1804 Operator Rules

In stating these rules, we will always assume that the functions involved are sufficiently differentiable, so that the operators can be applied to them.

Sum Rule. If $p(D)$ and $q(D)$ are polynomial operations, the any (sufficiently differentiable) function u, $$[P(D)+q(D)]u=p(D)u+q(D)u\tag{1}$$

Lineariry Rule. If $f$ and $g$ are functions and $c_1$ and $c_2$ are constants, $$p(D)(c_1f+c_2g)=c_1p(D)f+c_2P(D)g\tag{2}$$

Proof of the linearity rule: This rule follows from the linearity of differen­tiation. That is, $$D(c_1f+c_2g)=(c_1f+c_2g)'=c_1f'+c_2g'=c_1Du_1+c_2Du_2$$

Similarly taking the second or higher derivative also follows the linearity rule . That is, $$D^{(n)}(c_1f+c_2g)=\frac{d^n}{dt^n}(c_1f+c_2g)=c_1f^{(n)}+c_2g^{(n)}=c_1D^{(n)}f+c_2D^{(n)}g$$

Next, we can scale the linear operator $D^n$ by $a$ and it stays linear. That is, $$aD^{(n)}(c_1f+c_2g)=a\frac{d^n}{dt^n}(c_1f+c_2g)=c_1af^{(n)}+c_2ag^{(n)}=c_1aD^{(n)}f+c_2aD^{(n)}g$$

(Notice that a does not actually have to be a constant, it can be a function of t (or of whatever independent variable we're using). )
Finally we can combine these operators into a polynomial operator $$D^n+a_1D^{n-1}+...+a_{n-1}D^{n-1}+a_n$$ which clearly still obeys the linearity rule. □

Multiplication rule. If $p(D)=g(D)h(D)$ as polynomials in $D$, then $$p(D)u=g(D)(h(D)u)\tag{3}$$ The property is true when $h(D)$ is the simple operator $aD^k$, essentially because $$D^m(aD^ku)=aD^{m+k}u$$ It extends to general polynomial operators $h(D)$ by linearity. Note that here a must be a constant; it's false otherwise.

An important corollary of the multiplication property is that polynomial operators with constant coefficients commute; i.e., for every function $u(t)$, $$g(D)(h(D)u)=h(D)(g(D)u)\tag{4}$$ As polynomials, $g(D)h(D) = h(D)g(D) = p(D)$.

The remaining two rules are of a different type and are more concrete: they tell us how polynomial operators behave when applied to exponential functions and products involving exponential functions.

Substitution rule. $$p(D)e^{at}=p(a)e^{at}\tag{5}$$ Proof We have, by repeated differentiation, $$De^{at}=ae^{at},D^2e^{at}=a^2e^{at},...,D^ke^{at}=a^ke^{at}$$ therefore, $$(D^n+c_1D^{n-1}+...+c_n)e^{at}=(a^n+c_1a^{n-1}+...+c_n)e^{at}$$ which is the substitution rule. □

The exponential-shift rule This handles expressions such as $t^ke^{at}$ and $t^k\sin{at}$. Let $u=u(t)$. Then $$p(D)e^{at}u=e^{at}p(D+a)u\tag{6}$$ Proff. We prove it in successive stages. First, it is true when $P(D)=D$, since by the product rule for differentiation, $$De^{at}u(t)=e^{at}Du(t)+ae^{at}u(t)=e^{at}(D+a)u(t)\tag{7}$$ To show the rule is true for $D^k$,we apply $(7)$ to $D$ repeatedly: $$\begin{aligned} D^2e^{at}u=D(De^{at}u)&=D(e^{at}(D+a)u)\\ &=e^{at}(D+a)((D+a)u)\\ &=e^{at}(D+a)^2u \end{aligned}$$ In the same way, $$\begin{aligned} D^3e^{at}u=D(D^2e^{at}u)&=D(e^{at}(D+a)^2u)\\ &=e^{at}(D+a)((D+a)^2u)\\ &=e^{at}(D+a)^3u \end{aligned}$$ and so on. This shows that $(6)$ is true for an operator of the form $D^k$. To show it is true for a general operator $$P(D)=D^n+c_1D^{n-1}+...+c_n$$ we write $(6)$ for each $D^k(e^{at}u)$, multiply both sides by the coefficient $c_k$, and add up the resulting equations for the different values of $k$.

Example. Find $D^3e^{-t}\sin{t}$. Solution using the exponential-shift rule. $$\begin{aligned} D^3e^{-t}\sin{t}&=e^{-t}(D-1)^3\sin{t}\\ &=e^{-t}(D^3-3D^2+3D-1)\sin{t}\\ &=e^{-t}(-\cos{t}+3\sin{t}+3\cos{t}-\sin{t})\\ &=e^{-t}(2\cos{t}+2\sin{t}) \end{aligned}$$ Solution using the substitution rule. Write $e^{-t}\sin{t}=Img(e^{(-1+i)t}$. We have $$\begin{aligned} D^3e^{(-1+i)t}&=(-1+i)^3e^{(-1+i)t}\\ &=(2+2i)e^{-t}(\cos t+i\sin t) \end{aligned}$$ Take the imaginary part: $e^{-t}(2\cos{t}+2\sin{t})$