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?
differential-equations lipschitz-functions
edited Nov 17 at 14:06
Rebellos
12.5k21041
asked Nov 17 at 14:04
user52227
928412
add a comment |
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?
differential-equations lipschitz-functions
edited Nov 17 at 14:06
Rebellos
12.5k21041
asked Nov 17 at 14:04
user52227
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
add a comment |
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?
differential-equations lipschitz-functions
edited Nov 17 at 14:06
Rebellos
12.5k21041
asked Nov 17 at 14:04
user52227
928412
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
differential-equations lipschitz-functions
edited Nov 17 at 14:06
Rebellos
12.5k21041
asked Nov 17 at 14:04
user52227
928412
edited Nov 17 at 14:06
Rebellos
12.5k21041
asked Nov 17 at 14:04
user52227
928412
edited Nov 17 at 14:06
Rebellos
12.5k21041
edited Nov 17 at 14:06
Rebellos
12.5k21041
edited Nov 17 at 14:06
Rebellos
12.5k21041
12.5k21041
asked Nov 17 at 14:04
user52227
928412
asked Nov 17 at 14:04
user52227
928412
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
add a comment |
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
add a comment |
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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