Question

Consider the initial value problem $$ y^{\prime}=3 x(y-1)^{1 / 3}, \quad y\left(x_{0}\right)=y_{0} $$ (a) For what points \(\left(x_{0}, y_{0}\right)\) does Theorem 2.3 .1 imply that (A) has a solution? (b) For what points \(\left(x_{0}-y_{0}\right)\) does Theorem 2.3.1 imply that (A) has a unique solution on some open interval that contains \(x_{0} ?\)

Short answer

Based on Theorem 2.3.1, the initial value problem with the given first-order differential equation has a solution for all points (x₀, y₀) ≠ (x₀, 1). Furthermore, the problem has a unique solution on some open interval that contains x₀ for all points (x₀, y₀) ≠ (x₀, 1).

Step-by-step solution

Identify the given functions in the problem

We are given the differential equation: $$ y^{\prime} = 3x(y-1)^{1 / 3} $$ Here, the function is given by: $$ F(x, y) = 3x(y-1)^{1/3} $$

Determine the continuous function F in both x and y

To ensure the existence of a solution, Theorem 2.3.1 requires that the function \(F(x, y)\) is continuous in both \(x\) and \(y\). The function is a product of a linear function (\(3x\)) and a rational function (\((y-1)^{1/3}\)), which is continuous for all real numbers except when \(y = 1\). Thus, \(F(x, y)\) is continuous for all \((x, y) \neq (x_0, 1)\).

Determine the partial derivative of F with respect to y

To ensure the uniqueness of a solution, Theorem 2.3.1 also requires that the first partial derivative of \(F\) with respect to \(y\) is continuous. We'll calculate the partial derivative: $$ \frac{\partial F}{\partial y} = \frac{\partial}{\partial y}(3x(y-1)^{1/3}) = 3x \cdot \frac{1}{3}(y-1)^{-2/3} = x(y-1)^{-2/3} $$

Determine the continuous partial derivative in both x and y

Now, we need to determine for which \((x, y)\) values the partial derivative is continuous. The partial derivative is a product of a linear function (\(x\)) and a rational function (\((y-1)^{-2/3}\)), which is continuous for all real numbers except when \(y = 1\). Therefore, the first partial derivative is continuous for all \((x, y) \neq (x_0, 1)\).

Apply Theorem 2.3.1 to determine the existence and uniqueness of solutions

From Steps 2 and 4, we determined that both \(F(x, y)\) and its first partial derivative with respect to \(y\) are continuous for all \((x, y) \neq (x_0, 1)\). (a) Therefore, according to Theorem 2.3.1, the initial value problem has a solution for all points \((x_0, y_0) \neq (x_0, 1)\). (b) Similarly, the initial value problem has a unique solution on some open interval that contains \(x_0\) for all points \((x_0, y_0) \neq (x_0, 1)\).

Key concepts

Existence and Uniqueness Theorem

The Existence and Uniqueness Theorem is a cornerstone of differential equations. It helps us understand when a differential equation has a solution, and when that solution is unique in a given region. This theorem applies to initial value problems, where you're given a differential equation along with an initial condition that the solution must satisfy.
To use this theorem effectively, we need a function like the one in our exercise, which is expressed as \( F(x, y) \). For the initial value problem to have a solution, \( F(x, y) \) must be continuous around the initial condition value. The theorem essentially states that if both the function \( F(x, y) \) and its partial derivative with respect to \( y \) are continuous near the initial point, then the differential equation has not just a solution, but a unique one around that point.
In our problem, this means checking the continuity of \( 3x(y-1)^{1/3} \) and its partial derivative. As shown in the steps, it is clear that as long as \( y eq 1 \), the conditions are satisfied, ensuring both existence and uniqueness of the solution.

Differential Equation

A differential equation involves an unknown function and its derivatives, like the one in our problem: \( y' = 3x(y-1)^{1/3} \). Such equations are used in various scientific models to describe changes over time or space.
There are many types of differential equations, but in this case, we're dealing with an ordinary differential equation (ODE), which means the unknown function is dependent only on one variable. Solving a differential equation typically means finding a function that satisfies the equation.
The initial value problem specifies not only the differential equation but also an initial condition \( y(x_0) = y_0 \), which anchors the solution at a specific point. Our task is to find a function that fulfills both the differential equation and the initial condition, ensuring it meets real-world scenarios like starting positions or initial amounts.

Partial Derivative

The concept of a partial derivative is crucial in understanding changes in functions of multiple variables. For a function \( F(x, y) \), the partial derivative with respect to \( y \) is \( \frac{\partial F}{\partial y} \). In our exercise, calculating this derivative involves treating \( x \) as a constant while differentiating with respect to \( y \).
The computation results in \( x(y-1)^{-2/3} \), as demonstrated in the steps. This partial derivative is essential for evaluating the uniqueness of the solution in Theorem 2.3.1.
The need for the continuity of this partial derivative ensures that small changes in \( y \) around the initial point do not cause abrupt changes in the value of \( F \). Thus, checking where this partial derivative is continuous, namely where \( y eq 1 \), helps establish the uniqueness of solutions.

Continuity of Functions

Continuity is a fundamental concept in calculus that describes a function that does not have any abrupt changes, jumps, or gaps. It is vital in applying the Existence and Uniqueness Theorem.
The function \( F(x, y) = 3x(y-1)^{1/3} \) needs to be continuous in the region of interest for us to assure a solution exists for the initial value problem. This means predicting how the function behaves as you approach different values of \( x \) and \( y \).
In our problem, the function \( F(x, y) \) is shown to be continuous everywhere except where \( y = 1 \), due to the cube root becoming undefined for negative values. Similarly, its partial derivative must be continuous away from this singularity. These continuity conditions are pivotal for the Existence and Uniqueness Theorem, as they guarantee that small changes in inputs lead to small changes in outputs, effectively assuring that the problem's solution behaves predictably around the initial point.