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Let $A$ be a closed operator and let $V:D(A)\to H$ be compact when the domain carries its graph norm. If $A+V$ is closed on the same domain, compact perturbation invariance of Fredholm operators applied to the maps $D(A)\to H$ proves invariance of the Fredholm essential spectrum. For self-adjoint operators this is also the essential spectrum obtained by deleting isolated finite-multiplicity eigenvalues. A bounded potential tending to zero at spatial infinity is relatively compact with respect to a second-order constant-coefficient operator: local Rellich-Kondrachov compactness theorem gives compactness on bounded intervals and the small tails give compactness globally.

Let $L=a^{ij}(x)D_{ij}$ be uniformly elliptic with continuous coefficients. The coefficient-freezing interior second-derivative estimate extends from smooth functions to the graph norm closure of $L$ on $H^1(U)$: if $u_k\to u$ in $H^1(U)$ and $L{u}_k\to f$ in $L^2(U)$, then $(u_k)$ is Cauchy in $H^2(W)$ for every $W⋐U$. Hence $u\in H_{\mathrm{loc}}^2(U)$ and $Lu=f$. Equivalently, the standard local regularization argument gives the same conclusion for weak solutions.

We use $\Delta =\mathrm{div}\nabla$, whose spectrum on a compact manifold is nonpositive; write $P=-\Delta$ for the positive Laplace-Beltrami operator. This convention will also make the heat operator in Question 3 equal to ${\partial }_t-\Delta$.

Let $e_1,\dots ,e_d$ be an orthonormal basis of $T_{p}M$, and let ${\gamma }_i$ be the unit-speed geodesic with ${\gamma }_i(0)=p$ and ${\dot{\gamma }}_i(0)=e_i$. Since a geodesic has vanishing acceleration for the Levi-Civita connection, differentiation twice along it gives the Riemannian Hessian:Taking its metric trace yields the geodesic trace formula for the Laplace-Beltrami operatorThe positive-sign convention $P$ puts a minus sign in front of this sum.

For the unit sphere, take $p\in S^d$ and tangent unit vectors $e_i$ orthogonal to $p$. Its great circle geodesics are ${\gamma }_i(t)=p\cos t+e_i\sin t$. For a smooth ambient function $F$ near the sphere, the ordinary chain rule givesThe ambient orthonormal basis is $e_1,\dots ,e_d,p$. Thus, writing $F_r(p)={\partial }_{r}F(rp){|}_{r=1}$ and similarly for $F_{rr}$,This proves the ambient restriction formula for the spherical Laplacian and shows why ambient derivatives normal to the sphere must be subtracted. The equivalent formula in spherical coordinates is

Let ${\mathcal{P}}_{\ell }$ be the vector space of homogeneous polynomials of degree $\ell$ in $d+1$ variables, and ${\mathcal{H}}_{\ell }=\ker \tilde{\Delta }\cap {\mathcal{P}}_{\ell }$ its harmonic polynomials. If $H\in {\mathcal{H}}_{\ell }$, homogeneous polynomial scaling gives $H(r\omega )=r^{\ell }H(\omega )$. Substitution into the spherical coordinates formula givesThe restrictions are the degree-$\ell$ spherical harmonics. A homogeneous polynomial vanishing on the sphere vanishes on every nonzero ray and hence identically, so restriction is injective on ${\mathcal{H}}_{\ell }$.

