I Uniqueness and Existence Theorm

[Aldnoahz]

Consider y' = 1/sqrt(y)

I seem to be able to find a unique solution given the initial condition of the form y(c) = 0, but the theorem says I won't be able to do so, so I am kind of confused.

I just want some clarifications. Does the uniqueness and existence theorem say anything about the region outside the rectangle where y' and dy'/dy is continuous? Say in this example the theorem guarantees uniqueness when y > 0, but does it necessarily mean that when y = 0, there is no unique solution (i.e. you always find multiple solutions)?

 

[mfb]

You cannot plug in y(c) = 0 in the formula - the function cannot satisfy the differential equation there. You cannot apply a theorem using a differential equation in a region where this equation is not valid.

 

[Charles Link]

In this case, it appears the uniqueness theorem doesn't guarantee anything. There may be a unique solution, but from the uniqueness theorem you can not determine whether this is the case or not.

 

[Aldnoahz]

In this case, it appears the uniqueness theorem doesn't guarantee anything. There may be a unique solution, but from the uniqueness theorem you can not determine whether this is the case or not.

If I understand correctly, uniqueness and existence theorem only guarantees solutions in the region where y' and dy'/dy are continuous, so initial conditions outside this region may or may not result in multiple solutions?

 

[Aldnoahz]

You cannot plug in y(c) = 0 in the formula - the function cannot satisfy the differential equation there. You cannot apply a theorem using a differential equation in a region where this equation is not valid.

I see that the only reason we cannot substitute in y(c) = 0 as the solution is that y' goes to infinity as y approaches 0? But that only means the tangent at y = 0 tends to a vertical line. If I directly solve it and find the general solution, they still give me unique solutions that don't cross each other...

 

[mfb]

You need an applicable initial condition for a unique solution. y(c)=d for d>0 for example.

 

[Charles Link]

I see that the only reason we cannot substitute in y(c) = 0 as the solution is that y' goes to infinity as y approaches 0? But that only means the tangent at y = 0 tends to a vertical line. If I directly solve it and find the general solution, they still give me unique solutions that don't cross each other...

This reminds me of a similar problem that results when you do a Taylor expansion of $ln(1+x)$ about $x=0$. Because $x=-1$ clearly will diverge, the circle of convergence of this series in the complex plane has $r=1$. However, if you use $x=1$ you do get a converging series for $ln(2)=1-1/2+1/3-1/4...$.

 

[zwierz]

Actually it is easy to give sense to the original question. Just reformulate it for the system
$$\dot x=\sqrt y,\quad \dot y= 1$$