# A Rayleigh quotient Eigenvalues for a simple ODE

#### joshmccraney

Hi PF!

Given the ODE $$f'' = -\lambda f : f(0)=f(1)=0$$ we know $f_n = \sin (n\pi x), \lambda_n = (n\pi)^2$. Estimating eigenvalues via Rayleigh quotient implies $$\lambda_n \leq R_n \equiv -\frac{(\phi''_n,\phi_n)}{(\phi_n,\phi_n)}$$
where $\phi_n$ are the trial functions. Does the quotient hold for all $n\in\mathbb N$? It seems like it should (I haven't seen the proof so maybe not), but if I let $\phi_n=x(1-x^n)$ then $R_2 = 10.5$ which is larger than $(2\pi)^2$. What am I doing (understanding) wrong?

Also, the Rayleigh quotient only holds for admissible functions, right (i.e. functions satisfying $\phi(0)=\phi(1)=0$ which are sufficiently smooth)?

Last edited:
Related Differential Equations News on Phys.org

#### jasonRF

Gold Member
Hi PF!

Given the ODE $$f'' = -\lambda f : f(0)=f(1)=0$$ we know $f_n = \sin (n\pi x), \lambda_n = (n\pi)^2$. Estimating eigenvalues via Rayleigh quotient implies $$\lambda_n \leq R_n \equiv -\frac{(\phi''_n,\phi_n)}{(\phi_n,\phi_n)}$$
where $\phi_n$ are the trial functions. Does the quotient hold for all $n\in\mathbb N$? It seems like it should (I haven't seen the proof so maybe not), but if I let $\phi_n=x(1-x^n)$ then $R_2 = 10.5$ which is larger than $(2\pi)^2$. What am I doing (understanding) wrong?
The Rayleigh quotient doesn't quite work like that. Let $R[\phi] = - (\phi^{\prime\prime},\phi)/(\phi,\phi)$ be the Rayleigh quotient. If we order the eigenvalues $\lambda_1 < \lambda_2 < …$ then $$\lambda_1 = \min_{\phi} R[\phi],$$ where the minimization is over all admissible functions. The function $\phi_1$ that yields $\lambda_1 = R[\phi_1]$ is then the eigenfunction associated with the eigenvalue $\lambda_1$. For your example, you can use a trial function of $\phi = x(1-x^n)$ and compute $R[x(1-x^n)]$ which is just a function of $n$. The Rayleigh quotient just tells you $$\lambda_1 \leq \min_n R[x(1-x^n)].$$ You should do this and show us what you get.

If you want to find $\lambda_2$ then you cannot blindly minimize the Rayleigh quotient with a different trial function. The most straightforward option for estimating $\lambda_2$ is to use the fact that $$\lambda_2 = \min_{\phi, \, (\phi, \phi_1)=0} R[\phi],$$ where now the minimization is over all admissible functions that are orthogonal to the first eigenvector. Of course this requires you to know the first eigenvector. Another option is to use the min max theorem (see the section on self-adjoint operators at https://en.wikipedia.org/wiki/Min-max_theorem), which doesn't require you to know the first eigenvector but is not exactly simple.

Also, the Rayleigh quotient only holds for admissible functions, right (i.e. functions satisfying $\phi(0)=\phi(1)=0$ which are sufficiently smooth)?
yes

hope that helped,