We prove the harmonic decomposition of homogeneous polynomials that both counts these eigenfunctions and establishes completeness. For an ambient harmonic homogeneous polynomial $H_m$ of degree $m$ and $j\ge 1$, the spherical coordinates formula givesFor $d\ge 1$ its coefficient is nonzero. Inductively assume homogeneous polynomials of degree $\ell -2$ have been decomposed into sums of $r^{2j}{H}_m$. Apply this decomposition to $\tilde{\Delta }Q$ for $Q\in {\mathcal{P}}_{\ell }$, and lift each summand by multiplying by $r^2$ and dividing by the displayed nonzero coefficient. The resulting $R\in r^2{\mathcal{P}}_{\ell -2}$ satisfies $\tilde{\Delta }R=\tilde{\Delta }Q$, so $Q-R\in {\mathcal{H}}_{\ell }$. The same coefficient formula shows that no nonzero element of $r^2{\mathcal{P}}_{\ell -2}$ can be harmonic. Starting at degrees zero and one therefore proves the direct decompositionSince $\dim {\mathcal{P}}_{\ell }=\left(\genfrac{}{}{0ex}{}{\ell +d}{d}\right)$, the multiplicity iswhere the second term is zero for $\ell <2$.

On $r=1$, iterating the decomposition writes every polynomial restriction as a finite sum of spherical harmonics. The Stone-Weierstrass theorem makes these polynomial restrictions dense in $C(S^d)$, since coordinate functions separate points and constants are included; they are consequently dense in $L^2(S^d)$. Integration by parts makes $P$ symmetric on smooth functions and gives orthogonality between distinct eigenvalues. Normalizing within each finite-dimensional eigenspace gives a complete orthonormal eigenbasis $Y_{\ell a}$. To check the full operator spectrum, let $c_{\ell a}=⟨f,Y_{\ell a}⟩$. For smooth $f$, integration by parts gives the coefficients of $Pf$ as ${\lambda }_{\ell }{c}_{\ell a}$. Therefore finite harmonic partial sums approximate both $f$ and $Pf$ in $L^2$, hence approximate in the graph norm. Conversely every coefficient sequence with ${\sum }_{\ell ,a}{\lambda }_{\ell }^2|c_{\ell a}{|}^2<\infty$ is the graph norm limit of its finite harmonic sums. Thus the closure of $P$ has exactly this operator domain and is the real diagonal self-adjoint operator with entries ${\lambda }_{\ell }=\ell (\ell +d-1)$. If $z$ is not one of these entries, the inverse has entries $({\lambda }_{\ell }-z{)}^{-1}$ and is bounded because ${\lambda }_{\ell }\to \infty$. This proves that there are no missing spectral values. The spectrum of the Laplacian on a sphere iswith the above multiplicities; for $\Delta$, negate every eigenvalue. For example $S^1$ has eigenvalues ${\ell }^2$, with multiplicities one at zero and two for $\ell \ge 1$, and $S^2$ has multiplicity $2\ell +1$. The zero-dimensional sphere consists of two isolated points and has only the zero eigenvalue, with multiplicity two.

A sufficient set of assumptions is that $L$ is a densely defined closed operator on the Hilbert space $H$, generating an analytic semigroup $S(t)$, and that $f\in C^{0,\theta }([0,T],H)$ for some $0<\theta \le 1$. For an operator criterion, one may assume that $A=\omega I-L$ is a sectorial operator of angle less than $\pi /2$, with dense domain. The analytic semigroup then satisfies, on a fixed finite time interval,The Hölder continuity assumption means $\Vert f(t)-f(s){\Vert }_H\le K|t-s{|}^{\theta }$. It supplies time regularity of the forcing, not spatial membership of $f(t)$ in $D(L)$.

With initial datum $u_0\in H$, a classical solution of an abstract Cauchy problem meanswith the evolution identity holding in $H$ at positive times and $u(t)\to u_0$ in $H$. Equivalently, the positive-time domain continuity is in the graph norm. If the definition requires differentiability also at zero, assume additionally $u_0\in D(L)$; the proof below then gives $u\in C^1([0,T],H)\cap C([0,T],D(L))$.

The variation-of-constants formula gives the candidateThe integral is a Bochner integral. Boundedness of $S$ and continuity of $f$ make the expression continuous into $H$, and its integral term tends to zero with $t$. Thus it is a mild solution of an abstract Cauchy problem with the correct initial value. The issue is showing that its integral term lies in the domain of the unbounded linear operator $L$.

