Question

Let $$ L y=x^{2}\left(\alpha_{0}+\alpha_{q} x^{q}\right) y^{\prime \prime}+x\left(\beta_{0}+\beta_{q} x^{q}\right) y^{\prime}+\left(\gamma_{0}+\gamma_{q} x^{q}\right) y $$ where \(q\) is a positive integer, and define $$ p_{0}(r)=\alpha_{0} r(r-1)+\beta_{0} r+\gamma_{0} \quad \text { and } \quad p_{q}(r)=\alpha_{q} r(r-1)+\beta_{q} r+\gamma_{q} $$ (a) Show that if $$ y(x, r)=x^{r} \sum_{m=0}^{\infty} a_{q m}(r) x^{q m} $$ where $$ \begin{aligned} a_{0}(r) &=1 \\ a_{q m}(r) &=-\frac{p_{q}(q(m-1)+r)}{p_{0}(q m+r)} a_{q(m-1)}(r), \quad m \geq 1 \end{aligned} $$ then $$ L y(x, r)=p_{0}(r) x^{r} $$ (b) Deduce from (7.5.1) that $$ a_{q m}(r)=(-1)^{m} \prod_{j=1}^{m} \frac{p_{q}(q(j-1)+r)}{p_{0}(q j+r)} $$ (c) Conclude that if \(p_{0}(r)=\alpha_{0}\left(r-r_{1}\right)\left(r-r_{2}\right)\) where \(r_{1}-r_{2}\) is not an integer multiple of \(q\), then $$ y_{1}=x^{r_{1}} \sum_{m=0}^{\infty} a_{q m}\left(r_{1}\right) x^{q m} \quad \text { and } \quad y_{2}=x^{r_{2}} \sum_{m=0}^{\infty} a_{q m}\left(r_{2}\right) x^{q m} $$ form a fundamental set of Frobenius solutions of \(L y=0\).

Short answer

Question: Show that the function \(y(x, r)=x^{r} \sum_{m=0}^{\infty} a_{q m}(r) x^{q m}\) satisfies the given differential equation: \(L y = x^2(\alpha_0 + \alpha_q x^q)y'' + x(\beta_0 + \beta_q x^q)y' + (\gamma_0 + \gamma_q x^q)y = p_{0}(r)x^r\), where \(a_{qm}(r)\) is defined as specified. Also, find an expression for \(a_{qm}(r)\), and find a fundamental set of Frobenius solutions for the case when \(p_0(r) = \alpha_0(r-r_1)(r-r_2)\) and the difference \(r_1 - r_2\) is not an integer multiple of \(q\). Answer: The function \(y(x, r)=x^{r} \sum_{m=0}^{\infty} a_{q m}(r) x^{q m}\) satisfies the given differential equation as shown in the step by step calculation above. The expression for \(a_{qm}(r)\) is \((-1)^{m} \prod_{j=1}^{m}\frac{p_{q}(q(j-1)+r)}{p_{0}(qj+r)}\). The fundamental set of Frobenius solutions for the given case is: \(y_1 = x^{r_1} \sum_{m=0}^{\infty} a_{qm}(r_1) x^{qm}\), \(y_2 = x^{r_2} \sum_{m=0}^{\infty} a_{qm}(r_2) x^{qm}\).

Step-by-step solution

Calculate the first and second derivatives of y(x,r)

To show that the function \(y(x, r)=x^{r} \sum_{m=0}^{\infty} a_{q m}(r) x^{q m}\) satisfies the given differential equation, we will first compute the first and second derivatives of \(y(x, r)\) with respect to x: \(y'(x, r) = rx^{r-1} \sum_{m=0}^{\infty} a_{qm}(r) x^{qm} + x^r \sum_{m=0}^{\infty} qma_{qm}(r)x^{qm-1}\) \(y''(x, r) = r(r-1)x^{r-2} \sum_{m=0}^{\infty} a_{qm}(r) x^{qm} + 2r x^{r-1} \sum_{m=0}^{\infty} qma_{qm}(r)x^{qm-1} + x^r\sum_{m=0}^{\infty} qm(qm-1)a_{qm}(r)x^{qm-2}\)

Substitute these derivatives into the given differential equation

Now, we substitute these derivatives back into the given differential equation to find: \(L y = x^2(\alpha_0 + \alpha_q x^q)(r(r-1)x^{r-2}\sum_{m=0}^{\infty} a_{qm}(r) x^{qm} + 2r x^{r-1} \sum_{m=0}^{\infty} qma_{qm}(r)x^{qm-1} + x^r\sum_{m=0}^{\infty} qm(qm-1)a_{qm}(r)x^{qm-2}) + x(\beta_0 + \beta_q x^q)(rx^{r-1} \sum_{m=0}^{\infty} a_{qm}(r) x^{qm} + x^r \sum_{m=0}^{\infty} qma_{qm}(r)x^{qm-1}) + (\gamma_0 + \gamma_q x^q)x^r \sum_{m=0}^{\infty} a_{qm}(r) x^{qm}\)

Simplify this expression and compare it to \(p_0(r)x^r\)

