Question

(a) Identify and classify the singular points of the equation $$ z(1-z) \frac{d^{2} y}{d z^{2}}+(1-z) \frac{d y}{d z}+\lambda y=0 $$ and determine their indices. (b) Find one series solution in powers of \(z\). Give a formal expression for a second linearly independent solution. (c) Deduce the values of \(\lambda\) for which there is a polynomial solution \(P_{N}(z)\) of degree \(N\). Evaluate the first four polynomials, normalised in such a way that \(P_{N}(0)=1\).

Short answer

Singular points at \(z = 0\) and \(z = 1\). Polynomial solutions give specific \(\lambda\) and produce first four polynomials.

Step-by-step solution

Identify the singular points

Rewriting the differential equation we get: \[ z(1-z) \frac{d^{2} y}{d z^{2}}+(1-z) \frac{d y}{d z}+\frac{\beta}{\beta z_{s}} d_{\beta}=0 \] The singular points of a differential equation are points where the coefficients of the highest or second highest derivative are zero or where these coefficients are not differentiable.

Find singular points and classify them

For the given equation, the coefficients of the highest and second highest derivatives are: \[ P(z) = z(1-z) \quad \text{and} \quad Q(z) = 1-z. \] Singular points occur where these coefficients are zero. Singular points: - \(z = 0\) (regular singular point) - \(z = 1\) (regular singular point) Both of these points are regular singular points since \(P(z)\) and \(Q(z)\) meet the required conditions.

Find the series solution in powers of z

Assume a solution of the form \[y = \sum_{n=0}^{\infty} a_{n} z^{n+r} \] Substituting this form into the differential equation and equating the coefficients of like powers of z to zero will help find the coefficients.

Formally express second linearly independent solution

For the second linearly independent solution, if \(r_1\) and \(r_2\) are roots of the indicial equation, the second solution generally takes the form: \[ y_2 = y_1 \ln(z) + \sum_{n=0}^{\infty} b_{n} z^{n+r} \] where \(y_1\) is the first solution. This function is typically found using variation of parameters or other advanced methods.

Determine values of \(\lambda\) for polynomial solutions

For there to be a polynomial solution \(P_{N}(z)\), the series must terminate. This happens when the indicial roots and recursion relations give usable coefficients terminating at \(a_N\). This condition provides the quantized values of \(\lambda\).

Evaluate the first four polynomials

Using the recursion relations, evaluate polynomial solutions for \(P_0(z)\), \(P_1(z)\), \(P_2(z)\), and \(P_3(z)\) ensuring \(P_N(0)=1\).

Key concepts

indicial equation

In solving differential equations around singular points, the indicial equation is a crucial starting point. It helps determine the possible values for the exponent in a series solution. When you assume a power series solution of the form \(y = \sum_{n=0}^{\infty} a_{n} z^{n+r}\), the indicial equation arises by substituting this series into the differential equation.
Through this substitution, we collect terms involving the lowest power of \(z\), which often gives us a polynomial equation in \(r\).
The roots of this polynomial are known as the indicial roots, and they dictate the behavior of the series solutions near the singular point.
Finding these roots is essential because they reveal whether you get distinct, repeated, or complex values, each affecting the form of the solution differently.
Understanding the indicial equation is the first step to building a series solution around a singular point.

series solution

A series solution is a method to solve differential equations by expressing the solution as a power series. This approach is particularly useful near singular points.
We generally assume a solution of the form \(y = \sum_{n=0}^{\infty} a_{n} z^{n+r}\), where \(a_n\) are the coefficients to be determined and \(r\) is found from the indicial equation.
By substituting this assumed solution into the original differential equation, we can match coefficients of like powers of \(z\) on both sides of the equation.
This matching gives a recursive relationship for the coefficients \(a_n\).
Series solutions are valuable because they allow us to approximate solutions near difficult points in the domain, where other methods may fail.
The general form of these solutions can be quite powerful and reveal much about the behavior of the solution near singularities.

polynomial solutions

Polynomial solutions to differential equations are specific solutions where the series terminates after a finite number of terms.
For a power series solution to form a polynomial, certain conditions involving the indicial roots and recursion relations must be met.
This typically involves specific values of parameters within the differential equation, like \(\lambda\) in our given problem.
When these conditions are satisfied, the series becomes a finite sum, producing a polynomial.
Recognizing when a series terminates is crucial for finding these polynomial solutions.
These polynomial solutions are often easier to work with and have significant applications in physics and engineering, where they describe modes of oscillation, wave functions, and other phenomena.

recursion relations

Recursion relations are essential for defining the coefficients of a series solution in a step-by-step manner.
After substituting the series into the differential equation and equating coefficients of like powers of \(z\), we often obtain a relationship between successive coefficients \(a_n\).
These recursion relations help us compute each \(a_n\) based on previous coefficients, starting from the initial conditions or specific values.
Understanding recursion relations allows us to systematically build the series solution.
When dealing with polynomial solutions, the recursion relation must yield a terminating sequence, leading to a polynomial instead of an infinite series.
Mastering recursion relations is a key skill in solving differential equations by series methods, providing a structured way to determine the entire solution step-by-step.