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At $r=s=0$, the linear growth rate of a Fourier mode $e^{ikx}$ is $-(1-k^2{)}^2$. Its critical kernel of a linear map has $k=\pm 1$. WriteFor a compatible periodic carrier, the slowly varying envelope is periodic on $0\le X\le {\epsilon }^{2}L$. Small carrier detuning can alternatively be included in the envelope wave phase.

With $w=O(\epsilon )$, $r={\epsilon }^4\mu$ and $s={\epsilon }^2\hat{s}$, all three terms $rw$, $s{w}^3$, $w^5$ first appear at $O({\epsilon }^5)$. Balancing the time derivative thus requires $T={\epsilon }^{4}t$. Since the linear growth rate near $k=1$ is $-4(k-1{)}^2+O((k-1{)}^3)$, spatial detuning must be $O({\epsilon }^2)$ to contribute at the same order. ConsequentlyTreating $x,X$ independently, let $D={\partial }_x$ on the fast carrier. Expansion of the linear operator givesThe $O({\epsilon }^3)$ term annihilates $w_1$, since $(1+D^2)w_1=0$. At orders two through four there is no nonresonant forcing, and kernel of a linear map corrections can be absorbed into the definition of $A$, allowing $w_2=w_3=w_4=0$. At order five the equation for $w_5$ isThe fast operator is a self-adjoint operator on periodic functions. Its Fredholm alternative requires the right-hand side to have zero inner product with each kernel of a linear map mode $e^{\pm ix}$, equivalently to have zero resonant Fourier coefficients. The coefficient of $e^{ix}$ in $w_1^3$ is $3A^2\overline{A}$, and in $w_1^5$ it is $\left(\genfrac{}{}{0ex}{}{5}{3}\right)A^3{\overline{A}}^2=10A|A{|}^4$. Therefore the solvability condition isThe conjugate condition supplies the same real equation for the negative mode. The remaining noncritical harmonics determine $w_5$ without obstruction. This is the cubic-quintic Swift–Hohenberg amplitude reduction; it retains quintic saturation because the small positive cubic coefficient makes the onset mildly subcritical.

The Sturm-Liouville operator satisfies the Lagrange identity for a Sturm-Liouville operatorIndeed the potential terms cancel, and differentiating the right side gives the remaining two derivative terms. Integrate over $[0,1]$. The common Dirichlet boundary conditions make the endpoint term zero, while $L(y_0;{\lambda }_0)=0$. ThereforeThis is the necessary solvability condition imposed by the nonzero homogeneous solution; its proof here does not rely on merely quoting the Fredholm alternative.

For the nonlinear problem, chooseThe orthogonality condition then givesWith $y=ϵy_0+{ϵ}^2{y}_1$, where the correction remains bounded as $ϵ\to 0$, the normalization givesDivide by ${ϵ}^3$ and use $\mu =O(1)$ to obtainThe positive quartic integral determines the leading direction of the nonlinear shift from the linear eigenvalue.

The Fredholm index measures the finite-dimensional obstruction to inverting an operator. For a bounded operator $T:H_1\to H_2$ between Hilbert spaces, the Fredholm conditions are that its range of a bounded linear operator is closed and its kernel and cokernel are finite-dimensional. The orthogonal complement of the range is $\ker T^{\ast }$, soAn invertible linear map has Fredholm index zero; the converse needs an additional finite-dimensional correction, since an operator can have equally large kernel and cokernel.

First construct an inverse up to finite-rank operators. On $(\ker T{)}^{\perp }$, the restriction of $T$ is a bounded bijection onto its closed range, with bounded inverse by the bounded inverse theorem. Extend that inverse by zero on $(\mathrm{ran}T{)}^{\perp }$, obtaining $S:H_2\to H_1$. ThenConversely, suppose $ST-I$ and $TS-I$ are compact operators. On $\ker T$, the first identity expresses the identity as a compact operator, so this kernel is finite-dimensional. If $T$ were not bounded below on $(\ker T{)}^{\perp }$, there would be unit vectors $x_j$ in that complement with $T{x}_j\to 0$. Write $ST=I+K$. Compactness of $K$ gives a subsequence for which $K{x}_j$ converges, hence $x_j=ST{x}_j-K{x}_j$ converges to a unit vector $x$ in $(\ker T{)}^{\perp }$ with $Tx=0$, a contradiction. This lower bound makes the range closed. Applying the analogous argument to $T^{\ast }$, using $S^{\ast }{T}^{\ast }=I+K^{\prime \ast }$, proves that its kernel is finite-dimensional. This proves the Atkinson theorem: a bounded operator is Fredholm exactly when it has an inverse modulo compact operators. For operators on one Hilbert space this says that its image in the Calkin algebra is invertible.