Write $v(t)={\int }_0^{t}S(r)f(t-r)dr$. For $\epsilon >0$, the truncated integral $v_{\epsilon }(t)={\int }_{\epsilon }^{t}S(r)f(t-r)dr$ lies in $D(L)$. Subtract the current forcing before applying $L$:Here the last term follows by integrating $LS(r)f(t)=\frac{d}{dr}S(r)f(t)$ away from zero; it does not assume that $f(t)\in D(L)$. The first integrand satisfiesSince $v_{\epsilon }(t)\to v(t)$ in $H$ and the displayed images also converge, the closed operator property givesThis is the reason for the Hölder continuity assumption: the unsplit estimate would have the nonintegrable factor $r^{-1}$. Merely knowing that $f$ is continuous does not provide the integrable bound needed here. More generally, a modulus of continuity $\eta$ satisfying ${\int }_0^1\eta (r)dr/r<\infty$ suffices for this argument; a positive Hölder exponent is a convenient sufficient hypothesis.

The formula also proves continuity of $Lv$ at positive times. On compact positive-time intervals, split the integral at a small $r=\epsilon$: its short part is uniformly bounded by $C{\epsilon }^{\theta }$, while the remaining part is continuous by the strongly continuous semigroup property and ordinary dominated convergence. Together with analytic smoothing of $S(t)u_0$, this makes $Lu$ continuous at positive times. The semigroup property givesBecause $u(t)\in D(L)$, division of this identity by $h$ after subtracting $u(t)$ gives the continuous right derivative $Lu(t)+f(t)$. The left derivative agrees by the same identity and continuity of $Lu+f$. Thus $u^{\prime }=Lu+f$, proving existence of the classical solution of an abstract Cauchy problem on the entire given finite interval, and in particular local existence.

For uniqueness, the difference $h$ of two such classical solutions of an abstract Cauchy problem satisfies $h^{\prime }=Lh$ and $h(0)=0$. For $0<\epsilon <t$, differentiation of $S(t-s)h(s)$ on $[\epsilon ,t]$ gives zero, so $h(t)=S(t-\epsilon )h(\epsilon )$. Letting $\epsilon \downarrow 0$ yields $h(t)=0$. Finally, if $u_0\in D(L)$, then $LS(t)u_0=S(t)L{u}_0\to L{u}_0$, and the expression for $Lv(t)$ tends to zero. Hence $u^{\prime }(t)\to L{u}_0+f(0)$ and the stronger time-zero regularity follows. The solution is unique; the forcing regularity controls the singularity in the differentiated integral.

The natural expectation is orbital stability, since a spatial translation of a travelling wave is another travelling wave. Thus convergence for ordinary perturbations, when it holds, is to a suitably translated front. It also matters which perturbations are allowed in the leading edge: the zero state ahead of this front is linearly unstable, and a sufficiently shallow positive tail can select a different speed. One should not assert asymptotic stability for arbitrary disturbances without a norm and a class of initial data. The requested essential spectrum calculation is a necessary part of such an analysis, not by itself a complete nonlinear stability proof.

In the moving coordinate $z=x-ct$, write $u(x,t)=w_c(z)+v(z,t)$. Expansion of the reaction term givesThe first equation to study is its linearization,and the associated eigenvalue equation $\mathcal{L}\varphi =\lambda \varphi$. Differentiating the front equation gives $\mathcal{L}{w}_c^{\prime }=0$, the neutral translation mode whenever it belongs to the chosen function space. Ahead of the front the limiting coefficient is $k$, so in unweighted L2 space the limiting dispersion relation is $\lambda =-{\xi }^2+ic\xi +k$: its rightmost real part is positive. An exponential weight penalizes precisely these disturbances ahead of the front.

