We have a theorem that says: Let h : I → R and g : J → R be continuous functions, t 0 ∈ I and y 0 ∈ int(J), hen the differential equation y'(t)=g(y(t))h(t) has a unique solution in some surrounding of t 0 if g ( y 0 ) ≠ 0How I determine the magnitude of this surrounding? Especially, if I integrate the differential equation and solve it for y(t), am I only able to say: My solution is only unique in the surround of y 0 for which g(y(t)) is not zero, is this the condition that determines my surrounding? I mean, I could have determined a solution that gives me zero for some values of t, but is the only solution for my given problem? Am I correct, that in this case, my theorem is not able to say something about the uniqueness of this solution?

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We have a theorem that says: Let   h  :  I  →      R   and   g  :  J  →      R   be continuous functions,       t    0    ∈  I and       y    0    ∈  int(J), hen the differential equation y&#039;(t)=g(y(t))h(t) has a unique solution in some surrounding of       t    0   if   g  (      y    0    )  ≠  0How I determine the magnitude of this surrounding? Especially, if I integrate the differential equation and solve it for y(t), am I only able to say: My solution is only unique in the surround of       y    0   for which g(y(t)) is not zero, is this the condition that determines my surrounding? I mean, I could have determined a solution that gives me zero for some values of t, but is the only solution for my given problem? Am I correct, that in this case, my theorem is not able to say something about the uniqueness of this solution?

Mateo Carson · Accepted Answer

Short answer: local uniqueness + global existence     ⟹   global uniqueness.The solution is unique as long as it exists and stays out of the zeros of g; that is, as long as your theorem applies. Indeed, given two solutions y,z with the same initial value, consider the set   {  t  ∈  I  :  y  (  t  )  =  z  (  t  )  } where I is some interval in which y,z exist. This set is closed by continuity, and open by local uniqueness. Therefore, it coincides with I.The size of the interval of existence comes from the Peano existence theorem. Draw a rectangle  R  =  {  (  t  ,  y  )  :      |    t  −      t    0        |    ≤  c  ,      |    y  −      y    0        |    ≤  b  }around your initial point   (      t    0    ,      y    0    ). If   g  (  y  )  h  (  t  ) is continuous in this rectangle, the solution is safe there. Then you estimate for how long the solution is guaranteed to remain in the safe zone. Let M be the supremum of g(y)h(t) in R. The solution cannot exit through the horizontal side in time less than   b      /    M. Therefore, it exists for       |    t  −      t    0        |    ≤  min  (  c  ,  b      /    M  ). Of course, nothing here is specific to the particular structure of the right hand side that you have.

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