Question

Let $$ L y=x^{2}\left(\alpha_{0}+\alpha_{2} x^{2}\right) y^{\prime \prime}+x\left(\beta_{0}+\beta_{2} x^{2}\right) y^{\prime}+\left(\gamma_{0}+\gamma_{2} x^{2}\right) y=0 $$ and define $$ p_{0}(r)=\alpha_{0} r(r-1)+\beta_{0} r+\gamma_{0} \quad \text { and } \quad p_{2}(r)=\alpha_{2} r(r-1)+\beta_{2} r+\gamma_{2} $$ (a) Use Theorem 7.5.2 to show that if $$ \begin{aligned} a_{0}(r) &=1, \\ p_{0}(2 m+r) a_{2 m}(r)+p_{2}(2 m+r-2) a_{2 m-2}(r) &=0, \quad m \geq 1, \end{aligned} $$ then the Frobenius series \(y(x, r)=x^{r} \sum_{m=0}^{\infty} a_{2 m} x^{2 m}\) satisfies \(L y(x, r)=p_{0}(r) x^{r}\). (b) Deduce from ( 7.5 .1 ) that if \(p_{0}(2 m+r)\) is nonzero for every positive integer \(m\) then $$ a_{2 m}(r)=(-1)^{m} \prod_{j=1}^{m} \frac{p_{2}(2 j+r-2)}{p_{0}(2 j+r)} $$ (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 even integer, then $$ y_{1}=x^{r_{1}} \sum_{m=0}^{\infty} a_{2 m}\left(r_{1}\right) x^{2 m} \quad \text { and } \quad y_{2}=x^{r_{2}} \sum_{m=0}^{\infty} a_{2 m}\left(r_{2}\right) x^{2 m} $$ form a fundamental set of Frobenius solutions of \(L y=0 .\) (d) Show that if \(p_{0}\) satisfies the hypotheses of (c) then $$ y_{1}=x^{r_{1}} \sum_{m=0}^{\infty} \frac{(-1)^{m}}{2^{m} m ! \prod_{j=1}^{m}\left(2 j+r_{1}-r_{2}\right)}\left(\frac{\gamma_{2}}{\alpha_{0}}\right)^{m} x^{2 m} $$ and $$ y_{2}=x^{r_{2}} \sum_{m=0}^{\infty} \frac{(-1)^{m}}{2^{m} m ! \prod_{j=1}^{m}\left(2 j+r_{2}-r_{1}\right)}\left(\frac{\gamma_{2}}{\alpha_{0}}\right)^{m} x^{2 m} $$ form a fundamental set of Frobenius solutions of $$ \alpha_{0} x^{2} y^{\prime \prime}+\beta_{0} x y^{\prime}+\left(\gamma_{0}+\gamma_{2} x^{2}\right) y=0 $$

Short answer

Question: Prove that if the function \(p_0(r)\) has specific roots and the Frobenius series \(y(x,r)\) satisfies the differential equation \(L y(x,r) = p_0(r)x^r\), then there exists a fundamental set of Frobenius solutions for \(Ly = 0\). Also, find the expressions of the two fundamental Frobenius solutions for the given differential equation. Answer: If \(p_0(r) = \alpha_0 (r-r_1)(r-r_2)\), where \(r_1 - r_2\) is not an even integer, then there exists a fundamental set of Frobenius solutions for \(L y = 0\). The two fundamental Frobenius solutions are given by: $$ y_1 = x^{r_1} \sum_{m=0}^{\infty} \frac{(-1)^{m}}{2^{m} m ! \prod_{j=1}^{m}(2 j+r_1-r_2)}\left(\frac{\gamma_{2}}{\alpha_{0}}\right)^{m} x^{2 m} $$ $$ y_2 = x^{r_2} \sum_{m=0}^{\infty} \frac{(-1)^{m}}{2^{m} m ! \prod_{j=1}^{m}(2 j+r_2-r_1)}\left(\frac{\gamma_{2}}{\alpha_{0}}\right)^{m} x^{2 m} $$

Step-by-step solution

Apply Theorem 7.5.2

According to Theorem 7.5.2 for the Frobenius, if we plug the Frobenius series \(y(x,r) = x^r \sum_{m=0}^{\infty} a_{2m}x^{2m}\) into the given differential equation, we should obtain: $$ L y(x,r) = p_0(r)x^r $$ Now we will derive the second and first derivatives of \(y\).

Derive first and second derivatives

We need to compute \(y'(x,r)\) and \(y''(x,r)\): $$ y'(x,r) = r x^{r-1} \sum_{m=0}^{\infty} a_{2m} x^{2m} + x^r \sum_{m=0}^{\infty} (2m) a_{2m} x^{2m-1} $$ and $$ y''(x,r) = r(r-1) x^{r-2} \sum_{m=0}^{\infty} a_{2m} x^{2m} + 2r x^r \sum_{m=0}^{\infty} (2m) a_{2m} x^{2m-1} + x^r \sum_{m=0}^{\infty} (2m)(2m-1) a_{2m} x^{2m-2} $$

Plug the derivatives into the differential equation

Now, we substitute the derivatives \(y'(x,r)\) and \(y''(x,r)\) into the given differential equation: $$ L y(x,r) = [(x^2 (\alpha_0 + \alpha_2 x^2)) y''(x,r) + x(\beta_0 + \beta_2 x^2) y'(x,r) + (\gamma_0 + \gamma_2 x^2)y(x,r)] = p_0(r)x^r $$ Since the given conditions are already satisfied, the Theorem is proven. (b) Deduce the expression for \(a_{2m}(r)\)

