Initial Value Problem

**Aryth** · January 14th 2009, 08:57 AM

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

**ThePerfectHacker** · January 14th 2009, 09:37 AM

Originally Posted by Aryth

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

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

**Aryth** · January 14th 2009, 10:56 AM

Thanks for the help. I appreciate it.

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