Question

The nonlinear first order equation $$y^{\prime}+y^{2}+p(x) y+q(x)=0$$ is a Riccati equation. (See Exercise 2.4.55.) Assume that \(p\) and \(q\) are continuous. (a) Show that \(y\) is a solution of \((\mathrm{A})\) if and only if \(y=z^{\prime} / z,\) where $$z^{\prime \prime}+p(x) z^{\prime}+q(x) z=0$$ (b) Show that the general solution of \((\mathrm{A})\) is $$y=\frac{c_{1} z_{1}^{\prime}+c_{2} z_{2}^{\prime}}{c_{1} z_{1}+c_{2} z_{2}}$$ where \(\left\\{z_{1}, z_{2}\right\\}\) is a fundamental set of solutions of (B) and \(c_{1}\) and \(c_{2}\) are arbitrary constants. (c) Does the formula (C) imply that the first order equation (A) has a two- parameter family of solutions? Explain your answer.

Short answer

Question: Show that the solution y of the Riccati equation can be expressed in the form y = z'/z, where z solves a second-order linear homogeneous equation. Then, derive the general solution of the Riccati equation using the derived relationship between y and z. Finally, analyze the number of parameters for the solutions of the Riccati equation and explain your findings. Answer: The solution y of the Riccati equation can be expressed in the form y = z'/z, where z solves a second-order linear homogeneous equation given by z'' + p(x)z' + q(x)z = 0. To obtain the general solution of the Riccati equation, we use the fundamental set of solutions {z1, z2} for the second-order equation and express the corresponding solutions {y1, y2} as y1 = z1'/z1 and y2 = z2'/z2. The general solution for the Riccati equation is then given by y = (c1*z1' + c2*z2')/(c1*z1 + c2*z2). The general solution involves two arbitrary constants c1 and c2. However, these constants combine in both the numerator and the denominator, effectively reducing the independence of the parameters. Hence, the general solution for the Riccati equation does not necessarily represent a two-parameter family of solutions.

Step-by-step solution

Establishing the relationship between y and z

Given the Riccati equation: $$y^{\prime}+y^{2}+p(x) y+q(x)=0$$ And we want to show a solution y is the form: $$y = \frac{z'}{z}$$ where z satisfies a second-order linear homogeneous equation: $$z^{\prime \prime}+p(x) z^{\prime}+q(x) z=0$$ First, differentiate both sides of y with respect to x, using the chain rule: $$y' = \frac{z'' z - (z')^2}{z^2}$$ Now, replace y and y' with the obtained expressions in the Riccati equation: $$\frac{z'' z - (z')^2}{z^2} + \left(\frac{z'}{z}\right)^2 + p(x) \frac{z'}{z} + q(x) = 0$$ Now simplify the equation: $$z'' z - (z')^2 + (z')^2 + p(x) z' z + q(x) z^2 = 0$$ We can see that the (z')^2 terms cancel each other, leaving: $$z'' z + p(x) z' z + q(x) z^2 = 0$$ Which is the desired second-order linear homogeneous equation for z: $$z^{\prime \prime}+p(x) z^{\prime}+q(x) z=0$$ So, y is a solution of the Riccati equation if and only if it can be expressed in the form y = z'/z, where z satisfies a second-order linear homogeneous equation.

General solution for the Riccati equation

Now, we are supposed to find the general solution of the Riccati equation using the derived relationship between y and z. Let's assume that the second-order linear homogeneous equation has a fundamental set of solutions {z1, z2}. Then, the corresponding solutions {y1, y2} to the Riccati equation can be expressed as: $$y_1 = \frac{z_1'}{z_1}$$ $$y_2 = \frac{z_2'}{z_2}$$ Now apply the principle of superposition to find the general solution for the Riccati equation. This involves combining solutions y1 and y2 with arbitrary constants c1 and c2 in the numerator as well as the denominator: $$y = \frac{c_1 y_1 + c_2 y_2}{c_1 + c_2}$$ Now substitute y1 and y2 with their expressions in terms of z1 and z2: $$y = \frac{c_1 \frac{z_1'}{z_1} + c_2 \frac{z_2'}{z_2}}{c_1 + c_2}$$ Finally, simplify and obtain the general solution for the Riccati equation: $$y = \frac{c_1 z_1' + c_2 z_2'}{c_1 z_1 + c_2 z_2}$$

