I am reading the book "Ordinary Differential Equations" By Birkhoff and Rota (1989). I am a little confused about the meaning or definition of "adjoint operator" and "adjoint differential equation". In their Ch. 2.8, they define:
The second-order homogeneous linear differential equation (DE): $$ L[u] = p_0(x)u''(x)+ p_1(x) u'(x)+p_2(x)u(x) = 0\tag{1} $$
is said to be exact if and only if, for some $A(x),\,B(x)\in C^1$,
$$ p_0(x)u''(x)+ p_1(x) u'(x)+p_2(x)u(x) = \frac{\mathrm{d}}{\mathrm{d}x}\left[A(x)u'(x)+B(x)u(x)\right] $$
for all $u\in C^2$.
An integrating factor for the above DE is a function $v(x)$ such that $v(x) L[u]$ is exact.
It is easy to show that:
A function $v\in C^2$ is an integrating factor for the DE (1) if and only if it is a solution of the second order homogeneous linear DE:
$$ M[v] = \left[p_0(x) v(x)\right]'' - \left[p_1(x)v(x)\right]' + p_2(x)v(x) = 0\tag{2} $$
Now they define:
The operator $M$ in (2) is called adjoint of the linear operator $L$. The DE expanded to the DE:
$$ p_0v''+(2p_0'-p_1)v' + (p_0''-p_1'+p2)v = 0 $$
is called the adjoint of the DE (1).
The questions are:
- Is the adjoint stricly related to the existence of an integrating factor - can I talk or define an adjoint DE without the integrating factor $v$?
- Or else, if (1) is exact, and no integrating factor is necessary, does one also talk about "adjointment"?
- Or else, if in (2), I put instead of some integrating factor $v$ the same $u$ as in (1), that is, I would have instead of (2), $M[u]=0$, is this also called an adjoint of (1)?
- Or even, if I have an arbitrary function $w\in C^2$ that is not an integrating factor of (1) and basically has nothing to do with (1), is $M[w]=0$ called adjoint of (1)?
If you have a scalar product, the adjoint is just defined as $$ \langle Lu,v\rangle = \langle u,Mv\rangle $$ Taking the $L^2$ scalar product and integration-by-parts, this results in $$ M[v]=(p_0v)''-(p_1v)'+p_2v. $$ Expanded this gives exactly the claimed formula.
If the equation is exact with $A=p_0$, $B'=p_2$, then the adjoint is $$ M[v]=(Av')'-Bv'. $$
If with an integrating factor you can bring the equation to the form $$ S[u]=(q_0u')'+q_2u=0 $$ then $S$ is self-adjoint. In the original form this introduces a weight function in the scalar product. With this weighted scalar product $L$ is self-adjoint.