Exercise [21.03]

deant · 12 Aug 2012, 10:15

Simple expansion will show that

$\left(1-D^2+D^4-D^6+...\right)\left(1+D^2\right) = 1$

Thus,

$\left(1-D^2+D^4-D^6+...\right)\left(1+D^2\right)f(x) = f(x)$

Provided that there is no interval $x \in (a,b)$ on which $\left(1+D^2\right)f(x) = 0, \; \; \; f(x) \not\equiv 0$ .

If there is such an interval for some $f(x) \not\equiv 0$ , then the operator $\left(1+D^2\right)$ is not truly invertible; we have:

$\left(1-D^2+D^4-D^6+...\right)\left[\left(1+D^2\right)f(x)\right] = 0$

on this interval.

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Let us suppose that

is a linear operator, and that z(x)

is the most general solution of the equation

Pz = 0

(

is in general a parameterized family of solutions z(x,A,B,...)

). Then

$Py = f(x) \;\; \Leftrightarrow \;\; P(y-z) = f(x)$

Applying the (partial) inverse $P^{-1}$ to the right-hand equation then yields the most general possible solution:

$y = P^{-1}f(x) + z(x)$

In the particular case we are considering, $P = \left(1+D^2\right)$ , the general solution to Pz=0

is

$z = A\cos{x} + B\sin{x}$

Which leads directly to the same answer obtained in Exercise [21.02].