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Chapter 13: Q24E (page 561)

Use the recurrence relation in Problem 28 along with (26) and (27) to express \({J_{3/2}}(x),{J_{5/2}}(x),{J_{ - 3/2}}(x),{J_{ - 5/2}}(x)\) in terms of \(sinx,cosx\), and powers of \(x\).

Short Answer

Expert verified

The recurrence relation are,

\(\begin{array}{l}{J_{3/2}}(x) = \left( {\sqrt {\frac{2}{{\pi {x^3}}}} \sin x - \sqrt {\frac{2}{{\pi x}}} \cos x} \right.\\{J_{5/2}}(x) = \left( {\sqrt {\frac{{18}}{{\pi {x^5}}}} - \sqrt {\frac{2}{{\pi x}}} } \right)\sin x - \sqrt {\frac{{18}}{{\pi {x^3}}}} \cos x\\{J_{ - 3/2}}(x) = - \sqrt {\frac{2}{{\pi {x^3}}}} \cos x - \sqrt {\frac{2}{{\pi x}}} \sin x\\{J_{ - 5/2}}(x) = \left( {\sqrt {\frac{{18}}{{\pi {x^5}}}} - \sqrt {\frac{2}{{\pi x}}} } \right)\cos x + \sqrt {\frac{{18}}{{\pi {x^3}}}} \sin x\end{array}\)

Step by step solution

01

Define Recurrence relation.

Recurrence formulas that relate Bessel functions of different orders are important in theory and in applications.

\(\frac{d}{{dx}}\left( {{x^v}{J_v}(x)} \right) = {x^v}{J_{v - 1}}(x)\)

02

Determine the function for \(v = 1/2\).

Let the function be,

\(\begin{array}{c}x{J_{3/2}} = \frac{2}{2} \cdot \frac{1}{2}{J_{1/2}}(x) - x{J_{ - 1/2}}\\{J_{3/2}} = \frac{1}{x}\sqrt {\frac{2}{{\pi x}}} \sin x - x\sqrt {\frac{2}{{\pi x}}} \cos x\\ = \sqrt {\frac{1}{{{x^2}}}} \sqrt {\frac{2}{{\pi x}}} \sin x - \sqrt {\frac{2}{{\pi x}}} \cos x\\ = \sqrt {\frac{2}{{\pi x}}} \cos x\sqrt {\frac{2}{2}} \sin x - \sqrt {\frac{2}{{\pi x}}} \cos x\end{array}\)

03

Determine the function for \(v = 3/2\).

Let the function be,

\(\begin{array}{c}x{J_{5/2}}(x) = 2 \cdot \frac{3}{2}{J_{3/2}}(x) - x{J_{1/2}}\\ = 3\left( {\sqrt {\frac{2}{{\pi {x^3}}}} \sin x - \sqrt {\frac{2}{{\pi x}}} \cos x} \right) - x\sqrt {\frac{2}{{\pi x}}} \sin x\\{J_{5/2}}(x) = \frac{3}{x}\left( {\sqrt {\frac{2}{{\pi {x^3}}}} \sin x - \sqrt {\frac{2}{{\pi x}}} \cos x} \right) - \frac{1}{x} \cdot x\sqrt {\frac{2}{{\pi x}}} \sin x\\ = \sqrt {\frac{9}{{{x^2}}}} \left( {\sqrt {\frac{2}{{\pi {x^3}}}} \sin x - \sqrt {\frac{2}{{\pi x}}} \cos x} \right) - \sqrt {\frac{2}{{\pi x}}} \sin x = \sqrt {\frac{{18}}{{\pi {x^5}}}} \sin x - \sqrt {\frac{{18}}{{\pi {x^3}}}} \cos x - \sqrt {\frac{2}{{\pi x}}} \sin x\\ = \left( {\sqrt {\frac{{18}}{{\pi {x^5}}}} - \sqrt {\frac{2}{{\pi x}}} } \right)\sin x - \sqrt {\frac{{18}}{{\pi {x^3}}}} \cos x\end{array}\)

04

Determine the function for \(v =   - 1/2\).

