Explicit traveling wave solution for the diffusion equation

Consider the differential equation

$$ v'' + \sigma v' + f(v) = 0$$ as a system

$$ \eqalign{v' &= p\cr p' &= -f(v) - \sigma p\cr}$$ and consider the phase portrait as it depends on $\sigma$. The equilibria at $(v=0,p=0)$ and $(v=1,p=0)$ are both saddles. For $\sigma = 0$ the system is invariant under reflection about $v=1/2$, and there are heteroclinic orbits joining the two equilibria. For $\sigma \ne 0$ it looks to me like there are no heteroclinic orbits joining these two. Instead, for $\sigma > 0$ the trajectories coming out of $(0,0)$ going up and right, and out of $(1,0)$ going down and left, are attracted to the other equilibrium $(1/2,0)$. That equilibrium is a spiral for $0 < \sigma < \sqrt{2}$ and an attracting node for $\sigma > \sqrt{2}$. So you'll get travelling-wave solutions for $\sigma > 0$ that go to $0$ or $1$ at $-\infty$ and $1/2$ at $+\infty$. For example, with $\sigma = 3/2$ there is a closed-form solution

$$ v(x) = \dfrac{1}{4} \tanh(x/4)+\dfrac{1}{4} $$

HINT: you do not want this integral to be zero, because that means $v(s)$ is constant almost everywhere.

Hence, $\sigma = 0$.