Differential operators

Given the function $f(x,y)$, we have considered its partial derivatives, such as

$$\begin{array}{rcl}{f}_x(x,y)& =& \frac{\partial }{\partial x}f(x,y),\\ f_y(x,y)& =& \frac{\partial }{\partial y}f(x,y).\end{array}$$

We may interpret partial differentiation in the following manner:

The derivative $f_x(x,y)$ is obtained by applying $\frac{\partial }{\partial x}$ to the function $f(x,y)$ from the left.

$\frac{\partial }{\partial x}$ is neither a number nor a function. It's something different that we call a (partial) differential operator. The same argument applies to $\frac{\partial }{\partial y}$. This interpretation of differential operators turns out to be useful in many situations. For example, for any constants $a,b\in \mathbb{R}$, we may consider the following operator $D$:

$$D=a\frac{\partial }{\partial x}+b\frac{\partial }{\partial y}.$$

If we apply this operator to the function $f(x,y)$ from the left, we obtain $a{f}_x(x,y)+b{f}_y(x,y)$ as a result. That is,

$$Df(x,y)=a{f}_x(x,y)+b{f}_y(x,y).$$

Note that the result is different if we apply the operator to $f(x,y)$ from the right, which will be another differential operator rather than a function:

$$f(x,y)D=af(x,y)\frac{\partial }{\partial x}+bf(x,y)\frac{\partial }{\partial y}.$$

In other words, the "product" between an operator and a function is not commutative.

Let $D$ and $E$ be differential operators. We define their "product" (composition) $DE$ as follows: For the function $f(x,y)$,

$$DEf(x,y)=D(Ef(x,y)).$$

That is, first apply $E$ to $f(x,y)$, then apply $D$ to the result. When $D=E$, we write $D^2=DD$. For example,

$$\begin{array}{rcl}\frac{{\partial }^2}{\partial x^2}& =& {\left(\frac{\partial }{\partial x}\right)}^2,\\ \frac{{\partial }^2}{\partial x\partial y}& =& \frac{\partial }{\partial x}\cdot \frac{\partial }{\partial y}.\end{array}$$

Example ($\mathrm{\Delta }$, Laplacian). The differential operator defined by

$$\mathrm{\Delta }=\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}$$

is called the (two-variable) Laplacian after the French mathematician Pierre-Simon Laplace (1749-1827).

$$\mathrm{\Delta }f(x,y)=f_{xx}(x,y)+f_{yy}(x,y).$$

The Laplacian appears in many physical problems such as heat transfer and diffusion. □

Example ($\mathrm{\nabla }$, nabla). We can formally define an ordered pair of the partial differential operators:

$$\mathrm{\nabla }=(\frac{\partial }{\partial x},\frac{\partial }{\partial y}).$$

This operator is called the nabla. When this operator is applied to a function $f(x,y)$, we have an ordered pair of partial derivatives:

$$\mathrm{\nabla }f(x,y)=(\frac{\partial }{\partial x}f(x,y),\frac{\partial }{\partial y}f(x,y))=(f_x(x,y),f_y(x,y)).$$

The result is a vector function composed of the partial derivatives of $f(x,y)$.

The nabla may be regarded as a vector. Suppose we have a vector function $\mathbf{u}(x,y)=(u(x,y),v(x,y))$. Then, we can take their dot (scalar) product:

$$\mathrm{\nabla }\cdot \mathbf{u}(x,y)=\frac{\partial }{\partial x}u(x,y)+\frac{\partial }{\partial y}v(x,y)=u_x(x,y)+v_y(x,y).$$

We can also calculate the scalar product between the nabla and itself:

$$\mathrm{\nabla }\cdot \mathrm{\nabla }=\frac{\partial }{\partial x}\frac{\partial }{\partial x}+\frac{\partial }{\partial y}\frac{\partial }{\partial y}=\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}=\mathrm{\Delta }.$$

The result is the Laplacian. □

See also: Nabla symbol

Example. For $a,b\in \mathbb{R}$, let $a\frac{\partial }{\partial x}+b\frac{\partial }{\partial y}$ be a differential operator that operates on functions of class $C^{\mathrm{\infty }}$. Then, $\frac{{\partial }^2}{\partial x\partial y}=\frac{{\partial }^2}{\partial y\partial x}$. Accordingly, the following holds:

$${(a\frac{\partial }{\partial x}+b\frac{\partial }{\partial y})}^2=a^2\frac{{\partial }^2}{\partial x^2}+2ab\frac{{\partial }^2}{\partial x\partial y}+b^2\frac{{\partial }^2}{\partial y^2}.$$

More generally, we have

$$\begin{array}{}\text{(Eq:DObinom)}& {(a\frac{\partial }{\partial x}+b\frac{\partial }{\partial y})}^n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a^k{b}^{n-k}\frac{{\partial }^n}{\partial x^k\partial y^{n-k}}\end{array}$$

where $(\genfrac{}{}{0ex}{}{n}{k})$ is the binomial coefficient:

$$(\genfrac{}{}{0ex}{}{n}{k})=\frac{n!}{k!(n-k)!}.$$

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Exercise. Prove Eq. (Eq:DObinom). (Hint: Mathematical induction.)□