I am considering the following initial value problem:
$yy''=-(y')^2$ , $y(0)=y'(0)=1$
I am trying to show that this problem has a unique solution on an interval containing zero.
Firstly, I know that since y is continuous, and $y(0)=1$, so I can find some interval $[-\delta, \delta]$ on which $F(x,y(x),y''(x))=\frac{-(y'(x))^2}{y(x)}$ is well-defined (i.e. $|y(x)|> 0$).
Now, I would like to apply Picard's existence theorem for systems, which requires that I have a Lipschitz condition on F. That is, on the set $S = \{(x, u, v) : |x|\leq \delta, |u − 1| + |v − 1| ≤ k\}$ (where I think I need to choose k to guarantee the Lipschitz condition) we must have that at all points in S:
$|F(x, u, v) − F(x, a, b)|\leq L(|u − a| + |v − b|)|$ for some $L> 0$
$\iff$ $|\frac{b^2}{a}-\frac{v^2}{u}|\leq L(|u − a| + |v − b|)|$ for $L>0$
I have been trying to play around with this final inequality but have not been getting very far. Is this the right approach (I have only really covered Picard's theorem so far so I think I am meant to use it)? Any help would be much appreciated.
Best Answer
You need to correct the right-side function of your first-order system. It should read $$ [y'(x),y''(x)] = F(x,[y(x),y'(x)])=\left[y'(x),\,-\frac{y'(x)^2}{y(x)}\right]. $$ Then for the Lipschitz condition you get $$ \|F(x,[u,v])-F(x,[a,b])\|=\left\|\left[v-b,\,-\frac{v^2}{u}+\frac{b^2}{a}\right]\right\| \le |v-b|+\frac1{|a|}(|v|+|b|)|v-b|+\frac{v^2}{|au|}|u-a| $$ Thus you get $$ L = \max_{(u,v),(a,b)\in R}\max\left(1+\frac{|v|+|b|}{|a|},\frac{v^2}{|au|}\right) $$ which is a finite value for any rectangular domain $R$ around the initial point $(1,1)$ of radius smaller than $1$.