Ω-logic

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,

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

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 ($t<a$):

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