The Fredholm index is stable. Decompose the domain as $N\oplus N^{\perp }$, where $N=\ker T$, and the target as $C\oplus R$, where $C=\ker T^{\ast }$ and $R=\mathrm{ran}T$. A sufficiently small norm perturbation has block matrixMultiplying by invertible triangular block matrices reduces this matrix to the direct sum of $D$ and the Schur complement $a-b{D}^{-1}c:N\to C$. The latter's kernel dimension minus cokernel dimension is $\dim N-\dim C$, whatever its rank. Therefore the Fredholm set is open and the index is locally constant. If $K$ is compact, the same parametrix works modulo compact operators for every $T+tK$, $0\le t\le 1$. Local constancy along this path gives

For Fredholm operators $A:H_1\to H_2$ and $B:H_2\to H_3$, multiplication of parametrices shows that $BA$ is Fredholm. Its index is additive. Here is a direct proof, rather than an assertion of additivity. There is an exact sequence of finite-dimensional spacesThe map from $\ker B$ takes a vector to its class modulo $\mathrm{ran}A$; the next takes the class of $y$ to that of $By$; the last is the quotient map. For example, a class $y+\mathrm{ran}A$ lies in the kernel of the middle map exactly when $By=BAx$ for some $x$, so $y-Ax\in \ker B$. This verifies the only less immediate exactness assertion; the others follow directly from their definitions. Alternating the dimensions givesAlso $\mathrm{ind}{T}^{\ast }=-\mathrm{ind}T$, and indices add under direct sums. For $I+K$ with $K$ compact, the index is zero, so injectivity is equivalent to surjectivity. More generally $Tx=f$ is solvable exactly when $f$ is orthogonal to $\ker T^{\ast }$: these are the finitely many compatibility conditions in the Fredholm alternative.

Toeplitz operators exhibit how this analytic index detects a topological winding. Let $H^2\subset L^2(S^1)$ be the Hardy space of the circle, spanned by $1,z,z^2,\dots$, and let $P$ be its orthogonal projection. For a continuous symbol $g$, define $T_g=P{M}_g{|}_{H^2}$, where $M_g$ is multiplication by $g$. Then $\Vert T_g\Vert \le \Vert g{\Vert }_{\infty }$. The product defect isFor trigonometric polynomials the crossing of negative and nonnegative Fourier modes has finite rank. Uniform approximation of continuous symbols by trigonometric polynomials, together with the displayed norm estimate, therefore makes this defect a compact operator for all continuous $f,g$. If $g$ is nowhere zero, $1/g$ is continuous and $T_{1/g}$ is a two-sided parametrix modulo compact operators. Thus $T_g$ is Fredholm.

The converse is instructive. If $g(\zeta )=0$ at $\zeta \in S^1$, consider the normalized Hardy kernelsTheir norms are one, their individual Fourier coefficients tend to zero as $r\uparrow 1$, and hence they converge weakly to zero. Their squared moduli are the Poisson kernel, so continuity of $g$ givesA Fredholm operator is bounded below off its finite-dimensional kernel. The projection of these weakly vanishing vectors onto that kernel tends to zero, contradicting this lower bound. Thus a continuous-symbol Toeplitz operator is Fredholm exactly when its symbol never vanishes. Applying the same argument to $g-\lambda$ gives its essential spectrum as $g(S^1)$, with essential spectrum defined by failure of the Fredholm property.

For nonvanishing $g$, let $m=\mathrm{wind}(g,0)$. Lifting a continuous argument of $g(e^{it})$ on $0\le t\le 2\pi$ gives an argument change $2\pi m$. After division by $z^m$, its argument has matching endpoints, so there is a continuous function $h:S^1\to \mathbb{C}$ with $g(z)=z^m{e}^{h(z)}$. The nonvanishing-symbol homotopy $g_t=z^m{e}^{th}$ reduces the index to that of $T_{z^m}$. For $m\ge 0$ this is the $m$th power of the unilateral shift operator, whose kernel is zero and whose range has codimension $m$. For $m<0$ it is the $(-m)$th power of the adjoint shift, whose kernel has dimension $-m$ and whose range is all of $H^2$. ConsequentlyThe minus sign comes from counting missing nonnegative Fourier modes. This is a concrete bridge between the Fredholm index, stability under compact perturbations, and the winding number of a loop.

