Question

Find the indicial polynomial, characteristic exponents, and recursion relation at both of the regular singular points of the Legendre equation, $$ w^{\prime \prime}-\frac{2 z}{1-z^{2}} w^{\prime}+\frac{\alpha}{1-z^{2}} w=0 $$ What is \(a_{k}\), the coefficient of the Laurent expansion, for the point \(z=+1 ?\)

Short answer

The indicial polynomial at \(z=1\) will be \(r(r-1) - 2r + α + 2 = 0\), and recursion relation can be characterized as \[a_{k+2} = R_k(a_k, a_{k+1})\]. The coefficient of \(a_k\) in the Laurent expansion at \(z=1\) can be derived from corresponding coefficients in the expansion of \(v(z)\), the solution of the transformed Legendre equation.

Step-by-step solution

Identify Regular Singular Points

The first step is to identify the regular singular points of the given Legendre equation. This is done by pinpointing the values of z for which the coefficients of the differential equation become undefined or infinite. For the equation \(w''-2z(1-z^2)^{-1} w' + α(1-z^2)^{-1} w = 0\), it can be observed that singular points are \(z = ±1\).

Find Indicial Polynomial and Characteristic Exponents

Next, transform the differential equation in its normal form at \(z = +1\) by shifting z. With the transformation \(v(z) = w(z-1)\) you'll achieve \[v'' - 2(z+1)(2-z)^{-1}v' + (α+2)(2-z)^{-1}v = 0 \]The indicial polynomial is then derived by substituting the general power series \[v(z) = Σa_kz^k\] and considering the leading powers of z in the coefficient of the lowest power. This gives indicial equation \(r(r-1) - 2r + α + 2 = 0\), whose roots give the characteristic exponents.

Obtain Recursion Relation

The recursion relation can be derived by substituting the power series \[v(z) = Σa_kz^k\] into our transformed differential equation and equalizing the coefficients on both sides to zero. This gives us a recursion relation of the form \[a_{k+2} = R_k(a_k, a_{k+1})\] where \(R_k\) depends on the particular differential equation.

Derive the Coefficient of the Laurent Expansion

To solve for the coefficient \(a_k\) in the Laurent expansion at \(z=1\), take the general solution of the transformed differential equation, which is given by \[v(z) = C(z+1)^r(1 + Σb_kz^k)\]Expand this in a Taylor series around \(z=0\). The coefficient \(a_k\) in the expansion \[w(z) = C(z-1+1)^r(1 + Σb_k(z-1)^k) = Σa_kz^k\] will be directly related to the coefficients \(b_k\) of the solution \(v(z)\). By calculating these coefficients using the recursion relation and the initial conditions, we will obtain \(a_k\).

Key concepts

Regular Singular Points

In differential equations, identifying singular points is crucial for understanding the behavior of solutions near these points. A regular singular point is a type of singular point where the solution can be expressed in a series form like a power series or a Laurent series.
For the Legendre equation \[ w^{\prime \prime}-\frac{2 z}{1-z^{2}} w^{\prime}+\frac{\alpha}{1-z^{2}} w=0\]we observe that the coefficients become undefined at certain values of \(z\). Specifically, they are undefined when \(1-z^2=0\), so \(z=±1\) are the regular singular points.
This means both \(z=+1\) and \(z=-1\) are the points where the linearity and the form of the differential equation break down, yet around these points, solutions can still behave well enough to allow a series expansion.
Understanding these points helps in making use of special methods to find solutions, compared to points where the solution might be more erratic or complex.

Indicial Polynomial

The concept of the indicial polynomial is deeply linked to the method of Frobenius, which is a strategy for finding power series solutions near a regular singular point. This polynomial is essential to determine the starting behaviors of these solutions.
For a transformed differential equation, such as: \[v'' - 2(z+1)(2-z)^{-1}v' + (\alpha+2)(2-z)^{-1}v = 0 \]using the power series \(v(z) = \Sigma a_k z^k\), the indicial polynomial emerges by focusing on the lowest power of \(z\). In our case, substituting the series gives \[ r(r - 1) - 2r + \alpha + 2 = 0 \]This polynomial, derived from simplifying terms, provides the first clue to how solutions behave at these singular points. Solving this equation gives roots that are the characteristic exponents.
Each root indicates a distinct solution to the differential equation, setting the base for building converging series solutions around singular points. Recognizing the indicial polynomial's role ensures a steady start to solving complex differential equations.

Characteristic Exponents

Characteristic exponents are the roots obtained from the indicial polynomial; they play a critical role in determining the nature of series solutions to differential equations around singular points.
When the indicial polynomial is solved, say \[ r(r-1) - 2r + \alpha + 2 = 0 \]its roots, \(r_1\) and \(r_2\), are the characteristic exponents. These exponents define the leading term of the series solution, impacting how quickly or slowly the series converges near the singular point.
Multiple scenarios can arise:

• If \(r_1 eq r_2\), then two linearly independent solutions corresponding to each root exist.
• If \(r_1 = r_2\), special techniques might be required to find a second, independent solution often involving logarithmic terms.

Characteristic exponents essentially set the framework for the general solution of the differential equation in the vicinity of a regular singular point, helping predict solution behaviors effectively.

Recursion Relation

Recursion relations simplify the problem of finding coefficients in a power series solution of differential equations. They define a relation between a term and some previous terms in a series, crucial for determining successive terms systematically.
To find this relation, we substitute a series solution form, such as \(v(z) = \Sigma a_kz^k\), into the differential equation and match coefficients. This process yields a formula that connects terms, looking something like \[ a_{k+2} = R_k(a_k, a_{k+1}) \]where \(R_k\) is a function determined by the differential equation's coefficients.
Here's what makes recursion relations valuable:

• They provide a systematic way to generate any term based on previous ones.
• They make the job of writing a closed form solution simpler.
• They help identify convergent series solutions, ensuring you can build a usable series rather than an infinite one.

Mastering recursion relations offers insight into the structure and solvability of differential equations through series, paving the way for precise and practical solutions.