Question

Verify that the differential equation \(x^{4} y^{\prime \prime}+y=0\) is not Fuchsian; that it has the two independent solutions \(x \sin (1 / x)\) and \(x \cos (1 / x) ;\) and that these solutions are not expandable in Frobenius series.

Short answer

The equation is not Fuchsian. The solutions are \(x \sin(1/x)\) and \(x \cos(1/x)\). These solutions are not expandable in Frobenius series due to essential singularities at \(x = 0\).

Step-by-step solution

Identify the Differential Equation

The differential equation given is: \[ x^{4} y^{\text{''}} + y = 0 \]

Verify if the Equation is Fuchsian

A differential equation is Fuchsian if all its singular points are regular. To be regular, the coefficients of the lower-order terms divided by the leading coefficient must be analytic (holomorphic) at the singularity: Here, rewrite the equation in standard form:\[ y^{\text{''}} + \frac{1}{x^{4}} y = 0 \]The coefficient \(\frac{1}{x^{4}}\) is not analytic at \(x = 0\) (it has a pole of order 4). Therefore, the equation is not Fuchsian.

Identify the Given Solutions

The solutions given are:\[ y_1 = x \sin(1/x) \]\[ y_2 = x \cos(1/x) \]

Verify that the Given Solutions Satisfy the Differential Equation

To verify that these functions are solutions, find their second derivatives and substitute them back into the equation:For \(y_1 = x \sin(1/x)\):First derivative:\[ y_1' = \sin(1/x) - \frac{ \cos(1/x)}{x} \]Second derivative:\[ y_1'' = -2 \frac{\text{cos}(1/x)}{x^{2}}-\frac{ \sin(1/x)}{x^{3}} \]Substituting in the differential equation:\[ x^{4}y_1'' + y_1 = x^{4} \left( -2 \frac{\text{cos}(1/x)}{x^{2}}-\frac{ \sin(1/x)}{x^{3}} \right) + x \sin(1/x) = -2x^{2} \text{cos}(1/x) - x \sin(1/x) + x \sin(1/x) = 0\]Hence, \( y_1 \) is a solution.Similarly, for \(y_2 = x \cos(1/x)\):First derivative:\[ y_2' = \cos(1/x) + \frac{ \sin(1/x)}{x} \]Second derivative:\[ y_2'' = -2 \frac{\sin(1/x)}{x^{2}}+\frac{\text{cos}(1/x)}{x^{3}} \]Substituting in the differential equation:\[ x^{4}y_2'' + y_2 = x^{4} \left( -2 \frac{\sin(1/x)}{x^{2}}+\frac{\text{cos}(1/x)}{x^{3}} \right) + x \cos(1/x) = -2x^{2} \text{sin}(1/x)+ x \text{cos}(1/x) + x \text{cos}(1/x) = 0\]Hence, \( y_2 \) is a solution.

Determine Whether the Solutions are Expandable in Frobenius Series

Consider the form of Frobenius series expansion: \[ y(x) = \sum_{n=0}^{\infty}a_n x^{n+s} \]The series expansion must be analytic at x=0. However, both solutions involve trigonometric functions of \(1/x\), which have essential singularities at \(x = 0\), meaning they are not analytic at \(x = 0\) and thus cannot be represented as Frobenius series.

Key concepts

Fuchsian Differential Equations

A Fuchsian differential equation is a special type of linear differential equation characterized by its singular points. Singular points in differential equations are places where solutions may behave erratically or even become undefined. However, for a differential equation to be Fuchsian, all its singular points must be 'regular' or 'regular singular points'.
What does this mean? At a regular singular point, the coefficients of the differential equation, when divided by the leading term, must be analytic. Analyticity here means that the resulting function can be expressed as a power series around that point.
For example, consider the equation given in the exercise: \(x^{4} y^{\text{''}} + y = 0\). Rewrite it in standard form:\(y^{\text{''}} + \frac{1}{x^{4}} y = 0\). Notice that \(\frac{1}{x^{4}}\) isn't analytic at \(x = 0\) since it has a pole of order 4. The lack of analyticity at \(x = 0\) indicates that the equation is not Fuchsian.
This sets the stage for understanding the Frobenius method and how it pertains to solving such equations.

Frobenius Series

The Frobenius series is a type of power series used to solve linear differential equations, particularly around singular points. Unlike regular power series that only work at regular points, a Frobenius series incorporates a shift factor to handle singularities.
The general form of a Frobenius series is given by: \(y(x) = \sum_{n=0}^{\infty}a_n x^{n+s}\). Here, \(s\) is a constant that needs to be determined and \(a_n\) are the coefficients. The series converges around the singular point if \(s\) is chosen correctly.
In our example, the solutions provided are \(x \sin(1/x)\) and \(x \cos(1/x)\). These expressions involve trigonometric functions of \(1/x\), which exhibit essential singularities at \(x = 0\). Therefore, these solutions cannot be represented as Frobenius series because they are not analytic at \(x = 0\).
This difference in the nature of the singularity (essential vs. regular) is why the exercise concludes that these solutions are not expandable in Frobenius series.

Singular Points

Singular points of a differential equation are locations where the behavior of solutions changes dramatically. Singularities are classified into two categories: regular and irregular.
A regular singular point allows the differential equation to be manipulated such that the resulting coefficients are analytic. This property makes it possible to use methods like the Frobenius series to find solutions around these points.
Irregular singular points, on the other hand, involve coefficients that are not analytic (often having poles of higher orders). These types of singularities complicate the solutions and often require alternative methods for analysis.
In the context of the exercise, \(x = 0\) is an irregular singular point because the coefficient \(\frac{1}{x^{4}}\) isn't analytic. This irregularity is why the given solutions \(x \sin(1/x)\) and \(x \cos(1/x)\) are valid but do not conform to the Frobenius framework.
Understanding singular points is crucial for determining the appropriate method to solve differential equations and anticipate the behavior of solutions around these critical junctures.