The detailed proofs of the key properties of Bessel functions of the first kind \( J_n(x) \) and the second kind \( Y_n(x) \). We’ll focus on:
ODE derivation
Series expansion derivation for \( J_n(x) \)
Parity proof
Recurrence relations
Differentiation identities
Asymptotic behavior
Definition and properties of \( Y_n(x) \)
Bessel’s Differential Equation
We start from the ODE:
\[x^2 y” + x y’ + (x^2 – n^2)y = 0\]
This is derived naturally when separating variables in problems with cylindrical symmetry, such as solving Laplace’s or Helmholtz’s equation in cylindrical coordinates.
Power Series Representation of \( J_n(x) \)
Goal:
Find a series solution centered at \( x = 0 \) using the Frobenius method.
Assume:
\[y(x) = \sum_{m=0}^\infty a_m x^{m + s}, \]
Plug into the ODE:
\[x^2 y” + x y’ + (x^2 – n^2) y = 0\]
Compute:
\( y’ = \sum (m+s)a_m x^{m+s-1} \)
\( y” = \sum (m+s)(m+s-1)a_m x^{m+s-2} \)
Substituting:
\[\sum_{m=0}^{\infty} \left[ a_m (m+s)(m+s-1) + a_m (m+s) – n^2 a_m \right] x^{m+s} + \sum_{m=0}^{\infty} a_m x^{m+s+2} = 0\]
Rewriting and matching powers leads to a recurrence relation, whose solution (after choosing \( s = n \)) yields:
\[J_n(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m!(m+n)!} \left( \frac{x}{2} \right)^{2m + n}\] Parity Property: \( J_n(-x) = (-1)^n J_n(x) \)
From the power series:
\[J_n(-x) = \sum_{m=0}^\infty \frac{(-1)^m}{m!(m+n)!} \left( \frac{-x}{2} \right)^{2m + n}= (-1)^n J_n(x)\]
This uses \( (-x)^k = (-1)^k x^k \), and since \( 2m + n \) has the same parity as \( n \), we factor out \( (-1)^n \). Recurrence Relations
1: \( J_{n-1}(x) + J_{n+1}(x) = \frac{2n}{x} J_n(x) \)
Proof using series:
We start with:
\[J_n(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m!(m+n)!} \left( \frac{x}{2} \right)^{2m + n}\]
Now write \( J_{n-1}(x) \) and \( J_{n+1}(x) \), and combine:
\[J_{n-1}(x) + J_{n+1}(x) = \sum_{m=0}^\infty \left[ \frac{(-1)^m}{m!(m+n-1)!} \left( \frac{x}{2} \right)^{2m + n -1} + \frac{(-1)^m}{m!(m+n+1)!} \left( \frac{x}{2} \right)^{2m + n + 1} \right]\]
By adjusting indices and simplifying, this summation eventually leads to:
\[\frac{2n}{x} J_n(x)\]
More elegantly, this identity is derived using Rodrigues-type formulas, generating functions, or differential equation manipulations. Differentiation Identities
1: \( \frac{d}{dx} J_n(x) = \frac{1}{2}[J_{n-1}(x) – J_{n+1}(x)] \)
Proof:
Use the series:
\[J_n(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m!(m+n)!} \left( \frac{x}{2} \right)^{2m + n}\]
Differentiate term-by-term:
\[\frac{d}{dx} J_n(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m!(m+n)!} (2m+n) \left( \frac{x}{2} \right)^{2m+n – 1} \cdot \frac{1}{2} \]
Now rearrange and match with \( J_{n\pm1}(x) \). With some algebra, this leads to the stated result. 2: \( \frac{d}{dx} (x^n J_n(x)) = x^n J_{n-1}(x) \)
Proof:
Let’s define:
\[u(x) = x^n J_n(x)\]
Then:
\[u’ = n x^{n-1} J_n(x) + x^n J_n'(x)\]
Now use the earlier formula:
\[J_n'(x) = \frac{n}{x} J_n(x) – J_{n+1}(x)\]
Substitute:
\[u’ = n x^{n-1} J_n(x) + x^n \left( \frac{n}{x} J_n(x) – J_{n+1}(x) \right) = 2n x^{n-1} J_n(x) – x^n J_{n+1}(x)\]
But using the recurrence:
\[J_{n-1}(x) = \frac{2n}{x} J_n(x) – J_{n+1}(x)\]
we find:
\[u’ = x^n J_{n-1}(x)\]
QED. Asymptotic Behavior
As \( x \to 0 \):
From the series:
\[J_n(x) \approx \frac{1}{n!} \left( \frac{x}{2} \right)^n \quad \text{(leading term)}\]
This is immediate from the fact that all higher-order terms are of higher powers in \( x \). As \( x \to \infty \):
This requires steepest descent, stationary phase, or WKB method:
\[J_n(x) \approx \sqrt{\frac{2}{\pi x}} \cos\left( x – \frac{n\pi}{2} – \frac{\pi}{4} \right)\]
Sketch of derivation:
Start from the integral representation:
\[J_n(x) = \frac{1}{\pi} \int_0^\pi \cos(n \theta – x \sin \theta) \, d\theta\]
Using the method of stationary phase on this integral for large \( x \), we derive the stated asymptotic formula. Bessel Functions of the Second Kind \( Y_n(x) \)
Definition:
For integer \( n \), define:
\[Y_n(x) = \lim_{u \to n} \frac{J_u(x) \cos(u \pi) – J_{-u}(x)}{\sin(u \pi)}\]
This avoids the singularity in the general definition and gives a function linearly independent of \( J_n(x) \). Wronskian Identity:
\[W[J_n(x), Y_n(x)] = J_n(x) Y_n'(x) – J_n'(x) Y_n(x) = \frac{2}{\pi x}\]
Proof:
This follows from the general theory of second-order linear ODEs. Since \( J_n \) and \( Y_n \) form a fundamental set of solutions, the Wronskian is constant times \( 1/x \). The exact constant can be derived via asymptotics.
Bessel Function of the Second Kind: \( Y_n(x) \)
Definition and Series Representation
For non-integer \( u \), the second-kind Bessel function is defined as:
\[Y_u(x) = \frac{J_u(x)\cos(u\pi) – J_{-u}(x)}{\sin(u\pi)}\]
This definition fails when \( u \in \mathbb{Z} \), since the denominator vanishes. For integer order \( n \), \( Y_n(x) \) is defined by the limit:
\[Y_n(x) = \lim_{u \to n} \frac{J_u(x)\cos(u\pi) – J_{-u}(x)}{\sin(u\pi)}\]
Alternative expression for \( Y_n(x) \)
A standard expression valid for integer \( n \) is:
\[Y_n(x) = \frac{1}{\pi} \left( J_n(x) \ln \left( \frac{x}{2} \right) + \gamma_n(x) \right)\]
where \( \gamma_n(x) \) is an analytic function involving a series. For example, \( Y_0(x) \) has the expansion:
\[Y_0(x) = \frac{2}{\pi} \left[ \ln\left(\frac{x}{2}\right) J_0(x) + \sum_{k=1}^\infty \frac{(-1)^k}{(k!)^2} \left( \frac{x}{2} \right)^{2k} H_k \right]\]
with \( H_k = 1 + \frac{1}{2} + \dots + \frac{1}{k} \) the \( k \)-th harmonic number.
Differential Equation
Just like \( J_n(x) \), the function \( Y_n(x) \) satisfies Bessel’s differential equation:
\[x^2 y” + x y’ + (x^2 – n^2)y = 0\]
Thus, \( Y_n(x) \) and \( J_n(x) \) form a fundamental set of solutions.
Parity Property
Statement:
\[Y_n(-x) = (-1)^n Y_n(x) \quad \text{for integer } n\]
Proof:
From the definition:
\[Y_n(x) = \lim_{u \to n} \frac{J_u(x)\cos(u\pi) – J_{-u}(x)}{\sin(u\pi)}\]
Since \( J_u(-x) = (-1)^u J_u(x) \), and similarly for \( J_{-u} \), one finds:
\[Y_n(-x) = \lim_{u \to n} \frac{(-1)^u J_u(x)\cos(u\pi) – (-1)^{-u} J_{-u}(x)}{\sin(u\pi)}= (-1)^n Y_n(x)\]
Recurrence Relations
1:
\[2n Y_n(x) = x [Y_{n-1}(x) + Y_{n+1}(x)]\]
Proof:
This is identical in form to the recurrence for \( J_n(x) \), and is proved similarly.
