Let $t \to y(t)$ be a real-valued smooth function on any open interval containing $[0,1]$. Suppose that $y(t)$ satisfies the differential equation:
$$y''(t) + w(t)y(t) = \lambda y(t)$$
where $\lambda$ is some real constant and $w(t)$ is some function.
Suppose also that $y(t)>0$ for all $t \in [0,1]$ and that $y'(0)=y'(1)=0$.
If $\int_{0}^{1}\left(\frac{y'}{y}\right)^{2}\,dt = 10$, and $\int_{0}^{1}w(t)\,dt = 20$, what is the value of $\lambda$ ?
I have had some brief exposure to Sturm-Liouville Theory, but not enough to be able to handle such problems. How do we go about this one?
$$y''(t) + w(t)y(t) = \lambda y(t)$$ Since $y(t) \ne 0$: $$\dfrac {y''(t)}{y} + w(t) = \lambda $$ $$\left (\dfrac {y'(t)}{y}\right)'+\dfrac {y'^2}{y^2} + w(t) = \lambda $$ Integrate: $$\int_0^1\left (\dfrac {y'(t)}{y}\right)'dt+\int_0^1 \dfrac {y'^2}{y^2}dt + \int_0^1 w(t)dt = \lambda \int_0^1 dt $$ $$\lambda=\int_0^1\left (\dfrac {y'}{y}\right)'dt +30 $$ $$\lambda=\dfrac {y'}{y}\Bigg\rvert_0^1 +30 $$ Note that $y(0) \ne 0$ and $y(1) \ne 0$ (since suppose also that $y(t)>0$ for all $t \in [0,1])$ and $y'(0)=y'(1)=0$: $$\lambda=30$$