Consider the following IVP: $$ y'(x)=1+y^{2/3},\quad y(0)=0, $$ where the flux function is $f(x,y)=1+y^{2/3}$. According to Picard-Lindelöf Theorem, since $f_{y}$ is not continuous in any interval containing $x_{0}=0$, we can not guarantee existence of a solution for the above problem. On the other hand, using the separation of variables, we can find the following solution $$ 3\Big(y^{1/3}-\tan^{-1}(y^{1/3})\Big)=x. $$ It seems that this solution is unique for our ivp.

My Question: How we can find an interval of uniqueness without solving the problem?

  • $\begingroup$ See my answer for an explanation why the solution is defined on the whole of $\mathbb R$. $\endgroup$ – Yiorgos S. Smyrlis Jan 10 '14 at 20:03

The following fact is true:

Fact. Let $f:\mathbb R\to\mathbb R$ continuous and $f(\xi)\ne 0$, then the IVP $$ x'=f(x), \quad x(\tau)=\xi, \tag{1} $$ enjoys local uniqueness. If $f(x)\ne 0$, for all $x\in\mathbb R$, then the above
enjoys global uniqueness.

If $f(\xi)=0$, then even local uniqueness might be violated, as in the case of $x'=|x|^{1/2},\, x(0)=0$.

By global uniqueness we mean that any two solutions coincide on the intersection of their domains, while local uniqueness means that there exists an interval around the initial time where the any two solutions coincide. (Attention. The domain of the solution of an ODE is an interval, not a union of intervals.) For example, the IVP $x'=|x|^{1/2},\, x(0)=1$ enjoys local but not global uniqueness.

Proof of the Fact. If $\varphi: I \to\mathbb R$, where $\tau\in I$, satisfies $(1)$ and $$ F(x)=\int_\xi^x\frac{ds}{f(s)}, $$ then $F\big(\varphi(t)\big)=t-\tau$, since $$ F'\big(\varphi(t)\big)\varphi'(t)=\frac{\varphi'(t)}{f\big(\varphi(t)\big)}=1, $$ and $$ F\big(\varphi(\tau)\big)=\int_\xi^{\varphi(\tau)}\frac{ds}{f(s)}= \int_\xi^{\xi}\frac{ds}{f(s)}=0. $$ With the same argument, if $\psi: J\to\mathbb R$, where $\tau\in I$, is another solution of $(1)$, then $F\big(\psi(t)\big)=t-\tau$. But as $F$ is one-to-one, $\varphi\equiv\psi$ in $I\cap J$.

Your particular IVP $$ y'=1+|y|^{2/3},\quad y(0)=0, $$ has a global solution, i.e., $\varphi : \mathbb R\to \mathbb R$. This is not hard to prove. Let $$ F(x)=\int_0^x\frac{ds}{1+|s|^{2/3}}. $$ Clearly $F\in C^1(\mathbb R)$, $F$ is strictly increasing, with $F'>0$, and $$ \lim_{x\to\pm\infty}F(x)=\pm\infty. $$ Thus $F: \mathbb R\to\mathbb R$ is one-to-one and onto. Check that $\varphi=F^{-1}$ satisfies your IVP. Indeed, $F(0)=0$, and thus $\varphi(0)=F^{-1}(0)=0$, and $$ \varphi'(t)=(F^{-1})'(t)=\frac{1}{F'(F^{-1}(t))}=\frac{1}{\frac{1}{f(F^{-1}(t))}} =f\big(F^{-1}(t)\big)=f\big(\varphi(t)\big), $$ and it is unique.

  • $\begingroup$ Your answer is simpler than mine by quite a bit. My argument has the advantage that it could generalize to systems of ODE, but your method works for a much wider selection of 1D autonomous ODE than mine does. $\endgroup$ – Stephen Montgomery-Smith Jan 11 '14 at 17:59

I got some of the ideas from here: Counter-example to Cauchy-Peano-Arzela theorem, but I had to add a few more details.

By the Cauchy-Peano-Arzela theorem, we know that a solution exists on a small interval $[0,\epsilon]$. Let $y$ be a solution. Since $y$ is differentiable, using the definition of derivative at $x = 0$, we see $y(\epsilon) = \epsilon + o(\epsilon)$ as $\epsilon\to 0$.

Hence for sufficiently small $\epsilon$, we know that $y(\epsilon)$ is greater than zero. Now $y^{2/3}$ is Lipschitz on any domain of the form $[\delta,\infty)$, so this means that the $y(x)$ is uniquely determined and increasing for $x \ge \epsilon$. And on this interval $$ \frac{d}{dx} (3y^{1/3}) = \frac{y'}{y^{2/3}} \ge 1 ,$$ and hence for $x \ge \epsilon$ $$ 3(y(x))^{1/3} - 3(y(\epsilon))^{1/3} \ge x -\epsilon.$$ Since $(y(\epsilon))^{1/3} = \epsilon^{1/3} + o(\epsilon^{1/3})$, we have $$ y(x) \ge \left(\frac{x}3 + \epsilon^{1/3} + o(\epsilon^{1/3}) \right)^3 \ge \left(\frac{x + \epsilon^{1/3}}3 \right)^3 \tag1$$ if $\epsilon$ is sufficiently small.

Suppose that $y_1$ and $y_2$ are two solutions. Then by the mean value theorem $$ |y_1' - y_2'| = |y_1^{2/3} - y_2^{2/3}| \le |y_1 - y_2| \tfrac23 \min\{y_1,y_2\}^{-1/3} \le \frac2{x + \epsilon^{1/3}},$$ where at the end we applied equation (1). Applying Gronwall's inequality, we obtain $$ |y_1(x) - y_2(x)| \le |y_1(\epsilon) - y_2(\epsilon)| \exp\left(2 \int_{\epsilon}^x \frac{d\xi}{\xi+ \epsilon^{1/3}} \right) .$$ Now $$ \int_{\epsilon}^x \frac{d\xi}{\xi+ \epsilon^{1/3}} \le \int_\epsilon^{\epsilon^{1/3}} \frac{d\xi}{\epsilon^{1/3}} + \int_{\epsilon^{1/3}}^{x} \frac{d\xi}{\xi} \le 1 + \log(x) - \tfrac13\log(\epsilon) ,$$ and hence $$ |y_1(x) - y_2(x)| \le |y_1(\epsilon) - y_2(\epsilon)| \frac{(ex)^2}{\epsilon^{2/3}} = \frac {o(\epsilon)x^2}{\epsilon^{2/3}} .$$ Letting $\epsilon \to 0$, we obtain $y_1(x) = y_2(x)$.


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