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I teach myself to solve IVP and could use some help regarding this exercise I found:

Let $f:[0,+\infty) \to [0,+\infty)$ be continuous such that $0$ is the only zero of $f$. Consider the differential equation $\dot{x} = f(x)$ with initial value $x(0) = x_0 \in [0, +\infty)$.

I am asked to show that $\int_{0}^{1} \frac{1}{f(x)}\ \mathrm{d}x = +\infty$ implies that the IVP has a unique solution for $x_0 = 0$ . It's obvious that $x(t) = 0$ is a solution for $x_0 = 0$ but I have no idea how to proof the statement.

Are there any important theorems that might help here?

offline
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  • For how non-uniqueness might happen see https://math.stackexchange.com/questions/199086/how-can-i-show-that-y-sqrty-has-infinitely-many-solutions, for a proof in a special case that can be generalized see https://math.stackexchange.com/questions/2306047/show-ivp-ut-ut-lnut-has-unique-solution – Lutz Lehmann Jan 23 '21 at 20:57
  • I am reading this, as finding x(t) so that $$x'(t)=f(x(t))$$ with $x(0)=x_{0}$. Is that the right interpretation? – open problem Jan 23 '21 at 21:03
  • @openproblem Yes, that is right. (At least that's how I understand the problem.) – offline Jan 23 '21 at 22:18

1 Answers1

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Assume there is a solution with $x(t_1)=1$. Let $t_0=\sup\{t>0:x(t)=0\}$. Then for the time difference one gets $$ \frac{dx}{dt}=f(x)\implies t_1-t_0=\lim_{\varepsilon\to0}\int_ε^1\frac{dx}{f(x)}. $$ This of course only works if the limit of the integral has a finite value. If the integral diverges, then no such solution exists, the zero function is the unique solution.

Remarks: To properly apply existence theorems, one could extend $f$ as an odd function $f(-x)=-f(x)$, so that $f$ is defined and continuous on all of $\Bbb R$, and $(t_0,x_0)=(0,0)$ is an inner point of the domain.

For any $ε>0$, $f$ has a positive minimum on the interval $[ε,1]$, so that any solution that deviates from the zero solution will be strictly increasing after leaving the zero axis.

As a consequence, if a solution with $x(t_1)=1$ exists, then for every $ε>0$ there is a unique $δ>0$ so that $x(t_0+δ)=ε$. Now by the substitution rule one can integrate $$ t_1-(t_0+δ)=\int_{t_0+δ}^{t_1}dt=\int_{t_0+δ}^{t_1}\frac{\dot x(t)}{f(x(t))}dt=\int_{ε}^1\frac{dx}{f(x)}. $$ In view of the last two points, the emphasis for $t_0$ should perhaps be more that it is the start of the positive values, $t_0=\inf\{t>0:0<x(t)\}$.

If $f$ is Lipschitz on $[0,1]$, then any solution is unique. On the other side of the claim, $|f(x)|\le L|x|$ implies $$ \int_{ε}^1\frac{dx}{f(x)}\ge\frac{-\ln(ε)}{L}, $$ which is unbounded for $ε\to0$.

Lutz Lehmann
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  • Avoiding the messy $\frac{\mathrm{d}x}{f(x)}=\mathrm{d}t$, we can integrate over $[t_0+\epsilon,t_1]$: $$\int_{t_0+\epsilon}^{t_1}\frac{x'(t)}{f(x(t))}\mathrm{d}t=t_1-t_0+\varepsilon.$$ Now we rigourosly justify the change of variable $x(t)=:s$, $x'(t)\mathrm{d}t=\mathrm{d}s$ using the fact that $f$ is necessarily increasing near 0 as it is continuous, non-negative and only cancels at $0$ on $[0,1]$. – Nicolas Jan 23 '21 at 21:27
  • Yes, that is why I used the limit. For every $ε>0$ there is a unique $\delta>0$ so that $x(t_0+\delta)=ε$, which makes both approaches equivalent. – Lutz Lehmann Jan 23 '21 at 21:30
  • Maybe it is clear for you, but perhaps it is worth mentioning it in your answer? Also, this only works for sufficiently small $t_1>0$. – Nicolas Jan 23 '21 at 21:32
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    \int should be \inf when talking about $t_0$ near the end! – Nicolas Jan 23 '21 at 21:57
  • Addition done. And thanks. And no, it is not important where $t_1$ lies. It is the point with $x(t_1)=1$. If one has one branching solution, one can shift it around arbitrarily and thus construct an infinity of different solutions. – Lutz Lehmann Jan 23 '21 at 21:58
  • I meant for the change of variable: you have to ensure that $f>0$ so that $x'>0$ as well. Otherwise, I do not understand how you can change the variable? – Nicolas Jan 23 '21 at 22:01
  • That is why I define $t_0$ in such a way that with $t>t_0$ also $x(t)>0$ and thus also $f(x(t))>0$. – Lutz Lehmann Jan 23 '21 at 22:02
  • But $f$ has no reason for staying positive all the time, so that we should only consider low time for the change of variable? This is more than enough as we want to integrate near 0. – Nicolas Jan 23 '21 at 22:04
  • By the first assumption in the question, $0$ is the only root of $f$. – Lutz Lehmann Jan 23 '21 at 22:16
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    Oh silly me! Actually I had in mind "$f$ only cancels at $0$ on $[0,1]$". In that case there would be a restriction, but of course in the present situation there isn't! So good answer! +1 – Nicolas Jan 23 '21 at 22:30