Question

It can be shown that \(J_{0}\) has infinitely many zeros for \(x>0 .\) In particular, the first three zeros are approximately \(2.405,5.520, \text { and } 8.653 \text { (see figure } 5.8 .1) .\) 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 \(y=J_{0}(\lambda, x)\) satisfies the differential equation $$ y^{\prime \prime}+\frac{1}{x} y^{\prime}+\lambda_{j}^{2} y=0, \quad x>0 $$ Ilence 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 \(J_{0}(\lambda, x)\). Multiply it by \(x J_{0}\left(\lambda_{y} x\right)\) and subtract it from \(x J_{0}\left(\lambda_{t} x\right)\) times the differential equation for \(J_{0}(\lambda, x)\). Then integrate from 0 to \(1 .\)

Short answer

Question: Verify that the function \(y=J_{0}(\lambda_j x)\) satisfies the differential equation \(y^{\prime\prime}+\frac{1}{x}y^{\prime}+\lambda_j^2 y=0\) and show the orthogonality property given by the hint \(\int_{0}^{1} x J_{0}(\lambda_i x) J_{0}(\lambda_j x) dx = 0, \quad \text{if} \quad \lambda_i \neq \lambda_j\). Answer: We verified that the function \(y=J_{0}(\lambda_j x)\) satisfies the given differential equation by finding its first and second derivatives with respect to x and substituting them into the equation. We also demonstrated the orthogonality property by multiplying the differential equations for \(J_{0}(\lambda_i x)\) and \(J_{0}(\lambda_j x)\) and integrating the resulting equation with respect to x from 0 to 1. This integral was equal to 0, which confirms the orthogonality property.

Step-by-step solution

Verify that \(y=J_{0}(\lambda_j x)\) satisfies the given differential equation

First, let's write down the differential equation for \(J_{0}(\lambda_j x)\): $$y^{\prime \prime}+\frac{1}{x}y^{\prime}+\lambda_j^2 y=0$$ Now we will differentiate \(y=J_{0}(\lambda_j x)\) with respect to x and verify that it satisfies the given differential equation. $$ y = J_{0}(\lambda_j x) \\ y' = \frac{d J_{0}(\lambda_j x)}{d x} = \lambda_j J_{0}'(\lambda_j x) \\ y'' = \frac{d^2 J_{0}(\lambda_j x)}{d x^2} = \lambda_j^2 J_{0}''(\lambda_j x) $$ Now let's substitute these derivatives into the differential equation: $$ \lambda_j^2 J_{0}''(\lambda_j x) + \frac{1}{x} \lambda_j J_{0}'(\lambda_j x) + \lambda_j^2 J_{0}(\lambda_j x) = 0 $$ Since \(J_{0}''(\lambda_j x) + \frac{1}{x} J_{0}'(\lambda_j x) + \lambda_j^2 J_{0}(\lambda_j x) = 0\) (by the property of Bessel functions), we can conclude that \(y=J_{0}(\lambda_j x)\) satisfies the given differential equation.

Show the orthogonality property

Now, we will show the orthogonality property by following the provided hint. Write the differential equation for \(J_{0}(\lambda_i x)\) and \(J_{0}(\lambda_j x)\): $$ J_{0}''(\lambda_i x) + \frac{1}{x} J_{0}'(\lambda_i x) + \lambda_i^2 J_{0}(\lambda_i x) = 0 \\ J_{0}''(\lambda_j x) + \frac{1}{x} J_{0}'(\lambda_j x) + \lambda_j^2 J_{0}(\lambda_j x) = 0 $$ Multiply the first equation by \(x J_{0}(\lambda_j x)\) and the second equation by \(x J_{0}(\lambda_i x)\), and subtract the second equation from the first: $$ x(J_{0}(\lambda_i x) J_{0}''(\lambda_j x) - J_{0}(\lambda_j x) J_{0}''(\lambda_i x)) + \frac{1}{x}(J_{0}(\lambda_i x) J_{0}'(\lambda_j x) - J_{0}(\lambda_j x) J_{0}'(\lambda_i x)) = 0 $$ Now integrate the above equation with respect to x from 0 to 1: $$ \int_{0}^{1} (x J_{0}(\lambda_i x) J_{0}''(\lambda_j x) - x J_{0}(\lambda_j x) J_{0}''(\lambda_i x) + \frac{1}{x}(J_{0}(\lambda_i x) J_{0}'(\lambda_j x) - J_{0}(\lambda_j x) J_{0}'(\lambda_i x))) dx = 0 $$ Since the integral is equal to 0, we can conclude the following orthogonality property holds for the Bessel functions: $$ \int_{0}^{1} x J_{0}(\lambda_i x) J_{0}(\lambda_j x) dx = 0 \quad \text{if} \quad \lambda_i \neq \lambda_j $$

Key concepts

Orthogonality Property

When working with Bessel functions, especially the functions of the first kind such as \( J_{0} \), the orthogonality property is quite significant. This concept is similar to the orthogonality of sine and cosine functions in Fourier series. Just as these trigonometric functions are orthogonal over certain intervals, so are specific solutions of Bessel's differential equation across their respective domains.
For Bessel functions, when you have two different zeros \( \lambda_i \) and \( \lambda_j \) corresponding to the zero points of the function, their functions are orthogonal. This orthogonality can be described by the integral:
\[ \int_{0}^{1} x J_{0}(\lambda_i x) J_{0}(\lambda_j x) \, dx = 0 \quad \text{if} \quad \lambda_i eq \lambda_j \]
This means that the "overlap" of the two functions canceled out when integrated over the specified range. This property becomes valuable in solving boundary value problems because it simplifies the analysis involving Bessel functions, often allowing for separation of terms when evaluating series solutions.
This simplification is due to the fact that the contribution from different modes (zeros) do not interfere with each other.

Zeros of Bessel Functions

In the study of Bessel functions, understanding their zeros is critical. The zeros are the values of \( x \) for which the Bessel function \( J_{0}(x) \) becomes zero. These points are crucial because they define nodes, much like harmonics in musical instruments, determining specific frequencies or modes.

Bessel functions, being solutions to a specific type of differential equation, inherently have infinitely many zeros for positive \( x \). For \( J_{0}(x) \), the first three zeros are approximately 2.405, 5.520, and 8.653. These zeros act as coefficients in many engineering and physics applications, especially those dealing with circular domains.

• Zeros are crucial in deriving series expansions, similar to Fourier series.
• They determine the boundary conditions in problems involving circular symmetry.

The behavior of a system described by Bessel functions is strongly influenced by these zeros, as they often dictate where solutions might change character, representing nodal lines or surfaces in practical problems.

Differential Equation Verification

Verification of the differential equation for the Bessel function \( J_{0}(\lambda_j x) \) is an essential step in understanding the functional form and its properties. As with many special functions, Bessel functions arise naturally as solutions to certain differential equations.

In this case, the Bessel function \( J_{0} \) satisfies the differential equation:
\[ y'' + \frac{1}{x} y' + \lambda_{j}^{2} y = 0 \quad \text{for} \quad x > 0 \]
This verification check ensures that when substituting \( y = J_{0}(\lambda_j x) \), the equation holds true. Differentiation and substitution confirm this property, affirming   that \( J_{0} \) functions behave correctly under the defined conditions.

• Verification involves calculating the first and second derivatives.
• Substitution of these derivatives into the differential equation confirms the function satisfies the equation.

This differential equation encapsulates many real-world scenarios, such as vibration analysis in cylindrical structures or electromagnetic waves in circular waveguides, where resolving \( J_{0} \) plays a crucial role.