Fredholm solvability condition

At a simple eigenvalue of a self-adjoint linear operator, an inhomogeneous equation is solvable only when its forcing is orthogonal to the null eigenfunction. In weakly nonlinear expansions, this condition determines the amplitude equation.

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.

At the Maxwell balance for a scalar reaction-diffusion front, let $q=3\alpha /4$, $\beta =\sqrt{3}\alpha /2$, and perturb the linear growth rate by $\epsilon K\delta (x+cT)$, $T=\epsilon t$. Projection onto the front's translation mode gives $X_T=-K{\mathrm{sech}}^2(\beta Z/2)$, where $Z=X+cT+{\beta }^{-1}\log q$. Hence $Z_T=c-K{\mathrm{sech}}^2(\beta Z/2)$. Finite locked positions exist for $0<c/K\le 1$. For a strict inequality, the root with $KZ<0$ is stable and the other unstable; they meet in a saddle-node bifurcation at $c=K$. This is a leading-order solvability condition, not an exact finite-perturbation solution.