I have the following Sturm-Liouville problem \begin{cases} u''+\lambda u =0 & 0 < x < 1, \\ u(0)-u'(0)=0, & u(1)+u'(1)=0, \end{cases} and I am trying to show that all the eigenvalues $\lambda$ are positive.
I multiplied both sides by $u$, integrated with respect to $x$ from $0$ to $1$, applied initial conditions, and algebraically rearranged the result to obtain $$\lambda \int_0^1 u^2 \, dx = \int_0^1 (u')^2 \, dx + u(1)^2+u(0)^2.$$ This tells that $\lambda$ is nonnegative, or $\lambda \ge 0$. But I need to also show that $\lambda$ is positive, or $\lambda > 0$. To do this, I must show that $0$ cannot be an eigenvalue, or $\lambda \not=0$. But how can I show this? I'm aware this may be trivial, but I do not see it.
If you plug $\lambda = 0$ into your ODE, you get that $u'' = 0$, i.e. $u(x) = ax+b$ for some $a,b$. Using your initial conditions you get that:
$$u(0) - u'(0) = 0 \Longrightarrow b-a = 0$$
and
$$u(1) + u'(1) = 0 \Longrightarrow 2a+b = 0$$
The only solution to these two is if $a=0=b$, i.e. $u\equiv 0$. This is the trivial solution which is why we ignore it. $0$ is always a solution to a homogeneous ODE.