Question

In each exercise, (a) Rewrite the given \(n\)th order scalar initial value problem as \(\mathbf{y}^{\prime}=\mathbf{f}(t, \mathbf{y}), \mathbf{y}\left(t_{0}\right)=\mathbf{y}_{0}\), by defining \(y_{1}(t)=y(t), y_{2}(t)=y^{\prime}(t), \ldots, y_{n}(t)=y^{(n-1)}(t)\) and defining \(y_{1}(t)=y(t), y_{2}(t)=y(t), \ldots, y_{n}(t)=y^{\prime(n-t)}(t)\) \(\mathbf{y}(t)=\left[\begin{array}{c}y_{1}(t) \\ y_{2}(t) \\ \vdots \\\ y_{n}(t)\end{array}\right]\) (b) Compute the \(n^{2}\) partial derivatives \(\partial f_{i}\left(t, y_{1}, \ldots, y_{n}\right) / \partial y_{j}, i, j=1, \ldots, n\). (c) For the system obtained in part (a), determine where in \((n+1)\)-dimensional \(t \mathbf{y}\)-space the hypotheses of Theorem \(6.1\) are not satisfied. In other words, at what points \(\left(t, y_{1}, \ldots, y_{n}\right)\), if any, does at least one component function \(f_{i}\left(t, y_{1}, \ldots, y_{n}\right)\) and/or at least one partial derivative function \(\partial f_{i}\left(t, y_{1}, \ldots, y_{n}\right) / \partial y_{i}, i, j=1, \ldots, n\) fail to be continuous? What is the largest open rectangular region \(R\) where the hypotheses of Theorem \(6.1\) hold? $$ y^{\prime \prime}+t y=\sin y^{\prime}, \quad y(0)=0, \quad y^{\prime}(0)=1 $$

Short answer

Question: Rewrite the given 2nd order scalar initial value problem as a system of two first-order differential equations, compute the required partial derivatives, and find the largest open rectangular region where the hypotheses of Theorem 6.1 hold. Problem: \(y''(t) + ty(t) = \sin{y'(t)}\), \(y(0) = 0\), and \(y'(0) = 1\).

Step-by-step solution

Rewrite the given problem as a system of first-order differential equations

We define two functions: $$y_1(t) = y(t)$$ $$y_2(t) = y'(t)$$ Now we can rewrite the given problem: $$y''(t) + ty(t) = \sin{y'(t)}$$ as a first order system of equations: $$y_1'(t) = y_2(t)$$ $$y_2'(t) = \sin{y_2(t)} - ty_1(t)$$ Thus, the problem can be written in the vector form as: $$\mathbf{y}'(t) = \mathbf{f}(t, \mathbf{y}) = \begin{bmatrix} y_2(t) \\ \sin{y_2(t)} - ty_1(t) \end{bmatrix}, \quad \mathbf{y}(0) = \begin{bmatrix} y_1(0) \\ y_2(0)\end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

Compute the required partial derivatives

We need to compute the partial derivatives of \(\mathbf{f}(t, \mathbf{y})\): $$\frac{\partial f_{1}(t, y_{1}, y_{2})}{\partial y_{1}} = 0, \quad \frac{\partial f_{1}(t, y_{1}, y_{2})}{\partial y_{2}} = 1$$ $$\frac{\partial f_{2}(t, y_{1}, y_{2})}{\partial y_{1}} = -t, \quad \frac{\partial f_{2}(t, y_{1}, y_{2})}{\partial y_{2}} = \cos{y_2(t)}$$

Determine points where hypotheses of Theorem 6.1 are not satisfied

Theorem 6.1 requires that the component functions and the partial derivatives should be continuous everywhere in the \((n+1)\)-dimensional \(t\mathbf{y}\)-space. All the component functions and the partial derivatives we computed above are continuous everywhere. Therefore, the hypothesis of Theorem 6.1 is satisfied everywhere in the \((n+1)\)-dimensional \(t\mathbf{y}\)-space, and there are no points where they are not satisfied. The largest open rectangular region \(R\) where the hypotheses of Theorem 6.1 hold can be any open rectangle that contains the point \((t, y_1, y_2) = (0, 0, 1)\), since all the functions and their partial derivatives are continuous everywhere.

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) consists of a differential equation, usually represented as \( y'(t) = f(t, y(t)) \), together with an initial condition at a specific point, typically shown as \( y(t_0) = y_0 \). The aim is to find a function \( y(t) \), often referred to as the 'solution', which not only meets the differential equation but also satisfies the given initial condition at \( t_0 \). This type of problem is fundamental in the mathematical study because it mimics real-world situations where the state at an initial moment dictates future development, like the position and speed of a particle at the start dictating its future trajectory.

Solving an IVP often requires rewriting higher-order differential equations into a system of first-order equations. By doing this, we transform a complex problem into an assembly of simpler ones. Each first-order equation in a system like this represents a smaller piece of the IVP puzzle and can be integrated to obtain the solution to the full problem. As seen in the provided exercise, the original second-order equation is cleverly split into a pair of first-order equations that set the stage for analysis.

System of First-Order Differential Equations

A System of First-Order Differential Equations is a collection of equations involving derivatives of unknown functions, which are of the first order—in other words, the derivative is not taken more than once on any unknown function. The approach to transform a higher-order differential equation into a system of first-order differential equations, as seen in the exercise, is a valuable tool. This system can be visually represented as a matrix or a vector, providing a way to organize and systematically resolve the differential equations.

An important aspect is setting up these equations to ensure all derivatives are expressed as first-order ones, which may involve defining new 'intermediate' functions like \( y_1(t), y_2(t), \) etc. With this system, it is possible to apply a multitude of analytical and numerical techniques to find solutions, which widens the scope of solvable problems in physics, engineering, and other sciences.

Partial Derivatives

In mathematics, Partial Derivatives arise when dealing with functions of multiple variables, representing the rate at which the function changes with respect to one variable while holding the others constant. Calculating partial derivatives is an exercise critical to multidisciplinary fields, from economics to fluid dynamics, and is a fundamental tool in multivariable calculus.

The exercise provided demands computing the partial derivatives of functions that appear in a system of first-order differential equations. These derivatives inform us about the behavior and stability of the system under small changes in its variables. For instance, \( \frac{\partial f}{\partial y_1} \) signifies how the rate of change \( f \) with respect to \( y_1 \) while keeping \( y_2 \) constant. The calculation of these derivatives, as showcased, is essential for analyzing the system both theoretically and computationally and for applying relevant theorems about the system's nature and solutions.

Theorem 6.1 Continuity

Theorem 6.1 pertains to the continuity of the functions and their partial derivatives that are involved in a system of first-order differential equations, a condition critical for the existence and uniqueness of solutions in IVPs. For the theorem's hypotheses to hold, both the functions in the system and their respective partial derivatives must be continuous over a specified region of the domain in question. Continuity here means that there should be no 'jumps' or 'gaps' in the values of these functions.

In the context of the exercise, all functions and their calculated partial derivatives were continuous, thus satisfying the theorem's requirements. This is significant because it indicates the presence of a unique solution that extends throughout the domain, a reassurance that the IVP addressed is well-posed and solvable. Being able to identify regions where Theorem 6.1 is valid, or conversely, where it fails, is crucial for understanding the behavior and limits of the system under consideration.