TITLE: Sturm-Liouville differential equation eigenvalue problem
QUESTION [5 upvotes]: If we have a Sturm-Liouville differential equation of the form 
$$
\frac{d}{dx}[p(x)\frac{dy}{dx}]+q(x)y=-\lambda w(x)y
$$ 
and define the linear operator $L$ as 
$$L(u) = \frac{d}{dx}[p(x)\frac{du}{dx}]+q(x)u
$$ then we get the equation $L(y)=-\lambda w(x)y$ which defines what is called the eigenvalue problem of the Sturm-Liouville differential equation.
My question: why is it called that way despite the fact that there is still a function $w(x)$ in the equation? I thought an eigenvalue problem would have the form $L(y)=-\lambda y$.
What's happening here?

REPLY [5 votes]: The operator can be written as
$$
      Ly = \frac{1}{w}\left[-\frac{d}{dx}\left(p\frac{dy}{dx}\right)+qy\right].
$$
This operator is defined on weighted $L^2$ space $L^2_w[a,b]$. With the correct endpoint conditions, $L$ becomes selfadjoint, and the eigenvalue problem is $Ly=\lambda y$. Absording the negative sign into the second derivative term is standard because that has the best chance of making $L$ a positive operator. For example, in the simplest case of $Ly=-y''$ on $[a,b]$ with endpoint conditions at $x=a,b$,
$$
        \langle Ly,y\rangle = \int_a^b (-y'')y(x)dx = -y'y|_a^b + \int_a^b(y')^2dx
  = \int_a^b (y')^2dx \ge 0.
$$