$\displaystyle y''+\lambda y=0 \ \ \ \ \ \ y(0)=0, \ \ y(1)+y'(1)=0 $

has infinitely many eigenvalues $\displaystyle \lambda_{1}<\lambda_{2}<... $

and indicate the behaviour of lambda as n goes to infinity.

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- Nov 5th 2008, 06:12 PMszpengchaohow to show this DE has infinitely many eigenvalues
$\displaystyle y''+\lambda y=0 \ \ \ \ \ \ y(0)=0, \ \ y(1)+y'(1)=0 $

has infinitely many eigenvalues $\displaystyle \lambda_{1}<\lambda_{2}<... $

and indicate the behaviour of lambda as n goes to infinity. - Nov 6th 2008, 02:16 AMOpalg
You really ought not to need much of a hint in order to do this. You should recognise $\displaystyle y''+\lambda y=0$ as a simple harmonic motion equation (assuming that $\displaystyle \lambda>0$), with solution $\displaystyle y=A\cos\mu x + B\sin\mu x$, where $\displaystyle \mu = \pm\sqrt\lambda$. The initial conditions will give you a condition on $\displaystyle \mu$, of the form $\displaystyle \tan\mu=-\mu$. This is not an equation that you can solve explicitly, but by drawing a graph of the tan function, you can see that there is a doubly-infinite family of solutions, which for large values of $\displaystyle |\mu|$ will be close to odd multiples of $\displaystyle \pi/2$.