Week 12: Eigenvalues and eigenfunctions

Throughout this lecture, $\Omega$ is a connected bounded domain of class $C^{\infty}$.

The eigenvalue problem

We now turn our attention to the eigenvalue problem for $L$, i.e. the problem of finding $\lambda\in\mathbb{R}$ and $u$ not identically zero solving the following boundary value problem:

By Theorem 9.3, there are at most countably many eigenvalues $\lambda$ (and correponding eigenfunctions $u$). We now obtain more refined information on the solutions to this eigenvalue problem.

Assume $L$ is in divergence form with the same setup as in Week 8 and that $a_{ij}\in C^{\infty}(\overline{\Omega})$. We shall focus on the case where

in which case $L^{\ast}u\equiv Lu$, since $B_{L}(u,v)=\int_{\Omega}\sum_{i,j}a_{ij}\partial_{i}u\partial_{j}v=B_{L^{\ast}}(u,v)$ for any $u,v\in H_{0}^{1,2}(\Omega)$; we say that $L$ is symmetric. We shall require the following fundamental result from functional analysis.

Theorem 12.1 (eigenvectors of symmetric compact operators). If $X$ is a separable Hilbert space (i.e. $X$ has a countable dense subset) and $A:X\rightarrow X$ is a compact symmetric linear mapping (for all $x,y\in X$ $\left<x,Ay\right>=\left<Ax,y\right>$), then the eigenvalues $\{\mu{i}\}_{i=1}^{\infty}$ (counted with multiplicity) of $A$ are real and there exists an orthonormal basis $\{\varphi_{i}\}_{i=1}^{\infty}$ of $X$ such that for each $i\in\mathbb{N}$, $A \varphi_{i}=\mu_{i}\varphi_{i}$.

Theorem 12.2 (eigenvalues of symmetric elliptic operators). If $b\equiv 0$ and $c\equiv 0$, then each eigenvalue $\lambda$ of $L$ is real and positive. Moreover, writing $\{\lambda_{k}\}_{k=1}^{\infty}$ for the eigenvalues of $L$ counted with multiplicity, we have that $0<\lambda_{1}\leq \lambda_{2}\leq \dots$ with $\lambda_{k}\xrightarrow{k\rightarrow\infty}\infty$. Finally, there exists a sequence $\{w_{k}\}_{k=1}^{\infty}\subset H_{0}^{1,2}(\Omega)$ with $Lw_{k}=\lambda_{k}w_{k}$ for each $k\in\mathbb{N}$ such that $\{w_{k}\}_{k=1}^{\infty}$ is an orthonormal basis of $L^{2}(\Omega)$.

Proof. First note that if $Lu=\lambda u$ for some $\lambda\in \mathbb{C}$ and real-valued $u$ not equivalently zero, then

$B_{L}(u,u)=\left<Lu,u\right> = \lambda||u||_{2}^{2}\Leftrightarrow \lambda = \frac{B_{L}(u,u)}{||u||_{2}^{2}}=\frac{\int_{\Omega}\sum_{i,j}a_{ij}\cdot \partial_{i}u\partial_{j}u}{||u||_{2}^{2}}\geq \frac{||\D u||_{2}^{2}}{||u||_{2}^{2}}> 0.$

Now, since $B_{L}$ is coërcive, Lax-Milgram implies that there is a unique continuous linear mapping $A:L^{2}(\Omega)\rightarrow H_{0}^{1,2}(\Omega)\subset L^{2}(\Omega)$ such that

$B_{L}(Af,v)=\left<f,v\right>$

for all $f\in L^{2}(\Omega)$ and $v\in H_{0}^{1,2}(\Omega)$. Just as in Week 8, we deduce that $A:L^{2}(\Omega)\rightarrow L^{2}(\Omega)$ is compact, and we see that it is symmetric as well, since whenever $f\in L^{2}(\Omega)$ and $g\in H_{0}^{1,2}(\Omega)$,

$\left<f,Ag\right> = B_{L}(Af,Ag) = B_{L}(Ag,Af) = \left<g,Af\right>$

so that by approximation, $\left<f,Ag\right> = \left<Af,g\right>$ for all $f,g\in L^{2}(\Omega)$. Now,

