On the first eigenvalue of Laplace operator

[gugo82]

Exercise:

Let [tex]$a,b\in ]0,+\infty[$[/tex] and [tex]$\Omega (a,b):=]0,a[\times ]0,b[$[/tex].

1. Find the eigenvalues of the Laplace operator with homogeneous Dirichlet boundary conditions in [tex]$\Omega (a,b)$[/tex], i.e. the numbers [tex]$\lambda$[/tex] s.t. problem:

(*) [tex]$\begin{cases} -\Delta u=\lambda u &\text{, in } \Omega (a,b) \\ u=0 &\text{, on } \partial \Omega (a,b)\end{cases}$[/tex]

has some nontrivial solution.

2. Let [tex]$|E|$[/tex] denote the area of a measurable set [tex]$E\subseteq \mathbb{R}^2$[/tex] and set:

[tex]$\lambda_1(a,b):=\min \{ \lambda >0:\ \text{problem (*) has some nontrivial solution}\}$[/tex];

[tex]$\lambda_1(a,b)$[/tex] is called the first eigenvalue of the Laplace operator with Dirichlet homogeneous boundary conditions in [tex]$\Omega (a,b)$[/tex].
Prove that:

[tex]$\lambda_1 (l,l)=\min \{ \lambda_1 (a,b),\ |\Omega (a,b)|=l^2\}$[/tex],

that is, among all rectangles having area [tex]$l^2$[/tex], the square has the least possible first eigenvalue.

Hints: 1. Separate variables.
2. Keep in mind that [tex]$|\Omega (a,b)|=ab$[/tex] and impose the area constraint, solving for [tex]$b$[/tex]; let [tex]$b=b(a)$[/tex] in [tex]$\lambda_1(a,b)$[/tex] and minimize [tex]$\lambda_1(a):=\lambda_1(a,b(a))$[/tex] as a function of a single variable.

***

The result in 2 has a simple physical meaning: among all rectangular drums of given area, the square drum has the lowest principal tone.

[gugo82]

"gugo82":3. Compare the value [tex]$\lambda_1(l,l)$[/tex] with the first Dirichlet Laplacian's eigenvalue in the open disc having area [tex]$l^2$[/tex], i.e. with the number:

[tex]$\lambda_1^\star (l) := \min \{ \lambda >0:\ \text{problem (**) has nontrivial solutions}\}$[/tex]

where:

(**) [tex]$\begin{cases} -\Delta u=\lambda u &\text{, in } D(l) \\ u=0 &\text{, on } \partial D(l)\end{cases}$[/tex]

([tex]$D(l)$[/tex] is the open disc centered in the origin with [tex]$|D(l)|=l^2$[/tex]).

Here's the solution.

First of all, let us note that the radius of \(D(l)\) is exactly \(l/\sqrt{\pi}\).

