Construction of the Green's Function

The explicit construction of the Green's function is a very intuitive and mechanical process if one has available two independent solutions to the homogeneous (i.e. with zero on the right hand side) differential equation governing the physical system. Indeed, on the interval $(a,b)$ consider the two linearly independent solutions $u_1(x)$ and $u_2(x)$ which satisfy

$$Lu_1(x)=0$$

and

$$Lu_2(x)=0 ~~.$$

Let consider first the case where these two functions satisfy boundary conditions at each end point, $a$ and $b$ separately. We shall let these boundary conditions be the mixed Dirichlet-Neumann conditions at $a$ and $b$ respectively,

$$= B_1(u_1)\equiv \alpha u_1(a)+\alpha ' u_1'(a)$$ 0

$$= B_2(u_2)\equiv \beta u_2(b)+\beta ' u_2'(b) ~~.$$ 0

It is important to note that these boundary conditions do not determine these two functions uniquely. In fact, each one may be multiplied by its own multiplicative factor. Thus, one obtains two families of solutions,

$$c_1u_1(x)~:~~~~~B_1(c_1u_1)=0$$

and

$$c_2u_2(x)~:~~~~~B_2(c_2u_2)=0 ~~.$$

Because of the Fundamental Theorem, we must say (i) that the Green's function has the form

$$G(x;\xi)=\left\{ \begin{array}{cc} c_1u_1(x) & x < \xi \\ c_2u_2(x) & \xi < x \end{array} \right.$$ (416)

and that (ii) the constants $c_1$ and $c_2$ must be adjusted so that at $x=\xi$ the Green's function is continuous:

$$c_2u_2(\xi)- c_1u_1(\xi) =0$$ (417)

and has the prescribed jump in its slope:

$$c_2u'_2(\xi)- c_1u'_1(\xi) =\frac{-1}{p(\xi)} ~~.$$ (418)

These are two equations in the two unknowns $c_2$ and $c_1$ . Thus two unique members of each family of solutions have been determined. Figure 4.3 depicts how the graphs of the two solutions meet so as to fulfill the continuity requirement. Observe that, by itself, continuity at $x=\xi$ does not determine the amplitude at that point. Furthermore, at that point the graph has a kink, an abrupt change in its slope which depends entirely on the as-yet-indeterminate amplitude at that point.

Figure 4.3: Pictorial construction of the Green's function $G(x;\xi )$ . At $x=\xi$ , where the graphs of the two solutions meet, $G$ must have exactly that amplitude which guarantees that the jump in the slope equals precisely the requisite amount.

However, from Figure 4.3 one sees that by adjusting the amplitude $G(\xi;\xi)$ to an appropriate value, the magnitude of the change in the slope at $x=\xi$ can be made to equal the required amount, which is $-1/p(\xi)$ . This determines $G(x;\xi )$ uniquely.

Note, however, that there is one circumstance under which $G(x;\xi )$ does not exist, namely, when $u_1$ and $u_2$ form a linearly dependent set, i.e. when they are related by

$$u_1(x)=ku_2(x) ~~~~~~~~ a\le x \le b~~.$$

It is clear that in this circumstance the continuity condition at $x=\xi$ prevents the existence of any kink at $x=\xi$ : regardless how large an amplitude one chooses, the change in the slope will always be zero,

$$\left. \begin{array}{l} u_1(\xi)=ku_2(\xi)\\ c_2u_2(\xi)- c... ...d{array}\right\} \Longrightarrow c_2u'_2(\xi)- c_1u'_1(\xi) =0$$

Equation (4.18) will always be violated, and the Green's function does not exist.

If the Green's function does exist (i.e. when $u_1$ and $u_2$ form a linearly independent set) then it is given by Eq.(4.16), where $c_1$ and $c_2$ are determined uniquely by Eqs.(4.17) and (4.18). This circumstance is summarized by the following

Theorem 45.1 (Construction of $G(x;\xi )$ )
Given:  The functions $u_1(x)$ and $u_2(x)$ which satisfy

$$\begin{array}{rcc} Lu_1(x) & = & 0 \\ \alpha u_1(a)+\alpha ... ...(x)& = & 0\\ \beta u_2(b)+\beta ' u'_2(b) &=&0 ~~. \end{array}$$

Conclusion:  The Green's function for $L$ is

$$G(x;\xi) = \frac{-1}{c} \left\{ \begin{array}{ccl} u_1(x) u_2(\xi) & \te... ...\ u_1(\xi) u_2(x) & \textrm{for} & \xi < x \end{array} \right.$$

$$\equiv \frac{-1}{c} ~~u_1(x_<) u_2(x_>)$$ (419)

where

$$c=p(\xi)~[u_1(\xi) u'_2(\xi)-u'_1(\xi) u_2(\xi) ] ~~.$$

Remark. (i) The normalization constant $c$ is according to Abel's theorem (Section 3.3.3) always a constant.

(ii) It is evident that the notation introduced in Eq.(4.19),

$$G(x;\xi)= \frac{-1}{c} u_1(x_<) u_2(x_>)$$

is very suggestive. We shall use it repeatedly.

Proof. To verify that the formula given by Eq.(4.19) is the Green's function, simply check that properties (a)-(d) of the Fundamental Theorem are satisfied. Thus

(a)

$G(x;\xi )$ obviously satisfies the homogeneous differential equation $LG(x;\xi)=0$ whenever $x\ne \xi$ .

(b)

$G(x;\xi )$ does satisfy the given boundary conditions at each endpoint $a$ and $b$ .

(c)

$G(x;\xi )$ is obviously continuous.

(d)

The derivative $$\frac{dG}{dx}$$ satisfies the correct jump discontinuity at $x=\xi$ . Indeed,

$$\left.\frac{dG}{dx} \right\vert^{\xi^+} - \left.\frac{dG}{dx} \right\vert^{\xi^-} = \frac{-1}{c}[u_1(\xi) u'_2(\xi^+) -u'_1(\xi^-) u_2(\xi) ]$$

$$= \frac{-1}{c} \frac{c}{p(\xi)}$$

$$= \frac{-1}{p(\xi)}$$

Thus $G$ as given by formula (4.19) has all the identifying properties of the Green's function indeed.

Example (Response of a static string)

Consider the following boundary value problem:

$$\frac{d^2 u}{dx^2} = -f(x)$$

$$u(0) =$$ 0

$$u(1) =$$ 0

Find its Green's function and its solution.

Solution: There are three steps that lead up to the Green's function:

$$\begin{array}{rrcll} u''=0\Rightarrow~a)&u_1&=&x& \textrm{so... ...u'_1u_2]& \\ & &=&1~[x(-1)-1(1-x)] &\\ & &=&-1& \end{array}$$

Consequently,

$$d)~~~G(x;\xi) = \left\{ \begin{array}{lr} x(1-\xi )&~~~~~x<\xi \\ \xi (1-x)&~~~~~ \xi<x \end{array} \right.$$

$$\equiv x_< (1-x_>)$$

The solution is

$$e)~~~u(x) = \int_0^1 G(x;\xi)f(\xi)~d\xi$$

$$= \int_0^x \xi(1-x) f(\xi)~d\xi + \int_x^1 x(1-\xi) f(\xi)~d\xi$$

Lecture 31