Question

Show that a unique solution exists for each of the following initial-value problems. (a) $$ \begin{aligned} y^{\prime} &=x^{2}+y^{2} \\ y(0) &=0 \end{aligned} $$ on the interval \(|x|<\frac{1}{\sqrt{2}}\). (b) $$ y^{\prime}=1+y^{2} $$ $$ y(0)=0 $$ on the interval \(|x|<\frac{1}{2}\).

Short answer

Answer: A unique solution exists for the initial-value problem (a) on the interval \(|x|<\frac{1}{\sqrt{2}}\) and for the initial-value problem (b) on the interval \(|x|<\frac{1}{2}\).

Step-by-step solution

Checking the continuity of \(f(x,y)\) and \(\frac{\partial f}{\partial y}\)

Our function is given by \(f(x,y)=x^{2}+y^{2}\) and its partial derivative with respect to \(y\) can be found as follows: $$ \frac{\partial f}{\partial y} = 2y $$ Both \(f(x,y)\) and \(\frac{\partial f}{\partial y}\) are continuous for all values of \(x\) and \(y\). Thus, the Existence and Uniqueness Theorem guarantees the existence of a unique solution on the interval \(|x|<\frac{1}{\sqrt{2}}\). Problem (b) Initial-value problem: $$ y'=1+y^{2},\quad y(0)=0 $$

Checking the continuity of \(f(x,y)\) and \(\frac{\partial f}{\partial y}\)

Our function is given by \(f(x,y)=1+y^{2}\) and its partial derivative with respect to \(y\) can be found as follows: $$ \frac{\partial f}{\partial y} = 2y $$ Both \(f(x,y)\) and \(\frac{\partial f}{\partial y}\) are continuous for all values of \(x\) and \(y\). Thus, the Existence and Uniqueness Theorem guarantees the existence of a unique solution on the interval \(|x|<\frac{1}{2}\).

Key concepts

Initial-Value Problems

When solving differential equations, an initial-value problem combines an equation with an accompanying starting condition, specifying the value of the unknown function at a given point. It's akin to starting a journey with a known departure point. In the exercise, the initial-value problems are represented by equations like \( y' = x^2 + y^2 \) with the initial condition \( y(0) = 0 \), and \( y' = 1 + y^2 \) also with \( y(0) = 0 \). These problems seek the function \( y(x) \) that not only satisfies the differential equation but also passes through the specified point in the coordinate system, which in this case is the origin (0,0).

Correctly formulating the initial-value problem is crucial as it lays the foundation for finding a solution that is relevant to the context of the question posed, be it in mathematical theory or applications such as physics and engineering.

Continuity in Differential Equations

The concept of continuity is essential for understanding the behavior of solutions to differential equations. Ensuring the continuity of the functions involved in such equations is analogous to verifying that there are no unexpected jumps or breaks in the curve that represents the solution. In our case, the function \( f(x,y) = x^2 + y^2 \) in problem (a) and the function \( f(x,y) = 1 + y^2 \) in problem (b) must be continuous over the defined domain to assure that solutions can be effectively traced from the initial value without any discontinuities. The absence of any breaks or sharp turns in the graph of the function over the interval in question reinforces the idea that a solution will exist and will be smoothly traceable through the initial point given.

Partial Derivatives

When discussing multi-variable functions, such as those in our differential equations, the partial derivatives play a key role. These derivatives measure how the function changes as each variable changes, while the other variables are held constant. It's like examining the slope of the landscape in one direction at a time, in contrast to the total gradient that would consider all directions simultaneously. In both exercises, we calculate the partial derivative \( \frac{\partial f}{\partial y} \), which gives us \( 2y \) for the respective functions. We need this to apply the Existence and Uniqueness Theorem, ensuring that there is a single solution curve that touches our initial point, and that this curve does not suddenly split into multiple paths.

Unique Solution of ODE

The Existence and Uniqueness Theorem, a cornerstone of differential equations theory, addresses whether a given initial-value problem has a solution and whether that solution is unique. This theorem stipulates that if the function and its partial derivative with respect to the dependent variable (\( y \) in our cases) are continuous, then there is precisely one solution to the differential equation that passes through the specified initial value, much like having a single path that emerges from a crossroad. The workings of this theorem are illustrated in the step-by-step solutions for both problems. Since the conditions are met – our functions and their partial derivatives are continuous – we can state with confidence that in the specified intervals, our initial-value problems (a) and (b) have unique solutions that can be trusted to be the sole trajectory that satisfies the given conditions.