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.