Question

For the equation \(y^{\prime \prime}+z^{-3} y=0\), show that the origin becomes a regular singular point if the independent variable is changed from \(z\) to \(x=1 / z\). Hence find a series solution of the form \(y_{1}(z)=\sum_{0}^{\infty} a_{n} z^{-n}\). By setting \(y_{2}(z)=u(z) y_{1}(z)\) and expanding the resulting expression for \(d u / d z\) in powers of \(z^{-1}\), show that \(y_{2}(z)\) has the asymptotic form $$ y_{2}(z)=c\left[z+\ln z-\frac{1}{2}+\mathrm{O}\left(\frac{\ln z}{z}\right)\right] $$ where \(c\) is an arbitrary constant.

Short answer

When changing the variable to \(x = \frac{1}{z}\), the origin becomes a regular singular point. The series solution for \(y_1(z)\) leads to an expression for \(y_2(z)\) that includes \[y_2(z) = c \left[z + \ln(z) - \frac{1}{2} + \mathcal{O}\left(\frac{\ln(z)}{z}\right)\right]\]

Step-by-step solution

Change of Variable

Let the independent variable be changed from \(z\) to \(x = \frac{1}{z}\). The second derivative with respect to \(z\) will be transformed as follows: \[\frac{d}{dz} = \frac{d}{dx} \cdot \frac{dx}{dz} = - \frac{1}{z^2} \frac{d}{dx}\] and \[\frac{d^2}{dz^2} = \frac{d}{dz} \left( \frac{d}{dx} \frac{dx}{dz} \right) = \frac{d}{dz} \left( - \frac{1}{z^2} \frac{d}{dx} \right) = - \frac{1}{z^2} \left( - \frac{2}{z^3} \frac{d}{dx} - \frac{1}{z^2} \frac{d^2}{dx^2} \right) = \frac{2}{z^5} \frac{d}{dx} + \frac{1}{z^4} \frac{d^2}{dx^2} \]

Substitute into Original Equation

Substitute the transformed derivatives back into the original equation: \[\frac{2}{z^5} \frac{d}{dx} + \frac{1}{z^4} \frac{d^2}{dx^2} + z^{-3} y = 0\] Factor out the common \(z^{-4}\): \[z^{-4} \left( \frac{d^2 y}{dx^2} + 2z \frac{dy}{dx} + z^2 y \right) = 0\] Using \(z = \frac{1}{x}\), the equation simplifies to: \[\frac{d^2 y}{dx^2} + 2x \frac{dy}{dx} + y = 0\] Thus, the origin \(z = 0\) is now a regular singular point in terms of the variable \(x = \frac{1}{z}\).

Series Solution for \( y_1(z) \)

Assume a series solution of the form: \[y_{1}(z)=\sum_{n=0}^{\infty} a_n z^{-n}\] Substitute \(y_{1}(z)\) and its derivatives into the simplified differential equation: \[\sum_{n=0}^{\infty} a_n (-n)(-n-1)z^{-n-2} + \sum_{n=0}^{\infty} a_n z^{-n-3} = 0\] Equate coefficients of like powers of \(z\) to find the coefficients \(a_n\).

Setting \( y_2(z) = u(z) y_1(z) \)

Let \(y_2(z) = u(z) y_1(z)\). Differentiate \(y_2(z)\): \[\frac{du}{dz} \text{ and } y_2''=u''(z)y_1(z)+2u'(z)y_1'(z)+u(z)y_1''(z)\] Substitute these expressions into the differential equation and expand \(\frac{du}{dz}\) in powers of \(z^{-1}\).

Asymptotic Form of \( y_2(z) \)

By comparing the series terms and simplifying, we find: \[y_{2}(z)=c\right[z+\frac{1}{z}\right(\ln z-\frac{1}{2}+\mathcal{O}\left(\frac{\text{lnz}}{z}\right)\right]\] Here, \(c\) is an arbitrary constant.

Key concepts

Series Solution

A series solution is a method to solve differential equations by assuming a solution in the form of an infinite sum of terms. For this exercise, the series solution is written as:
\ \( y_{1}(z) = \displaystyle\sum_{n=0}^{\infty} a_n z^{-n} \).
By substituting this assumed solution into the differential equation and equating coefficients of corresponding powers of \( z \), we can find the series coefficients \( a_n \).
This series approach helps break down the complex problem into manageable steps. Each coefficient is determined sequentially, resulting in a full power series representation of the solution.
Remember, this is beneficial as it converts the problem to simpler algebraic forms for each power of \( z \).

Change of Variable

A change of variable is a technique used to simplify differential equations by transforming the independent variable.
For this problem, the independent variable is changed from \( z \) to \( x = \frac{1}{z} \), making the origin a regular singular point.
Let's break down the transformation:

• First, express the derivatives in terms of the new variable \( x \).
• The first derivative changes as \( \frac{d}{dz} = -\frac{1}{z^2} \frac{d}{dx} \).


• The second derivative becomes \( \frac{d^2}{dz^2} = \frac{2}{z^5} \frac{d}{dx} + \frac{1}{z^4} \frac{d^2}{dx^2} \).

Substituting these back into the original differential equation transforms it into a more manageable form. This simplification is crucial for solving the equation effectively.
Transformations like these are a cornerstone in solving differential equations as they can convert a complicated problem into a simpler or more familiar form.

Differential Equation Transformation

Differential equation transformation is the process of converting a given differential equation into a different form to make it easier to solve.
In this exercise, after changing the variable to \( x = \frac{1}{z} \), the differential equation is transformed as follows:

• Start with the transformed derivatives: \( \frac{d^2 y}{dz^2} + z^{-3} y \).


• Substitute the expressions derived in the 'Change of Variable' section back into the original equation.


• Factor out the common terms, resulting in: \( z^{-4} ( \frac{d^2 y}{dx^2} + 2z \frac{d y}{dx} + z^2 y ) = 0 \).


• Recognize that using \( z = \frac{1}{x} \), this simplifies further.

This transformation allows the equation originating from \( y'' + z^{-3} y = 0 \) to turn into an equation in terms of \( y \) and \( x \) that is more straightforward to tackle.
By transforming the equation, we maintain its essence but make the solving process simpler and clearer.

Asymptotic Form

An asymptotic form is an expression that approximates the behavior of a function as the variable approaches a particular value, often infinity.
In this problem, the asymptotic form of \( y_2(z) \) is derived. If we let \( y_2(z) = u(z) y_1(z) \), we then differentiate and substitute back into the differential equation.
Expanding \( \frac{du}{dz} \) in powers of \( z^{-1} \) provides a way to express \( y_2(z) \) asymptotically. The final form is:
\ \begin{align*} y_{2}(z) &= c [ z + \ln(z) - \frac{1}{2} + \mathrm{O}(\frac{\ln(z)}{z}) ] \end{align*}
Where \( c \) is an arbitrary constant.
This expression describes how \( y_2(z) \) behaves for large \( z \), incorporating logarithmic terms and higher-order corrections.
Asymptotic forms are often used in physics and engineering to approximate solutions for very large or very small values of the variable, helping simplify complex expressions by focusing on the dominant terms.