= Solution
Write the reaction term as $f(u;r)=u(1-u)(u-r)$ and assume $D>0$. At $r=1/2$, multiply the stationary <travelling-wave reduction of a reaction-diffusion system> $DU''+f(U;1/2)=0$ by $U'$ and integrate once. Since $U'=0$ at both asymptotic phases and $\int_0^U f(s;1/2)\,ds=-U^2(1-U)^2/4$, the integration constant is zero:
$$
\frac D2(U')^2-\frac14U^2(1-U)^2=0.
$$
The boundary orientation selects $U'=-U(1-U)/\sqrt{2D}$. Separating variables gives the <logistic front of a bistable cubic equation>, with arbitrary center $q_0$:
$$
\boxed{U(x)=\frac1{1+\exp[(x-q_0)/\sqrt{2D}]}=\frac12\left[1-\tanh\frac{x-q_0}{2\sqrt{2D}}\right].}
$$
The arbitrary <translation> is expected from <translation invariance>.
To develop the requested <perturbation theory>, let $r=1/2+\varepsilon a$ with signed $a=\pm1$, introduce slow time $T=\varepsilon t$, and put $z=(x-q(t))/\sqrt D$. Seek
$$
u=U_0(z)+\varepsilon U_1(z,T)+O(\varepsilon^2),\qquad \frac{\dot q}{\sqrt D}=\varepsilon c_1+O(\varepsilon^2),
$$
where $U_0(z)=(1+e^{z/\sqrt2})^{-1}$. The explicit $T$ derivative of $\varepsilon U_1$ is second order. Since $\partial_rf=-u(1-u)$, the first-order equation is
$$
LU_1=-c_1U_0'+aU_0(1-U_0),\qquad L=\frac{d^2}{dz^2}+f_u(U_0;1/2).
$$
Differentiation of $U_0''+f(U_0;1/2)=0$ gives $LU_0'=0$. The <self-adjoint differential operator> $L$ has this decaying <translation> mode, and $U_1\to0$ at both ends because the phases remain exactly $1$ and $0$. Multiplying the first-order equation by $U_0'$ and integrating by parts makes its left side zero. Therefore the <translation-mode solvability for a bistable front> requires
$$
0=-c_1\int_{-\infty}^{\infty}(U_0')^2\,dz+a\int_{-\infty}^{\infty}U_0'U_0(1-U_0)\,dz.
$$
Both integrals can be evaluated without a special-function formula. The decreasing profile takes $U$ from $1$ to $0$ and has $U_0'=-U(1-U)/\sqrt2$, so
$$
\int U_0'U_0(1-U_0)\,dz=\int_1^0U(1-U)\,dU=-\frac16,
$$
and
$$
\int(U_0')^2\,dz=\int_1^0\left[-\frac{U(1-U)}{\sqrt2}\right]dU=\frac1{6\sqrt2}.
$$
The resulting speed is
$$
\boxed{c_1=-\sqrt2a,\qquad \dot q=-\sqrt{2D}(r-1/2)+O(\varepsilon^2)=\sqrt{D/2}(1-2r)+O(\varepsilon^2).}
$$
This derives the velocity by the <Fredholm solvability condition>, rather than substituting a guessed speed.
With this value, the forcing actually cancels pointwise: $-c_1U_0'+aU_0(1-U_0)=0$. Thus $LU_1=0$. The decaying <kernel> is only the <translation> direction: an independent solution is $U_0'\int^z(U_0')^{-2}\,d\zeta$, which grows at at least one infinity. Fixing the center by $U_1(0)=0$ consequently gives $\boxed{U_1=0}$. In this particular <bistable cubic reaction-diffusion equation> the absence of shape deformation holds to all orders. Indeed $U_0''+f(U_0;r)=-(r-1/2)U_0(1-U_0)$, and the exact moving-frame equation has left side $-(\dot q/\sqrt D)U_0'$. Therefore
$$
\boxed{u(x,t)=\frac1{1+\exp[(x-q_0-ct)/\sqrt{2D}]},\qquad c=\sqrt{D/2}(1-2r)}
$$
is the exact front for every $0<r<1$. For $r<1/2$, $c>0$ and the $u=1$ phase advances to the right; for $r>1/2$ the $u=0$ phase advances to the left. The small-bias speed is consistent with the prescribed $\dot q/\sqrt D=O(\varepsilon)$ scaling.
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