Let the function be,

\(\begin{array}{c}x{J_{v - 1}}(x) = 2v{J_v}(x) - x{J_{v + 1}}\\x{J_{ - 3/2}} = 2 \cdot \frac{{ - 1}}{2}{J_{ - 1/2}}(x) - x{J_{1/2}}\\ = - \sqrt {\frac{2}{{\pi x}}} \cos x - x\sqrt {\frac{2}{{\pi x}}} \sin x\\{J_{ - 3/2}}(x) = - \sqrt {\frac{1}{{{x^2}}}} \sqrt {\frac{2}{{\pi x}}} \cos x - \frac{1}{x} \cdot x\sqrt {\frac{2}{{\pi x}}} \sin x\\ = - \sqrt {\frac{2}{{\pi {x^3}}}} \cos x - \sqrt {\frac{2}{{\pi x}}} \sin x\end{array}\)

05

Determine the function for \(v =   - 3/2\).

Let the function be,

\(\begin{array}{c}x{J_{ - 5/2}}(x) = - 3\left( { - \sqrt {\frac{2}{{\pi {x^3}}}} \cos x - \sqrt {\frac{2}{{\pi x}}} \sin x} \right) - x\sqrt {\frac{2}{{\pi x}}} \cos x\\{J_{ - 5/2}}(x) = - \frac{3}{x}\left( { - \sqrt {\frac{2}{{\pi {x^3}}}} \cos x - \sqrt {\frac{2}{{\pi x}}} \sin x} \right) - \frac{1}{x}x\sqrt {\frac{2}{{\pi x}}} \cos x\\ = - \sqrt {\frac{9}{{{x^2}}}} \left( { - \sqrt {\frac{2}{{\pi {x^3}}}} \cos x - \sqrt {\frac{2}{{\pi x}}} \sin x} \right) - \sqrt {\frac{2}{{\pi x}}} \cos x\\ = \sqrt {\frac{{18}}{{\pi {x^5}}}} \cos x + \sqrt {\frac{{18}}{{\pi {x^3}}}} \sin x - \sqrt {\frac{2}{{\pi x}}} \cos x\\ = \left( {\sqrt {\frac{{18}}{{\pi {x^5}}}} - \sqrt {\frac{2}{{\pi x}}} } \right)\cos x + \sqrt {\frac{{18}}{{\pi {x^3}}}} \sin x\end{array}\)

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Most popular questions from this chapter

Use the change of variables \(s = \frac{2}{\alpha }\sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}\) to show that the differential equation of the aging spring \(mx'' + k{e^{ - \alpha t}}x = 0\),\(\alpha > 0\) becomes \({s^2}\frac{{{d^2}x}}{{d{s^2}}} + s\frac{{dx}}{{ds}} + {s^2}x = 0\).

Use the change of variables \(s = \frac{2}{\alpha }\sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}\) to show that the differential equation of the aging spring \(mx'' + k{e^{ - \alpha t}}x = 0\),\(\alpha > 0\) becomes \({s^2}\frac{{{d^2}x}}{{d{s^2}}} + s\frac{{dx}}{{ds}} + {s^2}x = 0\).

Show that \(y = {x^{1/2}}w\left( {\frac{2}{3}\alpha {x^{3/2}}} \right)\) is a solution of the given form of Airy鈥檚 differential equation whenever w is a solution of the indicated Bessel鈥檚 equation. (Hint: After differentiating, substituting, and simplifying, then let \(t = \frac{2}{3}\alpha {x^{3/2}}\))

(a) \(y'' + {\alpha ^2}xy = 0,x > 0;{t^2}w'' + tw' + \left( {{t^2} - \frac{1}{9}w} \right) = 0,t > 0\)

(b) \(y'' - {\alpha ^2}xy = 0,x > 0;{t^2}w'' + tw' - \left( {{t^2} - \frac{1}{9}w} \right) = 0,t > 0\)

(a) Use the first formula in (30) and Problem 32 to find the spherical Bessel functions \({j_1}(x)\) and \({j_2}(x)\).

(b) Use a graphing utility to plot the graphs of \({j_1}(x)\) and \({j_2}(x)\) in the same coordinate plane.

(a) Use the general solution given in Example 5 to solve the IVP \(4x'' + {e^{ - 0.1t}}x = 0,x(0) = 1,x'(0) = - \frac{1}{2}\). Also use \(J_0^'(x) = - {J_1}(x)\) and \(Y_0^'(x) = - {Y_1}(x)\) along with Table 6.4.1 or a CAS to evaluate coefficients.

(b) Use a CAS to graph the solution obtained in part (a) for \(0 \le t < \infty \).
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