# Initial Value Problem

• Jan 14th 2009, 08:57 AM
Aryth
Initial Value Problem
Let $\displaystyle c > 0$. Show that the function $\displaystyle \phi(x) = (c^2 - x^2)^{-1}$ is a solution to the initial value problem $\displaystyle \frac{dy}{dx} = 2xy^2$, $\displaystyle y(0) = \frac{1}{c^2}$, on the interval $\displaystyle -c < x < c$. Note that this solution becomes unbounded as x approaches $\displaystyle \pm c$. Thus, the solution exists on the interval $\displaystyle (-\delta, \delta)$ with $\displaystyle \delta = c$, but not for larger $\displaystyle \delta$.
• Jan 14th 2009, 09:37 AM
ThePerfectHacker
Quote:

Originally Posted by Aryth
Let $\displaystyle c > 0$. Show that the function $\displaystyle \phi(x) = (c^2 - x^2)^{-1}$ is a solution to the initial value problem $\displaystyle \frac{dy}{dx} = 2xy^2$, $\displaystyle y(0) = \frac{1}{c^2}$, on the interval $\displaystyle -c < x < c$. Note that this solution becomes unbounded as x approaches $\displaystyle \pm c$. Thus, the solution exists on the interval $\displaystyle (-\delta, \delta)$ with $\displaystyle \delta = c$, but not for larger $\displaystyle \delta$.

If $\displaystyle \phi (x) = (c^2 - x^2)^{-1}$ for $\displaystyle |x| < c$ then $\displaystyle \phi '(x) = 2x(c^2 - x^2)^{-2}$ by the chain rule, therefore, $\displaystyle \phi '(x) = 2x \phi(x)^2$. Therefore, $\displaystyle \phi$ solves the equation $\displaystyle y' = 2xy^2$ with $\displaystyle y(0) = \tfrac{1}{c^2}$. This solution becomes unbounded because $\displaystyle \lim _{x\to c^-}\tfrac{1}{c^2-x^2} = \infty$ and $\displaystyle \lim_{x\to -c^+} \tfrac{1}{c^2 - x^2} = \infty$. Now say we want to solve $\displaystyle y' = 2xy^2, y(0) = \tfrac{1}{c^2}$ with $\displaystyle |x| < \delta$ (where $\displaystyle \delta > c$). Then by the existence and uniqueness theorem it means $\displaystyle y$ will coincide with the solution on $\displaystyle (-c,c)$ i.e. $\displaystyle y(x) = (c^2 - x^2)^{-1}$ where $\displaystyle |x| < c$. But this is a problem because $\displaystyle (c^2 - x^2)^{-1}$ becomes unbounded and so $\displaystyle y$ cannot be differenciable on $\displaystyle (-\delta,\delta)$ if $\displaystyle \delta > c$.
• Jan 14th 2009, 10:56 AM
Aryth
Thanks for the help. I appreciate it.