When I first got hold of a copy of Erich Kamke's classic book in 2002 I did not realise it would become the inspiration for two of my papers which to day I consider the best. One of them was an essay on lemniscate functions extracts from which I published in the first post here.

The other one was my dissertation for the MSc degree at the Open University. While working on the dissertation I had to remain within a rather restrictive framework of the general topic of variational methods in eigenvalue problems, with the emphasis on Ritz method and asymptotic properties of the eigenvalues.

Faced with a task of producing material which is both rigorous and original, in the area that is very well explored, and meeting the tight deadline, I decided to work through some of the less quoted results (not easily googled) and try to give an original account of them. Two of such results were given in Kamke's book without proof. In this post I discuss one of them.

The main focus of the dissertation was set on the asymptotic estimates for large eigenvalues. This topic was once brought to attention by the famous question of Mark Kac whether one can hear the shape of a drum. The estimates discussed in the course guide were all of the one-sided form, derived using the Rayleigh quotient. Kamke gives some examples of the two-sided estimates which I decided to put at the core of my project.

In particular, there is a theorem due to N. Krylov and N. Bogolyubov, published in 1929, which gives a two-sided estimate of the eigenvalues. In the Russian translation of Kamke's text a reference is made to a report for the Academy of Sciences of USSR which I was unable to get hold of. Other resources were not of much help either, until I found the book written by Lothar Kollatz. It gives the proof, however, the notational framework employed there seemed a bit obscure to me and was not compatible with the more modern setting I chose for my report. I decided to rewrite the proof.

Before I proceed I have to fix some definitions and cite some general results.

Let \(\Omega\) be a domain with a piecewise smooth boundary \(\partial\Omega\) and function \(\phi\in L_{2}\). Consider the following functional \(S:L_{2}\to\mathbb{R}\).

$$S\left\{ \phi\right\} =\int_{\Omega}\left(p\left|\nabla\phi\right|^{2}+q\phi^{2}\right)d\boldsymbol{x}+\int_{\partial\Omega}p\sigma\phi^{2}d\boldsymbol{s}$$

Where \(p>0\) and \(p\in C^{1}\left(\Omega\right)\), \(q\in C\left(\Omega\right)\). Now define the operator

$$L\left[u\right]=\frac{1}{\rho}\left(-\nabla\cdot\left(p\nabla u\right)+qu\right)$$

where \(\rho\in C^{1}\left(\Omega\right)\), \(\rho=\rho\left(\boldsymbol{x}\right)>0\) and demand that admissible functions satisfy the Robin boundary conditions

$$\left.\left(\frac{\partial u}{\partial n}+\sigma u\right)\right|_{\partial\Omega}=0$$

Define the scalar product \(L_{2}\times L_{2}\to\mathbb{C}\) as follows

$$\left\langle f,g\right\rangle =\int_{\Omega}\rho f\left(\boldsymbol{x}\right)g^{*}\left(\boldsymbol{x}\right)d\boldsymbol{x}$$

Denote the set of admissible functions

$$\mathfrak{A}=\left\{ \left. u \in L_{2} \right| \left. \left(\frac{\partial u}{\partial n}+\sigma u\right) \right|_{\partial\Omega}=0\right\}$$

and the subset set of normed admissible functions

$$\mathfrak{N}_{1}=\left\{ \left. u\in A\right| \int_{\Omega}\rho u^{2}d\boldsymbol{x}=1\right\}$$

It can be shown that the problem of minimization of the functional \(S\left\{\phi\right\} =\left\langle \phi,L\left[\phi\right]\right\rangle \to\min,\phi\in\mathfrak{N}_{1}\) is equivalent to the eigenvalue problem for the differential equation

$$L\left[u\right]=\lambda u$$

with boundary conditions

$$\left.\left(\frac{\partial u}{\partial n}+\sigma u\right)\right|_{\partial\Omega}=0$$

It can also be shown that \(L\) is self adjoint

$$\left\langle v,L\left[u\right]\right\rangle =\left\langle L\left[v\right],u\right\rangle$$

and that there exists a sequence an infinite and complete sequence of orthogonal functions (eigenfunctions) which satisfy boundary conditions and give rise to the corresponding sequence of the eigenvalues.

We can now state the theorem in question.

which does not vanish identically consider the following quantities

$$\alpha=\frac{\left\langle \phi,L\left[\phi\right]\right\rangle }{\left\langle \phi,\phi\right\rangle }\qquad\beta^{2}=\frac{\left\langle L\left[\phi\right],L\left[\phi\right]\right\rangle }{\left\langle \phi,\phi\right\rangle }$$

Then \(\beta^{2}\ge\alpha^{2}\) and in the interval between

$$\alpha-\sqrt{\beta^{2}-\alpha^{2}}\quad\text{and}\quad\alpha+\sqrt{\beta^{2}-\alpha^{2}}$$ there lies at least one eigenvalue.

