Explicit traveling wave solution for the diffusion equation Find explicit formulas for $v$ and $\sigma$ so that $u(x,t)=v(x-\sigma t)$ is a traveling wave solution of the nonlinear diffusion equation
$$u_t-u_{xx}=f(u)$$
where
$$f(z)=-2z^3+3z^2-z$$
and assume that $\lim_{s\to+\infty}v(s)=1$, $\lim_{s\to-\infty}v(s)=0$ and $\lim_{s\to\pm\infty}v'(s)=0$.
Here's what I got so far:
we have $u_t=-\sigma v'(x-\sigma t)$ and $u_{xx}=v''(x-\sigma t)$, so rewrite the equation we have
$$v''+\sigma v'+f(v)=0$$
This means
$$v''v'+\sigma(v')^2+v'f(v)=0$$
Hence
$$\sigma(v')^2=-v''v'-v'f(v)$$
Integrating both sides we have
$$\sigma\int_{-\infty}^{\infty}(v'(s))^2ds=-\int_{-\infty}^{\infty}v''(s)v'(s)ds-\int_{-\infty}^{\infty}v'(s)f(v(s))ds.$$
By using the assumptions $\lim_{s\to+\infty}v(s)=1$, $\lim_{s\to-\infty}v(s)=0$ and use the change of variable $y=v(s)$ we have
$$\int_{-\infty}^{\infty}v'(s)f(v(s))ds=\int_0^1f(y)dy=0$$
From $\lim_{s\to\pm\infty}v'(s)=0$ we have
$$\int_{-\infty}^{\infty}v''(s)v'(s)ds=0$$
Thus
$$\sigma\int_{-\infty}^{\infty}(v'(s))^2ds=0$$
At this point I cannot further proceed. Any further hints are welcomed. Thanks.
 A: 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
$$ v(x) = \dfrac{1}{4} \tanh(x/4)+\dfrac{1}{4} $$
A: HINT: you do not want this integral to be zero, because that means $v(s)$ is constant almost everywhere.
Hence,  $\sigma = 0$.
