- #1
Aldnoahz
- 37
- 1
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)?
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)?