Step 1. Now we turn to problem (**), looking for its eigenvalues and eigenfunctions.
Because \(D(l)\) is radially symmetric, it is convenient to look for solutions of (**) in polar coordinates.
So, let \(u(x,y)=u(r\cos \theta,r\sin \theta)=v(r,\theta)\) be a candidate solution; applying the chain rule we find:
\[\begin{align} v_r &= u_x \cos \theta +u_y \sin \theta \\ v_\theta &= -u_x r \sin \theta +u_y r \cos \theta \\
v_{rr} &= u_{xx} \sin^2 \theta +2u_{x,y} \sin \theta \cos \theta +u_{yy} \sin^2 \theta \\ v_{\theta \theta} &= r^2 (u_{xx} \sin^2 \theta -2u_{x,y} \sin \theta \cos \theta +u_{yy} \sin^2 \theta ) -r(u_x \cos \theta +u_y\sin \theta)\end{align}\]
therefore:
\[\Delta u(x,y) =v_{rr}(r,\theta)+\frac{1}{r}\ v_r (r,\theta)+\frac{1}{r^2}v_{\theta \theta} (r,\theta)\]
and the PDE in (**) transforms into:
\[r^2 v_{rr}(r,\theta)+r\ v_r (r,\theta)+v_{\theta \theta} (r,\theta) =\lambda r^2 v(r,\theta) \; ,\]
so that problem (**) rewrites:
\[\tag{***} \begin{cases} r^2 v_{rr}(r,\theta)+r\ v_r (r,\theta)+v_{\theta \theta} (r,\theta) =\lambda r^2 v(r,\theta) &\text{, in } ]0,l/\sqrt{\pi}[\times ]0,2\pi[ \\ v(r,0)=v(r,2\pi) &\text{, for } 0\leq r\leq l/\sqrt{\pi} \\ v(l/\sqrt{\pi},\theta) =0 &\text{, for } 0\leq \theta \leq 2\pi \; .\end{cases}\]
PDE in (***) separates variables, hence we can look for solutions in the form \(v(r,\theta)=R(r) \Theta (\theta)\); such a function solves the PDE iff there exists a constant \(\omega\) (which is usually called separation constant) such that:
\[\frac{r^2}{R}R^{\prime \prime} +\frac{r}{R} R^\prime +\lambda r^2 =\omega=\frac{1}{\Theta} \Theta^{\prime \prime}\]
that is iff \(R\) and \(\Theta\) solve:
\[\begin{align} r^2R^{\prime \prime} +rR^\prime +(\lambda r^2-\omega)R&=0 & \Theta^{\prime \prime} +\omega\ \Theta&=0 \; .\end{align}\]
Moreover, \(R(r)\Theta (\theta)\) solves (***) iff \(R\) and \(\Theta\) satisfy the following conditions:
\[\begin{align} &\begin{cases} R \text{ is continuous in } [0,l/\sqrt{\pi}] \\ R(l/\sqrt{\pi})=0 \end{cases} &, & \Theta(0)=\Theta(2\pi)\; ;\end{align}\]
therefore, problem (**) is transformed into a couple of BV problems for two ODEs, namely:
\[\tag{I} \begin{cases} \Theta^{\prime \prime} +\omega\ \Theta =0 &\text{, in } ]0,2\pi[ \\ \Theta (0)=\Theta (2\pi)\end{cases}\]
and:
\[\tag{II} \begin{cases} r^2\ R^{\prime \prime} +r\ R^\prime +(\lambda\ r^2-\omega)R =0 &\text{, in } ]0,l/\sqrt{2\pi}[ \\ R \text{ is continuous} &\text{, in } [0,l/\sqrt{\pi}] \\ R(l/\sqrt{\pi})=0\; . \end{cases}\]

Problem (I) can be easily solved: in fact, (I) has nontrivial solution iff \(\omega =n^2\) for some \(n\in \mathbb{N}\) and its general solution corresponding to \(\omega =n^2\) equals:
\[\Theta (\theta) =A_n \cos n\theta +B_n \sin n\theta\]
where \(A_n,B_n\) are arbitrary real constants.
Now we turn to (II). If we set \(\omega =n^2\) in the ODE, we find the Bessel-type equation:
\[r^2\ R^{\prime \prime} +r\ R^\prime +(\lambda r^2-n^2)R =0\]
which can be put into canonical form:
\[\rho^2\ \ddot{R}+\rho\ \dot{R}+(\rho^2-n^2)R=0\]
with the substitution \(\rho=\sqrt{\lambda}\ r\) (here the dot denotes derivative w.r.t. \(\rho\)); thus it becomes clear that the ODE has solution:
\[R(r)=C_n\ \text{J}_n(\sqrt{\lambda}\ r) +D_n\ \text{Y}_n (\sqrt{\lambda}\ r)\]
where \(C_n,D_n\) are arbitrary constants and \(\text{J}_n,\ \text{Y}_n\) are Bessel functions of order \(n\) of the first and second kind.
Such a function fulfill the continuity constraint in (II) iff \(D_n=0\) (for \(\text{Y}_n\) diverges when the argument approaches zero), and it satisfies \(R(l/\sqrt{\pi})=0\) iff \(\lambda\) is chosen among the solutions of equation \(J_n(\sqrt{\lambda}\ l/\sqrt{\pi})=0\). From the general theory of Bessel functions it follows that \(\lambda\) has to satisfy \( \sqrt{\lambda}\ l/\sqrt{\pi}=j_{n,k}\), where \(j_{n,k}>0\) is the \(k\)-th nontrivial zero of \(\text{J}_n\), therefore we have:
\[R(r)=C_k\ \text{J}_n (\sqrt{\lambda_{n,k}}\ r)\]
with \(\lambda_{n,k}=\pi\ j_{n,k}^2/l^2\).

