Existence and Uniqueness Theorem

In summary, the Existence and Uniqueness Theorem states that if the function and its y partial are continuous in a rectangle, then there exists a unique solution in some interval within that rectangle. Based on the wording of your question, it seems that you are trying to determine if a specific value, y_1, can exist in the same rectangle as another point (x_0,y_0) where a unique solution exists. However, if the function and its derivative are discontinuous at y_1, it cannot satisfy the first condition of the theorem and therefore cannot exist in the same rectangle.
  • #1
Bashyboy
1,421
5
Hello Everyone.

I have a question. Suppose I have a differential equation for which I want to find the values at which [itex]\displaystyle f(x,y)[/itex] and [itex]\displaystyle \frac{\partial f}{\partial y}[/itex] are discontinuous, that I might know the points at which more than one solution exists. Suppose that [itex]\displaystyle y_1[/itex] is such a value. Now suppose we want to find a unique solution at [itex]\displaystyle (x_0, y_0)[/itex], and that it exists. My question is, can the region that encloses [itex]\displaystyle (x_0,y_0)[/itex] also include [itex]\displaystyle y_1[/itex]?
 
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  • #2
Hi bashyboy,

Statement of the Existence and Uniqueness Theorem:

If ##f(x,y)## and ##f_y(x,y)## are continuous in a rectangle ##R: |t|<= a, |y| <= b##, then there is some interval ##|t| <= h <= a## in which there exists a unique solution ##y = \phi(t)## of the initial value problem.

I find the wording of your question a bit confusing, so I'm coming in with the assumption that you're trying to find out if ##y_1## is the value at which the function and/or its y partial are discontinuous if it can exist in the rectangle ##R##. If this assumption is correct then maybe I am the confusing one!

I'm also assuming here that you have created a line ##y_1## where the function and/or its derivative are continuous. Line, point, whatever :P

So if the function and its y partial are discontinuous at that point ##y_1##, then it does not satisfy the first condition of the theorem where it must be continuous in that rectangle. Therefore your points ##(x_0,y_0)## and your line ##y_1## cannot live in the same rectangle. Bad blood bro.

I hope this is what you were looking for!
 

1. What is the Existence and Uniqueness Theorem?

The Existence and Uniqueness Theorem is a fundamental theorem in mathematics that guarantees the existence and uniqueness of a solution to a certain type of mathematical problem. It is commonly used in the field of differential equations, but can also be applied to other areas of mathematics.

2. How does the Existence and Uniqueness Theorem work?

The theorem states that if certain conditions are met, such as continuity and differentiability of the function, then there exists one and only one solution to the problem. This solution can be found using various mathematical techniques, such as the Picard-Lindelöf method.

3. What are the conditions that must be met for the Existence and Uniqueness Theorem to hold?

The conditions vary depending on the type of problem, but generally include continuity and differentiability of the function, as well as a specified initial condition or boundary condition. These conditions ensure that the problem has a unique solution.

4. Can the Existence and Uniqueness Theorem be applied to all mathematical problems?

No, the theorem only applies to certain types of problems, particularly those involving differential equations. Other types of problems may have multiple solutions or no solutions at all.

5. Why is the Existence and Uniqueness Theorem important in mathematics?

The theorem provides a powerful tool for solving mathematical problems and proving the uniqueness of their solutions. It also helps to ensure the validity and reliability of mathematical models by guaranteeing the existence of a solution. This is crucial in various fields, such as physics, engineering, and economics.

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