Prove that $-cu'=u''+ku(1-u)$ has a unique solution given these hypotheses. I do not know how ti prove the following problem. Any help will be appreciated.
Prove that (if $c>2\sqrt{k}$ )  the equation
$$-cu'=u''+ku(1-u)$$ has a unique solution $u:\mathbb{R}\rightarrow [0,1]$ satisfying $u(-\infty)=0$ and $u(\infty)=1.$
 A: It is not hard to show $u\in[0,1]$. Let $u_i$ ($i=1,2$) satisfy 
$$-cu_i'=u_i''+ku_i(1-u_i), u_i(-\infty)=0,u_i(\infty)=1.$$
Then
$$ -(u_1''-u_2'')=k[u_1(1-u_1)-u_2(1-u_2)]+c(u_1'-u_2') $$
Let $U=u_1-u_2$. Multiplying both side by $U$ and integrating by parts give
\begin{eqnarray}
\int_{-\infty}^\infty|U'|^2dx=k\int_{-\infty}^\infty |U|^2dx-k\int_{-\infty}^\infty|U|^2(u_1+u_2)dx+c\int_{-\infty}^\infty U'Udx.
\end{eqnarray}
Using $ab\le \frac{\epsilon}{2}a^2+\frac{1}{2\epsilon}b^2$, one has
\begin{eqnarray}
\int_{-\infty}^\infty|U'|^2dx&=&k\int_{-\infty}^\infty |U|^2dx-k\int_{-\infty}^\infty|U|^2(u_1+u_2)dx+c\int_{-\infty}^\infty U'Udx\\
&\le&k\int_{-\infty}^\infty |U|^2dx-k\int_{-\infty}^\infty|U|^2(u_1+u_2)dx+\int_{-\infty}^\infty|U'|^2dx+\frac{c^2}{4}\int_{-\infty}^\infty|U|^2dx
\end{eqnarray}
which implies
$$ k\int_{-\infty}^\infty |U|^2dx+\frac{c^2}{4}\int_{-\infty}^\infty|U|^2dx \le k\int_{-\infty}^\infty|U|^2(u_1+u_2)dx\le2k\int_{-\infty}^\infty|U|^2dx.$$
Thus
$$\frac{c^2}{4}\int_{-\infty}^\infty|U|^2dx\le k\int_{-\infty}^\infty|U|^2dx.$$
or
$$（\frac{c^2}{4}-k)\int_{-\infty}^\infty|U|^2dx\le 0.$$
Therefore if $c>2\sqrt k$, one has
$$ \int_{-\infty}^\infty|U|^2dx\le 0 $$
which implies $U=0$ or $u_1=u_2$.
