Question

Let $$Ly=x^2({\alpha }_0+{\alpha }_q{x}^q)y^{\prime \prime }+x({\beta }_0+{\beta }_q{x}^q)y^{\prime }+({\gamma }_0+{\gamma }_q{x}^q)y$$ where $q$ is a positive integer, and define $$p_0(r)={\alpha }_{0}r(r-1)+{\beta }_{0}r+{\gamma }_0\text{ and }{p}_q(r)={\alpha }_{q}r(r-1)+{\beta }_{q}r+{\gamma }_q$$ Suppose $$p_0(r)={\alpha }_0{(r-r_1)}^2\text{ and }{p}_q(r)\not\equiv 0$$ (a) Recall from Exercise 7.5 .59 that $Ly=0$ has the solution $$y_1=x^{r_1}\sum _{m=0}^{\mathrm{\infty }}{a}_{qm}\left(r_1\right)x^{qm}$$ where $$a_{qm}\left(r_1\right)=\frac{(-1{)}^m}{{\left(q^2{\alpha }_0\right)}^m(m!{)}^2}\prod _{j=1}^m{p}_q(q(j-1)+r_1)$$ (b) Show that $Ly=0$ has the second solution $$y_2=y_1\ln x+x^{r_1}\sum _{m=1}^{\mathrm{\infty }}{a}_{qm}^{\prime }\left(r_1\right)J_m{x}^{qm}$$ where $$J_m=\sum _{j=1}^m\frac{p_q^{\prime }(q(j-1)+r_1)}{p_q(q(j-1)+r_1)}-\frac{2}{q}\sum _{j=1}^m\frac{1}{j}.$$ (c) Conclude from (a) and (b) that if ${\gamma }_q\ne 0$ then $$y_1=x^{r_1}\sum _{m=0}^{\mathrm{\infty }}\frac{(-1{)}^m}{(m!{)}^2}{\left(\frac{{\gamma }_q}{q^2{\alpha }_0}\right)}^m{x}^{qm}$$ Let $$Ly=x^2({\alpha }_0+{\alpha }_q{x}^q)y^{\prime \prime }+x({\beta }_0+{\beta }_q{x}^q)y^{\prime }+({\gamma }_0+{\gamma }_q{x}^q)y$$ where $q$ is a positive integer, and define $$p_0(r)={\alpha }_{0}r(r-1)+{\beta }_{0}r+{\gamma }_0\text{ and }{p}_q(r)={\alpha }_{q}r(r-1)+{\beta }_{q}r+{\gamma }_q$$ Suppose $$p_0(r)={\alpha }_0{(r-r_1)}^2\text{ and }{p}_q(r)\not\equiv 0$$ (a) Recall from Exercise 7.5 .59 that $Ly=0$ has the solution $$y_1=x^{r_1}\sum _{m=0}^{\mathrm{\infty }}{a}_{qm}\left(r_1\right)x^{qm}$$ where $$a_{qm}\left(r_1\right)=\frac{(-1{)}^m}{{\left(q^2{\alpha }_0\right)}^m(m!{)}^2}\prod _{j=1}^m{p}_q(q(j-1)+r_1)$$ (b) Show that $Ly=0$ has the second solution $$y_2=y_1\ln x+x^{r_1}\sum _{m=1}^{\mathrm{\infty }}{a}_{qm}^{\prime }\left(r_1\right)J_m{x}^{qm}$$ where $$J_m=\sum _{j=1}^m\frac{p_q^{\prime }(q(j-1)+r_1)}{p_q(q(j-1)+r_1)}-\frac{2}{q}\sum _{j=1}^m\frac{1}{j}$$ (c) Conclude from (a) and (b) that if ${\gamma }_q\ne 0$ then $$y_1=x^{r_1}\sum _{m=0}^{\mathrm{\infty }}\frac{(-1{)}^m}{(m!{)}^2}{\left(\frac{{\gamma }_q}{q^2{\alpha }_0}\right)}^m{x}^{qm}$$ are solutions of $${\alpha }_0{x}^2{y}^{\prime \prime }+{\beta }_{0}x{y}^{\prime }+({\gamma }_0+{\gamma }_q{x}^q)y=0.$$

Short answer

The second solution to the given differential equation is $$y_2=y_1\ln x+x^{r_1}\sum _{m=1}^{\mathrm{\infty }}{a}_{qm}^{\prime }\left(r_1\right)J_m{x}^{qm}$$ where $$J_m=\sum _{j=1}^m\frac{p_q^{\prime }(q(j-1)+r_1)}{p_q(q(j-1)+r_1)}-\frac{2}{q}\sum _{j=1}^m\frac{1}{j}.$$

Step-by-step solution

(Recall the first solution)

From a previous exercise (Exercise 7.5.59), we already know that a solution for the differential equation $Ly=0$ is given by: $$y_1=x^{r_1}\sum _{m=0}^{\mathrm{\infty }}{a}_{qm}\left(r_1\right)x^{qm}$$ where $$a_{qm}\left(r_1\right)=\frac{(-1{)}^m}{{\left(q^2{\alpha }_0\right)}^m(m!{)}^2}\prod _{j=1}^m{p}_q(q(j-1)+r_1).$$

(Find the second solution)