There is a small functional-analytic correction to the printed definition. Use the complete exponentially weighted L2 spaceThe additional requirement of unweighted L2 space membership in the printed set makes that set incomplete under its stated weighted norm. Indeed $v_N=1_{[-N,0]}$ belongs to the intersection and is a weighted Cauchy sequence, but its weighted limit $1_{(-\infty ,0]}$ does not belong to unweighted L2 space. The usual Hilbert space for this spectral argument is the completion $X_{\gamma }$. We use that completion explicitly, rather than treating the intersection as a complete space.

Multiplication $T_{\gamma }v=e^{\gamma z}v$ is a unitary map from $X_{\gamma }$ onto L2 space. Take the domain $T_{\gamma }^{-1}{H}^2(\mathbb{R})$ for $\mathcal{L}$. The transformed closed operator isThis follows from $T_{\gamma }{\partial }_z{T}_{\gamma }^{-1}={\partial }_z-\gamma$. The two limiting constant-coefficient dispersion relations, obtained by the Fourier transform, areIn particular the necessary strict leading-edge condition is ${\gamma }^2-c\gamma +k<0$. For $c>2\sqrt{k}$ this meansA particularly convenient choice isWith this choice there is no first derivative, and ${\mathcal{L}}_{\gamma }={\partial }_z^2+k-c^2/4-2w_c(z)$ is a self-adjoint differential operator on $H^2(\mathbb{R})$. This also avoids possible ambiguities between definitions of essential spectrum for nonnormal operators.

The relevant perturbation theorem is essential-spectrum invariance under relatively compact perturbations: if $A_0$ is a closed operator, $V:D(A_0)\to H$ is compact with respect to the graph norm, and $A_0+V$ is closed on the same domain, their Fredholm essential spectrum is the same. For self-adjoint operators this is the usual essential spectrum after removing isolated finite-multiplicity eigenvalues. To see why the theorem applies, regard $A_0-\lambda$ and $A_0+V-\lambda$ as bounded maps from the graph norm domain to $H$, and use compact perturbation invariance of Fredholm operators for each $\lambda$.

Define a step-potential comparison operatorBoth $A_0$ and ${\mathcal{L}}_{\gamma }$ are self-adjoint on $H^2(\mathbb{R})$. Their difference is multiplication byIt is relatively compact, not necessarily compact as a multiplication map on all of L2 space. To prove the needed compactness, take a sequence bounded in $H^2(\mathbb{R})$. On each bounded interval the Rellich-Kondrachov compactness theorem gives a subsequence converging in L2 space. A diagonal subsequence makes the products by $V$ converge locally. Outside a sufficiently large interval, $\Vert V{\Vert }_{\infty }$ is arbitrarily small and the sequence has a uniform L2 norm bound, so those tails are uniformly small. Thus the products converge globally in L2 space, proving graph-domain compactness.

Finally,Here is a direct verification. Its quadratic form is at most $(k-c^2/4)\Vert h{\Vert }_2^2$, so no spectrum lies above this endpoint. For every $\lambda \le k-c^2/4$, choose $\xi$ with $\lambda =k-c^2/4-{\xi }^2$. Normalized, increasingly long cutoffs of $e^{i\xi z}$, translated far to the right, have disjoint supports and satisfy $\Vert (A_0-\lambda )h_N{\Vert }_2\to 0$: differentiating the cutoff gives errors of orders $N^{-1}$ and $N^{-2}$. They converge weakly to zero. This is a singular sequence in the Weyl criterion, proving that every such $\lambda$ belongs to the essential spectrum. The relatively compact perturbation theorem therefore givesThis proves the weighted essential spectrum of a Fisher travelling front is strictly stable when $c>2\sqrt{k}$. In fact here the quadratic form gives more:Since $w_c\ge 0$, the full transformed spectrum lies below $-(c^2/4-k)$, and the evolution given by its linearization decays at least as $e^{-(c^2/4-k)t}$ in this weighted norm. This does not contradict translation invariance: the nonzero translation mode cannot lie in this weighted operator domain, since otherwise it would be a zero eigenfunction, contradicting the strictly negative quadratic-form bound. The weight restricts the permitted perturbations and excludes these neutral translations. Nor does a weighted L2 norm alone control the nonlinear remainder at $z\to -\infty$; a nonlinear theorem needs further boundedness or regularity assumptions.

