Question

Consider the differential equation \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\). In some cases, we may be able to find a power series solution of the form \(y(t)=\sum_{n=0}^{\infty} a_{n}\left(t-t_{0}\right)^{n}\) even when \(t_{0}\) is not an ordinary point. In other cases, there is no power series solution. (a) The point \(t=0\) is a singular point of \(t y^{\prime \prime}+y^{\prime}-y=0\). Nevertheless, find a nontrivial power series solution, \(y(t)=\sum_{n=0}^{\infty} a_{n} t^{n}\), of this equation. (b) The point \(t=0\) is a singular point of \(t^{2} y^{\prime \prime}+y=0\). Show that the only solution of this equation having the form \(y(t)=\sum_{n=0}^{\infty} a_{n} t^{n}\) is the trivial solution.

Short answer

Question: Find power series solutions for the given differential equations, even though their corresponding initial points are singular points. Differential equation 1: \(ty'' + y' - y = 0\) Differential equation 2: \(t^2y'' + y = 0\) Answer: Power series solution for the first differential equation: \(y(t) = a_0 + a_1 t + \sum_{n=2}^{\infty} a_n t^n,\) where \(a_n = \frac{(-1)^n}{(n-1)!a_0 + (n-2)!a_1\mathrm{, for\ } n\ge 2}.\) For the second differential equation, there is no nontrivial power series solution. The only solution is the trivial solution.

Step-by-step solution

Understanding singular points

A singular point is a point where the differential equation is not well-behaved. In other words, it's a point where the coefficients of the differential equation are not analytic. In the given examples, \(t=0\) is a singular point for both equations. Now, let's find power series solutions for the given differential equations.

Finding power series solution for the first equation

We are given the differential equation: \(ty'' + y' - y = 0\). We need to find the power series solution with the form: \(y(t) = \sum_{n=0}^{\infty} a_n t^n\). First, we find the first and second derivatives of y(t) with respect to t: \(y'(t) = \sum_{n=1}^{\infty} na_n t^{n-1}\) and \(y''(t) = \sum_{n=2}^{\infty} n(n-1)a_n t^{n-2}\). Now we can substitute these derivatives into our given equation: \(t\left(\sum_{n=2}^{\infty} n(n-1)a_n t^{n-2}\right) + \left(\sum_{n=1}^{\infty}na_n t^{n-1}\right) - \left(\sum_{n=0}^{\infty}a_n t^n\right) = 0.\) After multiplying each term, we have: \(\sum_{n=2}^{\infty} n(n-1)a_n t^{n-1} + \sum_{n=1}^{\infty}na_n t^{n-1} - \sum_{n=0}^{\infty}a_n t^{n+1} = 0.\) By looking at these series, we notice that they all have the same power for t as t^{n-1}, except for the last term. We need to adjust the index n as follows: \(\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2} t^n + \sum_{n=0}^{\infty}(n+1)a_{n+1} t^n - \sum_{n=-1}^{\infty}a_{n+1} t^{n+1} = 0.\) Now, we rewrite the series to have the same index n: \(\sum_{n=0}^{\infty} [(n+2)(n+1)a_{n+2} + (n+1)a_{n+1} - a_{n+1}]t^n = 0.\) Since this equation must hold for all n, we can find the recurrence relation: \((n+2)(n+1)a_{n+2} + (n+1)a_{n+1} - a_{n+1} = 0.\) We can solve for \(a_{n+2}\) and get: \(a_{n+2}=\frac{a_{n+1}-a_n(n+1)}{(n+2)(n+1)}.\) Now, we can use the recurrence relation to find the coefficients. For example: \(a_2 = \frac{a_1 - a_0(0+1)}{(0+2)(0+1)} = \frac{1}{2} a_0,\) \(a_3 = \frac{(a_2-a_1)(1)}{(1+2)(1+1)} = \frac{1}{6} (a_1 - a_2) = \frac{1}{6} (a_1 - \frac{1}{2} a_0),\) \(a_4 = \frac{a_3 - a_2(2)}{(3)(2)} = \frac{1}{12} (1 - a_0) a_0.\) Although we are not able to find a closed-form expression for each \(a_n\), we can express the power series solution as: \(y(t) = a_0 + a_1 t + \sum_{n=2}^{\infty} a_n t^n\), where \(a_n = \frac{(-1)^n}{(n-1)!a_0 + (n-2)!a_1\mathrm{, for \ } n\ge 2}.\) We have found a nontrivial power series solution for the first differential equation.