Finally, we can return to problem (**). Summing up what we wrote, the eigenvalues of (**) are:
\[\lambda_{n,k} = \pi\ j_{n,k}^2/l^2\]
(with \(n=0,1,2,\ldots \) and \(k=1,2,\ldots\))) and any eigenfunction \(u_{n,k} (x,y)\) corresponding to \(\lambda_{n,k}\) is given in polar coordinates by:
\[u_{n,k} (r,\theta) =(A_{n,k}\ \cos n\theta +B_{n,k}\ \sin n\theta )\ \text{J}_n \left( \frac{\sqrt{\pi}\ j_{n,k}}{l}\ r\right) \; .\]

Step 2. Now we determine the first eigenvalue \(\lambda_1^\star (l)\).
The following classical results about eigenfunctions corresponding to the first eigenvalue of the Dirichelet Laplacian will be useful:

Let \(\Omega \subset \mathbb{R}^N\) be a sufficiently smooth nonempty open set.
Then all the eigenfunctions of the homogeneous Dirichlet Laplacian in \(\Omega\) are of class \(C^\infty\).

Let \(B\subset \mathbb{R}^N\) be an open ball and \(u\) an eigenfunction corresponding to the first eigenvalue of the homogeneous Dirichlet Laplacian in \(B\).
If \(u\) is of class \(C^2\), then \(u\) is a radial function (in other words, \(u(x,y)=v(r)\) with \(r=\sqrt{x^2+y^2}\)).

[For a proof, which make use of the moving plane method (a geometric method based on the maximum principle), the reader may refer to § 9.5.2 in Evans, Partial Differential Equations - Second Edition, GSM-19, AMS, Providence.]

Let \(u_{n,k}\) be an eigenfunction corresponding to \(\lambda_1^\star (l)\); the angular part \(A_n \cos n\theta +B_n\sin n\theta\) of \(u_{n,k}\) must reduce to a constant (for \(u_{n,k}\) has to be radial), therefore \(n\) has to be zero and:
\[
u_{0,k}(r,\theta) =A_{0,k}\ \text{J}_0 (\sqrt{\pi}\ j_{0,k}/l\ r)\; .
\]
Thus the first eigenvalue we are looking for is the least element of the set \(\{\pi\ j_{0,k}^2/l^2\}_{k=1,2,\ldots}\), which is the one correspondig to \(k=1\), i.e.:
\[\lambda_1^\star (l)=\frac{\pi}{l^2}\ j_{0,1}^2 \approx 5.783\ \frac{\pi}{l^2} \; .\]

Step 3. Finally, we can compare \(\lambda_1(l,l)\) with \(\lambda_1^\star (l)\).
Remembering that \(\lambda_1(l,l)=2 \pi^2/l^2\), we easily find \(\lambda_1^\star (l)<\lambda_1(l,l)\).

[gugo82]

I provide the solution to the second question.

Using Camillo's argument, we find:

[tex]$\lambda_1 (a,b)=\pi^2 \left( \frac{1}{a^2}+\frac{1}{b^2}\right)$[/tex],

thus in order to answer question 2 we have to solve the following minimization problem with constraints:

(A) [tex]$\begin{cases} \text{minimize} & \lambda_1(a,b) \\ \text{under constraint} &ab=l^2\end{cases}$[/tex].

There are many way to solve this problem.
One, the most elementary perhaps, consists in plugging [tex]$b=\tfrac{l^2}{a}$[/tex] into [tex]$\lambda_1(a,b)$[/tex] so that problem (A) reduces to a free minimization problem for a function of the single positive real variable [tex]$a$[/tex].

On the other hand, the problem (A) can easily be solved using the properties of the [tex]$p$[/tex]-means (see here): in fact, problem (A) rewrites:

[tex]$\begin{cases} \text{minimize} & \frac{2\pi^2}{M_{-2}^2(a,b)} \\ \text{under constraint} &G(a,b)=l\end{cases}$[/tex],

where:

[tex]$M_{-2}(a,b) := \left\{ \frac{1}{2} (a^{-2}+b^{-2})\right\}^{-\frac{1}{2}} = \frac{\sqrt{2}}{\sqrt{\frac{1}{a^2} +\frac{1}{b^2}}}$[/tex],

for [tex]$\lambda_1 (a,b)=2\pi^2 M_{-2}^{-2}(a,b)$[/tex]; thus solving problem (A) is equivalent to solve:

(B) [tex]$\begin{cases} \text{maximize} & M_{-2}(a,b) \\ \text{under constraint} &G(a,b)=l\end{cases}$[/tex].

