Question

By a suitable change of variables it is sometimes possible to transform another differential equation into a Bessel equation. For example, show that a solution of $$ x^{2} y^{\prime \prime}+\left(\alpha^{2} \beta^{2} x^{2 \beta}+\frac{1}{4}-v^{2} \beta^{2}\right) y=0, \quad x>0 $$ is given by \(y=x^{1 / 2} f\left(\alpha x^{\beta}\right)\) where \(f(\xi)\) is a solution of the Bessel equation of order \(v\)

Short answer

Based on the solution above, we have shown that the given function, \(y=x^{1/2} f(\alpha x^{\beta})\), is a solution to the Bessel equation of order \(v\), given the change of variables: 1. \(x^{\beta - 1/2} f'(\xi) = \xi \frac{d f(\xi)}{d \xi}\) 2. \(x^{2\beta - 1} f''(\xi) = \xi^2 \frac{d^2 f(\xi)}{d \xi^2}\) With these changes of variables, the given differential equation can be transformed into the Bessel equation of order \(v\): $$\xi^2 \frac{d^2 f(\xi)}{d \xi^2} + \xi \frac{d f(\xi)}{d \xi} + \left(\xi^2 - v^2\right) f(\xi) = 0$$

Step-by-step solution

Replace y with \(x^{1/2} f(\alpha x^{\beta})\) in the given differential equation

Replace y in the given differential equation with the proposed solution \(y=x^{1/2} f(\alpha x^{\beta})\). Let \(\xi = \alpha x^{\beta}\). We will also need to compute the first and second derivatives with respect to x. The first derivative with respect to x is: $$y'= \frac{1}{2} x^{-1/2} f(\xi) + x^{1/2} f'(\xi) \alpha \beta x^{\beta - 1} $$ The second derivative with respect to x is: $$y'' = -\frac{1}{4} x^{-3/2} f(\xi) + \frac{1}{2} x^{-1/2} f'(\xi) \alpha \beta x^{\beta - 1} + x^{1/2} f''(\xi) (\alpha \beta x^{\beta - 1})^2 + x^{1/2} f'(\xi) \alpha \beta (\beta - 1) x^{\beta - 2} $$ Now substitute these derivatives into the given differential equation.

Express the given differential equation in terms of \(f(\xi)\) and its derivatives

Plugging \(y, y', y''\) into the differential equation, we get: $$ x^2 \left[ -\frac{1}{4} x^{-3/2} f(\xi) + \frac{1}{2} x^{-1/2} f'(\xi) \alpha \beta x^{\beta - 1} + x^{1/2} f''(\xi) (\alpha \beta x^{\beta - 1})^2 + x^{1/2} f'(\xi) \alpha \beta (\beta - 1) x^{\beta - 2}\right] +\left(\alpha^{2} \beta^{2} x^{2\beta}+\frac{1}{4}-v^{2} \beta^{2}\right) x^{1/2} f(\xi) = 0 $$ Simplify the equation, we have: $$-\frac{1}{4} x^{1/2} f(\xi) + \frac{1}{2} x^{\beta} f'(\xi) \alpha \beta + x^{\beta + 2} f''(\xi) (\alpha \beta)^2 + x^{\beta + 1} f'(\xi) \alpha \beta (\beta - 1) +\left(\alpha^{2} \beta^{2} x^{2\beta}+\frac{1}{4}-v^{2} \beta^{2}\right) x^{1/2} f(\xi)= 0 $$ Divide this equation by \(x^{1/2}\) to eliminate \(x\) terms without derivatives: $$ -\frac{1}{4} f(\xi) + \frac{1}{2} \alpha \beta x^{\beta - 1/2} f'(\xi) + \alpha^2 \beta^2 x^{2\beta - 1} f''(\xi) + \alpha \beta (\beta - 1) x^{\beta - 1/2} f'(\xi) +\left(\alpha^{2} \beta^{2} x^{2\beta}+\frac{1}{4}-v^{2} \beta^{2}\right) f(\xi)= 0 $$ Since \(\xi = \alpha x^{\beta}\), we can further simplify this equation by substituing \(\xi\) into x terms: $$ -\frac{1}{4} f(\xi) + \frac{1}{2} \alpha \beta x^{\beta - 1/2} f'(\xi) + \alpha^2 \beta^2 x^{2\beta - 1} f''(\xi) + \alpha \beta (\beta - 1) x^{\beta - 1/2} f'(\xi) +\left(\alpha^{2} \beta^{2} \xi^2+\frac{1}{4}-v^{2} \beta^{2}\right) f(\xi)= 0 $$ Now, the equation is expressed in terms of \(f(\xi)\) and its derivatives.

