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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?

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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 $p:q=2:1$.

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  • $\begingroup$ "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? $\endgroup$ Commented Dec 29, 2023 at 15:16
  • $\begingroup$ "In the case of the stochastic heat equation" -- What specific statement in Hairer's notes are you referring here to? $\endgroup$ Commented Dec 29, 2023 at 16:49
  • $\begingroup$ You have just found "where the quote in the OP comes from". However, the question was to explain why this is true. $\endgroup$ Commented Dec 31, 2023 at 1:26

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