Green's Function and Its Adjoint

Presently our task is to solve what in linear algebra corresponds to

$$(A-\lambda B)\vec u =\vec b~~.$$

In the framework of calculus this means that we must solve the inhomogeneous boundary value problem

$$\begin{array}{rclc} Lu(x) & = & -f(x) &~~~~ a<x<b \\ B_1(u)&=&0 & ~~~~\\ B_2(u)&=&0 & ~~~~ \end{array} ~~ ,$$ (46)

or, more generally, solve the problem

$$Lv(x) = -f(x) ~~~~ a<x<b$$

$$B_1(v) = d$$ (47)

$$B_2(v) = e~~,$$ (48)

Notation: the minus sign on the right hand sides is a convention which complies with the interpretation of $f$ as the force density on a simple string as discussed in the next section.
The operator $L$ is a second order linear differential operator, while $d$ and $e$ are constants. We shall first deal with the first problem where the boundary conditions are homogeneous ($d=e=0$ ). Once we have solved it, the solution to the second one is simply equal to the first solution augmented by that solution to the homogeneous differential equation which satisfies

$$\begin{array}{rclc} Lv_h(x) & = & 0 &~~~~ a<x<b \\ B_1(v_h)&=&d & ~~~~\\ B_2(v_h)&=&e & ~~~~ \end{array} ~~ .$$

Thus,

$$v(x) = u(x) +v_h (x)$$

$$= u(x) +c_1v_1 (x)+c_2 v_2 (x)~~.$$

Here $v_1$ and $v_2$ are any two independent solutions to the homogeneous differential equation, and the constants $c_1$ and $c_2$ are adjusted so that the two linear endpoint conditions, Eqs.(4.7) and (4.8), are satisfied. This means that $c_1$ and $c_2$ are determined by

$$c_1 B_1(v_1)+c_2 B_1(v_2) = d$$

$$c_1 B_2(v_1)+c_2 B_2(v_2) = e ~~.$$

Lecture 28

The solution to the inhomogeneous problem, Eqs.(4.6), is based on the corresponding Green's function. In the engineering sciences it is also known as the unit impulse response. It is given by the following

Definition. (Green's function and its adjoint)

Let $G(x;\xi )$ be a function with the property

$$\begin{array}{rclc} LG(x;\xi) & = & -\delta(x-\xi) &~~~~ a<x,\xi<b \\ B_1(G)&=&0 & ~~~~\\ B_2(G)&=&0 & ~~~~ \end{array} ~~ .$$ (49)

Such a function is the Green's function for the boundary value problem. The corresponding adjoint Green's function $H(x;\xi)$ is the function with the property

$$\begin{array}{rclc} L^*H(x;\xi) & = & -\delta(x-\xi) &~~~~ a<... ...b \\ B_1^*(H)&=&0 & ~~~~\\ B_2^*(H)&=&0 & ~~~~ \end{array} ~~ ,$$ (410)

where $L^*$ is the formal adjoint of the differential operator $L$ and $B_1^*(H)=0$ and $B_2^*(H)=0$ are the boundary conditions adjoint to $B_1(G)=0$ and $B_2(G)=0$ .

The adjoint Green's function is very useful because it allows us to solve the inhomogeneous boundary value problem, Eqs.(4.6). The solution is obtained with the help of Green's identity, Eq.(4.4),

$$\langle H,Lu \rangle -\langle L^* H,u \rangle =\int_a^b (\overline{H} L u-\overline{L^* H} u)~dx= P(\overline{H},u)\vert^b_a ~~.$$

Indeed, using the fact that the adjoint boundary conditions

$$\begin{array}{rcl} B_1^*(H)&=&0 ~~~~\\ B_2^*(H)&=&0 ~~~~ \end{array} ~~$$

have been constructed so as to guarantee that

$$P(\overline{H},u)\vert^b_a =0~,$$

we obtain with the help of the given Eq.(4.6), $Lu=-f$ , and with (4.10), $L^*H=-\delta(x-\xi )$ , the result

$$\int^b_a f(x)~\overline{H} (x;\xi)~dx =\int^b_a \delta (x-\xi)~u(x)~dx~~,$$

which yields the solution

$$\boxed{u(\xi) =\int^b_a \overline{H} (x;\xi)~f(x)~dx }$$

It turns out that the beauty of this result is that we don't even have to use the adjoint Green's function $H(x;\xi)$ . Instead, one may use the original Green's function $G(x;\xi )$ . This is based on the following

Theorem (Green's function and its adjoint)

$$\boxed{ \overline{H} (x;\xi)=G(\xi;x) }$$

The proof of this equation is given below.

Remark 1. This result says that in order to obtain the adjoint Green's function,

$$H(x;\xi)=\overline{G(\xi;x)}~~,$$

simply interchange the arguments $x$ and $\xi$ and then take the complex conjugate of the Green's function, the solution to Eq.(4.9). With the help of this result the solution to the inhomogeneous problem becomes simply

$$\boxed{u(\xi) =\int^b_a G(\xi;x)~f(x)~dx }$$ (411)

The advantage is clear: don't bother solving Eq.(). It is enough to find only the Green's function, i.e. the solution to Eq.(4.9).

Remark 2. The other noteworthy feature is algebraic. The process of interchanging the arguments $x$ and $\xi$ and then taking the complex conjugate is precisely the infinite-dimensional version of taking the Hermitian adjoint of a matrix. Moreover, the integration in Eq.(4.11) corresponds to the summation when a matrix acts on a vector and thereby yields a new vector.

Remark 3. If the boundary value problem is self-adjoint, i.e. $L=L^*$ , together with $B_1^*=B_1$ and $B_2^*=B_2$ , then $H(x;\xi)=G(x;\xi)$ and we have the result

$$\boxed{\overline{G (\xi;x) }=G(x;\xi) }~~.$$

This is generally known as the reciprocity relation. It says that $G(x;\xi )$ is what in linear algebra corresponds to a ``Hermitian matrix''.

Proof: (In three steps)

(i)

Again use Green's identity

$$\int_a^b (\overline{H} L G-\overline{L^* H} G)~dx= P(\overline{H},G )\vert^b_a$$

(ii)

The boundary conditions of the two boundary value problems (4.9) and () guarantee that the linear concomitant vanishes at the endpoints,

$$P(\overline{H},G )\vert^b_a =0~~.$$

(iii)

Inserting the two respective differential equations of (4.9) and () into the above Green's identity, one obtains

$$\int ^b_a \overline{H} (x;\xi ')\delta(x-\xi)~dx =\int ^b_a G(x;\xi)\delta(x-\xi ')~dx$$

or

$$\overline{H} (\xi ;\xi ')=G(\xi ';\xi)~~,$$

which is what had to be shown.