Fredholm Elliptic Problems and the Elliptic Spectrum — Examples

Membership in $H_0^1$. By [F2] and [F3], $u_k^{\mathrm{\prime }}=k\cos (kx)$ and $u_k^{\mathrm{\prime }\mathrm{\prime }}=-k^2{u}_k$ are weak derivatives. For $0<\epsilon <\pi \mathrm{/}4$ use the cutoff ${\eta }_{\epsilon }$ of [F3]; then ${\eta }_{\epsilon }{u}_k\in C_c^{\mathrm{\infty }}(0,\pi )$. On the boundary strips $A_{\epsilon }=(0,2\epsilon )\cup (\pi -2\epsilon ,\pi )$, $\mathrm{\mid }{u}_k\mathrm{\mid }\le k\mathrm{dist}(x,\{0,\pi \})\le 2k\epsilon$ by integration of $u_k^{\mathrm{\prime }}$ from the nearest endpoint, and $\mathrm{\mid }{u}_k^{\mathrm{\prime }}\mathrm{\mid }\le k$. Since $({\eta }_{\epsilon }{u}_k-u_k{)}^{\mathrm{\prime }}=({\eta }_{\epsilon }-1)u_k^{\mathrm{\prime }}+{\eta }_{\epsilon }^{\mathrm{\prime }}{u}_k$, $\mathrm{\parallel }{\eta }_{\epsilon }{u}_k-u_k{\mathrm{\parallel }}_{H^1}^2\le {\int }_{A_{\epsilon }}(\mathrm{\mid }{u}_k{\mathrm{\mid }}^2+2\mathrm{\mid }{u}_k^{\mathrm{\prime }}{\mathrm{\mid }}^2+2\mathrm{\mid }{\eta }_{\epsilon }^{\mathrm{\prime }}{\mathrm{\mid }}^2\mathrm{\mid }{u}_k{\mathrm{\mid }}^2)\le 4\epsilon (4k^2{\epsilon }^2+2k^2+8C^2{k}^2)⟶0.$ Thus $u_k\in H_0^1$ by the closure definition.

The weak eigenidentity. Let $v\in H_0^1(0,\pi )$ and choose $v_j\in C_c^{\mathrm{\infty }}(0,\pi )$ with $v_j\to v$ in $H^1$ (possible by [F2]). For each $j$, integration by parts on the compact support of $v_j$ has no boundary term and gives, since $u_k^{\mathrm{\prime }\mathrm{\prime }}=-k^2{u}_k$, ${\int }_0^{\pi }{u}_k^{\mathrm{\prime }}\overline{v_j^{\mathrm{\prime }}}=-{\int }_0^{\pi }{u}_k^{\mathrm{\prime }\mathrm{\prime }}\overline{v_j}=k^2{\int }_0^{\pi }{u}_k\overline{v_j}$; the left side differs from $\int u_k^{\mathrm{\prime }}\overline{v^{\mathrm{\prime }}}$ by at most $\mathrm{\parallel }{u}_k^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2}\mathrm{\parallel }{v}_j^{\mathrm{\prime }}-v^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2}$, and the right side from $k^2\int u_k\bar{v}$ by at most $k^2\mathrm{\parallel }{u}_k{\mathrm{\parallel }}_{L^2}\mathrm{\parallel }{v}_j-v{\mathrm{\parallel }}_{L^2}$, so passing to the limit gives the displayed identity; by [F1] and the weak eigenpair definition, $(k^2,u_k)$ is a Dirichlet eigenpair (Symmetric elliptic weak eigenpairs).

Orthogonality and norms. For integers $m,n\ge 1$ the addition formula [F4] gives $2\sin (mx)\sin (nx)=\cos ((m-n)x)-\cos ((m+n)x)$; integrating over $(0,\pi )$ with the second fundamental theorem gives $0$ when $m\mathrm{\ne }n$ (both cosine integrals vanish) and ${\int }_0^{\pi }{\sin }^2(nx)dx=\frac{1}{2}{\int }_0^{\pi }(1-\cos (2nx))dx=\pi \mathrm{/}2$. Hence the $u_k$ are pairwise $L^2$-orthogonal with squared norm $\pi \mathrm{/}2$, and the eigenvalues $k^2$ are pairwise distinct.

