Existence and Uniqueness Theorem

  • Thread starter Bashyboy
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  • #1
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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 [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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Answers and Replies

  • #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!
 

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