From Wikipedia, the free encyclopedia
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
a
x
y
)
≡
e
a
x
P
(
D
+
a
)
y
.
{\displaystyle 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
{\displaystyle 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
a
x
y
)
=
e
a
x
(
D
+
a
)
y
.
{\displaystyle D(e^{ax}y)=e^{ax}(D+a)y.\,}
Now suppose the result true for n = k , that is,
D
k
(
e
a
x
y
)
=
e
a
x
(
D
+
a
)
k
y
.
{\displaystyle D^{k}(e^{ax}y)=e^{ax}(D+a)^{k}y.\,}
Then,
D
k
+
1
(
e
a
x
y
)
≡
d
d
x
{
e
a
x
(
D
+
a
)
k
y
}
=
e
a
x
d
d
x
{
(
D
+
a
)
k
y
}
+
a
e
a
x
{
(
D
+
a
)
k
y
}
=
e
a
x
{
(
d
d
x
+
a
)
(
D
+
a
)
k
y
}
=
e
a
x
(
D
+
a
)
k
+
1
y
.
{\displaystyle {\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:
1
P
(
D
)
(
e
a
x
y
)
=
e
a
x
1
P
(
D
+
a
)
y
.
{\displaystyle {\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
{\displaystyle t<a}
L
(
e
a
t
f
(
t
)
)
=
L
(
f
(
t
−
a
)
)
.
{\displaystyle \scriptstyle {\mathcal {L}}(e^{at}f(t))=\scriptstyle {\mathcal {L}}(f(t-a)).\,}
