Suppose an ODE $$\frac{d}{dt} x(t) = f(x,t)$$ has a global solution for some initial condition $x(0) = x_0$.

I am interested in whether, for an arbitrary constant $k>0$, the proportional ODE $$\frac{d}{dt} x(t) = k \cdot f(x,t)$$ has a global solution for the same initial condition $x(0) = x_0$.

Does the existence of the solution to the first ODE weaken at all the conditions needed to ensure existence of the solution to the second ODE? In all of the examples and counterexamples to existence that I know of, multiplying $f$ by a constant does not affect existence. But I don't see a path to a general proof.


1 Answer 1


It's not true.

Let $$ f(x,t) = \cases{1 & if $x \le t + 1$\cr 1 + (x-t-1)^2 & otherwise\cr}$$

The solution of $x' = f(x,t)$ with initial condition $x(0)=0$ is $x=t$, defined for all real $t$.

The solution of $x' = 2 f(x,t)$ is $x = 2 t$ for $t \le 1$, but it enters the region $x > t+1$ at $(x,t) = (2,1)$ and then continues as $x(t) = t + 1 + \tan((t-1)\sqrt{2})/\sqrt{2}$ for $1 \le t < 1 + \pi/(2 \sqrt{2})$, going to $+\infty$ as $t$ approaches the right endpoint of that interval.

EDIT: Another example is $f(x,t) = 1 + (x-t)^2$, again with initial condition $x(0)=0$. Again $x' = f(x,t)$ with this initial condition has the globally defined solution $x(t) = t$, while for any $k > 1$, $x' = k f(x,t)$ with this initial condition has $x(t) = t + \sqrt{1-1/k}\; \tan(\sqrt{k^2-k}\; t)$ which blows up as $t \to \pi/(2 \sqrt{k^2-k})$.

EDIT: Consider a function
$$f(x,t) = \cases{\left(1 - x + \frac{1}{1-t}\right)\left(1 + x - \frac{1}{1-t}\right) \frac{x^2}{2} & if $-1 \le x - \frac{1}{1-t} \le 1$\cr 0 & otherwise}$$

Then for $k = 2$, the differential equation $x' = k f(x,t)$ with initial condition $x(0) = 1$ has a solution $x = 1/(1-t)$ that blows up as $t \to 1-$. But for $k < 2$, all initial conditions will result in global solutions, because $x(t)$ can't increase fast enough to stay in the region $-1 < x - 1/(1-t) < 1$.

  • $\begingroup$ Very nice examples! $\endgroup$ Apr 12 at 13:09
  • $\begingroup$ Thanks, this is very instructive. Both examples seem to exploit a particular choice of initial condition. What if the assumption were strengthened to "$x' = f(x,t)$ has a solution for all $x_0$"? $\endgroup$
    – John Sturm
    Apr 12 at 13:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .