I don't know what's happening with this exercise. I need a help becuase I'm quite puzzled.

Consider the Cauchy problem

\begin{cases} y'=\frac{2}{t} y + 2 t \sqrt{y} \\ y(1)=0 \end{cases}

Study the existence and uniqueness

Here $$f(t,y)=\frac{2}{t} y + 2 t \sqrt{y}$$ Since $y\geq0$ (I have the square root), I consider as open neigbourhood $K = \{t: |t-1|< r_1 \} \times \{y: 0 < y < r_2 \}$, but in this way I am in trouble with $$f_y(t,y)= \frac{2}{t} + \frac{t}{\sqrt{y}}$$ because it's discontinuous at $y=0$.

So I should look for a weaker condition as Lipschitz continuity: I take $(t,y_1)$ and $(t,y_2)$ in $K$:

$$|\frac{2}{t} \bigl(y_1 - y_2 \bigr) + 2t \bigl( \sqrt{y_1} - \sqrt{y_2} \bigr)| \leq |\frac{2}{t} \bigl(y_1 - y_2 \bigr)| + |2t \bigl( \sqrt{y_1} - \sqrt{y_2} \bigr)| $$

but the second term of the inequality is quite problematic: it is like proving that $x \mapsto \sqrt{x}$ is Lipschitz for $x\geq0$, which is known to be false.

So, I can't apply the theorem actually...Am I wrong? If so, what are my mistakes?

  • $\begingroup$ The r.h.s. is not Lipschitz continuous, hence you cannot apply the standard Cauchy-Picard theorem. By the way, your Cauchy problem admits more than one solution (one of them being $y(t) = 0$ for every $t>0$). $\endgroup$ – Rigel Aug 20 at 17:19
  • $\begingroup$ Thanks @Rigel. So I can't apply local existence... so I think that if I can't apply local existence, I can't also apply global existence, right? $\endgroup$ – andereBen Aug 20 at 17:21
  • $\begingroup$ You have local existence (Peano's theorem), but you don't really need it, since you already know one solution. You can find the other solutions looking for positive solutions of the equation using the substitution $z = \sqrt{y}$, and trying to prolong a positive solution joining it with the trivial ($y=0$) solution. $\endgroup$ – Rigel Aug 20 at 17:25
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    $\begingroup$ Yes sure, so when the local existence does not apply, it could be that I can still find global solutions, right? This is the case, as I have $y=0$ and also $t^2(1-t)^2$ @Rigel $\endgroup$ – andereBen Aug 20 at 18:00
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    $\begingroup$ No to the last comment, global existence always implies local existence, so when no solution exists locally, there can also be no global solution. // Set $y=t^2u^2$, then $y'=2t^2uu'+2tu^2=2tu^2+2t^2u$ so that $u'=1$ whenever $u\ne 0$. $\endgroup$ – Lutz Lehmann Aug 20 at 21:51

The r.h.s. $f(t,y)$, defined in $\Omega := (0, +\infty) \times [0,+\infty)$, is continuous in $\Omega$ but it is not locally Lipschitz continuous. Hence, Peano's theorem guarantees local existence, but uniqueness need not hold (and, indeed, in our case we have more than one solution).

Further, $f$ is sublinear in $y$, meaning that $|f(t,y)| \leq a(t) + b(t) |y|$ for some continuous functions $a, b \in C((0,+\infty))$, so that all solutions are global (meaning that every solution admits an extension on $(0,+\infty)$).

Let us compute the solutions of the given Cauchy problem. One solution is the constant function $y(t) = 0$, $t\in (0,+\infty)$.

Other solutions bifurcate from the constant solution at some time $\tau \geq 1$. In order to find them, let us first compute the strictly positive solutions of the differential equation. With the change of variable $z = \sqrt{y}$ we are let to the linear equation $z' = z/t + t$, whose solutions are of the form $z(t) = ct + t^2$, for some constant $c\in \mathbb{R}$. Recall that we are interested only in positive solutions defined in some subinterval of $(0,+\infty)$. The corresponding $y$ are then of the form $$ y_\tau (t) = t^2 (t- \tau)^2, \qquad t > \max\{\tau, 0\}, $$
where $\tau$ is a real parameter. It is easily seen that, if $\tau \geq 1$, then $y_\tau$ can be prolonged to the left with the $0$ solution, obtaining the global solution of the Cauchy problem $$ y_\tau(t) := \begin{cases} 0, & \text{if}\ t \in (0, \tau], \\ t^2(t-\tau)^2, & \text{if}\ t > \tau\,. \end{cases} $$ In conclusion, for every $\tau \geq 1$ the above function is a solution of the Cauchy problem. (This family of solutions is called the Peano brush.)

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  • $\begingroup$ Thanks so much @Rigel. Only one last question: could you show how $f$ is sublinear? I was thinking about $$|f(t,y)| \leq |2t \sqrt{y}| + |\frac{2}{t} y| $$ and now I would bound $\sqrt{y} \leq y$, which is true only for $y \geq 1$... how did you do that? $\endgroup$ – andereBen Aug 21 at 7:58
  • $\begingroup$ The idea is that the bound is needed only for large values of $y$. Anyway, you can use for example the estimate $\sqrt{y} \leq 1 + y$ (for $y\geq 0$). $\endgroup$ – Rigel Aug 21 at 8:07
  • $\begingroup$ thanks! But I think that the bound must hold also for small values, looking at the definition it seems that it has to hold for all the values of $(t,y)$ on a certain strip... I'm a bit puzzled $\endgroup$ – andereBen Aug 21 at 8:09
  • $\begingroup$ The bound $\sqrt{y} \leq 1 + y$ holds for every $y\geq 0$, so you can use it in the estimate for $f$ $\endgroup$ – Rigel Aug 21 at 9:05
  • $\begingroup$ Yes sure, but I was asking about "The idea is that the bound is needed only for large values of $y$". I don't think it's true, as the bound has to hold for all the values of $y \in \mathbb{R}$, not just for a subset of them, right? @Rigel $\endgroup$ – andereBen Aug 21 at 12:21

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