Shift theorem

In mathematics, the (exponential) shift theorem is a theorem about polynomial differential operators (D-operators) and exponential functions. It permits one to eliminate, in certain cases, the exponential from under the D-operators.

The theorem states that, if P(D) is a polynomial D-operator, then, for any sufficiently differentiable function y,

$P(D)(e^{{ax}}y)\equiv e^{{ax}}P(D+a)y.\,$

To prove the result, proceed by induction. Note that only the special case

$P(D)=D^{n}\,$

needs to be proved, since the general result then follows by linearity of D-operators.

The result is clearly true for n = 1 since

$D(e^{{ax}}y)=e^{{ax}}(D+a)y.\,$

Now suppose the result true for n = k, that is,

$D^{k}(e^{{ax}}y)=e^{{ax}}(D+a)^{k}y.\,$

Then,

${\begin{aligned}D^{{k+1}}(e^{{ax}}y)&\equiv {\frac {d}{dx}}\{e^{{ax}}(D+a)^{k}y\}\\&{}=e^{{ax}}{\frac {d}{dx}}\{(D+a)^{k}y\}+ae^{{ax}}\{(D+a)^{k}y\}\\&{}=e^{{ax}}\left\{\left({\frac {d}{dx}}+a\right)(D+a)^{k}y\right\}\\&{}=e^{{ax}}(D+a)^{{k+1}}y.\end{aligned}}$

This completes the proof.

The shift theorem applied equally well to inverse operators:

${\frac {1}{P(D)}}(e^{{ax}}y)=e^{{ax}}{\frac {1}{P(D+a)}}y.\,$

There is a similar version of the shift theorem for Laplace transforms ($t<a$):

$\mathcal{L}(e^{at} f(t))=\mathcal{L}(f(t-a)).\,$