Hi guys I have this problem

$$\frac{dy}{dt}= y^k$$

Such that $k >0$ and $y(0) \geq 0$. What we want to show is that the only value for which this equation has solutions bot unique and exist is $k=1$.

What I understand is that when $k=1$ we can use Picard's theorem and let $f(y,t)= y^1$ this is clearly continues implying existence, then we can see that $\frac{df}{dy}=1$ which is continues as well, thus this implies uniqueness. What I am stuck on is how to show in general that ode may not have solution or fail to exist.

My attempt. First we want to solve the ode and then view the cases : $0<k<1$ and $k>1$

Our solution $y= ((1-k)t+C)^{\frac{1}{1-k}}$

Using the initial condition $y(0)\geq 0$ we get that $C\geq 0$. Now I have no idea how to contradict uniqueness/ existence. I would really appreciate some input and also maybe a reference to a source to similar problems for my own practice. Thank you!


Your solution is wrong. It should be $y = ((1-k) t + C)^{1/(1-k)}$

For $0 < k < 1$, $C=0$ gives you a solution with $y(0) = 0$, but it is not unique (there is another, quite obvious, solution with this initial condition).

For $k > 1$, the solution blows up at a finite value of $t$.

  • $\begingroup$ Ugh, running out of votes is very frustrating when I find good answer. $\endgroup$ – Simply Beautiful Art Dec 30 '16 at 19:54
  • $\begingroup$ I fixed the typo $\endgroup$ – Kori Dec 30 '16 at 19:57
  • $\begingroup$ Just to check understanding the problem with $k>1$ is when $t =\frac{-C}{1-k}$, but if $t$ does not equal that value we are good. While in $0<k<1$ we may make a picewise function that is also a solution? $\endgroup$ – Kori Dec 30 '16 at 20:27
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    $\begingroup$ For $k>1$, the solution is valid for $ t < -C/(1-k)$. The values for $t > -C/(1-k)$ (which are not even real unless $1/(1-k)$ is a rational with odd denominator) constitute a different solution, not really connected to the solution in $t < -C/(1-k)$ despite having the same formula. $\endgroup$ – Robert Israel Dec 30 '16 at 22:35
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    $\begingroup$ For $k < 1$ you could simply take $y = 0$ for all $t$. $\endgroup$ – Robert Israel Dec 30 '16 at 22:36

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