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?


1 Answer 1


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 $p:q=2:1$.

  • $\begingroup$ "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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.