 Translation-mode solvability for a bistable front

Suppose $u_t=D{u}_{xx}+f_0(u)+\epsilon h(u)$ has a balanced stationary front $U_0(z)$, where $z=(x-q)/\sqrt{D}$ and $U_0^{\prime \prime }+f_0(U_0)=0$. Take $\dot{q}/\sqrt{D}=\epsilon c_1+\cdots$ and $u=U_0+\epsilon U_1+\cdots$. At first order, $L{U}_1=-c_1{U}_0^{\prime }-h(U_0)$ with $L=d^2/d{z}^2+f_0^{\prime }(U_0)$. Differentiating the front equation proves $L{U}_0^{\prime }=0$. Integration by parts against the decaying translation mode givesThis is the Fredholm solvability condition determining the velocity, derived from the self-adjoint differential operator rather than assumed. An orthogonality condition or a fixed midpoint removes the freedom to add a multiple of $U_0^{\prime }$ to $U_1$. For the bistable cubic reaction-diffusion equation, $h=-a{U}_0(1-U_0)$ gives $c_1=-\sqrt{2}a$, and its first-order forcing vanishes pointwise, so the phase-fixed shape correction is zero.