$$ with Gaussian noise $$W_s$$. If without the noise the map is $$C_t^1C_x^0$$, with the noise it becomes $$C_t^0 C_x^1$$, so the improvement of space regularity is obtained at the cost of loss of time regularity. See these lecture notes (pages 14+15).
In the case of the stochastic heat equation, the noise converts $$C_t^{1/2}C_x^0$$ regularity into $$C_t^{1/4}C_x^{1/2}$$ regularity, at a "rate of two powers of space to one power of time". This is derived in section 5.1 of these notes. The heuristics for the two-to-one conversion rate is based on dimensional analysis: The heat equation $$u_t=Du_{xx}$$ has one dimensionfull coefficient $$D$$ with dimensions length$$^2$$/time. This allows the conversion of $$p$$ powers of space into $$q$$ powers of time at a ratio $$p:q=2:1$$.
• "If without the noise the map is $C_t^1C_x^0$, with the noise it becomes $C_t^0 C_x^1$" -- (i) What map? $b$? (ii) What do you mean by "it becomes"? (iii) Why? Commented Dec 29, 2023 at 15:16