$u\in H_{0}^{1,2}(\Omega),\ Lu=\lambda u \Leftrightarrow u\in L^{2}(\Omega), Au = \lambda^{-1}u$

so that as in Weeks 8 and 9, we deduce that the set of all eigenvalues of $L$ may be written as monotone nondecreasing sequence $\{\lambda_{i}\}_{i=1}^{\infty}$ with $\lambda_{i}\xrightarrow{i\rightarrow\infty}\infty$. By Theorem 11.10, the (orthonormal) eigenfunctions $\{w_{i}\}_{i=1}^{\infty}$ associated with $\{\lambda_{i}\}_{i=1}^{\infty}$ span $L^{2}(\Omega)$.

Remark 12.3. By higher regularity, the $\{u_{k}\}_{k=1}^{\infty}$ are smooth on $\Omega$ (up to a set of measure zero), and if $\partial\Omega$ is smooth, they are smooth on $\overline\Omega$.

By Theorem 12.2, any $u\in L^{2}(\Omega)$ may be written in terms of the eigenfunctions $\{w_{k}\}_{k=1}^{\infty}$ as

where $u_{i}=\left<w_{i},u\right>$ for each $i\in\mathbb{N}$. We now turn our attention to the convergence of this series in $H_{0}^{1,2}(\Omega)$ for a given $u\in H_{0}^{1,2}(\Omega)$. We first note the following.

Lemma 12.4. The bilinear form $B_{L}$ associated to $L$ defines an inner product on $H_{0}^{1,2}(\Omega)$. Moreover, the associated norm is equivalent to $||\cdot||_{1,2}$, i.e. there exist constants $c_{0},c_{1}>0$ depending only on $\Omega$ such that for any $u\in H_{0}^{1,2}(\Omega)$,

Proof. $B_{L}$ is bilinear and symmetric by assumption. Moreover, by ellipticity,

$B_{L}(u,u)\geq \lambda_{0}\int_{\Omega}|\D u|^{2}\geq \frac{\lambda_{0}}{2}\min\{C_{\textup{Poinc}}^{-1},1\}||u||_{1,2}^{2}$

so that positive-definiteness is immediate. The equivalence claim now follows from this inequality and the boundedness of $(a_{ij})$:

$B_{L}(u,u)\leq\sup|(a_{ij})|\int_{\Omega}|\D u|^{2}\leq \sup|(a_{ij})|\cdot ||u||_{1,2}^{2}.$

Remark 12.5. $B_{L}$ is commonly known as the energetic inner product and the associated norm the energetic norm associated to $L$, the motivation being the Dirichlet-type energy

from which $B_{L}$ naturally arises (cf. Lemma 7.10).

Corollary 12.6. The eigenfunctions $\{w_{k}\}_{k=1}^{\infty}$ form a basis of $H_{0}^{1,2}(\Omega)$ and for any $u\in H_{0}^{1,2}(\Omega)$,

where $u_{i}=\left<w_{i},u\right>$ for each $i\in\mathbb{N}$.

Proof. We shall employ the energetic inner product. First of all, note that by definition,

$B_{L}(w_{i},w_{j})=\lambda_{i}\int_{\Omega}w_{i}w_{j}=\lambda_{i}\delta_{ij}\Leftrightarrow B_{L}(\frac{w_{i}}{\sqrt{\lambda_{i}}},\frac{w_{j}}{\sqrt{\lambda_{j}}})=\delta_{ij},$

i.e. the set of functions $\{\frac{w_i}{\sqrt{\lambda_{i}}}\}_{i=1}^{\infty}$ is $B_{L}$-O.N. To see that their span $X$ is all of $H_{0}^{1,2}(\Omega)$, note that if $u\in H_{0}^{1,2}(\Omega)$, we may write

$u=u_{0}+u_{1}$

with $u_{0}\in X$ and $u_{1}\in H_{0}^{1,2}(\Omega)$ such that $B_{L}(u_{1},v)=0$ for all $v\in X$. However, this implies that for each $i\in\mathbb{N}$,

