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)).

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