Is the unique solution of autonomous ODE infinitely differentiable











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Consider the ODE
begin{align}
y'&=f(y)\
y(0)&=y_0
end{align}

where $f$ is Lipschitz and, if it matters, always positive.
When restricted on an interval $[0,a]$, is the unique solution of such ODE infinitely differentiable? If not, under which assumptions would this be true?










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  • Assuming $y(t)$ is defined over an open interval of time, and $f$ is infinitely differentiable, we get $y'' = f'(y)y'=f'(y)f(y)$, $y''' = f''(y)y'f(y) + f'(y)f'(y)y' = f''(y)f(y)f(y)+f'(y)f'(y)f(y)$ and so on. In general the $n$th derivative of $y$ is a polynomial function of $f(y), f'(y), ... f^{(n-1)}(y)$. So $y$ is infinitely differentiable (over the open interval where it is defined) whenever $f$ is.
    – Michael
    Nov 17 at 17:40

















up vote
2
down vote

favorite












Consider the ODE
begin{align}
y'&=f(y)\
y(0)&=y_0
end{align}

where $f$ is Lipschitz and, if it matters, always positive.
When restricted on an interval $[0,a]$, is the unique solution of such ODE infinitely differentiable? If not, under which assumptions would this be true?










share|cite|improve this question
























  • Assuming $y(t)$ is defined over an open interval of time, and $f$ is infinitely differentiable, we get $y'' = f'(y)y'=f'(y)f(y)$, $y''' = f''(y)y'f(y) + f'(y)f'(y)y' = f''(y)f(y)f(y)+f'(y)f'(y)f(y)$ and so on. In general the $n$th derivative of $y$ is a polynomial function of $f(y), f'(y), ... f^{(n-1)}(y)$. So $y$ is infinitely differentiable (over the open interval where it is defined) whenever $f$ is.
    – Michael
    Nov 17 at 17:40















up vote
2
down vote

favorite









up vote
2
down vote

favorite











Consider the ODE
begin{align}
y'&=f(y)\
y(0)&=y_0
end{align}

where $f$ is Lipschitz and, if it matters, always positive.
When restricted on an interval $[0,a]$, is the unique solution of such ODE infinitely differentiable? If not, under which assumptions would this be true?










share|cite|improve this question















Consider the ODE
begin{align}
y'&=f(y)\
y(0)&=y_0
end{align}

where $f$ is Lipschitz and, if it matters, always positive.
When restricted on an interval $[0,a]$, is the unique solution of such ODE infinitely differentiable? If not, under which assumptions would this be true?







differential-equations lipschitz-functions






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edited Nov 17 at 14:06









Rebellos

12.5k21041




12.5k21041










asked Nov 17 at 14:04









user52227

928412




928412












  • Assuming $y(t)$ is defined over an open interval of time, and $f$ is infinitely differentiable, we get $y'' = f'(y)y'=f'(y)f(y)$, $y''' = f''(y)y'f(y) + f'(y)f'(y)y' = f''(y)f(y)f(y)+f'(y)f'(y)f(y)$ and so on. In general the $n$th derivative of $y$ is a polynomial function of $f(y), f'(y), ... f^{(n-1)}(y)$. So $y$ is infinitely differentiable (over the open interval where it is defined) whenever $f$ is.
    – Michael
    Nov 17 at 17:40




















  • Assuming $y(t)$ is defined over an open interval of time, and $f$ is infinitely differentiable, we get $y'' = f'(y)y'=f'(y)f(y)$, $y''' = f''(y)y'f(y) + f'(y)f'(y)y' = f''(y)f(y)f(y)+f'(y)f'(y)f(y)$ and so on. In general the $n$th derivative of $y$ is a polynomial function of $f(y), f'(y), ... f^{(n-1)}(y)$. So $y$ is infinitely differentiable (over the open interval where it is defined) whenever $f$ is.
    – Michael
    Nov 17 at 17:40


















Assuming $y(t)$ is defined over an open interval of time, and $f$ is infinitely differentiable, we get $y'' = f'(y)y'=f'(y)f(y)$, $y''' = f''(y)y'f(y) + f'(y)f'(y)y' = f''(y)f(y)f(y)+f'(y)f'(y)f(y)$ and so on. In general the $n$th derivative of $y$ is a polynomial function of $f(y), f'(y), ... f^{(n-1)}(y)$. So $y$ is infinitely differentiable (over the open interval where it is defined) whenever $f$ is.
– Michael
Nov 17 at 17:40






Assuming $y(t)$ is defined over an open interval of time, and $f$ is infinitely differentiable, we get $y'' = f'(y)y'=f'(y)f(y)$, $y''' = f''(y)y'f(y) + f'(y)f'(y)y' = f''(y)f(y)f(y)+f'(y)f'(y)f(y)$ and so on. In general the $n$th derivative of $y$ is a polynomial function of $f(y), f'(y), ... f^{(n-1)}(y)$. So $y$ is infinitely differentiable (over the open interval where it is defined) whenever $f$ is.
– Michael
Nov 17 at 17:40

















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