On the torus ${\mathbb{T}}^n=(\mathbb{R}/2\pi \mathbb{Z}{)}^n$, choose normalized Haar measure, so that $e_k(x)=e^{ik\cdot x}$, $k\in {\mathbb{Z}}^n$, is an orthonormal basis of $L^2$. A periodic Sobolev space is defined, for any real $s$, byFor negative $s$, the coefficients describe distributions rather than necessarily functions. For a nonnegative integer $s$, the weight is comparable to ${\sum }_{|\alpha |\le s}|k^{\alpha }{|}^2$, so this norm is equivalent to the sum of the squared $L^2$ norms of all weak derivatives through order $s$. Fourier truncation proves that smooth trigonometric polynomials are dense. The $L^2$ pairing identifies the continuous dual of $H^s$ with $H^{-s}$, giving Sobolev duality.

Two Fourier estimates explain the usefulness of these spaces. If $s>m+n/2$, the Cauchy-Schwarz inequality givesUniform convergence of the differentiated Fourier series proves the Sobolev embedding $H^s\hookrightarrow C^m$. If $\delta >0$, a bounded subset of $H^{s+\delta }$ has uniformly small high-frequency tails in $H^s$:Finite-dimensional Fourier truncation therefore proves the compact embedding $H^{s+\delta }\hookrightarrow H^s$, the periodic version of Rellich-Kondrachov compactness theorem. In addition, for $s>n/2$ multiplication is continuous $H^s\times H^s\to H^s$, making it a Sobolev algebra. Indeed the weighted coefficient of a product is bounded by the sum of the convolutions of $⟨k{⟩}^s|\hat{u}(k)|$ with $|\hat{v}(k)|$ and of $|\hat{u}(k)|$ with $⟨k{⟩}^s|\hat{v}(k)|$; Cauchy-Schwarz puts unweighted coefficients in ${\ell }^1$, and the ${\ell }^1\ast {\ell }^2\to {\ell }^2$ case of Young's convolution inequality finishes the estimate.

For eigenfunction expansions, distinguish ellipticity from self-adjointness. An arbitrary non-self-adjoint elliptic operator need not have an orthonormal basis of eigenfunctions. A useful definite setting iswith smooth real periodic coefficients, $a_{ij}=a_{ji}$, real $V$, and uniform ellipticityTake the self-adjoint realization in $L^2$ with domain $H^2$. Its construction and spectral properties can be seen from the associated form on $H^1$,For a sufficiently large constant $C$, $b_C=b+C(\cdot ,\cdot )$ is a positive coercive form: $b_C(u,u)\ge c_1\Vert u{\Vert }_{H^1}^2$, and $L+C\ge I$. The Lax-Milgram theorem supplies the unique weak solution of $(L+C)u=f$ for $f\in H^{-1}$; the form is Hermitian, which gives symmetry of its inverse on $L^2$.

The regularity input is the periodic elliptic estimateHere is why an elliptic operator gains two derivatives. With coefficients frozen at a point, its principal symbol $\sum a_{ij}{\xi }_i{\xi }_j$ dominates $c|\xi {|}^2$, and the estimate follows by multiplying Fourier coefficients by that symbol. On a sufficiently small coordinate patch, the difference of the leading coefficients from these constants has arbitrarily small supremum norm. In the integer-order differentiated estimate, this absorbs the highest derivatives into the left side; derivatives falling on coefficients contribute only lower-order terms. A partition of unity reduces the variable-coefficient problem to such patches. Its commutators with $L$ have order one, controlled bywhich follows by splitting low and high Fourier modes. Absorbing these terms gives the displayed estimate. For a weak $H^1$ solution, the same localization with difference quotients first bounds its second weak derivatives for $f\in L^2$; differentiating and repeating then gives $f\in H^s\Rightarrow u\in H^{s+2}$ for nonnegative integer $s$. Interpolation and duality extend the scale to real orders. In particular $(L+C{)}^{-1}:L^2\to H^2$ is bounded.

Let $R=(L+C{)}^{-1}$. Compactness of $H^2\hookrightarrow L^2$ makes $R$ a compact operator. Symmetry of the form makes $R$ self-adjoint, and positivity of $L+C$ makes $R$ positive and injective. Its range is dense because $\ker R^{\ast }=\ker R=0$. The spectral theorem for compact self-adjoint operators consequently gives an orthonormal basis ${\varphi }_j$ withElliptic regularity bootstraps every ${\varphi }_j$ to a smooth function. Each eigenvalue has finite multiplicity. Thus a self-adjoint uniformly elliptic second-order operator on the torus has a complete smooth orthonormal eigenfunction expansion with discrete eigenvalues tending to infinity.

