Assume we have a homogeneous solution $u_0$ of some pde that undergoes a finite-wavelength bifurcation at $\varepsilon=0$, say.
It is said this means that for the evolution of a Fourier mode $\exp(-i\Omega t+ ikx)$ the following holds:
For $\varepsilon <0$, the growth rate $Im(\Omega)<0$ for each Fouriermode, i.e. $u_0$ is stable.
For $\varepsilon >0$, we have that the growth rate of the Fouriermode, $Im(\Omega)$, is positive (for each Fouriermode), hence $u_0$ is unstable.
Two questions:
1) What is meant with that $u_0$ is homogeneous? Does this mean we can write it as an expression (integral or sum) of Fouriermodes $\exp(-i\Omega t+ kx)$, f.e. as an Fourier integral?
2) One naively question: Why is $Im(\Omega)$ the growth rate of $\exp(-i\Omega t+ikx)$?
Since you write "assume we have a homogeneous solution $u_0$ of some PDE" but then ask what this means I'm going to assume you're reading this somewhere. In this case I would venture to guess what what is really meant is that you have a homogeneous PDE, which essentially means you have a linear PDE with right hand side 0 (like homogeneous linear equations in linear algebra): $$ L u = 0. $$
Here if we write the differential operator as $$ L = \sum_{m=1}^M a_m \partial_t^m + \sum_{n=1}^N b_n \partial_x^n $$ then we can plug in pure waves of the form $u = \exp(-i \Omega t + i kx)$ to find that $L u =0$ if and only if $\Omega$ and $k$ obey the dispersion relation $$ \sum_{m=1}^M a_m (-i \Omega)^m + \sum_{n=1}^N b_n (ik)^n =0. $$
Now to address your second question. Suppose we have a pair $k,\Omega$ solving the dispersion relation, and $Im(\Omega) >0$, $k \in \mathbb{R}$. Then $\Omega = a + ib$ for $b>0$ and so our pure wave solution reads $$ u(x,t) = \exp(-i(a + ib)t + kx) = \exp(-i(at -kx) + bt) = e^{bt}\exp(-i(at -kx)), $$ which means that $$ \vert u(x,t)\vert = e^{bt}, $$ and so $b$ gives the growth rate in the sense that it tells how fast the solution grows pointwise in absolute value.