Nonlinear differential equation with imaginary number coefficient












0












$begingroup$


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 $t in mathbb R$)?










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$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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 $t in mathbb R$)?










share|cite|improve this question









$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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 $t in mathbb R$)?










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










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