I was considering this PDE $u_{t} = u_{xx} + u(1-u)(u-0.5)$; $0 \leq x \leq L$ with boundary condition $u_{x}(0,t) =0 ; u_{x}(L,t) =0$.
For the steady state we need $u_{t} = 0$ implying $u_{xx} + u(1-u)(u-0.5) = 0$.
I was trying to understand the phase portrait of the system: Let $v = u_{x}$, then $v_{x} + u(1-u)(u-0.5) = 0$ so we have two dimensional system :
$v_{x} = -u(1-u)(u-0.5) = f(u,v)$
$u_{x} = v = g(u,v)$
The fixed points of the system are $(u^*,v^*) = (0,0) ; (0.5,0) ; (1,0)$ in which $(1,0)$ is stable and the other two are unstable fixed points.
I am thinking of drawing the phase portrait, like i tried the online tool but seems that does not satisfy the boundary condition $v(0) = 0, v(L) = 0$.
I am not sure about the phase portrait of the system and to see the boundary conditions in the picture.