Question

In each exercise, the exponent \(n\) in the given differential equation is a nonnegative integer. Determine the possible values of \(n\) (if any) for which (a) \(t=0\) is a regular singular point. (b) \(t=0\) is an irregular singular point. $$ y^{\prime \prime}+\frac{1}{\sin t} y^{\prime}+\frac{1}{t^{n}} y=0 $$

Short answer

Question: Determine the possible values of the non-negative integer exponent $n$ for which, (a) $t=0$ is a regular singular point. (b) $t=0$ is an irregular singular point of the given differential equation: $$ y^{\prime \prime}+\frac{1}{\sin t} y^{\prime}+\frac{1}{t^{n}} y=0 $$ Answer: (a) There is no solution for part (a) as $t=0$ cannot be a regular singular point for any value of $n$. (b) For part (b), $t=0$ is an irregular singular point for all non-negative integer values of $n$.

Step-by-step solution

Recall what regular and irregular singular points are

A singular point of a differential equation is a point where the coefficients of the equation become infinite or undefined. For a second-order linear differential equation, $$ y^{\prime \prime} + p(t) y^{\prime} + q(t) y = 0 $$ A singular point is regular if \(tp(t)\) and \(t^2q(t)\) are analytic functions at the point. Otherwise, the singular point is irregular.

Calculate \(p(t)\) and \(q(t)\) and their products with powers of \(t\)

For our given differential equation, we have, $$ p(t) = \frac{1}{\sin t} $$ and $$ q(t) = \frac{1}{t^n} $$ Now let's calculate the products: $$ tp(t) = t\cdot\frac{1}{\sin t} = \frac{t}{\sin t} $$ and $$ t^2q(t) = t^2\cdot\frac{1}{t^n} = \frac{t^2}{t^n} = t^{2-n} $$

Analyze the regularity of the singular point at \(t=0\)

Now, we need to check if \(tp(t)\) and \(t^2q(t)\) are analytic functions at \(t=0\). We know that \(\sin t\) is an analytic function, but it has a zero at \(t=0\). Therefore, \(\frac{1}{\sin t}\) is NOT analytic at \(t=0\), so \(tp(t)\) is not analytic at \(t=0\). Thus, \(t=0\) cannot be a regular singular point for any value of \(n\), so there is no solution for part (a).

Check if \(t=0\) is an irregular singular point

Since \(t=0\) is not a regular singular point, it must be an irregular singular point. For irregular singular points, \(tp(t)\) and/or \(t^2q(t)\) are NOT analytic functions. Since we already showed \(tp(t)\) is not analytic at \(t=0\), there is no need to check for \(t^2q(t)\). Thus, for any non-negative integer value of \(n\), \(t=0\) is an irregular singular point. So, the solution for part (b) is all non-negative integer values of \(n\).

Key concepts

Regular Singular Point

Understanding a regular singular point is crucial when analyzing differential equations, especially second-order linear ones. In simple terms, a regular singular point is like a pothole in the otherwise smooth road of a function—it's a spot where normal methods hit a hiccup because our typical function behaviors (like having a finite value) break down.

Mathematically, if we have a second-order linear differential equation of the form $$y^{\textprime\textprime} + p(t) y^{\textprime} + q(t) y = 0$$ a point, say \(t = t_0\), is a regular singular point if when you multiply \(p(t)\) by \(t - t_0\) and \(q(t)\) by \((t - t_0)^2\), the resulting functions are analytic at \(t_0\). Analytic essentially means that the functions can be expressed as a power series, which is a basic, infinitely long polynomial like expansion.

For the equation given in the exercise, we examine \(t = 0\). To determine if it's a regular singular point, we need to see if both \(tp(t)\) and \(t^2q(t)\) can be neatly dressed in power series without any terms missing or misbehaving when \(t\) is zero. However, this is not the case here, as \(p(t)\) involves a \(\frac{1}{\text{sin } t}\), which is undefined at \(t = 0\).

Irregular Singular Point

On the other side, we have the irregular singular points—the wildcards of differential equations. Unlike their regular counterparts, irregular singular points don't play nice with analytic functions at all. When we approach an irregular singular point, either the function itself or its derivatives (or both!) go off the rails and no longer fit into the power series we like to use.

Returning to our example, since at \(t = 0\) the function \(p(t) = \frac{1}{\text{sin } t}\) becomes undefined, multiplying it by \(t\) doesn't resolve the issue. The result is non-analytic right at that point, making it impossible for \(t = 0\) to be a regular singular point. What we're left with is a confirmation that we're dealing with an irregular singular point instead for any value of \(n\), which is the hallmark of part (b) in our exercise. So, an irregular singular point is determined by what it can't do—allow for a smooth, series-based expression.

Analytic Functions

Analytic functions are the mathematicians best friends—well-behaved, predictable, and infinitely differentiable. Imagine drawing a curve without lifting your pencil; if you can do this and also describe each point on the curve with an infinite sum of its powers (a power series), you've got an analytic function!

To be deemed analytic, a function must not only be smooth but also posses a power series expansion around every point within its domain. These series are a super handy tool for approximating functions because they break them down into basic components that are much easier to work with, especially when solving differential equations.

However, not all functions are analytic. A function like \(1/\text{sin } t\) is not analytic at certain points, specifically where \(\text{sin } t = 0\) (such as at \(t = 0\)). As in our exercise, the absence of analyticity of \(p(t) = 1/\text{sin } t\) at \(t = 0\) has profound implications on the nature of the singular point for the differential equation in question.

Second-Order Linear Differential Equation

The second-order linear differential equation is a cornerstone of classical physics and engineering, describing everything from oscillations to electric circuits. These equations take on the form $$y^{\textprime\textprime} + p(t) y^{\textprime} + q(t) y = 0$$ where \(p(t)\) and \(q(t)\) are functions of \(t\). A solution to such an equation is a function \(y(t)\) that makes the equation true.

The 'second-order' bit of the name comes from the highest derivative present—\(y^{\textprime\textprime}\)—which is the second derivative of \(y\) with respect to \(t\). 'Linear' means the equation doesn't have any powers or products of \(y\) and its derivatives higher than one; everything is in a straight line, so to speak.

Solving these equations often involves finding the roots of a characteristic equation, which in turn dictate the general form of the solution. However, when singular points are present, as they are in our textbook exercise, we need to be careful, as the method of solution may need some adapting to accommodate the fact that standard power series solutions aren't going to cut it.