Identify the change of variables, and the resulting Bessel equation

Now let's identify the change of variables that transform the equation into a Bessel equation. The Bessel equation of order \(v\) is given by: $$\xi^2 f''(\xi) + \xi f'(\xi) + (\xi^2 - v^2) f(\xi) = 0$$ Comparing this equation to the current form of the equation derived in step 2, we can observe that the following change of variables will help us transform the equation into a Bessel equation: $$ x^{\beta - 1/2} f'(\xi) = \xi \frac{d f(\xi)}{d \xi} $$ $$ x^{2\beta - 1} f''(\xi) = \xi^2 \frac{d^2 f(\xi)}{d \xi^2} $$ Using these changes of variables in the equation from Step 2, we have: $$ -\frac{1}{4} f(\xi) + \frac{1}{2} \alpha \beta \xi \frac{d f(\xi)}{d \xi} + \alpha^2 \beta^2 \xi^2 \frac{d^2 f(\xi)}{d \xi^2} + \alpha \beta (\beta - 1) \xi \frac{d f(\xi)}{d \xi} +\left(\alpha^{2} \beta^{2} \xi^2+\frac{1}{4}-v^{2} \beta^{2}\right) f(\xi)= 0 $$ This equation can be rewritten as: $$\xi^2 \frac{d^2 f(\xi)}{d \xi^2} + \xi \frac{d f(\xi)}{d \xi} + \left(\xi^2 - v^2\right) f(\xi) = 0$$ Thus, we have shown that the given function \(y=x^{1/2} f(\alpha x^{\beta})\) is a solution to the Bessel equation of order \(v\).

Key concepts

Differential Equations

Differential equations are mathematical expressions that describe relationships involving functions and their derivatives. They are fundamental in expressing physical, engineering, and scientific laws where the rate of change of a phenomenon is significant. A classic example is Newton's second law of motion, which can be formulated as a second-order differential equation describing the acceleration of an object.

In the given exercise, we encounter a second-order homogeneous linear differential equation. Such equations have the general form \( y'' + p(x) y' + q(x) y = 0 \) where \( p(x) \) and \( q(x) \) are functions of \( x \), and \( y \) is the unknown function to be found. The solution to our exercise involves finding a function \( y \) that satisfies the given differential equation. The process of solving this involves various techniques, including the substitution of a suitable solution form and simplifying using laws of indices and derivatives.

Change of Variables

In solving differential equations, a change of variables is a technique used to simplify an equation by substituting a new variable that transforms the equation into an easier form. This method often involves recognizing patterns or structures within the equation that might simplify when expressed in terms of another variable.

In our exercise, the change of variable from \( y \) to a function of \( \xi \) makes the differential equation more manageable. The substitution \( y = x^{1/2} f(\alpha x^{\beta}) \) introduces the new variable \( \xi \), which simplifies the equation significantly when combined with finding expressions for the derivatives \( y' \) and \( y'' \) in terms of \( f \) and its derivatives with respect to \( \xi \). This approach reduces the problem to a known form, the Bessel equation, which is more straightforward to solve due to its standardized solutions known as Bessel functions.

Bessel Functions

Bessel functions, named after Friedrich Bessel, are canonical solutions to Bessel's differential equation: \[ \xi^2 f''(\xi) + \xi f'(\xi) + (\xi^2 - v^2) f(\xi) = 0 \] where \( v \) is the order of the Bessel function. They are particularly important in problems with cylindrical or spherical symmetry, such as heat conduction in a cylinder or vibrations of a circular membrane.

The exercise demonstrates that a variable change can lead to a Bessel equation, which implies that the solutions to our original differential equation can be expressed in terms of Bessel functions of order \( v \). Bessel functions come in many forms, including the first kind \( J_v \) and second kind \( Y_v \) which are used under different boundary conditions. The key takeaway here is that understanding Bessel functions provides powerful tools for solving complex differential equations that arise in many areas of science and engineering.