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Assume the IVP $$\frac{dy}{dx}=f(x,y), y(a)=b.$$ is there a way to know the nature of the solution without explicitly solving it?

what I tried if $f$ is Lipschitz, then it proves uniqueness. But there is no way to know about infinite solution or no solution (as I know it, without solving). Can any one suggest a good reference in this direction?

Thank you

Sine of the Time
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Hemanta Mandal
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  • Peano's theorem assures the existence – Sine of the Time Aug 28 '24 at 12:00
  • @SineoftheTime so is it like its continuous but not Lipschitz implies infinite solutions? – Hemanta Mandal Aug 28 '24 at 14:49
  • If $f(x,y)$ is continuous, there's at least a solution, but this does not mean there are infinitely many. – Sine of the Time Aug 28 '24 at 14:56
  • If $f$ is continuous and $y'=f(x,y)$, $y(a)=b$ is not uniquely solvable, then it has infinite many solutions, see H. Kneser's Theorem: https://en.wikipedia.org/wiki/Kneser%27s_theorem_(differential_equations) or here https://math.stackexchange.com/questions/657284/if-an-ivp-does-not-enjoy-uniqueness-then-it-possesses-infinitely-many-solutions/657301#657301 – Gerd Aug 28 '24 at 15:43
  • @Gerd, don't you have to first solve $y'=f(x,y)$ in order to determine whether $y(a)=b$ has more than one solution? – Gonçalo Aug 28 '24 at 15:56
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    @Gonçalo You are right, if you want to apply Kneser's Theorem you have to know that there are at least two different solutions of the IVP. Then you know that it has an infinite number of solutions. – Gerd Aug 28 '24 at 18:02

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