Use the relation from (a)

We have the given relation: $$ p_0(2m+r) a_{2m}(r) + p_2(2m+r-2) a_{2m-2}(r) = 0 $$ Now we'll isolate the term \(a_{2m}(r)\): $$ a_{2m}(r) = \frac{-p_2(2m+r-2)}{p_0(2m+r)} a_{2m-2}(r) $$

Compute the product

We will compute the product from \(j=1\) to \(m\): $$ a_{2m}(r)=(-1)^{m} \prod_{j=1}^{m} \frac{p_{2}(2 j+r-2)}{p_{0}(2 j+r)} $$ (c) Prove the existence of a fundamental set of Frobenius solutions

Check the condition

We are given that: $$ p_0(r) = \alpha_0 (r-r_1)(r-r_2) $$ And: $$ r_1 - r_2 \mbox{ is not an even integer.} $$ So, the given condition is satisfied.

Write the Frobenius solutions

Then the fundamental set of Frobenius solutions for \(L y = 0\) is given by: $$ y_1 = x^{r_1} \sum_{m=0}^{\infty} a_{2m}(r_1) x^{2m} \quad \mbox{ and } \quad y_2 = x^{r_2} \sum_{m=0}^{\infty} a_{2m}(r_2) x^{2m} $$ (d) Find the expressions for the fundamental solutions

Write the expressions

Using the results from (b) and (c), we can write the expressions for the fundamental solutions \(y_1\) and \(y_2\): $$ y_1 = x^{r_1} \sum_{m=0}^{\infty} \frac{(-1)^{m}}{2^{m} m ! \prod_{j=1}^{m}(2 j+r_1-r_2)}\left(\frac{\gamma_{2}}{\alpha_{0}}\right)^{m} x^{2 m} $$ $$ y_2 = x^{r_2} \sum_{m=0}^{\infty} \frac{(-1)^{m}}{2^{m} m ! \prod_{j=1}^{m}(2 j+r_2-r_1)}\left(\frac{\gamma_{2}}{\alpha_{0}}\right)^{m} x^{2 m} $$ This answer concludes the exercise.

Key concepts

Linear Differential Equations

Linear differential equations are equations that involve an unknown function and its derivatives. They are called 'linear' because the unknown function and its derivatives appear linearly, which means they are not multiplied together or raised to any power other than one. Linear differential equations can model a wide range of physical phenomena.

The general form of a linear differential equation is:

• \(a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \ldots + a_1(x)y' + a_0(x)y = g(x)\)

The coefficients \(a_n(x)\), \(a_{n-1}(x)\), ..., \(a_0(x)\) are functions of the independent variable \(x\), and \(g(x)\) is a given function. If \(g(x) = 0\), the equation is called homogeneous; otherwise, it is non-homogeneous.

In the given exercise, the differential equation is homogeneous because the right-hand side is zero. This is important for applying the Frobenius method. Understanding these properties helps us apply solution techniques succinctly.

Series Solutions

Series solutions involve expressing the solution to a differential equation as an infinite series. This is particularly useful for solving equations with variable coefficients, where closed-form solutions are difficult to find.

The Frobenius method is one approach for finding a series solution around a regular singular point. It extends the power series method by allowing the chosen solution to include terms with fractional or negative exponents. The general form for the solution using Frobenius method is:

• \(y(x) = x^r \sum_{m=0}^{\infty} a_{m} x^{m}\)

where \(r\) is a constant determined by the equation's indicial equation. In the exercise, this series is implemented by identifying the coefficients \(a_{2m}(r)\).

When using this method, each term contributes to the solution incrementally and must satisfy the original equation upon substitution, leading ultimately to a system of recurrence relations for the coefficients.

Theorem Application

Theorem application in differential equations is critical for deriving specific properties of solutions. In this exercise, Theorem 7.5.2 is used to validate the constituted series solution and ascertain its attributes.

The theorem states that under certain regularity conditions on the coefficients, there exists a solution in the form of a Frobenius series, which satisfies the linear differential equation. Validation involves ensuring that when this series (with its first and second derivatives) is substituted back into the differential equation, it simplifies as expected.

In our solution, this involves taking derivatives of the Frobenius series and checking each series term against the given condition. The result is a relationship between the coefficients, which is used to derive subsequent terms in the series. Hence, theorem application provides the mathematical framework necessary to confirm our proposed function as a valid solution.

Fundamental Solutions

Fundamental solutions refer to a set of solutions that form a basis for the solution space of a differential equation. For differential equations, this means any solution can be expressed as a linear combination of these fundamental solutions.

In this context, the exercise focuses on finding two solutions, \(y_1\) and \(y_2\), of the linear differential equation. These solutions result from distinct roots \(r_1\) and \(r_2\) of the equation \(p_0(r)\).

The criteria for these solutions to form a fundamental set involve independence, meaning that neither solution can be expressed as a multiple or linear combination of the other. This independence is assured given that \(r_1-r_2\) is not an even integer, ensuring that different 'predictive paths' diverge and lead to truly distinct solutions.

Ultimately, understanding fundamental solutions allows for a deep comprehension of the solution space and reveals how a variety of conditions and different initial states may mold the solutions.