By (pMqM) we know that [tex]$M_{-2}(a,b)\leq G(a,b)$[/tex] and that there is equality iff [tex]$a=b$[/tex], hence problem (B) is solved uniquely by [tex]$a=b=l$[/tex] with [tex]$M_{-2}(l,l)=l$[/tex].
Therefore this fact implies that:

[tex]$\frac{2\pi^2}{l^2} = \lambda_1(l,l) =\min \left\{ \lambda_1(a,b), |\Omega (a,b)|=l\right\}$[/tex],

so the square minimizes the first Dirichlet Laplacian's eigenvalue among all rectangles sharing the same measure.

***

3. Compare the value [tex]$\lambda_1(l,l)$[/tex] with the first Dirichlet Laplacian's eigenvalue in the open disc having area [tex]$l^2$[/tex], i.e. with the number:

[tex]$\lambda_1^\star (l) := \min \{ \lambda >0:\ \text{problem (**) has nontrivial solutions}\}$[/tex]

where:

(**) [tex]$\begin{cases} -\Delta u=\lambda u &\text{, in } D(l) \\ u=0 &\text{, on } \partial D(l)\end{cases}$[/tex]

([tex]$D(l)$[/tex] is the open disc centered in the origin with [tex]$|D(l)|=l^2$[/tex]).

***

For the ones interested in this subject.
There is a vast literature on the eigenvalues of the Laplacian. Good readings on various problems concerning the minimization of the eigenvalue of Laplace operator are this survey by Antoine Henrot and this survey by Rafael Benguria and Helmut Linde.

[gugo82]

@Camillo: There is a [tex]$+$[/tex] sign missing in your [tex]$\lambda_{n,m}(a,b)$[/tex]...

I mean, if you take [tex]$a=b$[/tex] ([tex]$\Omega (a,b)$[/tex] is a square) and [tex]$m=n=1$[/tex] (the first eigenvalue), then your general result yields [tex]$\lambda_{1,1}(a,a)=0$[/tex]; but this is impossible, for problem:

[tex]$\begin{cases} -\Delta u=0 &\text{, in } \Omega (a,a) \\ u=0 &\text{, on } \partial \Omega (a,a) \end{cases}$[/tex]

has only the trivial solution (in fact, there is only one harmonic function in [tex]$\Omega (a,a)$[/tex] which is zero on [tex]$\partial \Omega (a,a)$[/tex], i.e. the null function).

[Camillo]

1) Tentative solution

The problem is : $- u_(x x) -u_(yy)- lambda u(x,y)=0 $ with boundary conditions :
$u(x,0)=0 $ for $ 0<=x <=a $
$u(x,b) =0 $ for $ 0<=x <= a $
$u(0,y) =0 $ for $ 0<=y<=b $
$u(0,y) =0 $ for $ 0<=y<= b $

Considering the method of variable separation we write the solution in this way $ u(x,y)= X(x)Y(y)$ and the equation becomes

$-X''(x)Y(y) -X(x)Y''(y)= lambda X(x)Y(y) $.

Consequently we get :
$-(X''(x))/(X(x)) -(Y''(y))/(Y(y)) = lambda $

I put $ mu =-(X''(x))/(X(x)) =lambda +(Y''(y))/(Y(y)) $

Now I call $alpha = mu- lambda $ and we get two eigenvalues problems :

$X''(x) + mu X(x) =0 $ for $ 0< x< a ; X(0)=X(a ) =0 $.
$Y''(y) +alpha Y(y) =0 $ for $ 0
It is known that the relevant eigenfunctions and eigenvalues are :

$X_m(x)=A_m *sin(m*pi*x/a); mu_m=m^2 pi^2/a^2 , m=1,2,... $

$Y_n(y)= B_n sin( n*pi*y/b) ; alpha_n= m^2*pi^2/b^2 ; n= 1,2 ... $




Since $ lambda = mu-alpha $ the eigenvalues for the problem are :

$lambda_(m,n) = pi^2(m^2/a^2 -n^2/b^2) ; m,n = 1.2...$

and the corresponding eigenfunctions are $u_(m,n)= C _(m.n ) sin( m*pi*x/a)*sin( n*pi*y/b) m.n=1,2... $
The solution is $u(x,y) = sum _(m,n=1)^(oo) C_(m.n)sin( m*pi*x/a)*sin(n*pi*y/b) $ which is zero on the borders of the rectangle .