The Rayleigh bound. By step 1.2 applied with $k=1$, $u_1=\sin x$ is a weak eigenfunction with eigenvalue $1$; the Rayleigh principle The Rayleigh principle for the first Dirichlet eigenvalue then gives ${\lambda }_1\le 1$, since the Rayleigh quotient of $\sin x$ equals $\mathrm{\parallel }{u}_1^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2}^2\mathrm{/}\mathrm{\parallel }{u}_1{\mathrm{\parallel }}_{L^2}^2={\int }_0^{\pi }{\cos }^{2}xdx\mathrm{/}{\int }_0^{\pi }{\sin }^{2}xdx=1$ by [F4] and step 1.3. No completeness of the family $\{\sin (kx)\}$, no simplicity of the eigenvalues and no sharp Poincare constant is asserted.

The weak Neumann identities. By [F2], $u_k^{\mathrm{\prime }}$ is continuously differentiable on $[0,\pi ]$, hence absolutely continuous (its derivative is bounded); with [F1], The product of two absolutely continuous functions is absolutely continuous makes $u_k^{\mathrm{\prime }}{v}^{\ast }$ absolutely continuous with derivative $u_k^{\mathrm{\prime }\mathrm{\prime }}{v}^{\ast }+u_k^{\mathrm{\prime }}{v}^{\mathrm{\prime }}$ almost everywhere, and the fundamental theorem [F1] gives ${\int }_0^{\pi }{u}_k^{\mathrm{\prime }}{v}^{\mathrm{\prime }}=[u_k^{\mathrm{\prime }}{v}^{\ast }{]}_0^{\pi }-{\int }_0^{\pi }{u}_k^{\mathrm{\prime }\mathrm{\prime }}{v}^{\ast }=0+k^2{\int }_0^{\pi }{u}_{k}v,$ because the boundary term vanishes by [F2]. Thus $a_N(u_k,v)=k^2{\int }_0^{\pi }{u}_{k}v$ for every $v\in H^1(0,\pi ;\mathbb{R})$; for $k=0$ this reads $0=0$, giving the constant zero mode, and for $k\ge 1$ it says that $\cos (kx)$ is a weak Neumann eigenfunction with eigenvalue $k^2$.

Lowest eigenvalue on all of $H^1$. The form is $a_N(u,u)=\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2}^2\ge 0$ with $a_N(u_0,u_0)=0$ for the nonzero constant $u_0$, so the infimum of the Rayleigh quotient over $H^1(0,\pi )\setminus \{0\}$ is $0$, attained at the constants; in particular the lowest weak Neumann eigenvalue on $H^1(0,\pi )$ is $0$.

The mean-zero restriction. On the mean-zero subspace $V=\{u:{\int }_0^{\pi }u=0\}$ Poincare--Wirtinger [F4] gives a positive constant $C$ with $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2}\le C\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2}$, so the Rayleigh quotient on $V\setminus \{0\}$ is bounded below by $C^{-2}>0$, and by [F4] its infimum is the first positive Neumann Rayleigh value. Taking $v=\cos x$ in step 1.1 gives $a_N(\cos x,\cos x)={\int }_0^{\pi }{\sin }^{2}xdx=\pi \mathrm{/}2$ while ${\int }_0^{\pi }{\cos }^{2}xdx=\pi \mathrm{/}2$, so this value is at most $1$; the constant mode is exactly the element removed by the mean-zero restriction, in contrast with the Dirichlet problem where constants are excluded by the zero-trace domain.

Failure of coercivity. For $u=\sin x\in H_0^1(0,\pi )$ one has $u^{\mathrm{\prime }}=\cos x$, so by [F3] $a(u,u)={\int }_0^{\pi }({\cos }^{2}x-10{\sin }^{2}x)dx=\frac{\pi }{2}-10\cdot \frac{\pi }{2}=-\frac{9\pi }{2}<0.$ Hence $a$ is neither coercive nor nonnegative.

