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For spatially homogeneous solutions, the bistable cubic reaction-diffusion equation reduces to $\dot{u}=f(u)=u(1-u)(u-r)$. Its equilibria are $0,r,1$, andThus $0$ and $1$ are asymptotically stable, while $r$ is unstable. Small spatial Fourier perturbations have growth rates $f^{\prime }(u_{\ast })-m{k}^2$, so diffusion also leaves the two outer uniform states linearly stable.

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 $D{U}^{\prime \prime }+f(U;1/2)=0$ by $U^{\prime }$ and integrate once. Since $U^{\prime }=0$ at both asymptotic phases and ${\int }_0^{U}f(s;1/2)ds=-U^2(1-U{)}^2/4$, the integration constant is zero:The boundary orientation selects $U^{\prime }=-U(1-U)/\sqrt{2D}$. Separating variables gives the logistic front of a bistable cubic equation, with arbitrary center $q_0$:The arbitrary translation is expected from translation invariance.

To develop the requested perturbation theory, let $r=1/2+\epsilon a$ with signed $a=\pm 1$, introduce slow time $T=\epsilon t$, and put $z=(x-q(t))/\sqrt{D}$. Seekwhere $U_0(z)=(1+e^{z/\sqrt{2}}{)}^{-1}$. The explicit $T$ derivative of $\epsilon U_1$ is second order. Since ${\partial }_{r}f=-u(1-u)$, the first-order equation isDifferentiation of $U_0^{\prime \prime }+f(U_0;1/2)=0$ gives $L{U}_0^{\prime }=0$. The self-adjoint differential operator $L$ has this decaying translation mode, and $U_1\to 0$ at both ends because the phases remain exactly $1$ and $0$. Multiplying the first-order equation by $U_0^{\prime }$ and integrating by parts makes its left side zero. Therefore the translation-mode solvability for a bistable front requiresBoth integrals can be evaluated without a special-function formula. The decreasing profile takes $U$ from $1$ to $0$ and has $U_0^{\prime }=-U(1-U)/\sqrt{2}$, soandThe resulting speed isThis derives the velocity by the Fredholm solvability condition, rather than substituting a guessed speed.

With this value, the forcing actually cancels pointwise: $-c_1{U}_0^{\prime }+a{U}_0(1-U_0)=0$. Thus $L{U}_1=0$. The decaying kernel is only the translation direction: an independent solution is $U_0^{\prime }{\int }^z(U_0^{\prime }{)}^{-2}d\zeta$, 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 $U_0^{\prime \prime }+f(U_0;r)=-(r-1/2)U_0(1-U_0)$, and the exact moving-frame equation has left side $-(\dot{q}/\sqrt{D})U_0^{\prime }$. Thereforeis 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 $\dot{q}/\sqrt{D}=O(\epsilon )$ scaling.

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.