First we prove that \(\beta^{2}\ge\alpha^{2}\) using Cauchy-Schwarz inequality

$$\alpha^{2}=\left(\frac{\left\langle \phi,L\left[\phi\right]\right\rangle }{\left\langle \phi,\phi\right\rangle }\right)^{2}\le\frac{\left\langle L\left[\phi\right],L\left[\phi\right]\right\rangle \left\langle \phi,\phi\right\rangle }{\left\langle \phi,\phi\right\rangle ^{2}}=\frac{\left\langle L\left[\phi\right],L\left[\phi\right]\right\rangle }{\left\langle \phi,\phi\right\rangle }=\beta^{2}$$

Now consider the following expression

$$\begin{aligned}T\left[\phi\right] & =\frac{A\left[\phi\right]}{B\left[\phi\right]}=\frac{\left\langle \alpha\phi-L\left[\phi\right],\alpha\phi-L\left[\phi\right]\right\rangle }{\left\langle \phi,\phi\right\rangle }\\

& =\frac{\alpha^{2}\left\langle \phi,\phi\right\rangle -2\alpha\left\langle \phi,L\left[\phi\right]\right\rangle +\left\langle L\left[\phi\right],L\left[\phi\right]\right\rangle }{\left\langle \phi,\phi\right\rangle }\\

& =\alpha^{2}-2\alpha^{2}+\beta^{2}=\beta^{2}-\alpha^{2}

\end{aligned}$$

Denote

$$\psi=\alpha\phi-L\left[\phi\right]$$

Hence, by Bessel's inequality we obtain

$$A\left[\phi\right]=\left\langle \psi,\psi\right\rangle \ge\sum_{i=1}^{\infty}\left\langle \psi,\phi_{i}\right\rangle ^{2}$$

where \(\phi_{i}\) are the eigenfunctions of \(L\). Evaluating the coefficients

$$\begin{aligned}\left\langle \psi,\phi_{i}\right\rangle & =\left\langle \alpha\phi-L\left[\phi\right],\phi_{i}\right\rangle =\alpha\left\langle \phi,\phi_{i}\right\rangle -\left\langle L\left[\phi\right],\phi_{i}\right\rangle \\

& =\alpha\left\langle \phi,\phi_{i}\right\rangle -\left\langle \phi,L\left[\phi_{i}\right]\right\rangle =\left(\alpha-\lambda_{i}\right)\left\langle \phi,\phi_{i}\right\rangle

\end{aligned}$$

Therefore

$$A\left[\phi\right]\ge\left(\alpha-\lambda_{i}\right)^{2}\left\langle \phi,\phi_{i}\right\rangle ^{2}$$

Since \(\phi\) is an admissible function, by completeness theorem we obtain

$$\begin{aligned}B\left[\phi\right] & =\left\langle \phi,\phi\right\rangle =\sum_{i=1}^{\infty}\left\langle \phi,\phi_{i}\right\rangle ^{2}\end{aligned}$$

Recalling the expression for \(T\left[\phi\right]\) we obtain

$$T\left[\phi\right]=\beta^{2}-\alpha^{2}\ge\frac{\sum_{i=1}^{\infty}\left(\alpha-\lambda_{i}\right)^{2}\left\langle \phi,\phi_{i}\right\rangle ^{2}}{\sum_{i=1}^{\infty}\left\langle \phi,\phi_{i}\right\rangle ^{2}}$$

If \(\lambda_{n}\) is the nearest or one of the equidistant eigenvalues to \(\alpha\), then for all \(i\)

$$\left(\alpha-\lambda_{i}\right)^{2}\ge\left(\alpha-\lambda_{n}\right)^{2}$$

$$\frac{\sum_{i=1}^{\infty}\left(\alpha-\lambda_{i}\right)^{2}\left\langle \phi,\phi_{i}\right\rangle ^{2}}{\sum_{i=1}^{\infty}\left\langle \phi,\phi_{i}\right\rangle ^{2}}\ge\frac{\sum_{i=1}^{\infty}\left(\alpha-\lambda_{n}\right)^{2}\left\langle \phi,\phi_{i}\right\rangle ^{2}}{\sum_{i=1}^{\infty}\left\langle \phi,\phi_{i}\right\rangle ^{2}}=\left(\alpha-\lambda_{n}\right)^{2}$$

Hence

$$\left(\alpha-\lambda_{n}\right)^{2}\le\beta^{2}-\alpha^{2}$$

$$ \left|\alpha-\lambda_{n}\right|\le\sqrt{\beta^{2}-\alpha^{2}}$$

$$ \alpha-\sqrt{\beta^{2}-\alpha^{2}}\le\lambda_{n}\le\alpha+\sqrt{\beta^{2}-\alpha^{2}}$$

which completes the proof.

