Existence and Uniqueness Theorem

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SUMMARY

The discussion centers on the Existence and Uniqueness Theorem in the context of differential equations. It establishes that if the function \( f(x,y) \) and its partial derivative \( \frac{\partial f}{\partial y} \) are continuous within a rectangle \( R \), then a unique solution exists for the initial value problem. The participants clarify that if a point \( y_1 \) is where either \( f \) or \( \frac{\partial f}{\partial y} \) is discontinuous, then it cannot coexist with the point \( (x_0, y_0) \) in the same rectangle \( R \), as this violates the theorem's conditions.

PREREQUISITES
  • Understanding of differential equations
  • Familiarity with the Existence and Uniqueness Theorem
  • Knowledge of continuity and discontinuity in functions
  • Basic concepts of partial derivatives
NEXT STEPS
  • Study the implications of the Existence and Uniqueness Theorem in solving differential equations
  • Explore examples of functions that are continuous and discontinuous
  • Learn about the role of rectangles in defining regions of continuity
  • Investigate methods for identifying points of discontinuity in functions
USEFUL FOR

Mathematicians, students of calculus, and anyone involved in solving differential equations will benefit from this discussion, particularly those interested in the conditions for the existence of unique solutions.

Bashyboy
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Hello Everyone.

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

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