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?
Best Answer
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 $ε\to0$.