Second Order Operator and the Bilinear Concomitant

Let us extend our considerations from linear differential operators of first order to those of second order. To do this, let us find the adjoint of a second order operator. The given operator consists of

(i)

the differential operator

$$\boxed{L=\alpha(x)\frac{d^2}{dx^2} +\beta(x) \frac{d}{dx} +\gamma(x) }$$

(ii)

the domain $\mathcal {S} \subset \mathcal{H} =L^2(a,b)$ on which it operates,

$$\boxed{ \mathcal{S} = \{ ~u:~u\in L^2(a,b);~Lu \in L^2(a,b);~ B_1(u)=0;~B_2(u)=0 ~\} }$$

where $B_1$ and $B_2$ are two homogeneous boundary conditions,

$$\begin{array}{rcl} 0=B_1(u) &\equiv & \alpha_1 u(a)+\alpha_1'... ...2 u(a)+\alpha_2' u'(a) +\beta_2 u(b)+\beta_2' u'(b) \end{array}$$ (43)

The $\alpha_i's$ and $\beta_i's$ are given constants not to be confused with the functions $\alpha(x)$ and $\beta(x)$ . The task is to find the adjoint of the given operator, namely FIND

(i)

$L^*$

(ii)

$\mathcal{S}^* = \{ ~v\in L^2(a,b):~ B_1^*(v)=0;~B_2^*(v)=0 ~\}$

such that

$$\langle v,Lu \rangle =\langle L^* v,u \rangle$$

for all $u \in \mathcal{S}$ and all $v \in \mathcal{S^*}$ . The left-hand side of this equation is given, and it is

$$\langle v,Lu \rangle =\int^b_a\left( \alpha \overline{v} \frac{d^2u}{dx^2} +\beta \overline{v} \frac{du}{dx} + \overline{v}\gamma u \right)~dx ~~.$$

In order to have the derivatives act on the function $v$ , one does an integration by parts twice on the first term, and once on the second term. The result is

$$\langle v,Lu \rangle = \int^b_a \overline{ \underbrace{ \left( \frac{d^2}{dx^2}\overline... ...erline{\beta}v + \overline{\gamma }v \right) }_{\displaystyle L^* v} } u ~dx ~~$$

$$+ \underbrace{ \left[ \overline{v} \alpha u' -(\overline{v} \alpha)' u +\overline{v}\beta u \right]^b_a}_{\displaystyle P(\overline{v},u)\vert ^b_a}$$

The bilinear expression $P(\overline{v},u)$ is called the bilinear concomitant or the conjunct of $\overline{v}$ and $u$ . Thus we have

$$\boxed{ \langle v,Lu \rangle -\langle L^* v,u \rangle \equiv P(\overline{v},u)\vert^b_a }$$ (44)

This important integral identity is the one-dimensional version of Green's identity. Indeed, it relates the behavior of $v(x)$ and $u(x)$ in the interior of $[a,b]$ to their values on the boundary, here $x=a$ and $x=b$ . It is an extension of the integrated Lagrange identity, Eq.(3.16), from formally self-adjoint second order operators to generic second order operators. Observe that when

$$\beta =\alpha '~~,$$

$L$ becomes formally self-adjoint whenever the coefficient functions $\alpha ,\beta$ , and $\gamma$ are real. In this circumstance $L$ is the Sturm-Liouville operator and the bilinear concomitant reduces to

$$P(\overline{v},u)= \alpha [ \overline{v}u'-\overline{v}'u ]~~,$$

which is proportional to the Wronskian determinant of $\overline{v}$ and $u$ . The construction of $L^*$ from $L$ is based on the requirement that

$$\langle v,Lu \rangle -\langle L^* v,u \rangle =0~~.$$

This means that the bilinear concomitant evaluated at the endpoints must vanish,

$$P(\overline{v},u)\vert ^b_a =0 ~.\quad\quad \textrm{(\lq\lq compatibility~condition'')~~~~(I)}$$

This is a compatibility condition between the given boundary conditions, Eq.(4.3),

$$\begin{array}{rcl} B_1(u)&=&0\\ B_2(u)&=&0 \end{array}~, \quad \textrm{(\lq\lq given~boundary~ conditions'')~~~~(II)}$$

and the adjoint boundary conditions,

$$\begin{array}{rcl} B^*_1(u)&=&0\\ B^*_2(u)&=&0 \end{array}~. \quad \quad \textrm{(\lq\lq adjoint~boundary~ conditions'')~~~~(III)}$$

This means that any two of the three sets of conditions implies the third:

1. (II) and (III) imply (I).
2. (III) and (I) imply (II).
3. (I) and (II) imply (III).

The problem of obtaining the adjoint boundary conditions in explicit form,

$$\begin{array}{rcl} 0=B^*_1(v) &\equiv & \alpha^*_1 v(a)+{\alp... ...lpha^*_2}' v'(a) +\beta^*_2 v(b)+{\beta^*_2}' v'(b) \end{array}$$ (45)

is a problem in linear algebra. One must combine the given boundary conditions, Eq.(4.3) with the compatibility condition (I) to obtain the coefficients $\alpha^*_i,{\alpha^*_i}',\beta^*_i,{\beta^*_i}'$ in Eq.(4.5).