The general shift. With $\theta =1$, $M_b=0$, $M_c=10$ the Garding constants are $\alpha =1\mathrm{/}2$ and $\beta =1\mathrm{/}2+0+10=21\mathrm{/}2$, so [F2] gives $\mathrm{Re}a(u,u)\ge \frac{1}{2}\mathrm{\parallel }u{\mathrm{\parallel }}_{H_0^1}^2-\frac{21}{2}\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2}^2$ and makes $a_{\mu }$ bounded and coercive for every $\mu \ge 21\mathrm{/}2$. By Lax--Milgram the problem $-u^{\mathrm{\prime }\mathrm{\prime }}-10u+\mu u=f$ with zero boundary values has a unique weak solution for every $f\in L^2(0,\pi )$ at those shifts.

The sharper direct threshold. For every $u\in H_0^1(0,\pi )$ the shifted form is $a_{\mu }(u,u)={\int }_0^{\pi }(\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^2+(\mu -10)\mathrm{\mid }u{\mathrm{\mid }}^2)dx\ge \min (1,\mu -10)(\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2}^2+\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2}^2)=\min (1,\mu -10)\mathrm{\parallel }u{\mathrm{\parallel }}_{H_0^1}^2\mathrm{.}$ For $\mu >10$ the constant $\min (1,\mu -10)$ is strictly positive and $a_{\mu }$ is coercive with that constant, sharper than the general threshold $21\mathrm{/}2$ of step 1.2; the displayed inequality is positive for every nonzero $u$, so the shifted problem is uniquely solvable for every $\mu >10$ as well. No boundary regularity of $\mathrm{\Omega }$ was used.

The homogeneous space. By [F1] and the definition of $q_k$ in [F3], $u_k=\sin (kx)$ is a nonzero homogeneous solution. Conversely, if $u\in H_0^1$ is a weak homogeneous solution, compactly supported tests give $D(u^{\mathrm{\prime }})=-k^{2}u\in L^2$, so both $u$ and $u^{\mathrm{\prime }}$ have absolutely continuous representatives by [F2]. Their integral identities imply that $u$ is $C^1$ with derivative that representative of $u^{\mathrm{\prime }}$, and that derivative is $C^1$ with derivative $-k^{2}u$, since the latter is continuous. Hence $u$ is $C^2$ and $u^{\mathrm{\prime }\mathrm{\prime }}=-k^{2}u$ on $[0,\pi ]$, with $u(0)=u(\pi )=0$ by [F2]. Set $z=u-(u^{\mathrm{\prime }}(0)\mathrm{/}k)\sin (kx)$. It satisfies $z(0)=z^{\mathrm{\prime }}(0)=0$ and $z^{\mathrm{\prime }\mathrm{\prime }}=-k^{2}z$. The derivative of $\mathrm{\mid }{z}^{\mathrm{\prime }}{\mathrm{\mid }}^2+k^2\mathrm{\mid }z{\mathrm{\mid }}^2$ is zero, so [F2] makes that energy identically zero; thus $z=0$. This proves $N=\mathrm{span}\{\sin (kx)\}$, exactly one-dimensional.

Solvability fails on a nonzero datum. Since $q_k$ is symmetric, [F3] gives $N^{\ast }=N=\mathrm{span}\{\sin (kx)\}$, so the weak problem $-u^{\mathrm{\prime }\mathrm{\prime }}-k^{2}u=f$ with $u\in H_0^1(0,\pi )$ is solvable if and only if ${\int }_0^{\pi }f(x)\sin (kx)dx=0$; for $f=\sin (kx)$ this integral equals $\pi \mathrm{/}2\mathrm{\ne }0$ by [F4], so this datum admits no weak solution. Uniqueness also fails whenever a solution exists, because adding any multiple of the nonzero homogeneous solution $\sin (kx)$ produces another solution.

