0
$\begingroup$

For a linear non-homogeneous ODE $$ \dot{x}=f(t),\ \ x(0)=x_0, 0\leq t\leq 1, $$ if one has two different solutions then one has infinitely many solutions. The idea for showing such property is to use the fact that the solution space of $\dot{x}=0$ is a vector space.

Do we still have the same conclusion (locally) for the general nonlinear ODE, say $$ \dot{x}=g(x),\ \ x(0)=x_0, 0\leq t\leq 1, $$ where $g$ is assumed to be a continuous function?

The classical example $\dot{x}=\sqrt{|x|}$, which demonstrates that uniqueness can fail, seems relevant. But I don't know how to generalize it find infinitely many solutions in some $[0,\delta]\subset [0,1]$. I just have a vague guess the point at which the two different solutions have different values may give two new initial conditions, which may give two more new IVPs. I don't see how to go on.


[Added:]It seems that if $\varphi(t)$ is a solution, then $$ \varphi_c(t)=\begin{cases} x_0,\ &0\leq t\leq c;\\ \varphi(t-c),\ &c\leq t\leq 1. \end{cases} $$ gives a family of solutions where $0<c<1$. But how do I show that they must be different?

$\endgroup$
2
$\begingroup$

You essentially answered your own question under the cut. Here are the details.

Case 1: $g(x_0)\ne 0$. Then the solution of IVP is unique in some interval of time. Indeed, let $x(t)$ be a solution with $x(0)=x_0$. Then $x'(t)=g(x(t))$ is continuous and nonzero in a neighborhood of $0$. Therefore, $x(t)$ is invertible in this neighborhood; let $t(x)$ be its inverse. By the inverse function theorem, $t(x)$ is differentiable with $t'(x)=g(x)$. This (together with $t(0)=0$) determines $t(x)$ uniquely: two $C^1$ functions with the same derivative are equal.

Case 2: $g(x_0)=0$. If uniqueness fails, there is a nonconstant solution $\varphi$ starting at $x_0$. As you correctly observed, shifting the solution in time produces an uncountable family of solutions of IVP. They are different because the time at which the solution leaves the stationary point $x_0$ is different.

Conclusion: if IVP has two solutions, it has uncountably many solutions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.