Question

Let \(\phi\) be a real function defined on a rectangle \(R\) in the plane, given by \(a \leq x \leq b\), \(\alpha \leq y \leq \beta .\) A solution of the initial- value problem $$ y^{\prime}=\phi(x, y), \quad y(a)=c \quad(\alpha \leq c \leq \beta) $$ is, by definition, a differentiable function \(f\) on \([a, b]\) such that \(f(a)=c, \alpha \leq f(x) \leq \beta\), and $$ f^{\prime}(x)=\phi(x, f(x)) \quad(a \leq x \leq b) $$ Prove that such a problem has at most one solution if there is a constant \(A\) such that $$ \left|\phi\left(x, y_{2}\right)-\phi\left(x, y_{1}\right)\right| \leq A\left|y_{2}-y_{t}\right| $$ whenever \(\left(x, y_{1}\right) \in R\) and \(\left(x, y_{2}\right) \in R\) Hint: Apply Exercise 26 to the difference of two solutions. Note that this uniqueness theorem does not hold for the initial-value problem $$ y^{\prime}=y^{1 / 2}, \quad y(0)=0, $$ which has two solutions: \(f(x)=0\) and \(f(x)=x^{2} / 4\). Find all other solutions.

Short answer

Question: Prove the uniqueness of the solution of the given initial-value problem under the condition that the function satisfies \(\left|\phi\left(x, y_{2}\right)-\phi\left(x, y_{1}\right)\right| \leq A\left|y_{2}-y_{1}\right|\). Also, analyze the given initial-value problem \(y^{\prime}=y^{1 / 2}, y(0)=0\) for which the uniqueness theorem does not hold, and find all other solutions. Answer: Under the given condition, the initial-value problem has at most one solution. The uniqueness theorem does not hold for the given initial-value problem \(y^{\prime}=y^{1 / 2}, y(0)=0\), and the general solution is \(y = \frac{x^2}{4}\), along with the solution \(f(x) = 0\).

Step-by-step solution

Understand the given hints

First, we need to understand that the exercise wants us to study the uniqueness of the solution given the condition mentioned for the function \(\phi\). Exercise 26, referred in the hint, states the following: If \(f\) and \(g\) are continuous on \([a, b]\) and if \(|g(x) - f(x)| \leq K|\) for all \(x\) in \([a, b],\) where \(K\) is a constant, then there exists only one function whose derivative is equal to the difference of \(f(x)\) and \(g(x)\) and that is \(f(x) - g(x)\), provided their initial values are equal. Now, let's analyze both parts of the problem.

Prove the uniqueness of the solution

Suppose we have two different solutions \(f_1(x)\) and \(f_2(x)\) of the initial-value problem. We will analyze the difference between these two solutions, say, \(h(x) = f_1(x) - f_2(x)\). Now, let's find the derivative of \(h(x)\) using the chain rule: \(h'(x)= f_1'(x) - f_2'(x) = \phi(x, f_1(x)) - \phi(x, f_2(x))\) (since \(f_i'(x) = \phi(x, f_i(x)),\ i = 1, 2\)) Now, we are given that: \(\left|\phi\left(x, y_{2}\right)-\phi\left(x, y_{1}\right)\right| \leq A\left|y_{2}-y_{1}\right|\) Applying this to \(f_1\) and \(f_2\): \(|h'(x)| = \left|\phi\left(x, f_1(x)\right) - \phi\left(x, f_2(x)\right)\right| \leq A\left|f_1(x) - f_2(x)\right| = A|h(x)|\) By Exercise 26, there exists only one solution, when the constant \(K = A\) and their initial values are equal, which is \(h(x)\) here. Hence, the initial-value problem has at most one solution under the given condition.

Analyze the given initial-value problem and find all other solutions

Now, let's analyze the provided initial-value problem for which the uniqueness theorem does not hold: $$ y^{\prime}=y^{1 / 2}, \quad y(0)=0 $$ We have already been given two solutions: \(f(x)=0\) and \(f(x)=x^{2} / 4\). To find all other solutions, let's use separation of variables method: \(dy/dx = y^{\frac{1}{2}} \Rightarrow \int\frac{dy}{\sqrt{y}} = \int dx\) Integrating both sides, we get: \(2\sqrt{y} = x+C\), where \(C\) is the constant of integration. We have the initial condition \(y(0) = 0\): \(2\sqrt{0} = 0 + C \Rightarrow C=0\) So, the general solution is \(2\sqrt{y} = x\) or \(y = \frac{x^2}{4}\). This is the second given solution. However, the first solution \(f(x) = 0\) also satisfies the initial-value problem, so the uniqueness theorem does not hold for this problem.

