In the previous post I talked about one of the less quoted results in the variational theory of Sturm-Liouville eigenvalue problems which I owe to the book on differential equations by Erich Kamke and which I reworked for the purposes of my dissertation.

The other result that I also found in Kamke's book states an alternative variational principle which can be used to obtain estimates of the eigenvalues. This method gives a weaker upper bound than the Rayleigh quotient, however it requires smaller computational effort. In return, one can widen the class of trial functions without severely complicating the calculations.

We start with the Rayleigh quotient written in the abstract form:

$$J\left\{ \phi\right\}=\frac{\left\langle \phi,L\left[\phi\right]\right\rangle }{\left\langle \phi,\phi\right\rangle }$$

Then using Cauchy-Schwarz inequality we obtain.

$$\left(\left\langle \phi,L\left[\phi\right]\right\rangle \right)^{2}\leq\left\langle \phi,\phi\right\rangle \left\langle L\left[\phi\right],L\left[\phi\right]\right\rangle

\frac{\left\langle \phi,L\left[\phi\right]\right\rangle }{\left\langle \phi,\phi\right\rangle }\leq\frac{\left\langle L\left[\phi\right],L\left[\phi\right]\right\rangle }{\left\langle \phi,L\left[\phi\right]\right\rangle }$$

Written explicitly

$$\frac{\int_{\Omega}\rho\phi L\left[\phi\right]d\boldsymbol{x}}{\int_{\Omega}\rho\phi^{2}d\boldsymbol{x}}\le\frac{\int_{\Omega}\rho\left(L\left[\phi\right]\right)^{2}d\boldsymbol{x}}{\int_{\Omega}\rho\phi L\left[\phi\right]d\boldsymbol{x}}$$

Hence we can replace the original problem with the following one

$$K\left\{ \phi\right\} =\frac{\left\langle L\left[\phi\right],L\left[\phi\right]\right\rangle }{\left\langle \phi,L\left[\phi\right]\right\rangle }\to\min,\qquad\phi\in\mathfrak{A}$$

Consider the eigenvalue problem

$$-y''=\lambda y,\quad y\left(0\right)=y\left(1\right)=0$$

Take the trial function

$$y_{1}=\frac{x}{12}-\frac{x^{3}}{6}+\frac{x^{4}}{12}$$

First we perform the calculation using the Rayleigh quotient

$$J\left\{ y_{1}\right\} =\frac{\left\langle y_{1},L\left[y_{1}\right]\right\rangle }{\left\langle y_{1},y_{1}\right\rangle }=-\frac{\int_{0}^{1}y_{1}y_{1}''dx}{\int_{0}^{1}y_{1}^{2}dx}$$

$$\begin{aligned}\int_{0}^{1}y_{1}y_{1}''dx & =\int_{0}^{1}\left[\left(\frac{x}{12}-\frac{x^{3}}{6}+\frac{x^{4}}{12}\right)\left(-x+x^{2}\right)\right]dx\\

& =\int_{0}^{1}\left(\frac{x^{6}}{12}-\frac{x^{5}}{4}+\frac{x^{4}}{6}+\frac{x^{3}}{12}-\frac{x^{2}}{12}\right)dx\\

& =-\frac{17}{5040}

\end{aligned}$$

The factor evaluated above will be common to both variational quotients

$$\begin{aligned}\int_{0}^{1}y_{1}^{2}dx & =\int_{0}^{1}\left(\frac{x^{8}}{144}-\frac{x^{7}}{36}+\frac{x^{6}}{36}+\frac{x^{5}}{72}+\frac{x^{2}}{144}\right)dx\\

& =\frac{31}{90720}

\end{aligned}$$

$$ \lambda_{1}\le J\left\{ y_{1}\right\} =\frac{17\cdot90720}{5040\cdot31}\approx9.871$$

Now we use the alternative variational principle

$$K\left\{ y_{1}\right\} =\frac{\left\langle L\left[y_{1}\right],L\left[y_{1}\right]\right\rangle }{\left\langle y_{1},L\left[y_{1}\right]\right\rangle }=-\frac{\int_{0}^{1}y_{1}''^{2}dx}{\int_{0}^{1}y_{1}y_{1}''dx}$$

Obviously, the announced gain in computational efficiency comes from replacing \(y^{2}\) with \(\left(L\left[y\right]\right)^{2}\) which will be a polynomial of the order less by 4, than \(y^{2}\).

$$\int_{0}^{1}y_{1}''^{2}dx=\int_{0}^{1}\left(x^{4}-2x^{3}+x^{2}\right)dx=\frac{1}{30}$$

$$\lambda_{1}\le K\left\{ y_{1}\right\} =\frac{5040}{17\cdot30}\approx9.882$$

Hence

$$K\left\{ y_{1}\right\} >J\left\{ y_{1}\right\} >\lambda_{1}=\pi^{2}$$

as expected, however \(K\left\{ y_{1}\right\}\) takes less operations to evaluate.

