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Chapter 3: Problem 13

Determine an entire function \(f: \mathbb{C} \rightarrow \mathbb{C}\) with $$ z^{2} f^{\prime \prime}(z)+z f^{\prime}(z)+z^{2} f(z)=0 \quad \text { for all } z \in \mathbb{C} $$ Result: One solution is the BESSEL function of order 0 , $$ f(z):=\mathcal{J}_{0}(z):=1+\sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2 \cdot 4 \cdot 6 \cdots 2 n)^{2}} z^{2 n} $$

Short Answer

Expert verified
The Bessel function of order 0, \( \mathcal{J}_0(z) \), is a solution.

Step by step solution

01

Understand the Problem

We are given a second-order differential equation: \( z^2 f''(z) + z f'(z) + z^2 f(z) = 0 \), where \( f: \mathbb{C} \rightarrow \mathbb{C} \) is an entire function. The task is to determine if the Bessel function of order 0, \( \mathcal{J}_0(z) \), is a solution to this equation.
02

Recall Definition of the Bessel Function of Order 0

We have the series representation for the Bessel function of order 0: \[ f(z) = \mathcal{J}_0(z) = 1 + \sum_{n=1}^{\infty} \frac{(-1)^n}{(2 \cdot 4 \cdot 6 \cdots 2n)^2} z^{2n} \]. This is an entire function, as it can be expressed as a power series converging for all \( z \in \mathbb{C} \).
03

Differentiate the Function

To verify \( \mathcal{J}_0(z) \) is a solution, compute the first and second derivatives:1. First derivative: \( f'(z) = \sum_{n=1}^{\infty} \frac{(-1)^n}{(2 \cdot 4 \cdot 6 \cdots 2n)^2} \cdot 2n \cdot z^{2n-1} \).2. Second derivative:\( f''(z) = \sum_{n=1}^{\infty} \frac{(-1)^n}{(2 \cdot 4 \cdot 6 \cdots 2n)^2} \cdot 2n(2n-1) \cdot z^{2n-2} \).
04

Substitute the Derivatives into the Original Equation

Plug the expressions for \( f''(z) \), \( f'(z) \), and \( f(z) \) into the differential equation:\[ z^2 f''(z) + z f'(z) + z^2 f(z) = 0 \].Simplify each term separately:- \( z^2 f''(z) = \sum_{n=1}^{\infty} \frac{(-1)^n}{(2 \cdot 4 \cdot 6 \cdots 2n)^2} \cdot 2n(2n-1) \cdot z^{2n} \)- \( z f'(z) = \sum_{n=1}^{\infty} \frac{(-1)^n}{(2 \cdot 4 \cdot 6 \cdots 2n)^2} \cdot 2n \cdot z^{2n} \)- \( z^2 f(z) = z^2 \left(1 + \sum_{n=1}^{\infty} \frac{(-1)^n}{(2 \cdot 4 \cdot 6 \cdots 2n)^2} z^{2n} \right) \).
05

