I got a question which I'm quite stuck on...

Given the eigenvalue problem,

\(\displaystyle y''+\lambda y=0\)

\(\displaystyle hy(0)+y'(0) = 0, y(1) = 0\)

Show that if \(\displaystyle \lambda = 0\) is an eigenvalue, then \(\displaystyle h\) must have a specific value and find this value. If \(\displaystyle h \geq 1\), show that there is exactly one negative eigenvalue.

I believe I've done the first part, as my working is as follows:

Rearrange the HLDE into self adjoint form: \(\displaystyle -(-y')' = \lambda (-1)y\), and taking that \(\displaystyle \lambda = 0\), means the HLDE becomes \(\displaystyle y'' = 0\). The general solution to this should be of the form \(\displaystyle y(x)= Ax + B\). Using the boundary conditions,

\(\displaystyle hy(0)+y'(0) = hB + A = 0 \Rightarrow h = -\frac{A}{B}\)

\(\displaystyle y(1) = A + B = 0 \Rightarrow A = -B\)

Therefore, \(\displaystyle h = -\frac{(-B)}{B} = 1\)

Does this look correct?

The next part of the question I get stuck on. The general solution for \(\displaystyle -(-y')' = \lambda (-1)y\) is \(\displaystyle y(x)= Ccos (\sqrt{\lambda}x) + Dsin(\sqrt{\lambda}x) \).

Using the boundary conditions

\(\displaystyle hy(0)+y'(0) = h(Ccos (\sqrt{\lambda}0) + Dsin(\sqrt{\lambda}0)) + (-C\sqrt{\lambda}sin (\sqrt{\lambda}0) + D\sqrt{\lambda}cos(\sqrt{\lambda}0))\) \(\displaystyle = h*C +\sqrt{\lambda}D = 0\)

\(\displaystyle y(1) =Ccos (\sqrt{\lambda}) + Dsin(\sqrt{\lambda}) = 0 \)

From here, I not sure where to go... have I do the right thing?

Thanks for your time.