Spontaneous Symmetry Breaking In Spontaneous symmetry breaking we have got that, a field $$\phi=  \pm \sqrt{\frac{-m^2}{\lambda}}.$$ Now in order to get the unstable minima we need to guess the mass $m^2 <0 $.  But can mass be negative? Why are we guessing so? 

One more question is that, how this lead us to the solution of
  nonlinear equation?

 A: Sorry,I don't follow your notations in my answer, I hope it's clear enough nevertheless.
The mass can be supposed negative without much effort in the Ginzburg-Landau model: 
$$F=\int dr\left[\dfrac{\beta}{2}\left(\left|\Psi\right|^{2}-\dfrac{\alpha}{\beta}\right)^{2}+\dfrac{1}{2m}\left|-\mathbf{i}\hslash\nabla\Psi\right|^{2}\right]$$
with $\alpha\propto\left(T-T_{c}\right)$ negative for temperatures $T$ below the critical temperature $T_{c}$. You can call $\alpha$ a mass if you like, but it is a parameter of the model in fac, like $\beta$, ....
The minimum of the free energy $F$ above is $F=0$ given for
$$\Psi_{\infty}^{2}=\dfrac{\alpha}{\beta}\Rightarrow\Psi_{\infty}=\pm\sqrt{\dfrac{\alpha}{\beta}}$$
for positive $\alpha$ (so I choose $\alpha\propto\left|T-T_{c}\right|$ in fact). It corresponds to the space-independent field configuration for $x\rightarrow\infty$, such that $\nabla \Psi_{\infty} =0$. We nevertheless realise that there are two field configurations, $\pm\Psi_{\infty}$.
We next suppose a position dependent 1D problem, and real field for simplicity (it doesn't alter the main conclusion below, but it changes the exact solutions). We ask ourself: is there a space dependent solution tending to $\pm\Psi_{\infty}$ when $x\rightarrow \pm\infty$ ? Well, two possibilities: 


*

*$\Psi\left(x\rightarrow+\infty\right)=\Psi\left(x\rightarrow-\infty\right)=\Psi_{\infty}$ or $\Psi\left(x\rightarrow+\infty\right)=\Psi\left(x\rightarrow-\infty\right)=-\Psi_{\infty}$ in which case there is no space-dependent solution

*$\Psi\left(x\rightarrow+\infty\right)=-\Psi\left(x\rightarrow-\infty\right)=\Psi_{\infty} $ in which case there is a space-dependent solution (called an instanton, or a soliton) which connects the two non-trivial minima of the model
We now find the space-dependent solution. The equation of motion we should verify is 
$$\dfrac{\delta F}{\delta\Psi}=0\Rightarrow\xi^{2}\dfrac{d^{2}\Psi}{dx^{2}}+2\Psi\left(1-\dfrac{\alpha}{\beta}\Psi^{2}\right)=0$$
where we use that the field $\Psi$ is real (otherwise the factor 2 in front of the second term disappears, this is the simplifying hypothesis to find a simple instanton) and where $\xi=\hbar/\sqrt{2m\alpha}$ is the characteristic length of the problem (in the theory of superconductivity, it is called the coherence length). You see this is a non-linear equation, so solutions are complicated (when they are known). Here, you can nevertheless easily show that 
$$\Psi\left(x\right)=\pm\sqrt{\dfrac{\alpha}{\beta}}\tanh\dfrac{x-x_{0}}{\xi}=\pm\Psi_{\infty}\tanh\dfrac{x-x_{0}}{\xi}$$
is the solution we were looking at. You can choose $x_{0}=0$ for the initial condition. This solution smoothly connects the two non-trivial minima $\pm\Psi_{\infty}$ as required.
The local energy (the integrand of $F$) for this solution has a bump around $x_{0}$ of characteristic size $\xi$.
