Question

(Sturm-Liouville) I'm having trouble understanding to what extent I need to go to prove that for a given SLDE (more info below), a contradiction shows that for a given lambda, eigenvalues are not possible (due to contradictions or otherwise)

Answer 1

In the PDF below, on page 13, Problem 3, I'm not sure entirely how I show that it's not possible for lambda being negative that you can't have any eigenvalues or eigenfunctions. I'll post my working on it shortly. http://texas.math.ttu.edu/~gilliam/ttu/ode_pde_pdf/Ch5.pdf

Answer 2

\[\text{Let $\lambda=-\alpha^2$}, \ \ \ \alpha>0.\]\[y''-\alpha^2y=0; \ \ \ y''=\alpha^2y. \ \ \ m^2=\alpha^2\]

Answer 3

For real repeated roots solutions to a characteristic equation, you can have a solution of the form: \[y_h=c_1 \cosh(\alpha x)+c_2 \sinh(\alpha x).\]

Answer 4

\[y'_h=c_1 \alpha \cosh(\alpha x)+c_2 \alpha \sinh(\alpha x).\]
\[y'(0)=c_1\alpha \sinh(0)+c_2 \alpha \cosh(0)=c_2 \alpha \cosh(0)=c_2\alpha; \ \ \ c_2\alpha=0.\]

Answer 5

\[y'(l)=c_1\alpha \sinh(\alpha l)+c_2 \alpha \cosh(\alpha l)=c_1 \sinh(\alpha l). \ \ \ [c_2
\alpha=0]\]
\[c_1 \alpha \sinh(\alpha l)=0.\]

Now I don't know how to move forward or what I need to do to demonstrate that there is a contradiction or what. How do I do this?

Answer 6

@dan815

Answer 7

@SithsAndGiggles