Question

Find two power series solutions about \(z=0\) of the differential equation $$ \left(1-z^{2}\right) y^{\prime \prime}-3 z y^{\prime}+\lambda y=0 $$ Deduce that the value of \(\lambda\) for which the corresponding power series becomes an \(N\) th-degree polynomial \(U_{N}(z)\) is \(N(N+2)\). Construct \(U_{2}(z)\) and \(U_{3}(z)\).

Short answer

The value of \(\lambda\) for which the series becomes an Nth-degree polynomial is \(N(N+2)\). The polynomials are \(U_2(z) = a_0 + a_2 z^2\) and \(U_3(z) = a_0 + a_3 z^3\).

Step-by-step solution

Assume a Power Series Solution

Assume the solution is in the form of a power series: \[ y(z) = \sum_{n=0}^{\infty} a_n z^n \]

Compute the Derivatives

Calculate the first and second derivatives of the power series:\[ y'(z) = \sum_{n=1}^{\infty} n a_n z^{n-1} \]\[ y''(z) = \sum_{n=2}^{\infty} n(n-1) a_n z^{n-2} \]

Substitute into the Differential Equation

Substitute the power series and its derivatives into the differential equation:\[ (1-z^2) \sum_{n=2}^{\infty} n(n-1) a_n z^{n-2} - 3z \sum_{n=1}^{\infty} n a_n z^{n-1} + \lambda \sum_{n=0}^{\infty} a_n z^n = 0 \]

Simplify and Collect Like Terms

Expand and simplify each term, then collect like powers of z:\[ \sum_{n=2}^{\infty} n(n-1) a_n z^{n-2} - \sum_{n=2}^{\infty} n(n-1) a_n z^n - 3 \sum_{n=1}^{\infty} n a_n z^n + \lambda \sum_{n=0}^{\infty} a_n z^n = 0 \]Combining similar terms, we obtain a recurrence relation.

Relate Coefficients from the Series

Set the coefficients of like powers of z to zero, leading to a recurrence relation for \(a_n\):\[ (n+2)(n+1)a_{n+2} - n(n-1)a_n - 3na_n + \lambda a_n = 0 \]

Solve the Recurrence Relation

Simplify the recurrence relation to find a general relation for the coefficients:\[ a_{n+2} = \frac{(n(n-1)+3n-\lambda)}{(n+2)(n+1)} a_n \]

Find the Specific Value of \(\lambda\)

For the series to terminate and become a polynomial of degree N, the numerator must be zero for some value of n:\[ n(n+2)-\lambda = 0 \]Solving for \(\lambda\) we get:\[ \lambda = N(N+2) \]

Construct \(U_2(z)\)

For \(N=2\), \(\lambda=2(2+2)=8\). Using the recurrence relation, determine the coefficients:\[ U_2(z) = a_0 + a_2 z^2 \]Verify by substituting back into the original differential equation.

Construct \(U_3(z)\)

For \(N=3\), \(\lambda=3(3+2)=15\). Similarly, use the recurrence relation:\[ U_3(z) = a_0 + a_3 z^3 \]Verify by substituting into the differential equation.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve derivatives of a function. They describe how a particular quantity changes over time or space. In this exercise, we have a second-order linear differential equation involving the variable \( z \):\[ (1 - z^2) y'' - 3z y' + \lambda y = 0 \]. The goal is to find solutions for this equation in the form of power series. Understanding differential equations is crucial because they model many natural phenomena, such as motion, heat, and wave propagation.

Recurrence Relations

Recurrence relations are equations that define sequences using preceding terms. In the context of our power series solution, we come across a recurrence relation while equating coefficients of like terms. From the differential equation, substituting the power series \( y(z) = \sum_{n=0}^{\infty} a_n z^n \) and its derivatives, and then simplifying, we obtain: \[ a_{n+2} = \frac{n(n-1) + 3n - \lambda}{(n+2)(n+1)} a_n \]. This recurrence relation connects each coefficient \( a_{n+2} \) with \( a_n \), allowing us to calculate coefficients sequentially and determine the solution's form.

Polynomial Solutions

Polynomial solutions to differential equations are very important, especially when dealing with power series. In this exercise, the power series solution transforms into a polynomial when the series terminates after a finite number of terms. We achieve polynomial solutions by setting the numerator in the recurrence relation to zero: \( n(n+2) - \lambda = 0 \). Solving this yields: \( \lambda = N(N+2) \). For specific values of \( N \) (e.g., \( N = 2 \), \( N = 3 \)), we get polynomials \( U_2(z) = a_0 + a_2 z^2 \) and \( U_3(z) = a_0 + a_3 z^3 \), respectively.

Power Series

Power series are expressions in the form of \( \sum_{n=0}^{\infty} a_n z^n \), where \( a_n \) are constants called coefficients. They are powerful tools for solving differential equations, especially when exact solutions are difficult to find. By representing the solution as a power series, we can manipulate it term by term. In our exercise, we assumed the solution to be a power series and determined the coefficients using the differential equation. This method is particularly useful because it converts a differential equation into an algebraic problem, which is easier to handle.