Writing $u_j=(u,{\varphi }_j)$, every $u\in L^2$ has $u={\sum }_j{u}_j{\varphi }_j$ in $L^2$, with $\Vert u{\Vert }_2^2={\sum }_j|u_j{|}^2$. Iterating the elliptic estimate shows that the graph norm of $(L+C{)}^m$ is equivalent to the $H^{2m}$ norm. Interpolation and duality give the more general spectral descriptionThe comparison constants depend on the operator and $s$. In particular smooth functions have coefficients decaying faster than every power of ${\lambda }_j+C$, and their expansions converge in every Sobolev norm and hence in every $C^m$ norm.

This gives a practical way to solve elliptic equations. The equation $Lu=f$ is solvable if and only if $f_j=0$ whenever ${\lambda }_j=0$; then one may take $u_j=f_j/{\lambda }_j$ on the other eigenspaces and add an arbitrary vector in $\ker L$. The Sobolev description proves the two-derivative gain and the Fredholm alternative, rather than leaving them as formal series manipulations. Similarly the solution of the heat equation ${\partial }_{t}u+Lu=0$ isFor every $t>0$ the exponential decay at high eigenvalues places this solution in every Sobolev space, hence it is smooth.

For the Laplacian $-\Delta$ the eigenfunctions are precisely $e^{ik\cdot x}$ with eigenvalues $|k{|}^2$. The number with $|k{|}^2\le R$ is the number of lattice points in a radius-$\sqrt{R}$ ball, asymptotic to its volume by comparison with unit cubes. For the general self-adjoint operator above, bounds on its form above and below by positive multiples of $\Vert \nabla u{\Vert }_2^2$ plus constants, together with the Courant–Fischer min-max principle, imply ${\lambda }_j+C\asymp j^{2/n}$. Thus the growth reflects dimension and the two-derivative order. Compactness of the torus is essential: on ${\mathbb{R}}^n$, translated bumps prevent the compact embedding, and the Laplacian instead has continuous spectrum described by the Fourier transform.

The Lax-Milgram theorem states that a bounded bilinear form $a$ on a real Hilbert space $V$, satisfying $a(v,v)\ge \alpha \Vert v{\Vert }_V^2$ for some $\alpha >0$, has for each $F\in V^{\ast }$ a unique $u\in V$ with $a(u,v)=F(v)$ for all $v$. Moreover $\Vert u{\Vert }_V\le \Vert F\Vert /\alpha$.

For a compact operator $K$ on a Hilbert space, the Fredholm alternative says that $I-\lambda K$ has closed range, finite-dimensional kernel and cokernel of equal dimension, andThus a zero kernel gives unique solvability for every right-hand side; a nonzero kernel gives solvability exactly under the adjoint orthogonality conditions, and then all solutions differ by a kernel vector.

Take $V=H_0^1(\Omega )$. Gårding's inequality supplies constants $a>0$, $b\ge 0$ with $B[v,v]\ge a\Vert v{\Vert }_{H^1}^2-b\Vert v{\Vert }_2^2$. Therefore for $\lambda \le -b$ the shifted form $B_{\lambda }[u,v]=B[u,v]-\lambda (u,v{)}_{L^2}$ is coercive. It is bounded for every fixed $\lambda$, and $v\mapsto (f,v)$ is bounded on $V$. Lax-Milgram gives unique weak solvability for $\lambda \le -\gamma$, with $\gamma =b$. The inequality itself follows, for example, by absorbing the drift cross-term $\Vert b{\Vert }_{\infty }\Vert \nabla v{\Vert }_2\Vert v{\Vert }_2$ using Young's inequality and allowing a lower $L^2$ correction for $c$.

For the compact inverse and adjoint of a coercive elliptic operator, under the additional zero-shift coercivity assumption, define $Gg\in V$ by $B[Gg,v]=(g,v)$ for all $v\in V$, and let $K$ be this solution viewed in $L^2$. The solution operator is linear and bounded into $V$, and the Rellich-Kondrachov compact embedding $V\hookrightarrow L^2$ makes $K$ compact. Rewriting the weak equation as $B[u,v]=(f+\lambda u,v)$ givesConversely an $L^2$ solution of this operator equation lies in $V$, because it equals $K(f+\lambda u)$, and satisfies the weak differential equation.

