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Chapter 11: Problem 12

(a) Verify that $$ I_{0}(x)=\frac{1}{\pi} \int_{0}^{\pi} \cosh (x \cos \theta) d \theta $$ satisfies the modified Bessel equation, \(v=0\). (b) Show that this integral contains no admixture of \(K_{0}(x)\), the irregular second solution. (c) Verify the normalization factor \(1 / \pi\).

Short Answer

Expert verified
Yes, it satisfies the Bessel equation; it has no \(K_0(x)\) term, and the factor \(1/\pi\) is correct.

Step by step solution

01

Understanding the Modified Bessel Equation

The modified Bessel equation of order zero is given by \[x^2 y'' + x y' - x^2 y = 0.\]We need to verify that the given function \(I_0(x)\) satisfies this equation.
02

Differentiating the Function with Respect to x

Let's first differentiate \(I_0(x)\) with respect to \(x\). The function is:\[I_{0}(x)=\frac{1}{\pi} \int_{0}^{\pi} \cosh (x \cos \theta) d \theta.\]The derivative with respect to \(x\) is:\[I_0'(x) = \frac{1}{\pi} \int_{0}^{\pi} \frac{d}{dx}\cosh (x \cos \theta) d \theta = \frac{1}{\pi} \int_{0}^{\pi} \cos \theta \sinh (x \cos \theta) d \theta.\]
03

Second Derivative of the Function

Differentiate \(I_0'(x)\) once more to find the second derivative:\[I_0''(x) = \frac{1}{\pi} \int_{0}^{\pi} \left(\cos^2 \theta \cosh(x \cos \theta)\right)d \theta.\]
04

Substituting into the Bessel Equation

Now, plug \(I_0(x)\), \(I_0'(x)\), and \(I_0''(x)\) into the modified Bessel equation:\[x^2 I_0''(x) + x I_0'(x) - x^2 I_0(x) = 0.\]Substitute the expressions found in Steps 2 and 3, simplify, and verify that the equality holds. This confirms that \(I_0(x)\) satisfies the modified Bessel equation of order zero.
05

No Admixture of \(K_0(x)\)

The second solution, \(K_0(x)\), diverges logarithmically as \(x\to 0\), whereas \(I_0(x)\) remains finite at \(x=0\). Thus, a finite solution showing no diverging term means \(I_0(x)\) contains no admixture of \(K_0(x)\).
06

Verify the Normalization Factor 1/蟺

To verify the normalization factor, compute \(I_0(0)\). Simplify for \(x=0\):\[I_{0}(0) = \frac{1}{\pi} \int_{0}^{\pi} \cosh (0) d\theta = \frac{1}{\pi} \int_{0}^{\pi} 1 d\theta \ = 1.\]This confirms that the normalization factor \(1/\pi\) is correct, as \(I_0(0)=1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bessel Differential Equation
The Bessel differential equation is crucial in understanding patterns of waves, vibrations, and various physical systems. It is typically presented as:
  • Modified Bessel equation: \(x^2 y'' + x y' - x^2 y = 0\) for order zero.
  • It describes scenarios where radial symmetry is prominent in various fields like heat conduction and wave propagation.
The function \(I_0(x)\) represents the modified Bessel function of the first kind. To verify it satisfies the Bessel equation, you differentiate the function several times to obtain first and second derivatives. Subsequently, these derivatives are substituted back into the original equation. When doing so, if all terms consistently balance to zero, \(I_0(x)\) is confirmed to be a valid solution of the equation.
Normalization Factor
The normalization factor ensures that the Bessel functions maintain a consistent structure, especially as they are solutions to complex differential equations. In this exercise, the factor is \(\frac{1}{\pi}\).
  • It's vital for ensuring integral evaluations start or end at specific values.
  • Maintains the stability of the function even as variables reach limiting values.
To verify this factor, consider \(I_0(0)\). Here, the integral evaluates to 1 due to the integral of constants over a finite boundary. Checking this at \(x=0\), the integral becomes straightforward, and thus confirms the factor as correct, simplifying the form of any solutions derived.
Regular and Irregular Solutions
In the context of Bessel functions, regular and irregular solutions define behavior at critical points.
  • Regular solutions like \(I_0(x)\) remain finite as \(x\) approaches zero, meaning they don't blow up to infinity.
  • Irregular solutions, such as \(K_0(x)\), exhibit a logarithmic divergence as \(x\) nears zero.
This distinction is crucial when seeking physically meaningful solutions to differential equations, especially in applications like thermal diffusion or electromagnetic fields. Demonstrating that \(I_0(x)\) does not incorporate any part of \(K_0(x)\) assures us it remains finite and steady, without mixing behaviors indicative of irregular solutions.

