Answer: It can be shown that J0 has infinitely many zeros for x > 0. In particular, the

Chapter 5, Problem 14

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It can be shown that \(J_{0}\) has infinitely many zeros for \(x>0\). In particular, the first three zeros are approximately 2.405, 5.520, and 8.653 (see Figure 5.7.1). Let \(\lambda_{j}, j=1,2,3, \ldots\), denote the zeros of \(J_{0}\) ; it follows that

\(J_{0}\left(\lambda_{j} x\right)=\left\{\begin{array}{ll}
1, & x=0, \\
0, & x=1 .
\end{array}\right.\)

Verify that \(y=J_{0}\left(\lambda_{j} x\right)\) satisfies the differential equation

\(y^{\prime \prime}+\frac{1}{x} y^{\prime}+\lambda_{j}^{2} y=0, \quad x>0 .\)

Hence show that

\(\int_{0}^{1} x J_{0}\left(\lambda_{i} x\right) J_{0}\left(\lambda_{j} x\right) d x=0 \quad \text { if } \quad \lambda_{i} \neq \lambda_{j}\).

This important property of \(J_{0}\left(\lambda_{i} x\right)\), known as the orthogonality property,is useful in solving boundary value problems.

Hint: Write the differential equation for \(J_{0}\left(\lambda_{i} x\right)\). Multiply it by \(x J_{0}\left(\lambda_{j} x\right)\) and subtract it from \(x J_{0}\left(\lambda_{j} x\right)\) times the differential equation for \(J_{0}\left(\lambda_{i} x\right)\). Then integrate from 0 to 1.