= Solution
At $r=1/2$, write $F(u)=-u^2(1-u)^2/4$, so $F'=f$. Multiplying the stationary equation $mU''+F'(U)=0$ by $U'$ gives $m(U')^2/2+F(U)=C$. The endpoint values and vanishing endpoint derivatives make $C=0$. For the decreasing connection,
$$
U'=-\frac{U(1-U)}{\sqrt{2m}}.
$$
Separate variables and absorb the integration constant into a translation $x_0$. The <logistic front of a bistable cubic equation> is
$$
\boxed{U(x)=\frac{1}{1+e^{(x-x_0)/\sqrt{2m}}}.}
$$
It approaches $1$ on the left and $0$ on the right and satisfies the stationary differential equation directly.
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