The other one was my dissertation for the MSc degree at the Open University. While working on the dissertation I had to remain within a rather restrictive framework of the general topic of variational methods in eigenvalue problems, with the emphasis on Ritz method and asymptotic properties of the eigenvalues.

Faced with a task of producing material which is both rigorous and original, in the area that is very well explored, and meeting the tight deadline, I decided to work through some of the less quoted results (not easily googled) and try to give an original account of them. Two of such results were given in Kamke's book without proof. In this post I discuss one of them.

The main focus of the dissertation was set on the asymptotic estimates for large eigenvalues. This topic was once brought to attention by the famous question of Mark Kac whether one can hear the shape of a drum. The estimates discussed in the course guide were all of the one-sided form, derived using the Rayleigh quotient. Kamke gives some examples of the two-sided estimates which I decided to put at the core of my project.

In particular, there is a theorem due to N. Krylov and N. Bogolyubov, published in 1929, which gives a two-sided estimate of the eigenvalues. In the Russian translation of Kamke's text a reference is made to a report for the Academy of Sciences of USSR which I was unable to get hold of. Other resources were not of much help either, until I found the book written by Lothar Kollatz. It gives the proof, however, the notational framework employed there seemed a bit obscure to me and was not compatible with the more modern setting I chose for my report. I decided to rewrite the proof.

Before I proceed I have to fix some definitions and cite some general results.

Let \(\Omega\) be a domain with a piecewise smooth boundary \(\partial\Omega\) and function \(\phi\in L_{2}\). Consider the following functional \(S:L_{2}\to\mathbb{R}\).

$$S\left\{ \phi\right\} =\int_{\Omega}\left(p\left|\nabla\phi\right|^{2}+q\phi^{2}\right)d\boldsymbol{x}+\int_{\partial\Omega}p\sigma\phi^{2}d\boldsymbol{s}$$

Where \(p>0\) and \(p\in C^{1}\left(\Omega\right)\), \(q\in C\left(\Omega\right)\). Now define the operator

$$L\left[u\right]=\frac{1}{\rho}\left(-\nabla\cdot\left(p\nabla u\right)+qu\right)$$

where \(\rho\in C^{1}\left(\Omega\right)\), \(\rho=\rho\left(\boldsymbol{x}\right)>0\) and demand that admissible functions satisfy the Robin boundary conditions

$$\left.\left(\frac{\partial u}{\partial n}+\sigma u\right)\right|_{\partial\Omega}=0$$

Define the scalar product \(L_{2}\times L_{2}\to\mathbb{C}\) as follows

$$\left\langle f,g\right\rangle =\int_{\Omega}\rho f\left(\boldsymbol{x}\right)g^{*}\left(\boldsymbol{x}\right)d\boldsymbol{x}$$

Denote the set of admissible functions

$$\mathfrak{A}=\left\{ \left. u \in L_{2} \right| \left. \left(\frac{\partial u}{\partial n}+\sigma u\right) \right|_{\partial\Omega}=0\right\}$$

and the subset set of normed admissible functions

$$\mathfrak{N}_{1}=\left\{ \left. u\in A\right| \int_{\Omega}\rho u^{2}d\boldsymbol{x}=1\right\}$$

It can be shown that the problem of minimization of the functional \(S\left\{\phi\right\} =\left\langle \phi,L\left[\phi\right]\right\rangle \to\min,\phi\in\mathfrak{N}_{1}\) is equivalent to the eigenvalue problem for the differential equation

$$L\left[u\right]=\lambda u$$

with boundary conditions

$$\left.\left(\frac{\partial u}{\partial n}+\sigma u\right)\right|_{\partial\Omega}=0$$

It can also be shown that \(L\) is self adjoint

$$\left\langle v,L\left[u\right]\right\rangle =\left\langle L\left[v\right],u\right\rangle$$

and that there exists a sequence an infinite and complete sequence of orthogonal functions (eigenfunctions) which satisfy boundary conditions and give rise to the corresponding sequence of the eigenvalues.

We can now state the theorem in question.

**Theorem**. For an arbitrary admissible function \(\phi\)which does not vanish identically consider the following quantities

$$\alpha=\frac{\left\langle \phi,L\left[\phi\right]\right\rangle }{\left\langle \phi,\phi\right\rangle }\qquad\beta^{2}=\frac{\left\langle L\left[\phi\right],L\left[\phi\right]\right\rangle }{\left\langle \phi,\phi\right\rangle }$$

Then \(\beta^{2}\ge\alpha^{2}\) and in the interval between

$$\alpha-\sqrt{\beta^{2}-\alpha^{2}}\quad\text{and}\quad\alpha+\sqrt{\beta^{2}-\alpha^{2}}$$ there lies at least one eigenvalue.

