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A textbook asks

For which $p > 0$ is the solution of the IVP $$ \dot{x} = x^p, \quad x(0) = 1$$ unique and defined for all $t \leq 0$?

This is simple for $p \geq 1$, but otherwise we will have the solution $$ x \colon \mathbb{R} \to \mathbb{R}, x(t) = \begin{cases} \left( (1 - p)t + 1\right)^{1/(1 - p)} & t \geq \frac{1}{p - 1}\\ 0 & t < \frac{1}{p - 1} \end{cases}$$

which I think is only unique if

$$ x \colon \mathbb{R} \to \mathbb{R}, x(t) = \left( (1 - p)t + 1\right)^{1/(1 - p)}$$

is undefined or not a solution. At least in the case $p = 1/2$ it holds that $$ t/2 + 1 = \dot{x}(t) \neq \sqrt{x(t)} = \left| t/2 + 1 \right| \quad \left( t < \frac{1}{p - 1} \right) $$

Question: Suppose that $$ y^q, \qquad q := \frac{1}{1 - p} > 1 $$ is defined for negative $y$. Does the implication $$ y < 0 \quad\Rightarrow \quad y^q > 0$$ hold, hence leading to the same contradiction as for $p = 1/2$?

arney
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  • Sorry but the "Question" part of your post makes little sense and is quite unrelated to the question you want to see solved. Consider the case $p=\frac12$ since this case exhibits all the relevant phenomena for every $0<p<1$. You are supposed to note is that $\dot x=x^{1/2}$ has solutions $$x(t)=\tfrac14(t-t_0)^2\mathbf 1_{t>t_0}$$ for every $t_0$ (yes, you should check that these solve the differential equation on the whole real line, at the point $t_0$ included...). Asking that $x(0)=1$ then imposes that $t_0=-2$ hence indeed the solution ... – Did Jul 14 '16 at 10:31
  • ... to the IVP in your question is $$x(t)=\tfrac14(t+2)^2\mathbf 1_{t>-2}=(1+t+\tfrac14t^2)\mathbf 1_{t>-2}$$, this solution is unique and it is defined on the whole real line. – Did Jul 14 '16 at 10:32
  • @Did In the case of $p = \frac{2}{5}$, i.e. $q =5/3$, according to this answer, one would also have $$ x(t) = \left[sgn\left(\frac{3}{5}t + 1\right)\right]^5 \cdot \sqrt[3]{\left| \frac{3}{5}t + 1 \right|^5} $$ – arney Jul 14 '16 at 19:29
  • But this depends on the interpretation of $x^p$ when $p$ is in $(0,1)$ and $x<0$. The usual interpretation is that, then, $x^p$ does not exist while the formula in your comment considers that, for every real number $x$, $x^p=\mathrm{sign}(x)\cdot|x|^p$. – Did Jul 14 '16 at 19:53
  • @Did ... which is precisely my issue with the problem at hand. Did the original author consider that his problem is subject to convention ... – arney Jul 15 '16 at 13:48

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