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?
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 $\xi$.
