Show that $$Ly = -y''-\frac{y'}{x} + y \quad y'(0)=0;y(1) = 0$$ defines a symetric Sturm - Liouville operator for a suitable scalar product and determine the scalar product.
Determine eigenpairs.
So to show that the operator is symmetric we need to show that $$\langle Ly,z\rangle = \langle Lz,y\rangle$$
Hence let's start $$\langle Ly,z\rangle = \int(-y''-\frac{y'}{x} + y)z \, dx$$
On the other hand $$\langle Lz,y\rangle = \int(-z''-\frac{z'}{x} + z)y \, dx $$
Thus we will show that the operator is symmetric.
However how to show that operator is Sturm- Lioville operator?
Sturm- Liuville operator is of the form $Ly = p(x) y'' +q(x)y' +r(x)y$. Hence we can see that is Sturm Liuville right?
Question Could someone help please how to find a scalar product so that it will be positive? And how to find eigenpairs?
P.s I do not need solution to exactly this problem, but i would appretiate if someone would explain by steps how to do it
A symmetric form is $$ Ly = \frac{1}{x}\left[-(xy')'+xy\right] $$defined on $L^2_x[0,1]$ with inner product $$ \langle f,g\rangle = \int_{0}^{1}f(x)g(x)xdx. $$ Then, \begin{align} \langle Ly,w\rangle - \langle y,Lw\rangle &= \int_{0}^{1}-(xy')'w+y(xw')'dx \\ &=\int_{0}^{1}(-xy'w+xyw')'dx \\ &= -xy'w+xyw'|_{0}^{1}. \end{align} The solutions of $Ly-y=0$ are found by solving
$$ -y''-\frac{y'}{x}=0 \\ \frac{y''}{y'}+\frac{1}{x}=0 \\ \ln(xy')=C \\ y' = \frac{C}{x} \\ y = C\ln(x)+D $$ So $L$ is in the limit circle case at $x=0$. Setting $y'(0)=0$ imposes the condition $C=0$. The conditions $y'(0)=0=y(1)$ then lead to a well-posed singular Sturm-Liouville problem on $L^2_{x}[0,1]$. The eigenvalue problem is $$ -(xy')'+xy = \lambda x y,\;\; y'(0)=0=y(1). $$