Problem 19 - Differential Equations and Linear Algebra

Question

Let \(J_{p}(x)\) denote the Bessel function of the first kind of order \(p,\) and let \(\lambda\) be a positive real number. If \(u(x)=\) \(J_{p}(\lambda x),\) show that \(u\) satisfies the differential equation $$ \frac{d^{2} u}{d x^{2}}+\frac{1}{x} \frac{d u}{d x}+\left(\lambda^{2}-\frac{p^{2}}{x^{2}}\right) u=0 $$

Step-by-Step Solution

Answer

To show that \(u(x) = J_p(\lambda x)\) satisfies the given differential equation, we find the first and second derivatives of \(u(x)\) with respect to \(x\). We then substitute these derivatives along with the function \(u(x)\) into the given differential equation and verify that both sides match Bessel's differential equation, thereby demonstrating that \(u(x)\) indeed satisfies the given differential equation.

Step 1: Find the first derivative of u(x)

To find the first derivative of \(u(x) = J_p(\lambda x)\) with respect to x, we apply the chain rule as follows: $$ \frac{d u(x)}{d x}=\frac{d J_p(\lambda x)}{d (\lambda x)}\cdot\frac{d (\lambda x)}{d x} $$ The derivative of \(J_p(\lambda x)\) with respect to \(\lambda x\) is denoted as \(J_p'(\lambda x)\). And, the derivative of \(\lambda x\) with respect to x is simply \(\lambda\). Thus, the first derivative of u(x) is: $$\frac{d u(x)}{d x} = J_p'(\lambda x) \cdot \lambda$$

Step 2: Find the second derivative of u(x)

Now, we need to find the second derivative of \(u(x)\), which means we have to differentiate \(\frac{d u(x)}{d x}\) with respect to x: $$ \frac{d^2 u(x)}{d x^2} =\frac{d (J_p'(\lambda x) \cdot \lambda)}{d (\lambda x)}\cdot\frac{d (\lambda x)}{d x} $$ Now, the derivative of \(J_p'(\lambda x) \cdot \lambda\) with respect to \(\lambda x\) is denoted as \(J_p''(\lambda x)\). And, the derivative of \(\lambda x\) with respect to x is still \(\lambda\). Hence, the second derivative of u(x) is: $$\frac{d^2 u(x)}{d x^2} = J_p''(\lambda x) \cdot \lambda^2$$

Step 3: Plug the derivatives and u(x) into the given differential equation

Now, we have all the derivatives we need, and we can plug them into the given differential equation: $$ J_p''(\lambda x) \cdot \lambda^2 + \frac{1}{x} (J_p'(\lambda x) \cdot \lambda) + \left(\lambda^2 - \frac{p^2}{x^2}\right) J_p(\lambda x) = 0 $$

Step 4: Verify the Bessel's differential equation

We know that the Bessel function \(J_p(\lambda x)\) satisfies Bessel's differential equation of order p: $$ \frac{d^2 J_p(\lambda x)}{d (\lambda x)^2} + \frac{1}{\lambda x} \frac{d J_p(\lambda x)}{d (\lambda x)} + \left(1 - \frac{p^2}{(\lambda x)^2}\right) J_p(\lambda x) = 0 $$ Comparing this equation with our substituted equation from step 3, we can see that: $$ J_p''(\lambda x) \cdot \lambda^2 = \frac{d^2 J_p(\lambda x)}{d (\lambda x)^2} $$ $$ \frac{1}{x} (J_p'(\lambda x) \cdot \lambda) = \frac{1}{\lambda x} \frac{d J_p(\lambda x)}{d (\lambda x)} $$ $$ \left(\lambda^2 - \frac{p^2}{x^2}\right) J_p(\lambda x) = \left(1 - \frac{p^2}{(\lambda x)^2}\right) J_p(\lambda x) $$ Since both sides of the given differential equation match Bessel's differential equation, we have shown that the function \(u(x) = J_p(\lambda x)\) satisfies the given differential equation.

Key Concepts

Bessel Functions

Bessel functions are an important concept in many areas of mathematics and physics. These functions arise in solutions to differential equations with cylindrical symmetry, such as those describing wave propagation and heat conduction in circular geometries.

• The Bessel function of the first kind, denoted as \( J_p(x) \), is a common type of Bessel function where \( p \) represents the order. These functions are oscillatory and converge for all real \( x \).
• Bessel functions are used in various applications including signal processing, electromagnetics, and acoustics, because they naturally describe the modes of vibration in circular membranes and other cylindrical problems.

Recognizing their importance, Bessel functions like \( J_{p}(x) \) provide critical solutions to the associated Bessel's differential equation, which models many physical phenomena. Understanding how these functions behave, even outside the bounds of simple curves, is key to mastering these complex problems.

Differential Equations

A differential equation is a mathematical equation that relates a function to its derivatives. In simple terms, they express how a particular quantity changes in relation to changes in another.

• Differential equations can describe a wide range of phenomena, from population growth to mechanical vibrations.
• They can be classified into different types, such as ordinary differential equations (ODEs) and partial differential equations (PDEs).

Bessel's differential equation is a second-order ODE involving the second derivative and the first derivative of a function.
Understanding how to manipulate and solve these equations is essential to predicting systems' behavior in fields ranging from physics to engineering and beyond.

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions, i.e., where one function is applied to the result of another function. It's particularly useful for finding derivatives of nested functions where straightforward differentiation isn't possible.

• In the context of the exercise, we find the function \( u(x) = J_p(\lambda x) \). Here, the Bessel function is applied after scaling by \( \lambda \), necessitating the usage of the chain rule.
• The general formula for the chain rule is: if \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

To differentiate \( u(x) \), the chain rule allows us to take the derivative of the outer function and multiply it by the derivative of the inner function. This step is crucial in arriving at the correct derivative expressions needed for solving the differential equation associated with Bessel functions.

Derivative Calculation

Derivative calculation is the process of determining the rate at which a function is changing at any given point. With respect to Bessel functions and their associated differential equations, calculating derivatives accurately is important.

• For a function \( u(x) = J_p(\lambda x) \), finding the first and second derivatives involves applying the chain rule.
• The first derivative \( \frac{du}{dx} \) factors in the derivative \( J_p'(\lambda x) \) and the scaling factor \( \lambda \).
• The second derivative \( \frac{d^2u}{dx^2} \) builds on this result to compute \( J_p''(\lambda x) \cdot \lambda^2 \).

These derivative calculations are integrated into the differential equation to validate the solution as a valid Bessel function representation. Recognizing each derivative's role helps understand the function's behavior within the context of its differential equation formulation.