3
$\begingroup$

Let $f:\mathbb{R}^n \to \mathbb{R}^n $ be a continuously differentiable function. Prove that two distinct solutions of $x'(t)=f(x(t))$ (the ODE is autonomous) cannot intersect at any point, not even at different times.

Notes: Clearly there is a relationship between this question and the theorems for the existence and uniqueness of ODEs.

Theorem: Let A be an open subset of $\mathbb{R}^n$ and that $f \in C^1(A)$. Then $\forall x_0 \in A \ \exists \ \alpha>0$ such that the IVP $$x'(t)=f(x(t)), \ x(0)=x_0$$ has a unique solution $x(t)=x(t,x_0)$ on the interval $[-\alpha,\alpha]$.

As $f \in C^1$ the theorem applies so the solutions to $x'(t)=f(x(t))$ are unique (they will not be in the same position at the same time). However I do not know how to show that the two solutions cannot intersect at later times.

  • 0
    as long as it is autonomous, meaning that there is no explicit dependence on $t,$ only on $x(t)$ – 2017-02-07
  • 0
    @WillJagy could you clarify your comment – 2017-02-07
  • 1
    Maybe differentiating both sides could help. $x''(t) = x'(t)f'(x(t)) = f(x(t))f'(x(t))$ – 2017-02-07
  • 1
    @WillJagy is noting that you need the ODE to be autonomous, otherwise it is perfectly easy to construct trajectories that would intersect at some point. But then once you have that the ode is autonomous then you are considering solutions $x(t_0) = x_0 = y_0 = y(t_0)$ for some arbitrary $t_0$ and further $x^{\prime}(t_0) = y^{\prime}(t_0)$, and then try argue from the existence and uniqueness that $x(t)$ and $y(t)$ must also agree over some set – 2017-02-07
  • 1
    One idea: let $x,y$ be two solutions. If $x(t_1) = y(t_2)$ then $x(t_1) = z(t_1)$ where $z(t) = y(t - t_1 + t_2)$ and $z$ also satisfy $z' = f(z)$. Now imagine solving the $x,z$ ODE starting from $t = t_1$ with initial condition $x(t_1)$ and use uniqueness to get that $x(t) = z(t) = y(t - t_1 + t_2)$ in some suitable domain. – 2017-02-07

1 Answers 1

2

Assume there are two different solutions $y_1$ and $y_2$ of the ODE. Then we have points $a$ and $b$ with $y_1(a)=y_2(a)$ and $y_1(b) \neq y_2 (b)$. Therefore the set $$ M:= \left\{x \in [a,b] \mid y_1(x)=y_2(x) \right\} \subset [a,b] $$ is nonempty and bounded. Hence $s= \sup M$ exists and we can find a sequence $(x_n)_n$ with $x_n \to s$. $y_1$ and $y_2$ are continuous and therefore $$ 0 = \lim_n \left( y_1(x_n) - y_2(x_n)\right)=y_1(s)-y_2(s).$$ This shows $c:=y_1(s)=y_2(s)$. Further we have $s

  • 0
    How do we know the solutions are continuous? – 2017-02-07
  • 0
    @AzJ Basically because Picard's iteration method from uniqueness and existence theorem proof gives us continuous function as a solution of integral equation. – 2017-02-08
  • 0
    @Evgeny Okay makes sense – 2017-02-08