Nonlinear differential equation with imaginary number coefficient
$begingroup$
Let $p$ be a polynomial. Does the differential equation,
$dy/dt=i cdot p(y)$, with initial condition $y(0)=1$ always have a global solution (for all $t in mathbb R$)?
ordinary-differential-equations
$endgroup$
add a comment |
$begingroup$
Let $p$ be a polynomial. Does the differential equation,
$dy/dt=i cdot p(y)$, with initial condition $y(0)=1$ always have a global solution (for all $t in mathbb R$)?
ordinary-differential-equations
$endgroup$
$begingroup$
Does $p$ have complex coefficients or only real ones? You can split both $y$ and $p$ into their real and imaginary parts, to obtain a first-order system for $mathrm{Re}(y)$ and $mathrm{Im}(y)$. To this system you can apply Picard-Lindelöf.
$endgroup$
– Christoph
Jan 21 at 7:13
$begingroup$
I was thinking p has real coefficients.
$endgroup$
– Craig Feinstein
Jan 21 at 12:47
add a comment |
$begingroup$
Let $p$ be a polynomial. Does the differential equation,
$dy/dt=i cdot p(y)$, with initial condition $y(0)=1$ always have a global solution (for all $t in mathbb R$)?
ordinary-differential-equations
$endgroup$
Let $p$ be a polynomial. Does the differential equation,
$dy/dt=i cdot p(y)$, with initial condition $y(0)=1$ always have a global solution (for all $t in mathbb R$)?
ordinary-differential-equations
ordinary-differential-equations
asked Jan 21 at 0:34
Craig FeinsteinCraig Feinstein
387321
387321
$begingroup$
Does $p$ have complex coefficients or only real ones? You can split both $y$ and $p$ into their real and imaginary parts, to obtain a first-order system for $mathrm{Re}(y)$ and $mathrm{Im}(y)$. To this system you can apply Picard-Lindelöf.
$endgroup$
– Christoph
Jan 21 at 7:13
$begingroup$
I was thinking p has real coefficients.
$endgroup$
– Craig Feinstein
Jan 21 at 12:47
add a comment |
$begingroup$
Does $p$ have complex coefficients or only real ones? You can split both $y$ and $p$ into their real and imaginary parts, to obtain a first-order system for $mathrm{Re}(y)$ and $mathrm{Im}(y)$. To this system you can apply Picard-Lindelöf.
$endgroup$
– Christoph
Jan 21 at 7:13
$begingroup$
I was thinking p has real coefficients.
$endgroup$
– Craig Feinstein
Jan 21 at 12:47
$begingroup$
Does $p$ have complex coefficients or only real ones? You can split both $y$ and $p$ into their real and imaginary parts, to obtain a first-order system for $mathrm{Re}(y)$ and $mathrm{Im}(y)$. To this system you can apply Picard-Lindelöf.
$endgroup$
– Christoph
Jan 21 at 7:13
$begingroup$
Does $p$ have complex coefficients or only real ones? You can split both $y$ and $p$ into their real and imaginary parts, to obtain a first-order system for $mathrm{Re}(y)$ and $mathrm{Im}(y)$. To this system you can apply Picard-Lindelöf.
$endgroup$
– Christoph
Jan 21 at 7:13
$begingroup$
I was thinking p has real coefficients.
$endgroup$
– Craig Feinstein
Jan 21 at 12:47
$begingroup$
I was thinking p has real coefficients.
$endgroup$
– Craig Feinstein
Jan 21 at 12:47
add a comment |
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$begingroup$
Does $p$ have complex coefficients or only real ones? You can split both $y$ and $p$ into their real and imaginary parts, to obtain a first-order system for $mathrm{Re}(y)$ and $mathrm{Im}(y)$. To this system you can apply Picard-Lindelöf.
$endgroup$
– Christoph
Jan 21 at 7:13
$begingroup$
I was thinking p has real coefficients.
$endgroup$
– Craig Feinstein
Jan 21 at 12:47