Question Concerning Existence/Uniqueness Theorem

[Duderonimous]

Homework Statement

\frac{dy}{dt}=y(y-1)(y+3), y(0)=4

What does the exist/unique theorem say about the solution?

Homework Equations

Exist Theorem:

Suppose some continuous function f(t,y) goes through a rectangle. If (t_o,y_o)is a point in the rectanglein the ty-plane, then their exist a \epsilon>0 and func. y(t)
defined for t_o-\epsilon<t<t_o+\epsilon
that solves the init. val. prob.

\frac{dy}{dt}=f(t,y), y(t_o)=y_o

Unique Theorem:

Suppose f(t,y) and ∂f/∂y are continuous func. in a rectangle in the ty-plane. If (t_o,y_o) is a point in the rectangle and if y₁(t) and y₂(t) are two functions that solve the initial value prob.

\frac{dy}{dt}=f(t,y), y(t_o)=y_o

for all t in t_o-\epsilon<t<t_o+\epsilon then
y₁(t)=y₂(t) for that interval.

The Attempt at a Solution

Their are 3 equilibrium solutions namely y=0,1,-3

y(t)>3 because why? I know given my initial condition that if I am starting at y=4 when t=0 I can only approach y=3. Also in the book the solution includes that
y(t) goes to 3 as t decreases and y(t) goes to infinity as t increases. Why can't it be that y(t) goes to infinity as t increases and y(t) goes to 3 as t decreases. How do I know from which way my functions going if I just know the equilib. soln. and the initial cond.?

Also If I have an initial cond. of y(0)=2 how do I know which equilibrium soln it is approaching as t increases? Thanks

 

[HallsofIvy]

Yes, y= 0, y= 1, and y= -3 are equilibrium solutions and, by the existence and uniqueness theorem, no other solution can cross those lines. Now, for any y> 1, each of y, y- 1 and y+ 3 are positive so their product is positive. That means that dy/dt> 0 and so y is increasing in that region. If y(0)= 0, then, as t increases, y increases, going away from y= 1. But as t decreases, y also decreases- but can never cross y= 1. What does that mean?

Also in the book the solution includes that
y(t) goes to 3 as t decreases and y(t) goes to infinity as t increases. Why can't it be that y(t) goes to infinity as t increases and y(t) goes to 3 as t decreases

What? Those are the same thing. Grammatically "A and B" is the same as "B and A"! However, your book should say "1" rather than "3". There are equilibrium solutions at x= 1 and y= -3, not at y= 3.

If you meant "Why can't it be that y(t) goes to infinity as t decreases and y(t) goes to 1 as t decreases" it is because, as I said, for y> 1 dy/dt= y(y- 1)(y+ 3) is positive and so y is an increasing function not decreasing.