Originally Posted by

**Silverflow** Hi,

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? Yes.

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 $