How to obtain the formal scaling for the following PDE?

Question

In the paper by Mourrat and Weber( GLOBAL WELL-POSEDNESS OF THE DYNAMIC $\Phi^4_3$ MODEL ON THE TORUS) one reads:

Equation (1.1) is given by:

I couldn't arrive at (1.3)

Attempt:

note that $\partial_{\hat t} = \lambda^2 \partial_t $ and $\partial_{\hat{x_i}} = \partial_{x_i}$ yields $\hat \Delta = \lambda^2 \Delta$ so: $$\partial_{\hat t} \hat X = \lambda^2 \lambda^{\frac{2-d}{2}} \partial_t X = \lambda^2 \lambda^{\frac{2-d}{2}} \big(\Delta X - X^3 + mX + \xi\big) \\= \lambda^2 \lambda^{\frac{2-d}{2}}\Delta X - \lambda^2 \lambda^{\frac{2-d}{2}}X^3 + \lambda^2 \lambda^{\frac{2-d}{2}}mX + \lambda^2 \lambda^{\frac{2-d}{2}}\xi$$

now note that $$\lambda^2\lambda^{\frac{2-d}{2}}\Delta X = \hat \Delta \hat X $$ $$\lambda^2\lambda^{\frac{2-d}{2}}mX = \hat m \hat X $$

So we would like to see that

$$\lambda^2 \lambda^{\frac{2-d}{2}}X^3 = \hat X ^ 3 $$

$$ \lambda^2 \lambda^{\frac{2-d}{2}}\xi = \hat \xi $$

However $X^3 = \lambda^{-3\frac{2-d}{2}}\hat X^3$ so $$\lambda^2 \lambda^{\frac{2-d}{2}}X^3 =\lambda^2 \lambda^{\frac{2-d}{2}}\lambda^{-3\frac{2-d}{2}}\hat X^3 = \lambda^2 \lambda^{d-2}\hat X ^ 3 = \lambda ^d \hat X ^ 3$$

and $$ \lambda^2 \lambda^{\frac{2-d}{2}}\xi = \lambda^2 \lambda^{\frac{2-d}{2}}\lambda^{-\frac{2 + d}{2}}\hat \xi = \lambda ^ {2 - d} \hat \xi$$

So I can' t arrive at the desired result. I am missing something here?

Answer 1

Here's my attempt,

1. $\partial_tX$:

$$ \partial_tX = \frac{\partial X}{\partial t} = \frac{\lambda^2}{\lambda^{1 - d/2}}\frac{\partial \hat{X}}{\partial \hat{t}} = \lambda^{1+d/2}\partial_{\hat{t}}\hat{X} $$

2. $\Delta X$:

$$ \Delta X = \frac{\lambda^2}{\lambda^{1-d/2}}\frac{\partial \hat{X}}{\partial \mathbf{x}^2} = \lambda^{1+d/2}\hat{\Delta}\hat{X} $$

3. $X^3$:

$$ X^3 = \lambda^{3d/2 - 3}\hat{X}^3 $$

3. $mX$:

$$ mX = \frac{\lambda^{d/2-1}}{\lambda^2} \hat{m}\hat{X} = \lambda^{d/2 - 3}\hat{m}\hat{X} $$

4. $\xi$:

$$ \xi = \lambda^{-1 - d/2}\hat{\xi} $$

Putting all together we get

\begin{eqnarray} \lambda^{1+d/2}\partial_{\hat{t}}\hat{X} &=& \lambda^{1+d/2}\hat{\Delta}\hat{X} - \lambda^{3d/2-3}\hat{X}^3 + \lambda^{d/2 - 3}\hat{m}\hat{X} + \lambda^{-1 - d/2}\hat{\xi} \\ \Rightarrow \qquad \partial_{\hat{t}}\hat{X} &=& \hat{\Delta}\hat{X} - \lambda^{d-4}\hat{X}^3 + \lambda^{-4} \hat{m}\hat{X} + \lambda^{-2-d}\hat{\xi} \end{eqnarray}

Which is obviously not their Eq. (1.3). Unless there is something wrong with Eq. (1.3), I believe this should be the scaling:

$$ \hat{t} = \lambda^2t, \quad \hat{x} = \lambda x, \quad \hat{\xi} = \lambda^{(d-10)/4}\xi, \quad \hat{X} = \lambda^{(d-2)/4} \quad\mbox{and}\quad \hat{m} = \lambda^2m $$

This last set is obtained just by simply assuming $\hat{t} = \lambda^{\alpha}t$, $\hat{x} = \lambda^{\beta}x$, $\cdots$ and then finding the numbers $(\alpha,\cdots)$ that match their Eq. (1.3)