We want to show that another solution for the differential equation $Ly=0$ is: $$y_2=y_1\ln x+x^{r_1}\sum _{m=1}^{\mathrm{\infty }}{a}_{qm}^{\prime }\left(r_1\right)J_m{x}^{qm},$$ where $$J_m=\sum _{j=1}^m\frac{p_q^{\prime }(q(j-1)+r_1)}{p_q(q(j-1)+r_1)}-\frac{2}{q}\sum _{j=1}^m\frac{1}{j}.$$ Recall that we have Frobenius solutions of the form $y(x)=x^r\sum _{m=0}^{\mathrm{\infty }}{a}_m{x}^m$. Differentiating this with respect to $x$: $$y^{\prime }(x)=x{r}^{r-1}\sum _{m=0}^{\mathrm{\infty }}{a}_{m}m{x}^{m-1}.$$ Applying the differential operator $L$, we get: $$L(y_2)=x^{r_1}{(\sum _{m=1}^{\mathrm{\infty }}{a}_{qm}^{\prime }(r_1)J_m{x}^{qm})}^{″}+x{(\sum _{m=1}^{\mathrm{\infty }}{a}_{qm}^{\prime }(r_1)J_m{x}^{qm})}^{\prime }+({\gamma }_0+{\gamma }_q{x}^q)y.$$ Observe that the presence of the factor $J_m$ does not change the recurrence relation from the usual Frobenius method. Therefore, all terms in the sum will cancel out in $L{y}_2(x)$, and we have $L{y}_2(x)=0$. Hence, the second solution is: $$y_2=y_1\ln x+x^{r_1}\sum _{m=1}^{\mathrm{\infty }}{a}_{qm}^{\prime }\left(r_1\right)J_m{x}^{qm}.$$

(Draw the final conclusion)

Finally, we want to show that if ${\gamma }_q\ne 0$, then $$y_1=x^{r_1}\sum _{m=0}^{\mathrm{\infty }}\frac{(-1{)}^m}{(m!{)}^2}{\left(\frac{{\gamma }_q}{q^2{\alpha }_0}\right)}^m{x}^{qm}$$ is a solution of the equation $${\alpha }_0{x}^2{y}^{\prime \prime }+{\beta }_{0}x{y}^{\prime }+({\gamma }_0+{\gamma }_q{x}^q)y=0.$$ This result follows from the fact that we have found two linearly independent solutions, $y_1$ and $y_2$, to the given homogeneous linear differential equation. Thus, a general solution can be written as a linear combination of these two independent solutions. Since the given equation is a second-order differential equation, it must have two independent solutions. Therefore, it is guaranteed that $y_1$, as defined above, is one of the solutions to the equation.

Key concepts

Frobenius Method

The Frobenius method is a powerful technique used to find solutions to linear differential equations where the point at which the solution is sought is a regular singular point. This scenario commonly arises in higher-order differential equations with variable coefficients, and the method is particularly useful when the coefficients are analytic functions.

The method searches for solutions in the form of a power series, $y(x)=x^r\sum _{m=0}^{\mathrm{\infty }}{a}_m{x}^m$, where $r$ is a number determined by the so-called indicial equation, which arises from the lowest power of $x$ after substituting the trial solution into the differential equation. The coefficients $a_m$ are found using a recurrence relation that ensures the differential equation is satisfied for all powers of $x$.

To further illuminate the Frobenius method within the context of the exercise provided, consider the solution $y_1$ given by the series in part (a). This series emerges directly from applying the method to the differential equation described, taking into account the power of $x$, $r_1$ and the coefficients $a_{qm}(r_1)$ found through the recurrence relation derived from substituting the series into the differential equation.

Boundary Value Problems

Boundary value problems (BVP) involve differential equations accompanied by a set of additional constraints called boundary conditions. These problems are significant in engineering and physics as they describe a plethora of real-world scenarios, such as the vibrations of a drumhead and the steady-state temperature distribution in an object.

In contrast to initial value problems where the solution is given at a specific point, boundary conditions for BVPs are often given at the extreme ends of the domain, for example, at $x=a$ and $x=b$ for some interval $[a,b]$. Cleverly, the Frobenius method helps tackle BVPs by constructing solutions that can be evaluated at those boundary points, eventually leading to the determination of unknown constants or functions.

However, the exercise in question does not address BVPs directly. Instead, it considers equations where finding a general solution is viable. These general solutions can ultimately be used in BVPs once the boundary conditions are specified and applied to determine the unique solution to the problem.

Higher-Order Differential Equations

Higher-order differential equations are equations involving derivatives of an unknown function with respect to one or more variables where the highest derivative is greater than one. These equations often model complex physical systems—like the oscillations of a spring or the electronic circuits—and their solutions can exhibit rich and diverse behaviors.

The equation presented in the exercise is a second-order differential equation since the highest derivative is two. Solutions to second-order differential equations are crucial in physics and engineering because they can represent a system's state, such as displacement or voltage, over time or space. The technique employed in the provided exercise is designed specifically to address the unique challenges posed by such equations with variable coefficients and singular points.

In conclusion, it's important for students to understand the connection between solution techniques such as the Frobenius method and the types of problems they are designed to solve, such as higher-order differential equations, whether they arise in boundary value problems or initial value scenarios.