determine whether the given boundary value problem is self-adjoint. (1+x2)y''+2xy'+y=0,y'(0)=0,y(1)+2y'(1)=0

Solution

VerifiedGiven:

$\begin{align*} L[y]=(1+x^2)y''+2xy'+y&=0 \\ y'(0)&=0 \\ y(1)+2y'(1)&=0 \end{align*}$

First, we note that $2x$ is the derivative of $1+x^2$, which implies that we can rewrite the differential equation as:

$\begin{align*} L[y]&=(1+x^2)y''+2xy'+y \\ &=[(1+x^2)y']'+y \\ &=-[(-1-x^2)y']'+y \end{align*}$

A second-order differential equation is self-adjoint when the differential equation is of the form $L[y]=-(p(x)y')'+q(x)y$ and the boundary conditions are separated (that is, each condition involves only one boundary point).

$\begin{align*} p(x)&=-1-x^2 \\ q(x)&=1 \end{align*}$

Since $L[y]=-(p(x)y')'+q(x)y$ holds in this case and since the boundary conditions are separated (as one condition involves only $x=0$ and the other condition only involves $x=1$), the boundary value problem is $\textbf{self-adjoint}$.