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Magazine Chapter 8: Problem 1
Find two solutions of the initial-value problem $$ \left\\{\begin{array}{l} x^{\prime}=x^{1 / 3} \\ x(0)=0 \end{array}\right. $$
Short Answer
Expert verified
Solutions are \( x(t)=0 \) and \( x(t)=\left(\frac{2}{3}t\right)^{3/2} \).
Step by step solution
01
Identify the Differential Equation
The differential equation is given by \( x' = x^{1/3} \). We need to solve this equation to find \( x(t) \), the function that describes how \( x \) changes over time.
02
Separate Variables
Separate the variables to simplify the integration process. We rewrite the equation as \( dx/x^{1/3} = dt \).
03
Integrate Both Sides
Integrate both sides of the equation. The left side becomes \( \int x^{-1/3} \, dx \) and the right side becomes \( \int 1 \, dt \). This yields \( \frac{3}{2} x^{2/3} = t + C \), where \( C \) is the constant of integration.
04
Solve for \( x(t) \)
Solve the equation \( \frac{3}{2} x^{2/3} = t + C \) for \( x \):1. Multiply both sides by \( \frac{2}{3} \) to isolate \( x^{2/3} \): \( x^{2/3} = \frac{2}{3} (t + C) \).2. Raise both sides to the power of \( \frac{3}{2} \) to solve for \( x \): \( x = \left( \frac{2}{3} (t + C) \right)^{3/2} \).
05
Apply Initial Condition
Use the initial condition \( x(0) = 0 \) to find \( C \). Since \( x(t) = \left( \frac{2}{3} (t + C) \right)^{3/2} \), substituting \( t = 0 \) and \( x = 0 \) yields \( 0 = (\frac{2}{3} \times C)^{3/2} \). This implies \( C = 0 \).
06
Check for Alternative Solutions
Note that the differential equation is not Lipschitz continuous at \( x = 0 \). This may lead to more than one solution due to the initial condition. One obvious other solution is \( x(t) = 0 \). Therefore, there are two solutions: \( x(t) = 0 \) for all \( t \), and \( x(t) = (\frac{2}{3}t)^{3/2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial-value Problem
An initial-value problem in differential equations involves finding a function that satisfies a differential equation and meets an initial condition. In our example, the differential equation is \( x' = x^{1/3} \), and the initial condition is \( x(0)=0 \).
This means we're seeking a function \( x(t) \) that describes how \( x \) evolves with time, starting from a specific value when \( t=0 \).
Initial-value problems are common because they often represent real-world scenarios where a system changes over time, starting from a known state.
This means we're seeking a function \( x(t) \) that describes how \( x \) evolves with time, starting from a specific value when \( t=0 \).
Initial-value problems are common because they often represent real-world scenarios where a system changes over time, starting from a known state.
- The function that we derive must satisfy both the equation and the initial condition.
- Such problems require solving the equation and applying the initial condition to obtain constants of integration.
Variable Separation
Variable separation is a method to solve differential equations by separating variables on opposite sides of the equation.
In our exercise, this means putting all \( x \)-dependent terms on one side and all \( t \)-dependent terms on the other.
For \( x' = x^{1/3} \), variable separation transforms it into \( \frac{dx}{x^{1/3}} = dt \).
Here's how you proceed with it:
In our exercise, this means putting all \( x \)-dependent terms on one side and all \( t \)-dependent terms on the other.
For \( x' = x^{1/3} \), variable separation transforms it into \( \frac{dx}{x^{1/3}} = dt \).
Here's how you proceed with it:
- It allows for simple integration on both sides, leading to a more manageable form of the equation.
- It's useful when the equation can be directly partitioned into pure functions of each variable.
Integration
Integration is a fundamental operation in calculus that allows you to find the antiderivative of a function.
In the exercise, after separating the variables, the equation becomes \( \int x^{-1/3} \, dx = \int 1 \, dt \).
Integrating these leads to the expression \( \frac{3}{2} x^{2/3} = t + C \), where \( C \) is the constant of integration.
In the exercise, after separating the variables, the equation becomes \( \int x^{-1/3} \, dx = \int 1 \, dt \).
Integrating these leads to the expression \( \frac{3}{2} x^{2/3} = t + C \), where \( C \) is the constant of integration.
- Integration helps in transforming differential equations into algebraic equations.
- Simplified expressions allow us to solve for unknown functions like \( x(t) \).
Lipschitz Condition
The Lipschitz condition is a crucial concept in assuring the uniqueness of solutions to differential equations.
It's a type of constraint on the functions involved in the equation.
If a differential equation satisfies the Lipschitz condition in a region, then it guarantees uniqueness of the solution passing through a given point.
However, in our exercise, the equation \( x' = x^{1/3} \) fails to satisfy this condition at \( x = 0 \).
It's a type of constraint on the functions involved in the equation.
If a differential equation satisfies the Lipschitz condition in a region, then it guarantees uniqueness of the solution passing through a given point.
However, in our exercise, the equation \( x' = x^{1/3} \) fails to satisfy this condition at \( x = 0 \).
- This means multiple solutions could exist from the same initial condition.
- The lack of the Lipschitz condition at \( x = 0 \) implies the potential for solutions like \( x(t) = 0 \) and \( x(t) = \left( \frac{2}{3} t \right)^{3/2} \).
