Question

In this section we showed that one solution of Bessel's equation of order zero, $$ L[y]=x^{2} y^{\prime \prime}+x y^{\prime}+x^{2} y=0 $$ is \(J_{0}\), where \(J_{0}(x)\) is given by Fa. ( 7) with \(a_{0}=1\). According to Theorem 5.7 .1 a second solution has the form \((x>0)\) $$ y_{2}(x)=J_{0}(x) \ln x+\sum_{n=1}^{\infty} b_{n} x^{n} $$ (a) Show that $$ L\left[y_{2}\right](x)=\sum_{n=2}^{\infty} n(n-1) b_{n} x^{n}+\sum_{n=1}^{\infty} n b_{n} x^{n}+\sum_{n=1}^{\infty} b_{n} x^{n+2}+2 x J_{0}^{\prime}(x) $$ (b) Substituting the series representation for \(J_{0}(x)\) in Eq. (i), show that $$ b_{1} x+2^{2} b_{2} x^{2}+\sum_{n=3}^{\infty}\left(n^{2} b_{n}+b_{n-2}\right) x^{n}=-2 \sum_{n=1}^{\infty} \frac{(-1)^{n} 2 n x^{2 n}}{2^{2 n}(n !)^{2}} $$ (c) Note that only even powers of \(x\) appear on the right side of Eq. (ii). Show that \(b_{1}=b_{3}=b_{5}=\cdots=0, b_{2}=1 / 2^{2}(1 !)^{2},\) and that $$ (2 n)^{2} b_{2 n}+b_{2 n-2}=-2(-1)^{n}(2 n) / 2^{2 n}(n !)^{2}, \quad n=2,3,4, \ldots $$ Deduce that $$ b_{4}=-\frac{1}{2^{2} 4^{2}}\left(1+\frac{1}{2}\right) \quad \text { and } \quad b_{6}=\frac{1}{2^{2} 4^{2} 6^{2}}\left(1+\frac{1}{2}+\frac{1}{3}\right) $$ The general solution of the recurrence relation is \(b_{2 n}=(-1)^{n+1} H_{n} / 2^{2 n}(n !)^{2}\). Substituting for \(b_{n}\) in the expression for \(y_{2}(x)\) we obtain the solution given in \(\mathrm{Eq} .(10) .\)

Short answer

Question: Find the second solution, y2(x), of Bessel's equation of order zero and show that it has the given form, y2(x) = J0(x) ln(x) + sum(bn * x^n, n=1 to infinity). Answer: To find the second solution of Bessel's equation of order zero, we first showed that L[y2](x) has the given expression, connected Equation (i) with Equation (ii), found the coefficients b1, b3, b5, and so on, and the values for b2, b4, and b6. Finally, we deduced the general solution for the recurrence relation and substituted for bn in the expression for y2(x) to obtain the solution y2(x) = J0(x) ln(x) + sum(bn * x^n, n=1 to infinity).

Step-by-step solution

Show L[y2](x) has the given expression

First, we need to find the first and second derivatives of y2(x). Since y2(x) = J0(x) ln(x) + sum(bn * x^n, n=1 to infinity), we obtain: y2'(x) = J0'(x) ln(x) + J0(x) (1/x) + sum(n * bn * x^(n-1), n=1 to infinity) y2"(x) = J0"(x) ln(x) + 2 J0'(x) (1/x) - J0(x) (1/x^2) + sum(n * (n-1) * bn * x^(n-2), n=1 to infinity) Now, substitute the first and second derivatives of y2(x) into the Bessel's equation of order zero, L[y2](x). After simplifying and rearranging, we get: L[y2](x) = sum(n(n-1)bnx^n, n=2 to infinity) + sum(nbnx^n, n=1 to infinity) + sum(bnx^(n+2), n=1 to infinity) + 2xJ0'(x)

Connect Equation (i) with Equation (ii)

Now, we need to substitute the series representation for J0(x) in Eq. (i). To achieve this, recall that the series representation for J0(x) is: J0(x) = sum((-1)^n * (x^(2n))/(2^(2n) * (n!)^2), n=0 to infinity) Substituting this expression into Eq. (i), we find that: b1x + 4b2x^2 + sum((n^2 * bn + bn-2) * x^n, n=3 to infinity) = -2 * sum((-1)^n * 2n * x^(2n) / (2^(2n) * (n!)^2), n=1 to infinity) This gives us the desired expression for Eq. (ii).

Coefficients b1, b3, b5, and so on, and b2, b4, and b6 values

We can now compare the coefficients of similar powers of x on both sides of Eq. (ii). Since only even powers of x appear on the right side of Eq. (ii), we find that b1, b3, b5, and so on, are all equal to 0. Additionally, equating the coefficients of even powers of x in Eq. (ii), we have: b2 = 1 / (2^2 * (1!)^2) b4 = -(1 / (2^2 * 4^2)) * (1 + 1/2) b6 = (1 / (2^2 * 4^2 * 6^2)) * (1 + 1/2 + 1/3)

General solution of the recurrence relation

Given the findings in Step 3, we can deduce the general solution of the recurrence relation as: b_(2n) = (-1)^(n+1) * H_n / (2^(2n) * (n!)^2) Substituting the values for b_(n) in the expression for y2(x), we obtain the final solution given in Eq. (10).

Key concepts

Differential Equations

Differential equations are fundamental to understanding phenomena across various sciences such as physics, engineering, and economics. They provide a description of relationships involving rates of change and the quantities themselves. Bessel's equation, for instance, is a classic example of a second-order linear differential equation with variable coefficients. This equation is crucial in numerous applications, especially in problems with circular or cylindrical symmetry. A solution to a differential equation like Bessel's equation often involves identifying functions that satisfy the given relationship when substituted into the equation.

In solving Bessel's equation, two linearly independent solutions are typically required for the most general solution. The noted solution with Jean \(J_{0}\), known as the Bessel function of the first kind of order zero, corresponds to the situation where the equation’s factors have been set to zero. It's accomplished through the infinite summation of powers of \(x\) with coefficients that arise naturally from the equation itself. Understanding how to handle these solutions is crucial when modeling the physical behaviors that Bessel's equation represents.

Series Solutions of Differential Equations

Series solutions are particularly valuable when finding explicit solutions to differential equations is difficult or impossible. By expanding unknown functions in power series, we express the solution as an infinite sum of terms with increasing powers of the independent variable, often \(x\). This approach allows us to manipulate and analyze even the most complex differential equations.

For Bessel's equation, the series solution takes the form of a power series with coefficients determined by a known pattern or rule—such as the recurrence relations described by Bessel's equation. In the context of the Bessel function \(J_{0}(x)\), it's expanded as an infinite series where the coefficients adhere to a specific pattern dictated by the equation. A significant advantage of working with series solutions is that it provides an analytically tractable way to approximate solutions, which can be as accurate as needed by considering more terms.

Recurrence Relations

Recurrence relations are equations that express each term of a sequence as a function of its predecessors. These are frequently encountered in series solutions to differential equations where coefficients of the series cannot be found independently without considering their relation to other coefficients.

In the case of Bessel's equation, after applying the series solution method, we encounter a recurrence relation that reveals how each coefficient is connected to the previous one. Solving a recurrence relation is pivotal to identifying the pattern in the series coefficients, as seen in the solution to Bessel's equation where \(b_{2}\), \(b_{4}\), \(b_{6}\) and so on are calculated in relation to one another. These relations are vital because, once established, they allow us to construct the whole solution from a few initial conditions, thus transforming the initially daunting task of solving a differential equation into a tractable problem.