Straight forward ODE

• Feb 5th 2011, 03:13 PM
VonNemo19
Straight forward ODE
Hi again. Here's one.

(a) Without solving, explain why the intitial value problem

$\frac{dy}{dx}=\sqrt{y},$ $y(x_0)=y_0$

has no solution for $y_0<0$.

(b) Solve the initial value problem in part (a) for $y_0>0$ and find the largest interval $I$ on which the solution is defined.

My attempt:

(a) Because the only way to arrive at a radical through differentiation is to begin with one, and any expression involving a radical (even root) with nothing outside of the radical must always give a positive result over the reals.

(b) I get

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

If I set $y_0=\frac{1}{4}(x_0+c)^2$, I get $c=\pm2\sqrt{y}-x_0$.

But, I won't evfen attempt to find the interval of definition because the book doesn't include the plus or minus. They have $c=2\sqrt{y}-x_0$.

What's the deal here?
• Feb 5th 2011, 03:42 PM
zzzoak
$
\frac{dy}{dx}=\sqrt{y}
$

$
\frac{dy}{\sqrt{y}}=dx
$

$
2 \sqrt{y}=x+C
$

$
2 \sqrt{y_0}=x_0+C
$

Making square you add one more root.
• Feb 5th 2011, 04:32 PM
VonNemo19
OK, I unserstand that, but why is it wrong to find an explicit formula for y in this case? It seems clear to me that if $y=\frac{1}{4}(x+c)^2$, then $\frac{dy}{dx}=\frac{1}{2}(x+c)=\sqrt{y}$ which is the DE.
• Feb 6th 2011, 05:08 AM
VonNemo19
Quote:

Originally Posted by VonNemo19
OK, I unserstand that, but why is it wrong to find an explicit formula for y in this case? It seems clear to me that if $y=\frac{1}{4}(x+c)^2$, then $\frac{dy}{dx}=\frac{1}{2}(x+c)=\sqrt{y}$ which is the DE.

...and the interval $I$ on which the solution is defined would then be $I=(-\infty,\infty)$. I really don't understand what is wrong with this.
• Feb 6th 2011, 07:24 AM
TheEmptySet
Quote:

Originally Posted by VonNemo19
...and the interval $I$ on which the solution is defined would then be $I=(-\infty,\infty)$. I really don't understand what is wrong with this.

The key idea is that the domain of your solution does not have to be the same as the interval of existence and uniqueness of the solution to an ODE.

We know that if $\displaystyle \frac{dy}{dx}=f(x,y)$ and

$\displaystyle f(x,y) \text{ and } \frac{\partial x}{\partial y}$ are continuous $[a,b] \times [c,d]$ and $(x_0,y_0)$ is an interior point of the above rectangle.
Then there exists an interval $(x_0-h,x_0+h) \subset [a,b]$ such that a solution exists and is unique.

Now for your problem $\displaystyle \frac{\partial f}{\partial y}=\frac{1}{2\sqrt{y}}$ is discontinous at $(x,0)$

Now given an initial point $(x_0,y_0); y_0 > 0$ then the solution "is"

$\displaystyle y_0=\frac{1}{4}\left( x_0+c\right)^2 \implies c=-x_0 \pm 2y_0$

$\displaystyle y=\frac{1}{4}\left(x-x_0 +2y_0 \right)^2$

Things are already starting to look off because now we have a non unique solution and here is the kicker notice what happens when $x=x_0-2y_0$

This gives $y(x_0-2y_0)=0$ Now I can make another solution as follows

$\displaystyle y(x)=\begin{cases} \frac{1}{4}\left(x-x_0 +2y_0 \right)^2, \text{ if } x > x_0-2y_0 \\ 0, \text{ if } x \le x_0-2y_0\end{cases}$

Notice this new function is continuous and differentiable on all of $\mathbb{R}$ but is not unique. So we run into problems where $\displaystyle \frac{\partial f}{\partial y}$ is undefined.

So the largest interval with a unique solution is either $(-\infty,x_0-2y_0)\quad$ $(x_0-2y_0,x_0+2y_0) \quad$ $(-x_0+2y_0,\infty)\quad$