Differential operators

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

\[\begin{eqnarray*} f_x(x,y) &=&\frac{\partial}{\partial x}f(x,y),\\ f_y(x,y) &=&\frac{\partial}{\partial y}f(x,y). \end{eqnarray*}\]

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 \(af_x(x,y) + bf_y(x,y)\) as a result. That is,
\[Df(x,y) = af_x(x,y) + bf_y(x,y).\]
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,
\[\begin{eqnarray*} \frac{\partial^2}{\partial x^2} &=& \left(\frac{\partial}{\partial x}\right)^2,\\ \frac{\partial^2}{\partial x\partial y} &=& \frac{\partial}{\partial x}\cdot \frac{\partial}{\partial y}. \end{eqnarray*}\]

Example (\(\Delta\), Laplacian). The differential operator defined by
\[\Delta = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\]
is called the (two-variable) Laplacian after the French mathematician Pierre-Simon Laplace (1749-1827).
\[\Delta f(x,y) = f_{xx}(x,y) + f_{yy}(x,y).\]
The Laplacian appears in many physical problems such as heat transfer and diffusion. □

Example (\(\nabla\), nabla). We can formally define an ordered pair of the partial differential operators:
\[\nabla = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}\right).\]
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:
\[\nabla f(x,y) = \left(\frac{\partial}{\partial x}f(x,y), \frac{\partial}{\partial y}f(x,y)\right) = (f_x(x,y), f_y(x,y)).\]
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:
\[\nabla \cdot \mathbf{u}(x,y) = \frac{\partial}{\partial x}u(x,y) + \frac{\partial}{\partial y}v(x,y) = u_x(x,y) + v_y(x,y).\]
We can also calculate the scalar product between the nabla and itself:
\[\nabla\cdot\nabla = \frac{\partial}{\partial x}\frac{\partial}{\partial x} + \frac{\partial}{\partial y}\frac{\partial}{\partial y} = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} = \Delta.\]
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 \(C^{\infty}\). Then, \(\frac{\partial^2}{\partial x\partial y} = \frac{\partial^2}{\partial y\partial x}\). Accordingly, the following holds:
\[\begin{equation*} \left(a\frac{\partial}{\partial x} + b\frac{\partial}{\partial y}\right)^2 = a^2\frac{\partial^2}{\partial x^2} + 2ab\frac{\partial^2}{\partial x\partial y} + b^2\frac{\partial^2}{\partial y^2}. \end{equation*}\]
More generally, we have
\[\left(a\frac{\partial}{\partial x} + b\frac{\partial}{\partial y}\right)^n=\sum_{k=0}^{n}\binom{n}{k}a^kb^{n-k}\frac{\partial^n}{\partial x^k\partial y^{n-k}}\tag{Eq:DObinom} \]
where \(\binom{n}{k}\) is the binomial coefficient:
\[\binom{n}{k} = \frac{n!}{k!(n-k)!}.\]
□

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



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