Start from the Bessel differential equation and assume the recurrence holds. Use the known relation:
\[J_{n-1}(x) + J_{n+1}(x) = \frac{2n}{x} J_n(x)\]
and take derivatives of \( Y_n(x) \), or derive from Wronskian (see below) and a generating function approach. These derivations show both \( J_n \) and \( Y_n \) satisfy the same recurrence relations.
Differentiation Formula
Statement:
\[\frac{d}{dx} Y_n(x) = \frac{1}{2} \left[ Y_{n-1}(x) – Y_{n+1}(x) \right]\]
Proof:
This follows directly from differentiating the integral representation of \( Y_n(x) \):
\[Y_n(x) = \frac{1}{\pi} \int_0^\pi \sin(x \sin \theta – n\theta) \, d\theta – \frac{1}{\pi} \int_0^\infty \left[ e^{n t} + (-1)^n e^{-nt} \right] e^{-x \sinh t} dt\]
Differentiate under the integral and use trigonometric identities to arrive at the stated result.
Alternatively, one can prove it via the Rodrigues formula or series methods, using the identities for \( J_n(x) \), since:
\[Y_n(x) = \frac{J_n(x)\cos(n\pi) – J_{-n}(x)}{\sin(n\pi)}\]
and all the differentiation properties of \( J_n(x) \) are known.
Wronskian of \( J_n(x) \) and \( Y_n(x) \)
Statement:
\[W[J_n(x), Y_n(x)] = J_n(x) Y_n'(x) – J_n'(x) Y_n(x) = \frac{2}{\pi x}\]
Proof:
This is a general result for linearly independent solutions to second-order ODEs. For the Bessel equation:
\[x^2 y” + x y’ + (x^2 – n^2)y = 0\]
the Wronskian \( W(y_1, y_2) \) satisfies:
\[W(x) = \frac{C}{x}\]
To find the constant \( C \), take asymptotic forms:
\[J_n(x) \sim \sqrt{\frac{2}{\pi x}} \cos\left( x – \frac{n\pi}{2} – \frac{\pi}{4} \right)\]
\[Y_n(x) \sim \sqrt{\frac{2}{\pi x}} \sin\left( x – \frac{n\pi}{2} – \frac{\pi}{4} \right)\]
Then:
\[W \sim \left( \sqrt{\frac{2}{\pi x}} \cos(\dots) \right) \cdot \left( \text{derivative of } Y_n \right) – \text{vice versa}\]
Compute the determinant and simplify. The result is:
\[W[J_n, Y_n] = \frac{2}{\pi x}\]
Asymptotic Behavior
As \( x \to \infty \):
\[Y_n(x) \approx \sqrt{\frac{2}{\pi x}} \sin\left( x – \frac{n\pi}{2} – \frac{\pi}{4} \right)\]
Sketch of Proof:
Start from the integral representation:
\[Y_n(x) = \frac{1}{\pi} \int_0^\pi \sin(x \sin\theta – n\theta) d\theta – \frac{1}{\pi} \int_0^\infty \left[ e^{n t} + (-1)^n e^{-n t} \right] e^{-x \sinh t} dt\]
Use stationary phase or steepest descent on the oscillatory part of the integral. The dominant contribution comes from \( \theta \approx \pi/2 \). This yields the sinusoidal decay.
As \( x \to 0 \):
For \(n\gt0\):
\[Y_n(x) \sim -\frac{(n-1)!}{\pi} \left( \frac{2}{x} \right)^n\]
Proof:
Use the limiting definition involving \( J_u(x) \), and expand both \( J_u \) and \( J_{-u} \) for small \( x \):
\[J_u(x) \sim \frac{1}{\Gamma(u+1)} \left( \frac{x}{2} \right)^u \quad J_{-u}(x) \sim \frac{1}{\Gamma(1-u)} \left( \frac{x}{2} \right)^{-u}\]
Now take the limit \( u \to n \in \mathbb{Z} \) and compute:
\[Y_n(x) \sim \text{const} \cdot x^{-n}\]
which diverges as \( x \to 0 \), unlike \( J_n(x) \).