Conclusion. On $\mathrm{\Omega }=(0,\pi )$ with $\lambda =k^2$ the equation $-\mathrm{\Delta }u-\lambda u=f$ is neither uniquely solvable for every $f\in L^2$ (uniqueness fails at the eigenvalue) nor solvable for the particular datum $f=\sin (kx)$; hence the refuted statement fails, and the failure is exactly the one-dimensional orthogonality condition predicted by the Fredholm alternative.

Membership in $H_0^1$. For $\epsilon >0$ small let ${\chi }_{\epsilon }\in C_c^{\mathrm{\infty }}(0,\pi )$ satisfy $0\le {\chi }_{\epsilon }\le 1$, ${\chi }_{\epsilon }=1$ on $[2\epsilon ,\pi -2\epsilon ]$, ${\chi }_{\epsilon }=0$ outside $[\epsilon ,\pi -\epsilon ]$ and $\mathrm{\mid }{\chi }_{\epsilon }^{\mathrm{\prime }}\mathrm{\mid }\le C\mathrm{/}\epsilon$; put $w_{\epsilon }(x,y):={\chi }_{\epsilon }(x){\chi }_{\epsilon }(y)w(x,y)$ for $w=u$ or $w=v$. Then $w_{\epsilon }\in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$. Let $A_{\epsilon }$ be the union of the boundary strips where either cutoff differs from $1$; its area is $O(\epsilon )$. The sine factors give $\mathrm{\mid }w\mathrm{\mid }\le C_1\epsilon$ on $A_{\epsilon }$, while $\mathrm{\mid }\mathrm{\nabla }w\mathrm{\mid }\le C_2$ on the whole square. Thus $\mathrm{\parallel }{w}_{\epsilon }-w{\mathrm{\parallel }}_{L^2}^2=O({\epsilon }^3)$. For the gradient, $\mathrm{\nabla }(w_{\epsilon }-w)=({\chi }_{\epsilon }(x){\chi }_{\epsilon }(y)-1)\mathrm{\nabla }w+w\mathrm{\nabla }({\chi }_{\epsilon }(x){\chi }_{\epsilon }(y))$; the first term has squared $L^2$ norm $O(\epsilon )$, and on the support of the cutoff derivatives $\mathrm{\mid }w\mathrm{\mid }\le C_1\epsilon$, so the second term is bounded pointwise and supported on area $O(\epsilon )$, also giving squared $L^2$ norm $O(\epsilon )$. Hence $w_{\epsilon }\to w$ in $H^1(\mathrm{\Omega })$, and $u,v\in H_0^1(\mathrm{\Omega })$ by the closure definition [F1].

Weak eigenidentity. For every $\phi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$, integration by parts on the square has no boundary term and $-\mathrm{\Delta }u=5u$, $-\mathrm{\Delta }v=5v$ by [F1], so $a(w,\phi )={\int }_{\mathrm{\Omega }}\mathrm{\nabla }w\cdot \overline{\mathrm{\nabla }\phi }={\int }_{\mathrm{\Omega }}(-\mathrm{\Delta }w)\bar{\phi }=5{\int }_{\mathrm{\Omega }}w\bar{\phi }\mathrm{.}$ Given arbitrary $\phi \in H_0^1(\mathrm{\Omega })$, choose ${\phi }_k\in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ with ${\phi }_k\to \phi$ in $H^1$; both pairings are continuous in the $H^1$ norm, so passing to the limit extends the identity to all $\phi \in H_0^1(\mathrm{\Omega })$. By [F2], $u$ and $v$ are weak Dirichlet eigenfunctions with the common eigenvalue $5=1^2+2^2=2^2+1^2$.

Orthogonality and dimension. The inner product factors: $(u,v{)}_{L^2}=({\int }_0^{\pi }\sin x\sin 2xdx\left)\right({\int }_0^{\pi }\sin 2y\sin ydy)=0$ by [F5] (both one-dimensional integrals vanish), while the same factorization gives $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2}^2=\mathrm{\parallel }v{\mathrm{\parallel }}_{L^2}^2=(\pi \mathrm{/}2{)}^2>0$, so both are nonzero classes. Hence $u$ and $v$ are $L^2$-orthogonal nonzero classes, so they are linearly independent by [F3], and the eigenspace of the eigenvalue $5$ has dimension at least two.

