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Q8.3-31E

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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 444
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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Short Answer

(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) The values are \({j_1}(x) = \frac{1}{{{x^2}}}sinx - \frac{1}{x}cosx\) and \({j_2}(x) = \left( {\frac{3}{{{x^3}}} - \frac{1}{x}} \right)sinx - \frac{3}{{{x^2}}}cosx\).

(b) The graph has been plotted.

See the step by step solution

Step by Step Solution

Step 1:Define Spherical Bessel’s equation.

Bessel functions of half-integral order areused to dene two more important functions:

\(\begin{aligned}{j_n}(x) = \sqrt {\frac{\pi }{{2x}}} {J_{n + 1/2}}(x)\\{y_n}(x) = \sqrt {\frac{\pi }{{2x}}} {Y_{n + 1/2}}(x)\end{aligned}\)

The function \({j_n}(x)\) is called the spherical Bessel function of the first kind and \({y_n}(x)\)is the spherical Bessel function of the second kind.

Step 2: Find the value of \({j_1}(x)\).

Using the above formula and the result of Problem 32. Let,

\(\begin{aligned}{j_1}(x) &= \sqrt {\frac{\pi }{{2x}}} {J_{3/2}}\\ &= \sqrt {\frac{\pi }{{2x}}} \left( {\sqrt {\frac{2}{{\pi {x^3}}}} sinx - \sqrt {\frac{2}{{\pi x}}} cosx} \right)\\ &= \sqrt {\frac{\pi }{{2x}} \cdot \frac{2}{{\pi {x^3}}}} sinx - \sqrt {\frac{\pi }{{2x}} \cdot \frac{2}{{\pi x}}} cosx\\ &= \frac{1}{{{x^2}}}sinx - \frac{1}{x}cosx\end{aligned}\)

Step 3: Find the value of \({j_2}(x)\).

Let,

\(\begin{aligned}{j_2}(x) &= \sqrt {\frac{\pi }{{2x}}} {J_{5/2}}\\ &= \sqrt {\frac{\pi }{{2x}}} \left( {\left( {\sqrt {\frac{{18}}{{\pi {x^5}}}} - \sqrt {\frac{2}{{\pi x}}} } \right)sinx - \sqrt {\frac{{18}}{{\pi {x^3}}}} cosx} \right)\\ &= \left( {\sqrt {\frac{\pi }{{2x}} \cdot \frac{{18}}{{\pi {x^5}}}} - \sqrt {\frac{\pi }{{2x}} \cdot \frac{2}{{\pi x}}} } \right)sinx - \sqrt {\frac{\pi }{{2x}} \cdot \frac{{18}}{{\pi {x^3}}}} cosx\\ &= \left( {\sqrt {\frac{9}{{{x^6}}}} - \sqrt {\frac{1}{{{x^2}}}} } \right)sinx - \sqrt {\frac{9}{{{x^4}}}} cosx\\ &= \left( {\frac{3}{{{x^3}}} - \frac{1}{x}} \right)sinx - \frac{3}{{{x^2}}}cosx\end{aligned}\)

Step 4: Graph of the Bessel’s function.

Let the graph of \({j_1}(x)\) and \({j_2}(x)\) be,

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