Self-adjoint differential operator

A differential operator is self-adjoint when integration by parts makes its inner products symmetric on its domain.

For real coefficient functions, the fourth-order scalar operatoris formally self-adjoint exactly when $q=p^{\prime }$. Indeed, $(p{D}^2{)}^{\ast }=p{D}^2+2p^{\prime }D+p^{\prime \prime }$ and $(qD{)}^{\ast }=-qD-q^{\prime }$. Under clamped boundary conditions, its quadratic form is

For $L=D^4+q(x)$ where $q$ is a real continuous function on $[a,b]$, integration by parts givesThe potential terms cancel, and differentiating the boundary expression verifies the identity. Under clamped boundary conditions $u=u^{\prime }=v=v^{\prime }=0$ at both endpoints, this form vanishes. Conversely, requiring its vanishing against every clamped test function forces the adjoint function and its first derivative to vanish at both endpoints, because the second and third test-function derivatives are independent endpoint traces. Thus, on a regular finite interval with the usual fourth-order Sobolev domain, the clamped realization is self-adjoint. The equality of domains is part of this conclusion; formal self-adjointness of the expression alone does not supply it.

A second-order scalar differential equation is in self-adjoint form when it can be writtenMultiplication by $y$ and integration by parts then gives an energy identity.

If $L(y_0;{\lambda }_0)=0$ and $y_0,y$ satisfy homogeneous separated boundary conditions, the Lagrange identity givesThus $L(y;{\lambda }_0)=f$ can be solved only if $f$ is orthogonal to the null mode $y_0$.

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.

For $L(y;\lambda )=y^{m+1}$ near a normalized simple null mode $y_0$, put $y=\epsilon y_0+O({\epsilon }^2)$ and $\lambda -{\lambda }_0={\epsilon }^m\mu$. Orthogonality to $y_0$ yields

Suppose a real simple leading eigenfunction is $y_0=C\varphi$ and a first-order correction satisfies $L_{0}v=-\nu w{y}_0-y_0^p$, where $p\ge 2$ is an integer, $L_0$ is self-adjoint with null space spanned by $\varphi$, the boundary conditions are homogeneous and independent of the parameter, and $w>0$. The Fredholm solvability condition givesIndeed, orthogonality of the forcing to $\varphi$ gives $\nu C\int w{\varphi }^2+C^p\int {\varphi }^{p+1}=0$, and $C\ne 0$ permits division. Thus the numerical eigenvalue correction depends on the chosen leading amplitude; fixing an eigenfunction normalization is part of defining the nonlinear branch. This calculation assumes a perturbative branch and does not by itself establish that branch's existence.