---
title: Adjoint and self-adjoint
tags: Applied_math_method
GA: G-77TT93X4N1
---

# Adjoint and self-adjoint

Consider the second-order linear differential operator
$$
L = a_0(x)\frac{d^2}{dx^2} + a_1(x)\frac{d}{dx} + a_2(x),
$$
with inner product defined as
$$
<f, g> = \int^b_a f(x) g(x)\,dx.
$$
We have
$$
<v, L[u]> = \int^b_a v L[u]\,dx = \int^b_a v\left( a_0 u'' + a_1 u' + a_2 u\right)\,dx \\
= \int^b_a \left[ a_0 v'' + (2a'_0 -a_1) v' + (a''_0-a'_1+a_2) v\right]u\,dx +\left.\left[a_0(vu'-uv')+(a_1-a'_0)uv\right]\right|^b_a \\
=<L^*[v], u> + J(u,v)|^b_a,
$$
where
$$
L^*[v] = a_0 v'' + (2a'_0 -a_1) v' + (a''_0-a'_1+a_2) v,
$$
and
$$
J(u,v) = a_0(vu'-uv')+(a_1-a'_0)uv.
$$

## Formally self-adjoint
### Definition:
We say that $L$ is **formally self-adjoint** if $L^*=L$.

If $a'_0=a_1$, then
$$
L = a_0 \frac{d^2}{dx^2} + a'_0 \frac{d}{dx} +a_2 = \frac{d}{dx}\left(a_0\frac{d}{dx}\right)+a_2 = L^*,
$$
that is, $L$ is formally self-adjoint.

## Self-adjoint

Let
$$
D = \{u \,|\, B_1(u) = B_2(u)=0\},
$$
where $B_1(u)=0$ and $B_2(u)=0$ are homogeneous boundary conditions. We can define
$$
D^* = \{v \,|\, J(u,v)|^b_a = 0, \quad \forall u\in D\} = \{u \,|\, B^*_1(u) = B^*_2(u)=0\}.
$$

### Definition:
The system $Lu=f$, $B_1(u)=B_2(u)=0$ is said to be self-adjoint if $L=L^*$ and $D=D^*$.