0

Consider $$g(x) = \cases{ 1 & $x < 0$ \\ 2 & $x\geq0$}$$

and the differential equation $x'=g(x).$ Prove that there is no solution if we set $x(0)=0$.

My idea is that the differential equation $\frac{dx}{dt}=g(x)$ can be solved by $\displaystyle{\int\frac{1}{g(x)}dx}=t$. Now suppose $x(0)=0$; we set $t=0$ in the former equation and looks like it must be $\displaystyle{\int\frac{1}{g(x)}dx}=0$, but this can't be because $g(x)>0$ which implies $\displaystyle{\int\frac{1}{g(x)}dx}>0$.

Is this a correct argument?

Cure
  • 4,179
  • 1
    Is $g(0)$ equal to $1$ or $2$...? Anyway you probably want a differentiable solution to write $x'$, do you see how that could be a problem? – Najib Idrissi Oct 27 '14 at 12:53
  • Hi. Is $g(0)=2$, I just fixed it. – Cure Oct 27 '14 at 12:55
  • Then there is a solution: x(t)=2t for every nonnegative t. – Did Oct 27 '14 at 12:56
  • What would be wrong with the solution $x(t)=t.g(t)$, except that it is not differentiable at $t=0$ ? –  Oct 27 '14 at 12:59
  • You need to swap one of the inequality signs in the definition of $g$. I cleaned up the source code, but i cannot tell offhand which is supposed to be which way. – Arthur Oct 27 '14 at 13:00
  • @YvesDaoust This is one of the reasons I asked this question. The problem was copied as it was stated, and the first thing I thought was that I could define $x(t)=t$ if $t<0$ and $x(t)=2t$ if $t\geq0$. Assuming the question was right in what it asked to prove, I thought I may be wrong so I tried what I did above. Maybe the fact that $x$ is not diff. at $t=0$ is enough to say that there is no general solution if $x(0)=0$?. – Cure Oct 27 '14 at 13:07
  • Yep, $x'(0)=2$ is incompatible with $x(t)=t$; the derivative is not just discontinuous, it is non-existent at $0$. –  Oct 27 '14 at 13:58
  • @YvesDaoust Thanks. One last question: My approach is not good because of the $C$ right?. Couldn't I "save" it arguing that it must be $C=0$?. Since the left hand side has only factors in terms of $x$ without independent variables, it is gonna be $x(0)=0$ can't be $C\neq 0$. Still, I believe there would be problems if some factor of the kind $x^{-k}$ with positive $k$ appears... :( – Cure Oct 27 '14 at 15:38
  • No. Specify the integration bounds and you'll stumble on a problem in your argument. –  Oct 27 '14 at 15:51

2 Answers2

2

Your argument is not correct as the solution is $$\int\frac{dx}{g(x)}=t+C.$$ Anyway, $x'(t)$ is positive so that $x(t)$ is strictly increasing. Given $x(0)=0$, we have $x(-h)<0$ and $x(h)>0$, so that by the equation $x'(-h)=1$ and $x'(h)=2$, then integrating, $x(-h)=-h$ and $x(h)=2h$.

Taking the limit $h\to0$, $x'(t)$ is not defined at $t=0$, though we expected it to be $2$, a contradiction.

Actually, whatever the initial condition there is always a $t$ such that $x(t)=0$ and a solution never exists.

  • You're right, I messed it up. But is what I was asked to prove possible?. If it is, then the problem to find such solution is that is not possible to find an $x$ differentiable for every $t$?. – Cure Oct 27 '14 at 13:13
1

The non-differentiability of $g$ doesn't enter in to it, all Cure's argument depends on is that $g$ is positive everywhere. The error comes from

$\displaystyle{\int\frac{1}{g(x)}dx}=t$

which is only true up to a (usually implicitly assumed) arbitrary constant. By letting that slide and simply substituting in $t=0$, you've invalidated the actual solution.

Julia Hayward
  • 690