Differential equation dimension analysis

[Forhad3097]

A differential equation of solitary wave oscillons is defined by,
$$ \Delta S -S +S^3=0 $$
**How can we write this equation as,**
\begin{equation}
\langle(\vec{\nabla}S)^2\rangle+\langle S^2\rangle-\langle S^4\rangle=0 \tag{1}
\end{equation}
where $\langle f\rangle:=\int d^Dx f(x)$. Furthermore, another virial identity
can be found
from the scaling transformation ($$\vec{x}\to \mu \vec{x}$$)
by extremizing the scaled ($$\vec{x}\to\mu \vec{x}$$)
of the action corresponding to
$$ \int d^Dx[(\vec{\nabla}S)^2+S^2-S^4/2]$$:
\begin{equation}
(D-2)\langle(\vec{\nabla}S)^2\rangle+D\langle S^2\rangle-\frac{D}{2}\langle S^4\rangle=0 \tag{2}
\end{equation}
From Eqs. (1) and (2) one immediately finds
\begin{equation}
2\langle S^2\rangle+\frac{1}{2}(D-4)\langle S^4\rangle=0\,,
\end{equation}
which equality can only be satisfied if $D<4$.
D= Refers dimension.

If you have any Query then ask me please.
Thanks in advance.
To see details, please check the paper here in equations (21), (41)and (42)

 

[AndyW]

Thank you for your post about the differential equation of solitary wave oscillons. It is an interesting topic that I am familiar with. I would be happy to help you understand how the equation can be written in the form given in equations (1) and (2).

First, let's consider the original equation:
$$ \Delta S -S +S^3=0 $$

We can write this in terms of the Laplacian operator $\nabla^2$ as:
$$ \nabla^2S -S +S^3=0 $$

Now, to get to equation (1), we can take the average of both sides of the equation over all space, denoted by $\langle \rangle$. This gives us:
$$ \langle \nabla^2S \rangle -\langle S \rangle +\langle S^3 \rangle =0 $$

Using the definition of $\langle \rangle$ given in the post, we can write this as:
$$ \langle (\vec{\nabla}S)^2 \rangle +\langle S^2 \rangle -\langle S^4 \rangle =0 \tag{1}$$

This is the form given in equation (1). Now, let's move on to equation (2).

To get to equation (2), we can use a scaling transformation, as mentioned in the post. This transformation is given by:
$$ \vec{x} \to \mu \vec{x} $$

This means that all distances are scaled by a factor of $\mu$. Using this transformation, the Laplacian operator becomes:
$$ \nabla^2 \to \frac{1}{\mu^2} \nabla^2 $$

We can then apply this transformation to the original equation and take the average over all space, giving us:
$$ \langle \frac{1}{\mu^2} \nabla^2S \rangle -\langle S \rangle +\langle \frac{1}{\mu^4} S^3 \rangle =0 $$

Using the definition of $\langle \rangle$, we get:
$$ \frac{1}{\mu^2} \langle (\vec{\nabla}S)^2 \rangle -\langle S \rangle +\frac{1}{\mu^4} \langle S^3 \rangle =0 $$

Now, we can