Conclusion. The eigenspace of dimension at least two established in step 2.1 contains the linearly independent eigenfunctions $u,v$, so the eigenvalue $5$ occurs with multiplicity at least two in the list of [F4]; the refuted statement, that all eigenvalues of the Dirichlet Laplacian on a bounded domain are simple (no repetitions in the list with multiplicity), therefore fails on the square.

The Hermitian part of $M$ is $H:=\frac{1}{2}(M+M^{\ast })=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$ and $\mathrm{Re}a(u,u)=(Hu,u)$ for $u=(u_1,u_2)$. Writing $x=\mathrm{\mid }{u}_1\mathrm{\mid }$ and $y=\mathrm{\mid }{u}_2\mathrm{\mid }$, the elementary bound $2\mathrm{Re}(u_2\overline{u_1})\ge -2xy$ gives $\mathrm{Re}a(u,u)\ge 2x^2-2xy+y^2$. With $\phi :=\frac{1+\sqrt{5}}{2}$ one has $\alpha :=\frac{3-\sqrt{5}}{2}=2-\phi >0$ and ${\phi }^{-1}=\phi -1=1-\alpha$, hence $2x^2-2xy+y^2-\alpha (x^2+y^2)=\phi (x-y\mathrm{/}\phi {)}^2\ge 0$. Therefore $\mathrm{Re}a(u,u)\ge \alpha \mathrm{\parallel }u{\mathrm{\parallel }}^2$, so $a$ is coercive with constant $\alpha$.

Boundedness: for all $u,v$ one has $\mathrm{\mid }a(u,v)\mathrm{\mid }=\mathrm{\mid }(Mu,v)\mathrm{\mid }\le \mathrm{\parallel }Mu\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$ by [F4], and the coordinate estimate $\mathrm{\parallel }Mu{\mathrm{\parallel }}^2=\mathrm{\mid }2u_1+2u_2{\mathrm{\mid }}^2+\mathrm{\mid }{u}_2{\mathrm{\mid }}^2\le (4+4+1)(\mathrm{\mid }{u}_1{\mathrm{\mid }}^2+\mathrm{\mid }{u}_2{\mathrm{\mid }}^2)=9\mathrm{\parallel }u{\mathrm{\parallel }}^2$ obtained from Cauchy--Schwarz in the two-dimensional index gives $\mathrm{\parallel }Mu\mathrm{\parallel }\le 3\mathrm{\parallel }u\mathrm{\parallel }$; hence $\mathrm{\mid }a(u,v)\mathrm{\mid }\le 3\mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$ and $a$ is bounded.

Eigenvalues and eigenvectors: the characteristic polynomial of $M$ is ${\chi }_M(\lambda )=(2-\lambda )(1-\lambda )$, whose roots are $1$ and $2$; by [F3] the spectrum is $\{1,2\}$, both roots are simple, and each eigenspace is one-dimensional. Solving $(M-2I)u=0$ gives $u_2=0$, so $E_2=\mathbb{K}(1,0)$, and solving $(M-I)u=0$ gives $u_1+2u_2=0$, so $E_1=\mathbb{K}(2,-1)$. The two exhibited eigenvectors satisfy $((2,-1),(1,0))=2\mathrm{\ne }0$.

No orthonormal eigenbasis exists. Suppose $u,v$ were an orthogonal pair of nonzero eigenvectors; since $E_1$ and $E_2$ are one-dimensional and distinct, after relabelling $u\in E_2$ and $v\in E_1$, so $u=c(1,0)$ and $v=d(2,-1)$ with $c,d\mathrm{\ne }0$, and step 1.3 gives $(u,v)=2c\bar{d}\mathrm{\ne }0$, a contradiction. Thus no orthogonal pair of eigenvectors exists, although $a$ is bounded and coercive by steps 1.1 and 1.2 and all its eigenvalues $1,2$ are real. The displayed form is therefore a counterexample to the refuted statement: the symmetry hypothesis of the symmetric elliptic spectral theorem is not redundant.