Question

By initially writing $y(x)$ as $x^{1/2}f(x)$ and then making subsequent changes of variable, reduce $$\frac{d^{2}y}{d{x}^2}+\lambda xy=0$$ to Bessel's equation. Hence show that a solution that is finite at $x=0$ is a multiple of $x^{1/2}{J}_{1/3}\left(\frac{2}{3}\sqrt{\lambda x^3}\right)$.

Short answer

The solution that is finite at $x=0$ is $x^{1/2}{J}_{1/3}(\frac{2}{3}\sqrt{\lambda x^3})$.

Step-by-step solution

Substitute the given form of y(x)

Write the given function as: $$y(x)=x^{1/2}f(x)$$ Now, calculate the first and second derivatives of this function with respect to x.

Calculate the first derivative

First, apply the chain rule of differentiation: $$\frac{dy}{dx}=\frac{d}{dx}(x^{1/2}f(x))=x^{-1/2}f(x)+x^{1/2}\frac{df}{dx}$$

Calculate the second derivative

Again, apply the chain rule: $$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}(x^{-1/2}f(x)+x^{1/2}\frac{df}{dx})=-\frac{1}{2}{x}^{-3/2}f(x)+2x^{-1/2}\frac{df}{dx}+x^{1/2}\frac{d^{2}f}{d{x}^2}$$

Substitute back into the original differential equation

Now, substitute $y(x)=x^{1/2}f(x)$ and its second derivative into the original differential equation: $$\frac{d^{2}y}{d{x}^2}+\text{bml}xy=0$$ This becomes: $$-\frac{1}{2}{x}^{-3/2}f(x)+2x^{-1/2}\frac{df}{dx}+x^{1/2}\frac{d^{2}f}{d{x}^2}+\lambda x^{3/2}f(x)=0$$

Simplify the equation

Multiply through by $x^{3/2}$ to clear denominators: $$-\frac{1}{2}f(x)+2x\frac{df}{dx}+x^2\frac{d^{2}f}{d{x}^2}+\lambda x^{3}f(x)=0$$ Collect terms involving $f(x)$: $$x^2\frac{d^{2}f}{d{x}^2}+2x\frac{df}{dx}+(\lambda x^3-\frac{1}{2})f(x)=0$$

Introduce a change of variable

Let $t=\sqrt{\lambda }{x}^{3/2}$, hence: $$t=\frac{3}{2}\sqrt{\lambda }{x}^{1/2}$$ Then, rewrite the differential equation in terms of t.

Rewrite and identify Bessel's equation

Using chain rule, derivatives transform as follows: $$\frac{d}{dx}=\frac{dt}{dx}\frac{d}{dt}=\frac{3}{2}\sqrt{\lambda }{x}^{-1/2}\frac{d}{dt}$$ Thus, the differential equation becomes Bessel's equation: $$t^2\frac{d^{2}f}{d{t}^2}+t\frac{df}{dt}+(t^2-(\frac{1}{3}{)}^2)f=0$$ This is the standard form of Bessel’s equation with order $\frac{1}{3}$.

Show the solution in terms of Bessel function

The general solution is given by Bessel function of the first kind, for $\frac{1}{3}$, which is finite at zero: $$f(t)=J_{1/3}(t/\sqrt{\lambda })=J_{1/3}(\frac{2}{3}\sqrt{\lambda }{x}^{3/2})$$ Therefore, $$y(x)=x^{1/2}f(x)=x^{1/2}{J}_{1/3}(\frac{2}{3}\sqrt{\lambda }{x}^{3/2})$$

Key concepts

Differential Equations

Differential equations are mathematical equations that involve derivatives of an unknown function. These derivatives represent rates of change, and the aim is to find the function that satisfies the equation. One specific type we deal with here is a second-order linear differential equation: $$\frac{d^{2}y}{d{x}^2}+\text{bml}xy=0$$This equation involves the function y(x) and its second derivative with respect to x. To solve such equations, we might often use techniques like substitution or changes of variables to simplify them into forms we can recognize and solve more easily. Generally, these equations appear in various fields like physics and engineering to model phenomena.

Change of Variables

The 'change of variables' technique is used to transform a difficult equation into a simpler or more familiar form. In this specific problem, we start by assuming a form: $$y(x)=x^{1/2}f(x)$$ This helps us separate the power of x from the unknown function f(x). After substituting y(x) into the original differential equation and simplifying, we introduce a new variable: $$t=\frac{3}{2}\text{bml}{x}^{3/2}$$ This transformation makes it easier to recognize the modified expression as a known type of differential equation, such as Bessel's equation. By rewriting our derivatives with respect to t, we match the standard forms we are familiar with.

Bessel Function

Bessel functions are solutions to Bessel's differential equation, which appears frequently in problems with cylindrical or spherical symmetry, such as heat conduction, vibrations, and electromagnetic waves. In our transformed equation, we recognize it as a form of Bessel's equation: $$t^2\frac{d^{2}f}{d{t}^2}+t\frac{df}{dt}+(t^2-(\frac{1}{3}{)}^2)f=0$$ The solution to such an equation is given by Bessel functions of the first kind.For our specific order, this is: $$y(x)=x^{1/2}{J}_{1/3}(\frac{2}{3}\text{bml}{x}^{3/2})$$Where $J_{1/3}$ is the Bessel function of order $\frac{1}{3}$.

Chain Rule

The chain rule is a fundamental rule in calculus used to differentiate compositions of functions. When we have a function of another function, the chain rule helps us take the derivative.For our first substitution: $$y(x)=x^{1/2}f(x)$$ we need to apply the chain rule to find the first and second derivatives. For the first derivative: $$\frac{dy}{dx}=x^{-1/2}f(x)+x^{1/2}\frac{df}{dx}$$ For the second derivative: $$\frac{d^{2}y}{d{x}^2}=-\frac{1}{2}{x}^{-3/2}f(x)+2x^{-1/2}\frac{df}{dx}+x^{1/2}\frac{d^{2}f}{d{x}^2}$$ Each term is derived by differentiating the product or composition of functions and applying the chain rule.

Mathematical Methods in Physics

Mathematical methods, like the ones used in this problem, are critical in physics to model and solve complex physical phenomena. Differential equations, changes of variables, and special functions like Bessel functions all provide tools to describe and analyze systems. In fields such as quantum mechanics, electromagnetism, and thermal dynamics, these methods help bridge the gap between abstract mathematics and real-world physical systems. Solving such equations often requires a mix of intuitive understanding and technical manipulation, demonstrating the deep overlap between math and physics.