Let $\mathcal{L}$ be the second-order linear differential operator

$$\mathcal{L}=p_0(x)\frac{d^2}{dx^2}+p_1(x)\frac{d}{dx}+p_2(x)$$

which acts on a function $u(x)$ as

\begin{equation}\label{eq:ldo}\mathcal{L}u(x)=p_0(x)\frac{d^2u(x)}{dx^2}+p_1(x)\frac{du(x)}{dx}+p_2(x)u(x).\end{equation}

Define an *adjoint* operator $\bar{\mathcal{L}}$ by

\begin{align*}

\bar{\mathcal{L}}&:=\frac{d^2}{dx^2}[p_0u]-\frac{d}{dx}[p_1u]+p_2u\\

&=p_0\frac{d^2u}{dx^2}+(2p_0^\prime-p_1)\frac{du}{dx}+(p_0^{\prime\prime}-p_1^\prime+p_2)u.

\end{align*}

If $\mathcal{L}=\bar{\mathcal{L}}$, $\mathcal{L}$ is said to be *self-adjoint*. One can immediately see that $\mathcal{L}=\bar{\mathcal{L}}$ if and only if \begin{equation}\label{eq:self-adjoint}p_0^\prime=p_1.\end{equation} Let $p(x)=p_0(x)$ and $q(x)=p_2(x)$. Then

\begin{align*}

\mathcal{L}=\bar{\mathcal{L}}&=p\frac{d^2u}{dx^2}+\frac{dp}{dx}\frac{du}{dx}+qu\\

&=\frac{d}{dx}\left[p(x)\frac{du(x)}{dx}\right]+qu(x).

\end{align*}

Note that one can transform a non-self-adjoint 2nd-order linear differential operator to a self-adjoint one. The idea is similar to that of finding a integrating factor to transform a non-separable first-order linear differential equation to a separable one.

Suppose that \eqref{eq:ldo} is not self-adjoint, i.e. $p_1\ne p_0′$. Multiply $\mathcal{L}$ by $\frac{f(x)}{p_0(x)}$. Then

$$\mathcal{L}’:=\frac{f}{p_0}\mathcal{L}=f\frac{d^2u}{dx^2}+f\frac{p_1}{p_0}\frac{du}{dx}+f\frac{p_2}{p_0}u.$$

Suppose $\mathcal{L}’$ is self-adjoint. Then by \eqref{eq:self-adjoint}

$$f’=f\frac{p_1}{p_0}.$$

That is,

$$f(x)=\exp\left[\int^x\frac{p_(t)}{p_0(t)}dt\right].$$

If $p_1=p_0′$, then

\begin{align*}

\frac{f(x)}{p_0}&=\frac{1}{p_0}\exp\left[\int^x\frac{p_1}{p_0}dt\right]\\

&=\frac{1}{p_0}\exp\left[\int^x\frac{p_0^\prime}{p_0}dt\right]\\

&=\frac{1}{p_0}\exp(\ln p_0(x))\\

&=\frac{1}{p_0(x)}\cdot p_0\\

&=1

\end{align*}

i.e. $f(x)=p_0(x)$ as expected.

**Eigenfunctions**, **Eigenvalues**

From separation of variables or directly from a physical problem, we have second-order linear differential equation of the form

\begin{equation}\label{eq:sl}\mathcal{L}u(x)+\lambda w(x)u(x)=0,\end{equation}

where $\lambda$ is a constant and $w(x)>0$ is a function called a density or weighting function. The constant $\lambda$ is called an eigenvalue and $u(x)$ is called an eigenfunction.

Example. [Schrödinger Equation]

The Schrödinger equation

$$H\psi=E\psi$$

is of the form \eqref{eq:sl}. Recall that $H$ is the Hamiltonian operator

$$H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)$$

where $V(x)$ is a potential. So $H$ is a second-order linear differential operator. The weight function $w(x)=-1$ and $E$ is energy as an eigenvalue. Clearly Schrödinger equation is self-adjoint.

Example. [Legendre's Equations]

Legendre’s equation

$$(1-x^2)y^{\prime\prime}-2xy’+n(n+1)y=0$$ is of the form \eqref{eq:sl}, where $\mathcal{L}y=(1-x^2)y^{\prime\prime}-2xy’$, $w(x)=1$, and $\lambda=n(n+1)$. Since $p_0^\prime=-2x=p_1$, Legendre’s equations are self-adjoint.

Masoud MirzaeiThanks This Really helped.

Ian LeslieI believe there is a typo in the adjoint operator definition. In the second line of that definition the p1 multiplying u(x) should be primed.

John LeePost authorIan,

You are right about the typo. It’s fixed now. Thanks a lot for pointing it out.

John

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