Key concepts

Mathematical Analysis

Mathematical analysis is a branch of mathematics that deals with limits and related theories, such as differentiation, integration, measure, sequences, and series. It is the rigorous underpinning of calculus, and its study involves taking an in-depth look into the behavior of real functions and how they can be described with mathematical operations. In the context of differential equations, analysis investigates the existence, uniqueness, and characteristics of solutions. The exercise given centers around understanding the uniqueness of solutions to initial-value problems – a fundamental concept in mathematical analysis. When a uniqueness theorem applies, it assures that under certain conditions, an initial-value problem cannot have more than one solution. This provides certainty in physical sciences and engineering, where differential equations often model real-world phenomena.

In the step-by-step solution provided, mathematical analysis comes into play to establish that for a given function \( \phi \), if it meets certain conditions—specifically one involving a constant \( A \) that bounds the difference between the function's values at two points—then the initial-value problem has at most one solution. Such an analysis involves understanding the behavior of \( \phi \) and employing the properties of continuity and differentiability to draw conclusions about the solution's behavior.

Differential Equations

Differential equations are mathematical equations that describe the relationship between functions and their derivatives. They are fundamental in modeling the behavior of systems where change is continuous, such as fluid dynamics, thermodynamics, and population growth. The initial-value problem is a specific type of differential equation where the solution is determined not only by the differential equation itself but also by the value the solution takes at a starting point, known as the initial condition.

In the provided exercise, we encounter an initial-value problem that asks for the proof of the uniqueness of its solutions. By applying the condition \( |\phi(x, y_{2}) - \phi(x, y_{1})| \leq A|y_{2} - y_{1}| \) and involving the concept of mathematical analysis, one needs to prove that the solution to the differential equation is singular. This creates a direct link between the abstract concepts of mathematical analysis and the practical application of solving differential equations. The method involved in the solution, which assumed two solutions and compared them, serves to illustrate the direct application of theoretical concepts to problem-solving.

Real Functions

Real functions are mappings from a set of real numbers to real numbers. These functions are central to calculus and mathematical analysis and often are the subject of differential equations. In our case, the function \( \phi \) is a real function that takes two real number inputs and produces a real number output. It is this function that dictates the rate of change, or the derivative, of the solution to our differential equation.

Understanding the properties of real functions like \( \phi \) is crucial for solving and making assertions about differential equations. Real functions can exhibit behaviors like continuity, boundedness, and differentiability which play a role in the existence and uniqueness of solutions to differential equations. The exercise provided demonstrates how the nature of \( \phi \) directly affects the solution to the initial-value problem. If \( \phi \) satisfies certain properties—the Lipschitz condition in this case—then we can guarantee the uniqueness of the solution of the differential equation. This is an example of how investigating the characteristics of real functions is paramount in the study of differential equations.

Continuity

Continuity is a fundamental concept in mathematical analysis and describes a property of functions in which small changes in the input lead to small changes in the output. This concept is essential when dealing with differential equations, as the behavior of solutions over time is often dependent on the continuity of the function that represents the change. For a function to be continuous at a point, the limit of the function as it approaches the point must exist and be equal to the function's value at that point.

In the context of this exercise, we need a function \( \phi \) that is continuous. Continuity ensures the function does not have abrupt changes, which allows us to deduce certain behaviors about potential solutions of a differential equation. When a function is continuous and satisfies the given condition that involves the constant \( A \) in the exercise, it helps to set a foundation upon which the uniqueness theorem relies. The step-by-step solution uses this notion of continuity implicitly when utilizing the properties of \( \phi \) to determine that the solution to the initial-value problem is unique. Thus, continuity is not just an abstract mathematical concept but a practical tool that guarantees the predictability and stability of solutions in differential equations.