Problem 18
Question
Determine a region of the \(x y\) -plane for which the given differential equation would have a unique solution whose graph passes through a point \(\left(x_{0}, y_{0}\right)\) in the region. $$ \frac{d y}{d x}=\sqrt{x y} $$
Step-by-Step Solution
Verified Answer
The region is where \( x > 0 \) and \( y > 0 \).
1Step 1: Identify the differential equation
The given differential equation is \( \frac{d y}{d x} = \sqrt{x y} \). This equation expresses how \( y \) changes as \( x \) changes.
2Step 2: Examine the expression for the derivative
The derivative \( \sqrt{x y} \) involves the square root of the product \( x y \). For this expression to be defined, \( x y \) must be non-negative, meaning \( x y \geq 0 \).
3Step 3: Determine where the expression is continuous and differentiable
To satisfy the initial value problem conditions, \( \sqrt{x y} \) must be continuous and differentiable. Therefore, neither \( x \) nor \( y \) can be negative simultaneously, which implies the region includes where \( x > 0 \) and \( y > 0 \).
4Step 4: Identify the region for a unique solution
By the existence and uniqueness theorem for differential equations (Picard-Lindelöf theorem), a unique solution passing through a point \((x_0, y_0)\) exists when \( \frac{d y}{d x} \) is Lipschitz continuous in \( y \). In this case, this condition is automatically satisfied in any non-negative region due to the continuity of \( \sqrt{x y} \).
5Step 5: Conclude the region
The region \( x > 0 \) and \( y > 0 \) ensures both continuity and differentiability, providing a unique solution to the differential equation that passes through any point \((x_0, y_0)\) in this quadrant.
Key Concepts
Existence and Uniqueness TheoremInitial Value ProblemLipschitz Continuity
Existence and Uniqueness Theorem
The Existence and Uniqueness Theorem, also known as the Picard-Lindelöf theorem, is a vital concept in differential equations. It addresses whether a differential equation has a solution and if that solution is unique. This theorem applies to first-order ordinary differential equations of the form \( \frac{dy}{dx} = f(x, y) \).
According to the theorem, if the function \( f(x, y) \) is continuous in a region that contains the initial value point \((x_0, y_0)\), then there exists at least one solution that passes through \( (x_0, y_0) \).
Moreover, if \( f(x, y) \) satisfies an additional condition known as Lipschitz continuity with respect to \( y \), then this solution is unique within the region.
In our given problem, \( \frac{dy}{dx} = \sqrt{x y} \), both continuity and Lipschitz conditions must be examined. Since \( \sqrt{x y} \) is continuous in places where \( x > 0 \) and \( y > 0 \), the Existence and Uniqueness Theorem assures us of a unique solution in this quadrant.
According to the theorem, if the function \( f(x, y) \) is continuous in a region that contains the initial value point \((x_0, y_0)\), then there exists at least one solution that passes through \( (x_0, y_0) \).
Moreover, if \( f(x, y) \) satisfies an additional condition known as Lipschitz continuity with respect to \( y \), then this solution is unique within the region.
In our given problem, \( \frac{dy}{dx} = \sqrt{x y} \), both continuity and Lipschitz conditions must be examined. Since \( \sqrt{x y} \) is continuous in places where \( x > 0 \) and \( y > 0 \), the Existence and Uniqueness Theorem assures us of a unique solution in this quadrant.
Initial Value Problem
An Initial Value Problem (IVP) requires us to find a function that not only satisfies the differential equation but also passes through a specific point called the initial condition.
The general form for an initial value problem is:
In the exercise, you are given the differential equation \( \frac{dy}{dx} = \sqrt{x y} \) and must find the region where solutions exist for given initial points.
Solving this requires ensuring that the derivative function \( \sqrt{x y} \) remains valid, which it does for \( x > 0 \) and \( y > 0 \). Thus, any initial value within this region leads to a valid IVP solution.
The general form for an initial value problem is:
- A differential equation: \( \frac{dy}{dx} = f(x, y) \)
- An initial condition: \( y(x_0) = y_0 \)
In the exercise, you are given the differential equation \( \frac{dy}{dx} = \sqrt{x y} \) and must find the region where solutions exist for given initial points.
Solving this requires ensuring that the derivative function \( \sqrt{x y} \) remains valid, which it does for \( x > 0 \) and \( y > 0 \). Thus, any initial value within this region leads to a valid IVP solution.
Lipschitz Continuity
Lipschitz Continuity is a condition that enhances the uniqueness aspect of the Existence and Uniqueness Theorem. It provides a more stringent requirement than mere continuity, ensuring that solutions do not diverge from each other.
A function \( f(x, y) \) is Lipschitz continuous with respect to \( y \) if there exists a constant \( L \) such that for all points \( (x, y_1) \) and \( (x, y_2) \) in the region:
\[ |f(x, y_1) - f(x, y_2)| \leq L |y_1 - y_2| \]
This essentially means that the rate of change of \( f \) with respect to \( y \) is bounded by \( L \), preventing solutions from drastically departing from one another.
In our problem of \( \frac{dy}{dx} = \sqrt{x y} \), analyzing Lipschitz continuity is crucial to guarantee a single, unique solution. Given that both \( x \) and \( y \) are positive within the region considered, and since the complexity of the square root does not markedly increase differences between values, the function inherently satisfies Lipschitz conditions, supporting unique solution paths.
A function \( f(x, y) \) is Lipschitz continuous with respect to \( y \) if there exists a constant \( L \) such that for all points \( (x, y_1) \) and \( (x, y_2) \) in the region:
\[ |f(x, y_1) - f(x, y_2)| \leq L |y_1 - y_2| \]
This essentially means that the rate of change of \( f \) with respect to \( y \) is bounded by \( L \), preventing solutions from drastically departing from one another.
In our problem of \( \frac{dy}{dx} = \sqrt{x y} \), analyzing Lipschitz continuity is crucial to guarantee a single, unique solution. Given that both \( x \) and \( y \) are positive within the region considered, and since the complexity of the square root does not markedly increase differences between values, the function inherently satisfies Lipschitz conditions, supporting unique solution paths.
Other exercises in this chapter
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