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,

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

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 ($t<a$):

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