$0=B_{L}(u_{1},w_{i})=B_{L}(w_{i},u_{1})=\lambda_{i}\int_{\Omega}w_{i}u_{1}\Rightarrow\int_{\Omega}w_{i}u_{1}=0$

so that $u_{1}=0$ in $L^{2}(\Omega)$, thus also in $H_{0}^{1,2}(\Omega)$. Therefore, $u\in X$ and we may expand $u$ in terms of the orthonormal basis $\{\frac{w_{i}}{\sqrt{\lambda_{i}}}\}_{i=1}^{\infty}$ (limit in $W^{1,2}(\Omega)$):

$u= \sum_{i=1}^{\infty}B_{L}(u,\frac{w_{i}}{\sqrt{\lambda_{i}}})\cdot\frac{w_{i}}{\sqrt{\lambda_{i}}}=\sum_{i=1}^{\infty}\frac{B_{L}(u,w_{i})}{\lambda_{i}}\cdot w_{i}=\sum_{i=1}^{\infty}\left<w_{i},u\right>\cdot w_{i}.$

Example 12.7. A ubiquitous concrete instance of the above considerations is $n=1$, $\Omega=]0,1[$ and $Lu:=-\Delta u = -u''$. In this case, the eigenvalue problem boils down to the problem of solving the ODE

for $u$ and $\lambda>0$ subject to the boundary conditions $u(0)=u(1)=0$. From ODE theory, we know that any $C^{2}$ solution to this equation is of the form

where $A,B\in\mathbb{R}$. Using the boundary conditions, we obtain constraints on $A$, $B$ and $\lambda$:

The last condition implies that either $B=0$, in which case $u$ is the trivial solution, or $\sin\sqrt\lambda=0\Leftrightarrow \sqrt{\lambda}=j\pi$ for $j\in\mathbb{N}$. Therefore, we see that the eigenvalues of $L$ are given by $\{\lambda_{j}=j^{2}\pi^{2}\}_{j=1}^{\infty}$ with corresponding orthonormal eigenfunctions $\{x\mapsto w_{j}(x)=\sqrt{2}\sin(j\pi x)\}_{j=1}^{\infty}$ so that any $u\in L^{2}(\left]0,1\right[)$ may be expanded in the form

such a series is called a Fourier series.

The first eigenvalue

We now take a closer look at the smallest eigenvalue $\lambda_{1}>0$ of $Lu=-\sum_{i,j=1}^{n}\partial_{i}(a_{ij}\partial_{j}u)$, which we call the principal eigenvalue. It turns out it may be characterised variationally.

Theorem 12.8 (variational principle for principal eigenvalue). The principal eigenvalue of $L$ is characterised by the condition

moreover, the infimum is attained at a positive function $w_{1}\in H_{0}^{1,2}(\Omega)\cap C^{\infty}(\Omega)$ solving the equation $Lw_{1}=\lambda_{1}w_{1}$ , and if $u\in H_{0}^{1,2}(\Omega)$ is any weak solution to the equation $Lu=\lambda_{1}u$, then $u=k\cdot w_{1}$ a.e. for some $k\in\mathbb{R}$.

Remark 12.9. This theorem states in particular that the eigenvalue $\lambda_{1}$ is simple so that the eigenvalues of $L$ satisfy the inequality

Remark 12.10. (*) is known as Rayleigh's formula and may be written as

If in particular $L=-\Delta$, then the lowest eigenvalue is given by

cf. Exercise 4 on tutorial sheet 3.

Proof of Theorem 12.8. Suppose $u\in H_{0}^{1,2}(\Omega)$ with $||u||_{2}=1$. Since by Corollary 12.6, $u$ may be expanded as

$u=\sum_{j=1}^{\infty}\left<w_{j},u\right>w_{j}$

in $H_{0}^{1,2}(\Omega)$, we see that

$B_{L}(u,u) = \sum_{j,k=1}^{\infty}\left<w_{j},u\right>\cdot \left<w_{k},u\right>\cdot B_{L}(w_{j},w_{k}) = \sum_{j=1}^{\infty}\lambda_{j}\cdot |\left<w_{j},u\right>|^{2}\geq \sum_{j=1}^{\infty}\lambda_{1}\cdot |\left<w_{j},u\right>|^{2} = \lambda_{1}||u||_{2}^{2}=\lambda_{1},$

with equality iff $\left<w_{j},u\right>=0$ for all $j>1$, i.e. iff $u=w_{1}$. This establishes the first claim. For the second claim, we shall show that if $u\in H_{0}^{1,2}(\Omega)$ is a nontrivial weak solution to $Lu=\lambda_{1}u$, then it must be either positive or negative on $\Omega$. To this end, write $u_{+}(x):=\max\{u(x),0\}$ and $u_{-}(x):=\max\{-u(x),0\}$ so that $u$ may be written as

