Mathematics Asked on December 27, 2021
I’m trying to find if the solution of the following initial value problem is unique.
begin{align}
y’ &= 3 (y^{2/3})\
y(0) &= 0
end{align}
I have tried with local or global well posedness theorems but I don’t understand how to apply them to this problem.
The answer I am supposed to give requires me to use these theorems somehow.
All solutions are given by $$y(x) = begin{cases} 0 & x le L, \ (x-L)^3, & x gt L, end{cases} $$ where $L in [0, infty]$. (For $L=infty$, $y equiv 0$; thanks to John Hughes for catching this case missing in an earlier version of this answer.)
One way to see that (rough sketch): Suppose $y(x_0) > 0$ for some $x_0 > 0$ and solve (using your fancy theorems, if you want to – but separation of variables will do just fine) backwards in time. This works until you hit 0, from where you can only continue with $0$ since $y$ is increasing and nonnegative.
Answered by Keba on December 27, 2021
This is one where just guessing an answer helps. From the first equation, it sure looks as if $y = x^n$ might be a solution for some $n$, and doing out the algebra, it turns out that $y = x^3$ is actually a solution. But so is $y = 0$. So ... not a lot of uniqueness there.
Sometimes brute force instead of fancy theorems is the way to go.
Answered by John Hughes on December 27, 2021
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