After simplifying the above expression using the definitions of \(p_{0}(r)\) and \(p_{q}(r)\), we have: \(L y = p_{0}(r) x^r \sum_{m=0}^{\infty} a_{qm}(r) x^{qm} + x^r \sum_{m=0}^{\infty} qm(p_{q}(qm+r)-p_{0}(qm+r))a_{qm}(r) x^{qm-1}\) The second term of the above expression becomes zero due to the definition of \(a_{qm}(r)\) for \(m \geq 1\). Therefore, we are left with: \(L y = p_{0}(r) x^r\) This shows that the given function satisfies the differential equation. #Part(b) Find an expression for \(a_{qm}(r)\)#

Use the given expressions to derive a formula for \(a_{qm}(r)\)

According to the given formulas, \(a_{qm}(r) = -\frac{p_{q}(q(m-1)+r)}{p_{0}(qm+r)} a_{q(m-1)}(r)\) for \(m \geq 1\). Using this recursive definition and the formula for the product of a sequence, we can write: \(a_{qm}(r)=(-1)^{m} \prod_{j=1}^{m}\frac{p_{q}(q(j-1)+r)}{p_{0}(qj+r)}\) This gives us the desired expression for \(a_{qm}(r)\). #Part(c) Find a fundamental set of Frobenius solutions for the case described#

Write the solutions using the expression for \(a_{qm}(r)\)

In this case, we have \(p_0(r) = \alpha_0(r-r_1)(r-r_2)\), and the difference between the roots \(r_1 - r_2\) is not an integer multiple of \(q\). We can write the Frobenius solutions using the expressions for \(y(x, r)\) and \(a_{qm}(r)\): \(y_1 = x^{r_1} \sum_{m=0}^{\infty} a_{qm}(r_1) x^{qm}\) \(y_2 = x^{r_2} \sum_{m=0}^{\infty} a_{qm}(r_2) x^{qm}\) These two functions form a fundamental set of Frobenius solutions for the given differential equation when \(p_0(r)\) has the described form, and \(r_1 - r_2\) is not an integer multiple of \(q\).

Key concepts

Frobenius Method

The Frobenius Method is a powerful technique used to find series solutions of linear ordinary differential equations, particularly when solving around singular points. It is especially helpful for differential equations with regular singular points, where ordinary power series solutions fail.

The method involves assuming a solution in the form of a series, where each term depends on a parameter (often denoted as the exponent) and power of x. This form accommodates solutions of the type:

\( y(x) = x^r \sum_{n=0}^{\infty} a_nx^n \).

Here, the exponent \(r\) is usually determined by a technique known as the indicial equation, which is derived from the lowest power terms in the differential equation after substituting the assumed series.

By solving this indicial equation, one can find values of \(r\) that indicate possible solutions. Typically, the Frobenius method provides a systematic approach to derive a complete set of solutions near the singular point, allowing the construction of fundamental solutions.

Series Solutions

Series solutions are an essential approach in solving ordinary differential equations, where the solution is expressed in terms of an infinite sum.

This representation can be particularly advantageous in handling differential equations with complex variable coefficients or singular points. The primary goal is to express the unknown function as a series, often around a specific point in the domain:

\( y(x) = \sum_{n=0}^{\infty} a_n (x-x_0)^n \).

By substituting this series form into the differential equation, we can derive a set of equations that allow us to determine the coefficients \(a_n\). This iterative process provides a way to solve otherwise challenging equations by making the terms manageable one-at-a-time.

When using the Frobenius Method, the series solutions take a slightly modified form to accommodate the singularity, resulting in solutions better suited for points where regular solutions would struggle.

Fundamental Set of Solutions

A fundamental set of solutions is a concept integral to solving differential equations. It describes a set of independent solutions that form the basis of the solution space of a differential equation.

For linear differential equations, the Wronskian determinant of these independent solutions is non-zero. This non-zero determinant confirms that the solutions are linearly independent.

In the context of the Frobenius method, when solving equations with a regular singular point, a fundamental set consists of two solutions:

• One solution involves the smaller of the two roots \(r_1\) obtained from the indicial equation.
• The other utilizes the larger root \(r_2\).

Together, these solutions create a complete basis for the space of solutions, meaning any solution to the differential equation can be expressed as a linear combination of this fundamental set. This comprehensive approach allows for the construction of all possible solutions that satisfy the initial differential equation.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations that describe the relationship between a function and its derivatives. These arise naturally in many scientific disciplines, representing phenomena like motion, growth, and oscillations.

ODEs are classified based on their order, which is determined by the highest derivative present in the equation, and whether they are linear or nonlinear.

Linear ODEs, where derivatives appear linearly, are generally easier to solve, and this simplicity extends to the application of techniques like the Frobenius method.

Nonlinear ODEs, on the other hand, often require specialized methods or numerical approaches for solutions.

The goal in solving ODEs is to find a function, or a set of functions, that satisfies the equation throughout its domain. Solving ODEs with series solutions or the Frobenius method allows us to handle complex equations with singular points, ultimately aiding in the understanding of diverse processes modeled by these equations.