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I just need a bit of help with this question.

If I know that $dg/dx = g^2$, and that $g(0) = g_0$, then I can solve:

$$ dg/dx = g^2\\ \frac{1}{g^2} dg = dx \\ -\frac{1}{g} = x + \hat c \\ -g = \frac {1}{x + \hat c} \\ g = \frac{1}{-x - \hat c}. \\ $$

Solving for the specific solution, $$ g_0 = \frac{1}{0 - \hat c} \\ \hat c = -\frac{1}{g_0} $$

This means that $g(x) = \frac {g_0}{1-g_0x}$.

My question is: How do I determine the maximum interval on which the solution is defined?

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1 Answer 1

Firstly note that there might not exist a maximum interval in which a solution is defined, in general you can only be sure about maximal intervals.

It's easy to check that the maximal intervals in which a solution to $g'=g^2$ is defined are $\left]-\infty, \dfrac{1}{g_0}\right[$ and $\left]\dfrac{1}{g_0}, +\infty\right[$, if $g_0\neq 0$.

Now regarding the I.V.P. $g'=g^2, g(0)=g_0$ you can indeed assure the existence of a maximum interval in which the solution is defined, because $0$ can only be in one of the above intervals. Do a discussion by cases:

  1. If $g_0=0$, clearly $g=\bf 0$ and it is defined over $\Bbb R$.
  2. If $g_0>0$, then $g(x)=g(x) = \dfrac {g_0}{1-g_0x},$ for all $x\in \left]-\infty, \dfrac{1}{g_0}\right[$ and this is a maximum interval of definition.
  3. If $g_0<0$, then $g(x)=g(x)=g(x) = \dfrac {g_0}{1-g_0x}$, for all $x\in \left]\dfrac{1}{g_0}, +\infty\right[$ and this is a maximum interval of definition.

This takes care of existence.

To justify uniqueness for each of the cases note that the I.V.P. is $g'(x)=f(x,g(x)), g(0)=g_0$, where $f(a,b)=b^2$ for all $(a,b)\in \Bbb R^2$ and $f$ is locally Lipschitz continuous and thus, by the Picard–Lindelöf theorem, there's uniqueness of solutions in each compact interval contained in one of the intervals above. Since this happens for any compact interval, it guarantees existence in the whole interval (one of the above).

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