Differential operators

-


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.)□



Comments

Post a Comment

Popular posts from this blog

Birth process

-
Image
  Defining the birth process  Consider a colony of bacteria that never dies. We study the following process known as  the birth process , also known as  the Yule process . The colony starts with \(n_0\) cells at time \(t = 0\). Assume that the probability that any individual cell divides in the time interval \((t, t + \delta t)\) is proportional to \(\delta t\) for small \(\delta t\). Further assume that each cell division is independent of others. Let \(\lambda\) be the  birth rate.  The probability of a cell division for a population of \(n\) cells during \(\delta t\) is \(\lambda n \delta t\). We assume that the probability that two or more births take place in the time interval \(\delta t\) is \(o(\delta t)\). That is, it can be ignored. Consequently, the probability that no cell divides during \(\delta t\) is \(1 - \lambda n \delta t - o(\delta t)\). Note that this process is an example of the Markov chain with states \({n_0}, {n_0 + 1}, {n_0 + 2}...
Read more

Informal introduction to formal logic

-
Image
 In mathematics, we must prove (almost) everything and the proofs must be done logically and rigorously. Therefore, we need some understanding of basic logic. Here, I will informally explain some rudimentary formal logic. Definitions (Proposition): A proposition is a statement that is either true or false. "True" and "false" are called the truth values, and are often denoted \(\top\) and \(\bot\). Here is an example. "Dr. Akira teaches at UBD." is a statement that is either true or false (we understand the existence of Dr. Akira and UBD), hence a proposition. The following statement is also a proposition, although we don't know if it's true or false (yet): Any even number greater than or equal to 4 is equal to a sum of two primes. See also: Goldbach's conjecture   Next, we define several operations on propositions. Note that propositions combined with these operations are again propositions.   (Conjunction, logical "and"):  Let \(P\)...
Read more

Computing integrals (3): Integration by parts

-
Image
 Sometimes, we may simplify integration by using the product rule of differentiation. This technique is called integration by parts. Theorem (Integration by parts) Let \(f(x)\) and \(g(x)\) be differentiable functions on an open interval \(I\). Then,  \(\int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx\); For any \(a, b \in I\), \[\int_a^bf(x)g'(x)dx = \left[f(x)g(x)\right]_a^b - \int_a^bf'(x)g(x)dx.\] Proof . By the product rule, \[[f(x)g(x)]' = f'(x)g(x) + f(x)g'(x)\] so \[f(x)g'(x) = [f(x)g(x)]' - f'(x)g(x).\] By integrating both sides, we have the desired results. ■ Example . Let us find \(\int x\cosh x dx\). \[ \begin{eqnarray*} \int x\cosh x dx &=& \int x(\sinh x)'dx \\ &=& x \sinh x - \int 1 \cdot \sinh x dx\\ &=& x \sinh x - \cosh x + C. \end{eqnarray*} \] Example (eg:recur) . Let us study how we can compute \[I_n = \int \frac{dx}{(x^2 + 1)^n}\] for \(n\in \mathbb{N}\). Note \[I_{n} = \int \fr...
Read more