# Thread: Uniqueness / Lipschitz condition

1. ## Uniqueness / Lipschitz condition

Hi,

I'm have some trouble understanding what my lecturer did.

The question is:

For the IVP, $y' = \frac{x+y}{2y}$, $0 \le x \le 1$, y(0)=1,

find an appropriate Lipschitz constant L in the domain $\{(x,y) : 0\le x \le 1, y \ge 1\}$. Hence determine whether or not the soln y(x) is unique.

We've found L = 1/2 (by finding $f_y$ and maximizing in our domain, but doesn't this function have a singularity at y=0? Which means the uniqueness thm doesn't apply?

Can someone explain?

EDIT: Also, he did things like:

y(0)=1 => (x+y)/2y > 0 => y' > 0
=> y increases
=> y >= 1

Which I don't understand.

2. The DE can be written as...

$y^{'} = \frac{1}{2} + \frac{x}{2y}, x \in [0,1], y(0)=1$ (1)

This type of equation has been solved in eighteenth century by the Italian mathematician Gabriele Manfredi with the substitution...

$\frac{y}{x} = t \rightarrow dy= t\cdot dx + x\cdot dt$ (2)

Applying (2) the (1) becomes...

$x\cdot \frac{dt}{dx} = \frac {1+t-2t^{2}}{2t} \rightarrow \frac{t\cdot dt}{(t-1)\cdot (t+\frac{1}{2})} = - \frac{dx}{x}$ (3)

In (3) ther variables are separated, so that we can integrate both terms and obtain the general solution of (1)...

$\{(y-x)^{2}\cdot (y+\frac{x}{2}\}^{\frac{1}{3}} = c$ (4)

Imposing the condition $y(0)=1$ we obtain finally the solution in implicit form...

$(y-x)^{2} \cdot (y + \frac{x}{2}) -1 =0$ (5)

... that is represented here...

The Cauchy-Lipschitz condition in $[0,1]$ are satisfied and the (1) has only one solution for $y(0)=1$. However in the case $y(0)=0$ that isn't and [easy to verify...] we have two different solution of (1)...

$y=x$

$y=-\frac{x}{2}$ (6)

Kind regards

$\chi$ $\sigma$