The other result that I also found in Kamke's book states an alternative variational principle which can be used to obtain estimates of the eigenvalues. This method gives a weaker upper bound than the Rayleigh quotient, however it requires smaller computational effort. In return, one can widen the class of trial functions without severely complicating the calculations.

We start with the Rayleigh quotient written in the abstract form:

$$J\left\{ \phi\right\}=\frac{\left\langle \phi,L\left[\phi\right]\right\rangle }{\left\langle \phi,\phi\right\rangle }$$

Then using Cauchy-Schwarz inequality we obtain.

$$\left(\left\langle \phi,L\left[\phi\right]\right\rangle \right)^{2}\leq\left\langle \phi,\phi\right\rangle \left\langle L\left[\phi\right],L\left[\phi\right]\right\rangle

\frac{\left\langle \phi,L\left[\phi\right]\right\rangle }{\left\langle \phi,\phi\right\rangle }\leq\frac{\left\langle L\left[\phi\right],L\left[\phi\right]\right\rangle }{\left\langle \phi,L\left[\phi\right]\right\rangle }$$

Written explicitly

$$\frac{\int_{\Omega}\rho\phi L\left[\phi\right]d\boldsymbol{x}}{\int_{\Omega}\rho\phi^{2}d\boldsymbol{x}}\le\frac{\int_{\Omega}\rho\left(L\left[\phi\right]\right)^{2}d\boldsymbol{x}}{\int_{\Omega}\rho\phi L\left[\phi\right]d\boldsymbol{x}}$$

Hence we can replace the original problem with the following one

$$K\left\{ \phi\right\} =\frac{\left\langle L\left[\phi\right],L\left[\phi\right]\right\rangle }{\left\langle \phi,L\left[\phi\right]\right\rangle }\to\min,\qquad\phi\in\mathfrak{A}$$

**Example**.Consider the eigenvalue problem

$$-y''=\lambda y,\quad y\left(0\right)=y\left(1\right)=0$$

Take the trial function

$$y_{1}=\frac{x}{12}-\frac{x^{3}}{6}+\frac{x^{4}}{12}$$

First we perform the calculation using the Rayleigh quotient

$$J\left\{ y_{1}\right\} =\frac{\left\langle y_{1},L\left[y_{1}\right]\right\rangle }{\left\langle y_{1},y_{1}\right\rangle }=-\frac{\int_{0}^{1}y_{1}y_{1}''dx}{\int_{0}^{1}y_{1}^{2}dx}$$

$$\begin{aligned}\int_{0}^{1}y_{1}y_{1}''dx & =\int_{0}^{1}\left[\left(\frac{x}{12}-\frac{x^{3}}{6}+\frac{x^{4}}{12}\right)\left(-x+x^{2}\right)\right]dx\\

& =\int_{0}^{1}\left(\frac{x^{6}}{12}-\frac{x^{5}}{4}+\frac{x^{4}}{6}+\frac{x^{3}}{12}-\frac{x^{2}}{12}\right)dx\\

& =-\frac{17}{5040}

\end{aligned}$$

The factor evaluated above will be common to both variational quotients

$$\begin{aligned}\int_{0}^{1}y_{1}^{2}dx & =\int_{0}^{1}\left(\frac{x^{8}}{144}-\frac{x^{7}}{36}+\frac{x^{6}}{36}+\frac{x^{5}}{72}+\frac{x^{2}}{144}\right)dx\\

& =\frac{31}{90720}

\end{aligned}$$

$$ \lambda_{1}\le J\left\{ y_{1}\right\} =\frac{17\cdot90720}{5040\cdot31}\approx9.871$$

Now we use the alternative variational principle

$$K\left\{ y_{1}\right\} =\frac{\left\langle L\left[y_{1}\right],L\left[y_{1}\right]\right\rangle }{\left\langle y_{1},L\left[y_{1}\right]\right\rangle }=-\frac{\int_{0}^{1}y_{1}''^{2}dx}{\int_{0}^{1}y_{1}y_{1}''dx}$$

Obviously, the announced gain in computational efficiency comes from replacing \(y^{2}\) with \(\left(L\left[y\right]\right)^{2}\) which will be a polynomial of the order less by 4, than \(y^{2}\).

$$\int_{0}^{1}y_{1}''^{2}dx=\int_{0}^{1}\left(x^{4}-2x^{3}+x^{2}\right)dx=\frac{1}{30}$$

$$\lambda_{1}\le K\left\{ y_{1}\right\} =\frac{5040}{17\cdot30}\approx9.882$$

Hence

$$K\left\{ y_{1}\right\} >J\left\{ y_{1}\right\} >\lambda_{1}=\pi^{2}$$

as expected, however \(K\left\{ y_{1}\right\}\) takes less operations to evaluate.