Potential in Relativistic Scalar Field Theory My intention is to establish a Soliton equation. I have cropped a page from  Mark Srednicki page no 576.
I have understand the equation (92.1) but don't understand that how  they guessed the  potential   in equation (92.2). 
EDIT:
Contrast the above potential the author  used this potential in equation (2)  $$U(\phi)= \frac{1}{8} \phi^2 (\phi -2)^2.$$
 My query is , are we using different potential just for our convenient using? or I need to redefine the $\phi^4$ theory to get the potential ?
 A: Any term in the action that does not transform under any symmetries is allowed. This means for a scalar field you can have any power $\phi^n$ in your Lagrangian. In equation (92.2) Scrednicki just choses a particular potential (that is, he chose (1) a polynomial in the scalar field along with (2)particular coefficents) that additionally gives you a vev for the scalar field of $\langle \phi \rangle = \pm v$. There are many potentials that will give your scalar field a vev, this is just the simplest. 
EDIT: Like I said before 'There are many potentials that will give your scalar field a vev, this is just the simplest.' A bit more detail:
You can start with a general $\phi^4$ theory:
$V(\phi) = a_1 \phi + a_2 \phi^2 + a_3 \phi^3 +a_4 \phi^4$
and go from there. For example if you impose a discrete symmetry $\phi \rightarrow - \phi$ then $a_1 = a_3 = 0$. Moreover if you want $\phi$ to get a vev, one of the ways to do this is to choose $a_2 < 0$. But there are other potentials that will give the field a vev. 
Also, judging from this and other questions to have posted, you seem to be obsessing over the 2 potentials 
$V(\phi) = \frac{1}{8} \lambda (\phi^2 - v^2)^2$
and
$U(\phi) = \frac{1}{8}\phi^2 (\phi - v)^2$.
This is a bit like obsessing over whether vanilla or chocolate ice cream is better. It depends on your taste and what you are going to serve the ice cream with. 