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Most popular questions from this chapter

A right circular cylinder has an electrostatic potential of \(\psi(\rho, \varphi)\) on both ends. The potential on the curved cylindrical surface is zero. Find the potential at all interior points. Hint. Choose your coordinate system and adjust your \(z\) dependence to exploit the symmetry of your potential.

The quantum mechanical radial wavefunction for a scattered wave is given by $$ \psi_{k}=\frac{\sin \left(k r+\delta_{0}\right)}{k r} $$ where \(k\) is the wavenumber, \(k=\sqrt{2 m E / \hbar}\), and \(\delta_{0}\) is the scattering phase shift. Show that the normalization integral is $$ \int_{0}^{\infty} \psi(r) \psi_{k^{\prime}}(r) r^{2} d r=\frac{\pi}{2 k} \delta\left(k-k^{\prime}\right) . $$

The differential cross section in a nuclear scattering experiment is given by \(d \sigma / d \Omega=|f(\theta)|^{2}\). An approximate treatment leads to $$ f(\theta)=\frac{-i k}{2 \pi} \int_{0}^{2 \pi} \int_{0}^{R} \exp [i k \rho \sin \theta \sin \varphi] \rho d \rho d \varphi . $$ Here, \(\theta\) is an angle through which the scattered particle is scattered, and \(R\) is the nuclear radius. Show that $$ \frac{d \sigma}{d \Omega}=\left(\pi R^{2}\right) \frac{1}{\pi}\left[\frac{J_{1}(k R \sin \theta)}{\sin \theta}\right]^{2} $$

(a) Show by direct differentiation and substitution that $$ J_{\nu}(x)=\frac{1}{2 \pi i} \int_{C} e^{(x / 2)(t-1 / t)} t^{-\nu-1} d t $$ or that the equivalent equation $$ J_{v}(x)=\frac{1}{2 \pi i}\left(\frac{x}{2}\right)^{v} \int e^{s-x^{2} / 4 s} s^{-v-1} d s $$ satisfies Bessel's equation; \(C\) is the contour shown in Fig. 11.4. The negative real axis is cut line. Hint. Show that the total integrand (after substituting in Bessel's differential equation) may be written as a total derivative: $$ \frac{d}{d t}\left\\{\exp \left[\frac{x}{2}\left(t-\frac{1}{t}\right)\right] t^{-v}\left[v+\frac{x}{2}\left(t+\frac{1}{t}\right)\right]\right\\} . $$ (b) Show that the first integral (with \(n\) an integer) may be transformed into $$ J_{n}(x)=\frac{1}{2 \pi} \int_{0}^{2 \pi} e^{i(x \sin \theta-n \theta)} d \theta=\frac{i^{-n}}{2 \pi} \int_{0}^{2 \pi} e^{i(x \cos \theta+n \theta)} d \theta . $$

If \(r=\left(x^{2}+y^{2}\right)^{1 / 2}\), prove that $$ \frac{1}{r}=\frac{2}{\pi} \int_{0}^{\infty} \cos (x t) K_{0}(y t) d t . $$ This is a Fourier cosine transform of \(K_{0}\).
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