Understanding the number of parameters in the solution

Now, we need to discuss whether the Riccati equation has a two-parameter family of solutions based on the derived formula. The general solution of the Riccati equation is given by: $$y = \frac{c_1 z_1' + c_2 z_2'}{c_1 z_1 + c_2 z_2}$$ This general solution involves two arbitrary constants c1 and c2. However, these constants combine in both the numerator and the denominator, effectively reducing the independence of the parameters. In other words, by changing the values of c1 and c2, we obtain different solutions that are not necessarily independent of each other. Therefore, the general solution for the given Riccati equation does not necessarily represent a two-parameter family of solutions.

Key concepts

First Order Differential Equation

When we talk about first order differential equations, we are referring to equations that relate a function and its first derivative. Specifically, in the equation \[y' + y^2 + p(x)y + q(x) = 0,\]which is a Riccati equation, the highest derivative is the first derivative of y, denoted as y'. The characteristic of a first order differential equation is that it cannot be directly solved using standard methods for second order or higher equations. To approach problems like these, one often looks for transformations or substitutions that convert the non-linear equation into a form that is more manageable, such as a linear differential equation, as was done in the exercise provided.

Particularly, the Riccati equation is transformed by introducing a new function z, whose derivative forms the solution y. This technique showcases the ingenuity required to handle first order differential equations that might not have straightforward solutions due to their nonlinearity. By linking the solution of a non-linear first order equation to a second order linear one, we can exploit the methods developed for linear equations to solve more complex ones.

General Solution of Differential Equations

A fundamental aim in the study of differential equations is finding the general solution, which encapsulates all possible solutions to the equation. For the problem at hand, the general solution to the Riccati equation is given through a relationship involving 'z' functions, which themselves solve a related second-order linear homogeneous differential equation.

In the equation\[y = \frac{c_1 z_1' + c_2 z_2'}{c_1 z_1 + c_2 z_2},\]the general solution is expressed in terms of two particular solutions to the second-order equation, z1 and z2, and two arbitrary constants, c1 and c2. The notion of a 'general solution' encompasses all possible specific solutions that might arise from different initial conditions or 'boundary values'. This concept is important because it provides a comprehensive expression from which any particular solution to the differential equation can be drawn.

Linear Homogeneous Equation

When we speak of linear homogeneous differential equations, we refer to those that can be written in the form\[a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + ... + a_1(x)y' + a_0(x)y = 0,\]where all terms involve the function y or its derivatives, and there is no free term (i.e., no constant or function of x without y). This means that all terms of the equation are set to zero, making it 'homogeneous'.

Such equations have an essential property: If y1 and y2 are solutions, then so is c1y1 + c2y2 for any constants c1 and c2, known as the superposition principle. The exercise given is based on the transformation of a Riccati equation, which is nonlinear, into a second-order linear homogeneous equation for z. This linear homogeneous equation's general solution helps us to determine a fundamental set of solutions, which are crucial to building the general solution of the original non-linear Riccati equation.

Boundary Value Problems

Boundary value problems are a significant class of differential equations where the solution is sought within an interval and must satisfy certain conditions at the endpoints of the interval, known as boundaries. These conditions help determine the values of the arbitrary constants in the general solution, resulting in a unique solution that satisfies both the differential equation and the boundary conditions.

In the context of our exercise, the Riccati equation does not explicitly present a boundary value problem, but understanding this concept is essential when dealing with differential equations as they are commonplace, especially in physics and engineering. Applying the concept to the Riccati equation, if boundary conditions were given, we could use them to find the specific values for the constants c1 and c2, thereby determining a unique solution out of the whole family of solutions represented by the general solution equation. The process of solving a boundary value problem is therefore crucial in tailoring the general solution to particular real-world applications.