Finding the power series solution for the second equation

We are given the differential equation: \(t^2 y'' + y = 0\). We need to find the power series solution with the form: \(y(t) = \sum_{n=0}^{\infty} a_n t^n.\) Again, we find the first and second derivatives of y(t) with respect to t: \(y'(t) = \sum_{n=1}^{\infty} na_n t^{n-1}\) and \(y''(t) = \sum_{n=2}^{\infty} n(n-1)a_n t^{n-2}.\) Now we substitute these derivatives into our given equation: \(t^2\left(\sum_{n=2}^{\infty}n(n-1)a_n t^{n-2}\right) + \left(\sum_{n=0}^{\infty}a_n t^n\right) = 0.\) After we multiply each term and equate the coefficients for each power of t, we get the following: \(\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2} t^n + \sum_{n=0}^{\infty} a_n t^n = 0.\) Comparing coefficients, we get a recurrence relation: \((n+2)(n+1)a_{n+2} + a_n = 0.\) Solving for \(a_{n+2}\), we get: \(a_{n+2}= - \frac{a_n}{(n+1)(n+2)}.\) Now we use the recurrence relation to find the coefficients. For example: \(a_2 = -\frac{a_0}{2!},\) \(a_4 = \frac{a_0}{4!},\) \(a_6 = -\frac{a_0}{6!},\) \(a_3 = -\frac{a_1}{3!},\) \(a_5 = \frac{a_1}{5!}.\) As we can see, all odd coefficients are multiples of \(a_1\), and all even coefficients are multiples of \(a_0\). Now we can express the power series solution as: \(y(t) = a_0\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}t^{2n}+ a_1\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}t^{2n+1}.\) But the above result is the same as the power series solution for cosine and sine functions with \(a_0=\cos(1)\) and\(a_1=\sin(1) \). Therefore, there is no nontrivial power series solution for the second differential equation; the only solution is the trivial solution.

Key concepts

Singular Points

In the realm of differential equations, understanding singular points is crucial as they represent locations where the behavior of the equation changes. More formally, a singular point is where the coefficients of a differential equation fail to be analytic. This simply means that the function that describes the differential equation has difficulties behaving normally at that point.

For instance, in the differential equation given in the exercise, power series solutions are sought at the point \(t = 0\). The exercise focuses on two equations with singular behavior at this point:

• \(t y^{\prime\prime} + y^{\prime} - y = 0\)
• \(t^2 y^{\prime\prime} + y = 0\)

The point \(t = 0\) is called a singular point for these equations because plugging \(t = 0\) in results in undefined behavior, typically causing either coefficients to become zero or making them non-existent. Understanding these points is key because it determines whether a power series solution might exist or not.

In scenarios involving singular points, sometimes ingenious manipulations such as series expansions can still yield meaningful solutions, as shown in this problem. But the presence of a singular point often complicates the analysis, influencing whether non-trivial solutions are attainable.

Recurrence Relations

A significant part of solving differential equations using power series lies in finding and using recurrence relations. These are equations that relate various terms in the series to each other. In the context of the given exercise, the recurrence relation forms the backbone of constructing the power series solution.

For the differential equation \(t y^{\prime\prime} + y^{\prime} - y = 0\), the power series approach involves:

• Expressing the solution as \(y(t) = \sum_{n=0}^{\infty} a_n t^n\)
• Calculating derivatives and substituting back into the original equation
• Rearranging terms to establish a consistent pattern across the power series terms

This process results in a recurrence relation such as \((n+2)(n+1)a_{n+2} + (n+1)a_{n+1} - a_{n+1} = 0\) which links a term in the series to the preceding terms, allowing computation of coefficients iteratively. Recurrence relations are powerful because once established, they enable the further calculation of any desired number of terms in the series.

The relation tells you how to find \(a_{n+2}\) if you know \(a_n\) and possibly other previous terms. They thus often form the heart of finding both the existence and the form of the solutions for differential equations, especially around singular points.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations involving functions and their derivatives. They serve as essential tools in modeling the behavior of complex systems in fields like physics, engineering, and beyond. Understanding how to find solutions to ODEs is a major component of studying them.

In our problem, the ODEs provided are in the second order and linear. The general form of second-order linear ODEs can be represented as:\[ y^{\prime\prime} + p(t) y^{\prime} + q(t) y = 0 \]The task of finding power series solutions becomes interesting when dealing with singular points, as not all ODEs have solutions that are expressible as power series at such points. The methods used often involve expressing the function \(y(t)\) as a series and accounting for each term's differentiation based on recurrence relations.

In the context of this exercise, the challenge was to determine when these methods yield non-trivial solutions. For instance, in \(t y^{\prime\prime} + y^{\prime} - y = 0\), a power series solution was found even though \(t=0\) was a singular point. In the second ODE, \(t^2 y^{\prime\prime} + y = 0\), we see that the equation only admits the trivial solution, showcasing that singular points can affect whether meaningful solutions exist. Exploring such dynamics helps build a deeper understanding of the underlying nature of ODEs and their solutions.