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Consider the initial value problem $$\frac{dy}{dx} = x^2 + y^2 \\ y(0) = 0$$

on $[-0.5, 0.5]\times [-0.5, 0.5]$

Find $[-a, a]$ such that the solution exists and is unique so there are two steps involved apparently.

1) verify the Lipschitz continuous condition

2) Fix the parameters M and h

I am able to get $$M = \sup(x^2 + y^2) = 0.25 + 0.25 = 0.5$$ and $$h = \min\{a, \frac{b}{M}\} = \min[0.5, \frac{0.5}{0.5}] = 0.5$$

But the problem I have is with step $1$. I do not know how to verify the Lipschitz condition. If someone can help explain it, that would be awesome.

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  • $\begingroup$ You can explicitly solve the differential equation and take it from there. Or is the idea to obtain the Lipschitz condition without doing so? $\endgroup$
    – Urgje
    Commented Mar 27, 2014 at 11:49
  • $\begingroup$ Could you please explain how to solve the equation explicitely? $\endgroup$ Commented Mar 27, 2014 at 11:52

1 Answer 1

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You can find the Lipschitz costant as $$ L=\sup\Bigl|\frac{\partial}{\partial y}(x^2+y^2)\Bigr|. $$

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  • $\begingroup$ why am I taking the derivative w.r.t to y and not x? $\endgroup$
    – Kennan
    Commented Mar 28, 2014 at 12:26
  • $\begingroup$ Because the theorem of existence and uniqueness for the equation $y'=f(x,y)$ states that $f(x,y)$ should be continuous (as a function of two variables) an Lipschitz with respect to $y$. $\endgroup$ Commented Mar 28, 2014 at 15:01
  • $\begingroup$ oh okay, I understand now. So in this case the Lipschitz constant will be L = sup|2y|? $\endgroup$
    – Kennan
    Commented Mar 29, 2014 at 9:58
  • $\begingroup$ That is correct. $\endgroup$ Commented Mar 29, 2014 at 13:57

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