When does an initial value problem have exactly two solutions.

Question

My book have examples of an initial value problem which have unique solution and some which have infinite solutions.But I want to know when an initial value problem have two solutions. Is possible or not. Thanks in advance

Answer 1

The generic situation is that if there is no uniqueness of solutions, then there are infinitely may solutions. Under reasonable conditions on the equation $y'=f(x,y)$, $y(x_0)=y_0$, it can be shown that there is a unique solution or there exist a maximal solution $\phi(x)$ and a minimal solution $\psi(x)$ such that $\phi(x_0)=\psi(x_0)=y_0$ and $\phi(x)<\psi(x)$ for $x\ne x_0$ in the interval of existence. In the last case, the region $\{(x,y):\psi(x)<y<\phi(x)\}$ is covered by the graphs of infinitely many solutions of the initial value problem. This is the situation for the well known example $y'=\sqrt{y}$, $y(0)=0$.

Answer 2

An IVP can only have either a unique solution, no solutions or infintely many solutions. If an IVP, with f is continuous, has more than one solution, then it has uncountably many solutions. This is also known as Kneser’s Theorem.