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I am to show that solutions to the ODE:

$x'(t) = -\frac{1}{2} \left(x(t)^2 - 3 c \right)$,

are not unique. I don't know why this is problematic, since, the ODE clearly has a solution that seems to work for all $t$:

$x(t) = \sqrt{3} \sqrt{c} \tanh \left(\frac{1}{2} \left(\sqrt{3} \sqrt{c} t+2 \sqrt{3} \sqrt{c} c_1\right)\right)$,

where $c_1$ is a constant. But, my professor insists that this is not a unique solution!

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I think the way to go about this is to use a result from Arnold's classical ODE book, which says that a sufficient condition for uniqueness of a first-order ODE $x'(t) = v(x)$, with initial data $(x_0, t_0)$ is that the integral

$\int_{x_{0}}^{x} \frac{d\xi}{v(\xi)}$ diverges at $x_{0}$.

So, let us consider the IVP related to your problem:

$x'(t) = -\frac{1}{2}(x(t)^2 - 3c), \quad x(0) = x_{0}$.

Evaluating the integral above, we find that it is equal to:

$\frac{2 \tanh ^{-1}\left(\frac{x}{\sqrt{3} \sqrt{c}}\right)}{\sqrt{3} \sqrt{c}}-\frac{2 \tanh ^{-1}\left(\frac{x_{0}}{\sqrt{3} \sqrt{c}}\right)}{\sqrt{3} \sqrt{c}}$

We see that the above integral diverges for when

$\frac{x_{0}}{\sqrt{3 c}} = \pm 1$.

Therefore, ONLY for the initial condition of $x_{0} = \pm \sqrt{3 c}$ does one have a unique solution, or when $c = 0$ (where the integral diverges), and hence, a unique solution.

In all other cases, the integral converges, and uniqueness fails.

I hope that helps!

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  • $\begingroup$ Hi. Thanks, that is very helpful! $\endgroup$ – Thomas Moore Sep 9 '16 at 15:31
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Your solution is not unique because of the constant $c_1$. You have actually written down infinitely many different solutions, since you have a different solution for each possible value of $c_1$.

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