$u=u_{+}-u_{-},$

and we have that $u_{+}\cdot u_{-}=0$ so that $||u||^{2}= ||u_{+}||^{2}+ ||u_{-}||^{2}$. Moreover, it may be shown that $u_{\pm}\in H_{0}^{1,2}(\Omega)$ and

$\D u_{\pm}(x)=\begin{cases}\pm \D u(x),& x\in \{\pm u \geq 0\}\\ 0,& x\in \{\pm u\leq 0\} \end{cases}$

a.e. so that $\partial_{i}u_{+}\cdot \partial_{j}u_{-}\equiv 0$ a.e., which in turn implies that $B_{L}(u_{+},u_{-})=0$ so that $B_{L}(u,u) = B_{L}(u_{+},u_{+}) + B_{L}(u_{-},u_{-})$. Now, by the proof of the first part, $B_{L}(u_{\pm},u_{\pm})\geq \lambda_{1}||u_{\pm}||_{2}^{2}$. It turns out that equality holds in both cases, for if we had a strict inequality, then

$\lambda_{1}||u||_{2}^{2} = B_{L}(u,u) = B_{L}(u_{+},u_+) + B_{L}(u_-,u_-)> \lambda_{1}\left(||u_{+}||_{2}^{2} + ||u_{-}||_{2}^{2}\right) = \lambda_{1}||u||_{2}^{2},$

which is impossible. Therefore, by the first part, $u_{+}\in H_{0}^{1,2}(\Omega)$ is a weak solution to $Lu = \lambda_{1}u$. By regularity theory, $u_{+}\in C^{\infty}(\overline\Omega)$, and since

$Lu_{+}=\lambda_{1} u_{+}\geq 0,$

$u_{+}$ is a supersolution to the equation $Lu=0$ so that by the strong maximum principle, since $\left.u_{+}\right|_{\partial\Omega}\equiv 0$ and $u_{+}\geq 0$, either $u_{+}\equiv 0$ on $\Omega$ or $u_{+}>0$, in which case $u_{-}\equiv 0$. A similar argument applies to $u_{-}$ as well, thus establishing the second claim. To establish the last claim, let $u,v\in H_{0}^{1,2}(\Omega)$ be any nontrivial solutions to $Lu=\lambda_{1}u$. Then by the preceding part, both $\int_{\Omega}u$ and $\int_{\Omega}v$ are nonzero so that we may find an $\alpha\in \mathbb{R}$ such that

$\int_{\Omega} u - \alpha v=0.$

But since $u-\alpha v$ is a solution to the same equation, the preceding part now implies that $u-\alpha v\equiv 0$.

All of the above considerations were for the symmetric elliptic operator

More generally, consider the more general nondivergence form elliptic operator

with $a_{ij}, b^{i},c\in C^{\infty}(\overline\Omega)$ for all $i,j\in\{1,\dots,n\}$ and $c\geq 0$. This operator is not symmetric in general; therefore, its eigenvalues will not necessarily be real. However, we can still say something about its principal eigenvalue, which is both real and positive.

Theorem 12.11. There exists a $\lambda_{1}>0$ such that

• $\lambda_{1}$ is an eigenvalue of $L$: There exists a $u\in H_{0}^{1,2}(\Omega)\backslash\{0\}$ such that $Lu=\lambda_{1}u$ weakly.

• $\lambda_{1}$ is simple: If $v\in H_{0}^{1,2}(\Omega)$ is any weak solution to $Lv=\lambda_{1}v$, then $v=\alpha\cdot u$ for some $\alpha\in\mathbb{R}$.

• If $\lambda\in\mathbb{C}$ is any other eigenvalue of $L$, then $\Re\lambda\geq\lambda_{1}$.