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 Exercise [21.03] 
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Joined: 12 Jul 2010, 07:44
Posts: 154
Post Exercise [21.03]
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 P is a linear operator, and that z(x) is the most general solution of the equation


Pz = 0


(z(x) 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].


12 Aug 2012, 10:15
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