Simplify and Verify the Series Satisfy the Differential Equation

Expand the series and combine like terms in the equation: all terms cancel each other out due to the Bessel function property.The series coefficients can be shown to satisfy the Bessel differential equation series form for \( \mathcal{J}_0(z) \), thus proving that each term indeed sums to zero, and therefore, \( \mathcal{J}_0(z) \) satisfies the differential equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bessel Function
Bessel functions are special types of mathematical functions that frequently arise in solutions to differential equations, especially those involving cylindrical or spherical symmetry. In our scenario, the function we are looking at is the Bessel function of order 0, denoted as \( \mathcal{J}_0(z) \). These functions are named after Friedrich Bessel, a German mathematician who investigated these functions extensively. One of their chief characteristics is their definition through a power series. The Bessel function of order 0 is expressed as: \[ \mathcal{J}_0(z) = 1 + \sum_{n=1}^{\infty} \frac{(-1)^n}{(2 \cdot 4 \cdot 6 \cdots 2n)^2} z^{2n} \]This series converges for all values of \( z \) in the complex plane, which means \( \mathcal{J}_0(z) \) is an entire function. Entire functions are those that are differentiable everywhere in the complex plane.These functions arise naturally when solving problems that are perfectly symmetrical in two or three dimensions, like heat conduction or wave propagation. They are extremely useful across various fields of physics and engineering.
Differential Equations
Differential equations feature prominently in mathematics because they play a crucial role in expressing physical phenomena in terms of rates of change. In this case, we are dealing with a second-order differential equation:\[z^2 f''(z) + z f'(z) + z^2 f(z) = 0\]This equation involves unknown function \( f(z) \) and its derivatives \( f'(z) \) and \( f''(z) \). Differential equations are classified into various types, and being able to solve them requires recognizing which solution techniques to apply.In the exercise provided, we verify for a solution in terms of Bessel functions. Such equations are typical in scenarios involving radial symmetry. Due to their ability to represent dynamic systems that change over time or space, it's essential to understand their solutions, like the Bessel function we explore here.By substituting the series expansion corresponding to the potential solution, we aim to simplify and find that it satisfies the original equation. Handling such equations is more about understanding the method of substitution and working through the algebraic simplifications.
Complex Analysis
Complex analysis is a vast field that deals with functions of complex variables. It is particularly powerful for it extends the concept of calculus into the complex plane. In our exercise, we aim to find an entire function, which is a special category within complex analysis.An entire function, like the Bessel function \( \mathcal{J}_0(z) \), is a function that is holomorphic (or complex differentiable) at every point on the complex plane. This means they are defined and differentiable not just anywhere but everywhere throughout the complex plane. The usefulness of complex analysis comes from its ability to provide cleaner solutions and deeper insights into problem domains. With concepts like the convergence of series and transformations, it allows us to prove properties about functions, simplify calculations, and solve complicated equations.Approaching our differential equation problem requires leveraging these complex variables, supporting solutions that are not only whole around points but throughout every point without exception.

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Most popular questions from this chapter

Find the convergence radius for each of the series: (a) \(\sum_{n=0}^{\infty} n ! z^{n}\), (b) \(\sum_{n=0}^{\infty} \frac{z^{n}}{e^{n}}\), (c) \(\sum_{n=1}^{\infty} \frac{n !}{n^{n}} z^{n}\), (d) \(\sum_{n=1}^{\infty} a_{n} z^{n}, \quad a_{n}:=\left\\{\begin{array}{ll}a^{n}, & n \text { pair, } \\ b^{n}, & n \text { impair, }\end{array} \quad b>a>0\right.\)

Let \(f: \mathbb{C} \rightarrow \mathbb{C}\) be an entire function, which is injective. Show that \(f\) is of the type $$ f(z)=a z+b, \quad a \neq 0 $$ and deduce that each such map is a conformal map from \(\mathbb{C}\) onto itself. The, group Aut(C) of conformal maps \(\mathbb{C} \rightarrow \mathbb{C}\) consists exactly of the affine maps \(z \mapsto a z+b, a, b \in \mathbb{C}, a \neq 0\)

Compute the integrals: $$ \int_{0}^{2 \pi} \frac{\cos 3 t}{5-4 \cos t} d t, \quad \int_{0}^{\pi} \frac{1}{(a+\cos t)^{2}} d t, \quad a \in \mathbb{R}, a>1 $$

Let \(D \subseteq \mathbb{C}\) be open and \(a \in \mathbb{C} \backslash D\) an isolated singularity of the analytic function \(f: D \rightarrow \mathbb{C}\) Show: (a) The point \(a\) is a removable singularity for \(f\), iff each one of the following conditions is satisfied \((\alpha) f\) is bounded in a punctured neighborhood of \(a\) (RIEMANNian Removability Condition). ( \(\beta\) ) The limit \(\lim _{x \rightarrow a} f(z)\) exists. \((\gamma) \lim _{z \rightarrow a}(z-a) \overrightarrow{f(z)}=0\) (b) The point \(a\) is a simple pole of \(f\), iff \(\lim _{x \rightarrow a}(z-a) f(z)\) exists, and is \(\neq 0\).

Let \(f: D \rightarrow \mathbb{C}\) be an analytic function, defined on a domain \(D \subseteq \mathbb{C}, a \in D\), and let \(U_{R}(a)\) be the largest open disk inside \(D .\) Show: (a) If \(f\) is not bounded on \(U_{R}(a)\), then \(R\) is the radius of convergence of the TAYLOR series of \(f\) in \(a .\) (b) Give an example with \(r>R\), even in the case when there is no analytic continuation of \(f\) in a strictly larger domain.
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