Existence and uniqueness

  • #1
Using the existence and uniqueness criteria, give the region (call it D) in the x-y plane consisting of all points (xo, yo) such that there is a unique solution. Choose a point in D as your initial condition, show that the equation is exact, then use the fact to solve the associated initial value problem.



2x+4y+(4x-2y)(dy/dx)=0, y(xo)=yo



I know how to solve of exactness, by proving that (dP/dy)=(dQ/dx), but what I don't quite get is how do I figure out what the initial value is, do I just let the inital values of x and y be just some random value and prove for uniqueness or is there a way that I can find what the initial values are?
 
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Answers and Replies

  • #2
136
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I know you've already "solved" that part, but personally I don't like thinking of exactness in terms of "P's" and "Q's", unless those things have been carefully defined.

What you're really trying to do w/ exactness suppose you have a solution to the D.E. given by:

[tex]u(x,y) = c[/tex]

Then you must have that:

[tex]du = 0[/tex]

as the differential equation for which u is the solution (i.e., primitive). But we can write:

[tex]du = \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy = 0[/tex]

Taking

[tex]P(x,y) = \frac{\partial u}{\partial x}[/tex]

[tex]Q(x,y) = \frac{\partial u}{\partial y}[/tex]

we have your condition for exactness, which follows from the continuity condition:

[tex]\frac{\partial^2u}{\partial x\partial y} = \frac{\partial^2u}{\partial y\partial x}[/tex]

Which is a long-winded way of saying something you already know, but I think it's important to understand the reasons why we want exactness, and what it means, rather than just memorizing rote formulas w/ Ps and Qs in them.

Rewriting your equation as a total differential, we have:

[tex](4y + 2x) dx + (4x- 2y) dy = 0[/tex]

So yes, clearly the equation is exact.

In answer to your (real) question, I think the problem is just asking you to find the domain on which this equation has a solution (is it the whole real plane R^2? ... not all equations have solutions at all points in the real plane). At any rate, find out the domain in which the equation is valid (it could be the whole real plane), then pick any point in that domain, and solve the associated IVP.

I think that's what the question is asking.
 
  • #3
thanks psholtz for the help... The Q makes more sense now.
 
  • #4
97
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I am confused as to how you find the domain to which the equation has a solution and then how to determine an appropriate point to solve the IVP...
 

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