For the equation

$$y'' = \frac{(y')^2}{y} - 1$$

there are two solutions $y_1 = 1+ \sin x$ and $y_2 = (\frac{x}{\sqrt{2}} + 1)^2$ passing through the point $(0,1)$. Does the fact contradict the Picard-Lindelof uniqueness theorem?

I am guessing that the equation does not satisfy the conditions for the theorem to hold!

that is

Let $t_0 \in \mathbb{R},a>0,R>0,y_0 \in \mathbb{R}^n$ and $f:[t_0,t_0 + a] \times B_R(y_0) \rightarrow \mathbb{R}^n$ (jointly) continuous, $|f(t,y)| \leq M (t\in [t_0,t_0+a] y \in B_R(y_0)) $ and $\alpha = \min(a,\frac{R}{M})$ Assume also that there is an $L>0$ such that the Lipschitz condition

$$|f(t,y)-f(t,z)|\leq|y-z| (t\in [t_0,t_0+a] y \in B_R(y_0))$$ holds, then the IVP

$$u'(t) = f(t,u(t)) ,u(t_0) = y_0$$ has a unique solution on $[t_0,t_0+a]$


$$f(t,y) = y'' - \frac{(y')^2}{y} + 1 $$

The odd thing is that I am not given any range of $y$ to work with.

Hence have little idea how to deal with.

$$f(t,y) - f(t,z) = |y'' - z'' - \frac{(y')^2}{y} +\frac{(z')^2}{z}|$$

Would the answer be that it is not jointly continuous at the origin? y = 0? but we can choose a B such that this is not the case, by the way my balls are closed balls.

  • $\begingroup$ Picard-Lindelöf talks about Cauchy ivp's, i.e., in the form $y'=f(y,t)$. $\endgroup$ – Nigel Overmars May 2 '16 at 19:21
  • $\begingroup$ To be more specific, the usual form of Picard-Lindelof can be adapted to apply to an $n$th order ODE with sufficient regularity provided you give $n$ initial conditions. If you provide $1$ initial condition to a $2$nd order ODE, you will almost never have uniqueness. $\endgroup$ – Ian May 2 '16 at 19:43
  • $\begingroup$ I am sorry, I do not understand, so how does it fail in this case? specifically? $\endgroup$ – user197848 May 2 '16 at 21:03
  • $\begingroup$ One expects uniqueness for a second degree ODE when, at some initial time value $t_0,$ both $y(t_0)$ AND $y'(t_0)$ are specified. For example, if I said $u'' + u = 0$ and $u(0) = 0,$ there would still be infinitely many choices with different derivatives at $0,$ namely $u = A \sin t$ for constant $A.$ $\endgroup$ – Will Jagy May 2 '16 at 22:37

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