At the minimal speed $c_{\ast }=2\sqrt{k}$, the best choice is $\gamma =\sqrt{k}$. Now ${\gamma }^2-c\gamma +k=(\gamma -\sqrt{k}{)}^2$, andThe essential spectrum has no positive part but touches zero; no positive exponential weight makes it strictly negative. Thus the critical front has no spectral gap and this argument gives no exponential decay rate there. Strict weighted spectral stability holds above the minimal speed; at the minimal speed it is marginal.

Fix $W⋐V⋐U$. The standard local regularization of the maximal graph norm domain supplies smooth compactly supported approximants $u_k$ on $V$ such thatApply part 4(c), with $V$ as the outer domain, to $u_k-u_m$. It givesThus $(u_k)$ is Cauchy in $H^2(W)$. Its $H^1$ limit is $u$, so $u\in H^2(W)$. Since $W⋐U$ was arbitrary, the definition of a Local Sobolev space givesThis is the Interior H2 regularity for continuous nondivergence coefficients.

Let $(u_n)$ be a Cauchy sequence in the graph normThen $(u_n)$ and $(A{u}_n)$ are Cauchy sequences in the Banach space $H$, so for some $u,v\in H$,Because $A$ is a closed linear operator, $u\in D(A)$ and $Au=v$. It follows thatThus every Cauchy sequence converges in $Y=(D(A),\Vert \cdot {\Vert }_Y)$, and

For $y\in D(A)$, the semigroup property givesHence $U(t)y\in D(A)$ andThe generator domain is therefore an invariant subspace, and the operator norm bound givesThusMoreover, strong continuity applied separately to $y$ and $Ay$ givesTherefore the restrictions form the semigroup restricted to its generator domain. Its derivative at zero exists in the graph norm exactly when $y\in D(A)$ and $Ay\in D(A)$, namely when $y\in D(A^2)$, and then the derivative is $Ay$. Hence its generator is

A solution operator for a nonautonomous evolution equation is an evolution family $U(t,s)$ satisfyingand, on a suitable common domain $\mathcal{D}$,

One applicable nonautonomous generation theorem is the following. Suppose $\mathcal{D}$ is a dense linear subspace of $H$, each $A(t)$ has domain $\mathcal{D}$, the family is a stable family of semigroup generators with constants $M,0$, and $t\mapsto A(t)$ is continuously differentiable as a map from $[0,T]$ to $\mathcal{B}(\mathcal{D},H)$, where $\mathcal{D}$ carries one of the uniformly equivalent graph norms. Then there is a unique evolution family such that:

For the uniform partition $t_j=s+j(t-s)/N$, the frozen-generator product approximation isAs $N\to \infty$, $U_N(t,s)u\to U(t,s)u$ in the norm of $H$ for every $u\in H$, uniformly for $(t,s)$ in the compact time triangle $0\le s\le t\le T$. This is convergence in the strong operator topology, rather than convergence in the operator norm.

It remains to verify the second differential equation. The evolution family law gives, for $h>0$,Divide by $h$. Since $u\in \mathcal{D}$,while strong continuity gives $U(t,s+h)A(s)u\to U(t,s)A(s)u$. ThereforeThe left derivative follows in the same way, so $s\mapsto U(t,s)u$ is differentiable.

Equip $D(A)$ with its graph norm. The restrictions $U(t){|}_{D(A)}$ form a C0-semigroup on this Banach space; its generator is $A$ restricted to $D(A^2)$.