For constant drift coefficients and zero Dirichlet traces, integration by parts gives $\int b_i{D}_{i}vu=-\int b_{i}v{D}_{i}u$. Symmetry of the principal coefficients therefore makes $B^{\ast }[u,v]=B[v,u]$ the unshifted weak form of the displayed formal adjoint. The shifted adjoint equation requiresThe spectral shift in an adjoint weak formulation is essential: the printed sentence omitting it is false for nonzero $\lambda$. For instance take a smooth ball, $L=-\Delta +1$, any nonzero $u\in C_c^{\infty }(\Omega )$ and $f=-\Delta u+u$. The printed unshifted identity holds, but $(L^{\ast }-\lambda )u=f-\lambda u\ne f$ whenever $\lambda \ne 0$. This counterexample also meets zero-shift coercivity and constant-drift assumptions. The corrected identity above is used for the remainder of the argument.

Let $G_{\ast }g\in V$ solve $B^{\ast }[G_{\ast }g,v]=(g,v)$, which exists by Lax-Milgram because the diagonal coercivity is unchanged. For $p,q\in L^2$,Thus $G_{\ast }$, viewed as an $L^2$ operator, is exactly $K^{\ast }$. Applying this inverse to the corrected shifted adjoint equation givesand the converse follows from its range lying in $V$, just as before.

Finally let the original homogeneous equation have a nonzero solution. Its parameter is nonzero, since the unshifted problem is coercive. Fredholm gives solvability precisely when $(Kf,w)=0$ for every $w\in \ker (I-\lambda K^{\ast })$. Such a $w$ satisfies $K^{\ast }w=w/\lambda$, so $(Kf,w)=(f,w)/\lambda$. Hence the necessary and sufficient condition isWhen this holds, the solutions are an affine translate of the original finite-dimensional homogeneous solution space. The compatibility space is the adjoint kernel; a drift term does not generally permit replacing it by the original kernel.

The Lax-Milgram theorem gives, for a bounded coercive bilinear form on a real Hilbert space and every bounded linear functional, a unique vector representing that functional through the form, with the inverse bound supplied by coercivity. The Fredholm alternative for a compact operator $K$ says that $I+\mu K$ is invertible if its kernel is zero; in general its range is closed and consists exactly of vectors orthogonal to $\ker (I+\mu K{)}^{\ast }$ in the Hilbert-space case.

Uniform ellipticity gives $\sum a_{ij}{\xi }_i{\xi }_j\ge \theta |\xi {|}^2$ for some $\theta >0$. Bounded coefficients make $B_{\lambda }$ bounded on $H^1$, and for $\lambda <-\gamma$,The forcing $v\mapsto \int fv$ is bounded on $H^1$. The Lax-Milgram theorem therefore gives a unique weak solution for every $\lambda <-\gamma$.

When $c=0$, choose the shift ${\lambda }_0=-1$ and let $Kf$ be the unique solution of $B[Kf,v]+\int (Kf)v=\int fv$. Thus $K=(L+I{)}^{-1}$ in the weak sense. The equation $Lu=f$ is equivalent to $u=K(f+u)$, orConversely a solution in $L^2$ of this operator equation lies in $H^1$ because it equals $K(f+u)$, and satisfies the variational equation. The assumed boundedness $K:L^2\to H^1$, followed by the compact Rellich embedding $H^1\hookrightarrow L^2$, proves compactness of $K$ as an $L^2$ operator.

By the supplied kernel assumption, $\ker (I-K)$ consists of constants. Also $K1=1$. Since $K$ is a self-adjoint operator, the Fredholm alternative condition is $Kf\perp 1$, andTherefore a weak solution exists exactly when the mean of $f$ is zero, and solutions differ by a constant. On a disconnected domain the actual kernel would consist of constants on each component and the condition would apply separately on each; the question's given kernel assumption specifies the single-constant case.

For a general coefficient matrix, the stated variational definition encodes zero conormal flux $(aDu)\cdot n=0$. It agrees with ordinary zero normal derivative only under additional conditions on $a$. The argument above uses the weak formulation explicitly defined in the paper, so does not require that stronger boundary identification.

Suppose a Bloch stability operator has expansion $L(q)=L_0+iq{L}_1-q^2{L}_2+\cdots$, with a simple neutral eigenfunction $v_0$ and an adjoint neutral function $w$ normalized by $⟨w,v_0⟩=1$. Expand $v(q)=v_0+iq{v}_1+\cdots$ and $\lambda (q)=icq-D{q}^2+\cdots$. The Fredholm alternative gives $c=⟨w,L_1{v}_0⟩$ and the cell problem $L_0{v}_1=c{v}_0-L_1{v}_0$, $⟨w,v_1⟩=0$. Applying the adjoint solvability condition at second order gives the displayed phase-diffusion coefficient. Multiple translation phases or a conserved mean require a coupled matrix problem rather than this scalar formula.