**Proof.**First we prove that \(\beta^{2}\ge\alpha^{2}\) using Cauchy-Schwarz inequality

$$\alpha^{2}=\left(\frac{\left\langle \phi,L\left[\phi\right]\right\rangle }{\left\langle \phi,\phi\right\rangle }\right)^{2}\le\frac{\left\langle L\left[\phi\right],L\left[\phi\right]\right\rangle \left\langle \phi,\phi\right\rangle }{\left\langle \phi,\phi\right\rangle ^{2}}=\frac{\left\langle L\left[\phi\right],L\left[\phi\right]\right\rangle }{\left\langle \phi,\phi\right\rangle }=\beta^{2}$$

Now consider the following expression

$$\begin{aligned}T\left[\phi\right] & =\frac{A\left[\phi\right]}{B\left[\phi\right]}=\frac{\left\langle \alpha\phi-L\left[\phi\right],\alpha\phi-L\left[\phi\right]\right\rangle }{\left\langle \phi,\phi\right\rangle }\\

& =\frac{\alpha^{2}\left\langle \phi,\phi\right\rangle -2\alpha\left\langle \phi,L\left[\phi\right]\right\rangle +\left\langle L\left[\phi\right],L\left[\phi\right]\right\rangle }{\left\langle \phi,\phi\right\rangle }\\

& =\alpha^{2}-2\alpha^{2}+\beta^{2}=\beta^{2}-\alpha^{2}

\end{aligned}$$

Denote

$$\psi=\alpha\phi-L\left[\phi\right]$$

Hence, by Bessel's inequality we obtain

$$A\left[\phi\right]=\left\langle \psi,\psi\right\rangle \ge\sum_{i=1}^{\infty}\left\langle \psi,\phi_{i}\right\rangle ^{2}$$

where \(\phi_{i}\) are the eigenfunctions of \(L\). Evaluating the coefficients

$$\begin{aligned}\left\langle \psi,\phi_{i}\right\rangle & =\left\langle \alpha\phi-L\left[\phi\right],\phi_{i}\right\rangle =\alpha\left\langle \phi,\phi_{i}\right\rangle -\left\langle L\left[\phi\right],\phi_{i}\right\rangle \\

& =\alpha\left\langle \phi,\phi_{i}\right\rangle -\left\langle \phi,L\left[\phi_{i}\right]\right\rangle =\left(\alpha-\lambda_{i}\right)\left\langle \phi,\phi_{i}\right\rangle

\end{aligned}$$

Therefore

$$A\left[\phi\right]\ge\left(\alpha-\lambda_{i}\right)^{2}\left\langle \phi,\phi_{i}\right\rangle ^{2}$$

Since \(\phi\) is an admissible function, by completeness theorem we obtain

$$\begin{aligned}B\left[\phi\right] & =\left\langle \phi,\phi\right\rangle =\sum_{i=1}^{\infty}\left\langle \phi,\phi_{i}\right\rangle ^{2}\end{aligned}$$

Recalling the expression for \(T\left[\phi\right]\) we obtain

$$T\left[\phi\right]=\beta^{2}-\alpha^{2}\ge\frac{\sum_{i=1}^{\infty}\left(\alpha-\lambda_{i}\right)^{2}\left\langle \phi,\phi_{i}\right\rangle ^{2}}{\sum_{i=1}^{\infty}\left\langle \phi,\phi_{i}\right\rangle ^{2}}$$

If \(\lambda_{n}\) is the nearest or one of the equidistant eigenvalues to \(\alpha\), then for all \(i\)

$$\left(\alpha-\lambda_{i}\right)^{2}\ge\left(\alpha-\lambda_{n}\right)^{2}$$

$$\frac{\sum_{i=1}^{\infty}\left(\alpha-\lambda_{i}\right)^{2}\left\langle \phi,\phi_{i}\right\rangle ^{2}}{\sum_{i=1}^{\infty}\left\langle \phi,\phi_{i}\right\rangle ^{2}}\ge\frac{\sum_{i=1}^{\infty}\left(\alpha-\lambda_{n}\right)^{2}\left\langle \phi,\phi_{i}\right\rangle ^{2}}{\sum_{i=1}^{\infty}\left\langle \phi,\phi_{i}\right\rangle ^{2}}=\left(\alpha-\lambda_{n}\right)^{2}$$

Hence

$$\left(\alpha-\lambda_{n}\right)^{2}\le\beta^{2}-\alpha^{2}$$

$$ \left|\alpha-\lambda_{n}\right|\le\sqrt{\beta^{2}-\alpha^{2}}$$

$$ \alpha-\sqrt{\beta^{2}-\alpha^{2}}\le\lambda_{n}\le\alpha+\sqrt{\beta^{2}-\alpha^{2}}$$

which completes the proof.

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