Self-Adjoint Differential Equations I

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.

4 thoughts on “Self-Adjoint Differential Equations I

    1. Ian Leslie

      I 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.

      Reply
  1. Pingback: Self-Adjoint Differential Equations II: Hermitian Operators | MathPhys Archive

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