Question

Prove: (a) If $$ f\left(x, y_{0}\right)=0, \quad a

Short answer

Based on the given information, explain why the constant function y ≡ y0 is a unique solution of the differential equation y' = f(x, y) on the interval (a, b), given that f(x, y0) = 0 for all x in (a, b) and that f and its partial derivative with respect to y are continuous on an open rectangle containing (x0 + y0). The constant function y ≡ y0 is a unique solution of the differential equation y' = f(x, y) on the interval (a, b) because: 1. It satisfies the differential equation since y' = 0 = f(x, y0) for all a < x < b, given that f(x, y0) = 0 for all x in (a, b). 2. There can be no other solution in the interval (a, b) that equals y0 at some point. This is proven by assuming the existence of such a solution and using the Mean Value Theorem and the uniqueness theorem for differential equations, which rely on the continuity of f and its partial derivative with respect to y. The proof concludes that the only possible solution is y ≡ y0, making this solution unique.

Step-by-step solution

Verify if the constant function \(y \equiv y_{0}\) satisfies the given differential equation

To check whether the function \(y \equiv y_{0}\) is a solution to the differential equation \(y'=f(x,y)\), we need to calculate its derivative and check if it's equal to \(f(x, y_0)\) for all \(a < x < b\), given that \(f(x,y_0)=0\) for all \(x\) in \((a, b)\). Since \(y \equiv y_{0}\) is a constant function, its derivative is: $$ y^\prime = \frac{d}{dx}(y_0) = 0 $$ It satisfies the given differential equation because \(y^\prime = 0 = f(x, y_0)\) for all \(a < x < b\). So, \(y \equiv y_{0}\) solves the differential equation in the interval \((a, b)\).

Prove that no other solution can equal \(y_{0}\) at any point in \((a, b)\)

Assume that there exists another solution, \(y(x)\), of \(y'=f(x,y)\) that is equal to \(y_0\) at some point \(c\) in \((a, b)\). Then, $$ y(c) = y_0 $$ Now, consider the function \(g(x) = y(x) - y_0\). From the above assumption, we have \(g(c) = 0\). Taking the derivative of \(g(x)\), we get: $$ g'(x) = \frac{d}{dx} (y(x) - y_0) = \frac{dy}{dx} = f(x, y) $$ We know that \(f(x, y_0) = 0\) by the given condition. By the Mean Value Theorem (since \(f\) and \(f_y\) are continuous on an open rectangle containing \((x_{0}+y_{0})\)), there exists a \(y^* = y(c) + k (y(x) - y(c))\), where \(0 < k < 1\), such that: $$ f_x(x, y^*) = \frac{f(x, y(x)) - f(x,y_0)}{y(x) - y_0} = \frac{f(x, y) - 0}{g(x)} = \frac{g'(x)}{g(x)} $$ Since \(f_y\) is continuous, \(f_x(x, y^*)\) is continuous in \((a, b)\). Furthermore, \(g'(x) = 0 = g(c)\). So on applying the uniqueness theorem, we conclude that the only possible solution \(y(x)\) is \(y \equiv y_0\), as there can be no other solution in the interval \((a, b)\) that equals \(y_0\) at some point. Hence, the proof is complete.

Key concepts

Constant Function Solution

A constant function solution involves finding if a function that remains constant over an interval solves a given differential equation.
For instance, given a differential equation \( y'=f(x,y) \) and a solution \( y \equiv y_{0} \), we verify if \( y_{0} \) solves the equation.
Since \( y \equiv y_{0} \) is a constant function, its derivative is \( 0 \). In this scenario, we attempt to see if \( 0 = f(x, y_{0}) \) holds true for all \( x \) in an interval \((a, b)\).
If so, \( y \equiv y_{0} \) is indeed a solution of the differential equation.

Uniqueness Theorem

The Uniqueness Theorem is a vital part of differential equations, stating that under certain conditions, a differential equation has a unique solution.
Typically, if the function \( f \) and its partial derivative concerning \( y \), denoted \( f_{y} \), are continuous in a region surrounding the initial conditions, then the differential equation has a unique solution.
This principle asserts that if a solution passes through a specific point, no other solution can intersect this point within the domain, given that all conditions are met.
This becomes imperative in proving that no other solution than the constant function can achieve a specified outcome in scenarios like the exercise provided.

Mean Value Theorem

The Mean Value Theorem (MVT) is a fundamental concept used in calculus to estimate changes and differences.
Suppose \( f \) is continuous on a closed interval and differentiable on the open interval within the same boundaries. The theorem claims there exists a point \( c \) in the interval where the derivative \( f'(c) \) equals the average rate of change over the entire interval.
In the context of differential equations, MVT helps relate changes in function values at two points accounting for continuity and differentiability assumptions, as applied in proving uniqueness in solutions.

Initial Value Problem

An Initial Value Problem (IVP) is a type of differential equation along with a set of initial conditions.
Solving an IVP involves finding a function \( y(x) \) that not only satisfies the differential equation but also meets the predefined initial value conditions.
This process usually looks for a continuous solution that conforms to the given initial condition \( y(x_{0})=y_{0} \).
Initial value problems are common in physics and engineering, where initial states are known, and the objective is to predict the behavior over time.

Continuous Functions

Continuous functions are those that have no gaps, jumps, or breaks in their graphs over a specific interval.
These functions uphold the fundamental property that small changes in input lead to small changes in output, making them predictable and manageable in calculations.
In differential equations, continuity is crucial, as it assures that the functions involved are well-behaved and amenable to techniques like differentiation or integration.
Ensuring that \( f \) and its derivatives are continuous is often a prerequisite for applying theorems such as the Mean Value or Uniqueness Theorems.