If an IVP does not enjoy uniqueness, then it possesses infinitely many solutions

I am trying to prove that when an IVP in ODEs does not enjoy uniqueness, then it has infinitely many different solutions. I know that when the Lipschitz condition is satisfied, then there is a unique solution.

I guess, I am trying to show that if the IVP possesses two solutions, then we can create infinitely many solutions using these two. However, I am slightly lost as where to start the proof.

Solution 1:

This is not something totally trivial.

It is Helmut Kneser's Theorem.

Let the following IVP $$ x'=f(t,x), \quad x(\tau)=\xi,\tag{1} $$ where $x,f,\xi\in\mathbb R^n$. Then the set for any $s\in\mathbb R$, for which there exists a solution of the above in the interval $[\tau,s]$ the set $$ S=\{x(s): x\,\, \text{is a solution of $(1)$}\}, $$ is connected.

In the one-dimensional case, the proof is rather simple. Assume that $\varphi,\psi:[\tau,\sigma]\to\mathbb R$ are solutions of $(1)$ and $\varphi(\sigma)<\eta<\psi(\sigma)$. Then solve "backwards" the IVP $$ x'=f(t,x), \quad x(\sigma)=\eta,\tag{2} $$ which means that you obtain a solution $\zeta$ for $t\le \sigma$. As soon as the graph of $\zeta$ hits the graph of $\varphi$ or $\psi$, say at $t_0\in(\tau,\sigma)$ you have $\zeta(t_0)=\psi(t_0)$, then you can define a new solution $\tilde\zeta$ as $$ \tilde\zeta(t)=\left\{ \begin{array}{lll} \psi(t) & \text{if} & t\in[\tau,t_0], \\ \zeta(t) & \text{if} & t\in[t_0,\sigma]. \end{array} \right. $$ Clearly $\tilde\zeta$ satisfies both $(1)$ and $(2)$, and that's how you obtain a continuum of solutions.

For a more general theorem, see Philip Hartman, Ordinary Differential Equations, page 15.

Solution 2:

You can start like this. Let the IVP be \begin{eqnarray} L(D)y=f,y(0)=y_0. \end{eqnarray} Let $y_1,y_2$ be two different solutions; namely, $y_1,y_2$ satisfy $$ L(D)y_i=f, y_i(0)=y_0, i=1,2. $$ Now let $y_\alpha=(1-\alpha)y_1+\alpha y_2$ ($\alpha$ is a constant) and you can check that $y_\alpha$ is a solution of the IVP for any $\alpha$. Hence your argument is proved.