- 1,797

- 74

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)?

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: