Consider the initial value problem (IVP) 

Consider the following statements: 

S1: There is an ε > 0 such that for all y₀ ∈ ℝ, the IVP has more than one solution.

S2: There is a  such that for all ε > 0, the IVP has more than one solution.  

Then

Concept:

 Lipschitz condition:  A function   satisfies the Lipschitz condition with respect to  in a domain

 if there exists a constant  such that for any  and  in  D , the following inequality holds

The constant  is called the Lipschitz constant.

Explanation:



, where  is a constant, and  .

S1: There is an  such that for all  , the IVP has more than one solution.
 

S2: There is a   such that for all  , the IVP has more than one solution.

The differential equation is  . Since  , the right-hand side of the

equation is always positive and well-defined for any  .

In general, the uniqueness of solutions to IVPs can often be determined by the Lipschitz condition.

For a function , we need to check if the function satisfies the Lipschitz condition with respect to y .

This function is continuous and bounded for all  because  ensures no singularity at y = 0 .

Therefore, the function satisfies the Lipschitz condition, ensuring that the IVP has a unique solution for each  when  .

S1: This statement claims that there is some  for which the IVP has more than one solution for all  .

From our uniqueness analysis (Lipschitz condition), the IVP actually has a unique solution for all  and  .

Therefore, S1 is false.

S2: This statement claims that for some  and for all , the IVP has more than one solution.

Based on the same reasoning (Lipschitz condition and continuity of the derivative), the solution remains

unique for all  and for any  . Therefore, S2 is also false.

Both statements S1 and S2 are false.

Thus, the correct option is 4).