What is a Differential Operator in Physics?
Explanation
One of the methods to solve differential equations is to use the differential operator. Let’s define the differential operator $D$ as follows.
$$ D:= \frac{d}{dx} $$
When explicitly expressing the variable being differentiated, it is also denoted as $D_{x}$. For partial differentiation, it is represented as follows.
$$ \partial _{x}:=\frac{ \partial }{ \partial x},\quad \partial_{y}=\frac{ \partial }{ \partial y} $$
Using the differential operator, the differential equation is expressed as follows.
$$ \begin{align*} y^{\prime \prime}+4y^{\prime}-y=0 && \implies&& D^{2}y+4Dy-y=0 \\ && && (D^{2}+4D-1)y=0 \end{align*} $$ Here, the solution $y=0$ is physically meaningless. Therefore, solving the differential equation changes to solving a quadratic equation for the constant $r$ that satisfies $Dy=ry$ $$ r^{2}+4r-1=0 $$ Solving $Dy=ry$ is essentially an eigenvalue problem, so solving the eigenvalue problem is virtually the same as solving the differential equation. Since the differential operator includes differentiation, special attention must be paid to the order of operations. For example, $D$ and $x$ do not commute, hence $Dx\ne xD$ is true. Given that $y$ is a function concerning $x$, $$ Dxy=D(xy)=\frac{ d }{ d x }(xy)=y+xy^{\prime}=y+xDy=(xD+1)y $$ thus $$ Dx=xD+1 $$ is true. There are useful properties regarding the differential operator as follows.
Properties
$$ \begin{align*} D(D+x) &= D^{2}+xD+1 \tag{a} \\ (D-a)(D-b)=(D-b)(D-a) &= D^{2}-(a+b)D+ab \tag{b} \\ (D+1)(D^{2}-D+1) &= D^{3}+1 \tag{c} \\ Dx &= xD+1 \tag{d} \\ (D-x)(D+x) &=D^{2}-x^{2}+1 \tag{e} \\ (D+x)(D-x) &= D^{2}-x^{2}-1 \tag{f} \end{align*} $$
Proof
Since the proof method is the same, some proof processes are omitted.
$(a)$
$$ \begin{align*} D(D+x)y &= D(y^{\prime}+xy) \\ &= y^{\prime \prime}+y+xy^{\prime} \\ &= D^{2}y+xDy+y \\ &= (D^{2}+xD+1)y \end{align*} $$ Therefore, $$ D(D+x) = D^{2}+xD+1 $$
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$(b)$
$$ \begin{align*} (D-a)(D-b)y &=(D-a)(y^{\prime}-by) \\ &= y^{\prime \prime}-ay^{\prime}-by^{\prime}+aby \\ &= D^{2}y-(a+b)Dy+aby \\ &=[D^{2}-(a+b)D+ab]y \\ &=[D^{2}-(b+a)D+ba]y \\ &=(D-b)(D-a)y \end{align*} $$ Therefore, $$ (D-a)(D-b)=(D-b)(D-a) = D^{2}-(a+b)D+ab $$
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$(e)$
$$ \begin{align*} (D-x)(D+x)y &= (D-x)(y^{\prime}+xy) \\ &= y^{\prime \prime} -xy^{\prime} +y+xy^{\prime}-x^{2}y \\ &= D^{2}y+(1-x^{2})y \\ &= (D^{2}-x^{2}+1)y \end{align*} $$ Therefore $$ (D-x)(D+x)=D^{2}-x^{2}+1 $$
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