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How do I solve $$x'=e^{it}\overline{x}?$$

This is a complex differential equation, but I don't see how to solve it.

Edit:
the original ODE is given by $$(x', y')=\begin{pmatrix}\cos t& \sin t\\ \sin t&-\cos t\end{pmatrix}(x,y)$$ I want to show that solutions of this ODE do not remain bounded for all $t$, and the idea was to solve the complex equation.

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  • $\begingroup$ Convert it to a system of two ODE. $\endgroup$ Nov 21, 2017 at 11:51
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    $\begingroup$ @Martín-BlasPérezPinilla You mean by setting $z=x+iy$ and setting $(x',y')=\begin{pmatrix}\cos t& \sin t\\\sin t& -\cos t\end{pmatrix}(x,y)$? Because that was the original question and the hint was to convert it to a complex ode $\endgroup$
    – Rafaelo
    Nov 21, 2017 at 11:59
  • $\begingroup$ I don't think this works $\endgroup$
    – mercio
    Nov 21, 2017 at 13:03
  • $\begingroup$ @mercio Why not? $\endgroup$
    – Rafaelo
    Nov 21, 2017 at 13:03
  • $\begingroup$ @Rafaelo Could you also show the original real ode? $\endgroup$
    – lisyarus
    Nov 21, 2017 at 13:08

1 Answer 1

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With $z'=e^{it}\bar z$ you also get by conjugation $\bar z'=e^{-it}z$ and thus for the second derivative $$z'' = ie^{it}\bar z+e^{it}\bar z'=iz'+z.$$ This second order linear ODE has as characteristic polynomial $$ \lambda^2-iλ-1 =\left(λ-\frac i2\right)^2-\frac34 =\left(λ-\frac{i-\sqrt3}2\right)\left(λ-\frac{i+\sqrt3}2\right) $$ which allows you to construct the solution. $$ z=e^{it/2}\left(c_1e^{\sqrt3 t/2}+c_2e^{-\sqrt3 t/2}\right)\\ $$

You have to find a relation between the integration constants that restricts the general solution to the solutions of the original equation. \begin{align} z'&=e^{it/2}\left(\frac12(i+\sqrt3)c_1e^{\sqrt3 t/2}+\frac12(i-\sqrt3)c_2e^{-\sqrt3 t/2}\right) \\ z'-e^{it}\bar z&=e^{it/2}\left(\frac12\Bigl[(i+\sqrt3)c_1-2\bar c_1\Bigr]e^{\sqrt3 t/2}+\frac12\Bigl[(i-\sqrt3)c_2-2\bar c_2\Bigr]e^{-\sqrt3 t/2}\right) \end{align} This implies $$\Bigl[(\sqrt3+1)+(\sqrt3-1)i\Bigr]c_1=\Bigl[(\sqrt3+1)-(\sqrt3-1)i\Bigr]\bar c_1$$ where the right side is the conjugate of the left and thus both sides are real, so that $$c_1=A\,\Bigl[(\sqrt3+1)-(\sqrt3-1)i\Bigr]$$ and similarly $$c_2=B\,\Bigl[(\sqrt3-1)-(\sqrt3+1)i\Bigr]$$ with real constants $A,B\in \Bbb R$.

As the components have absolute values $e^{\pm\sqrt3/2\,t}$, you get that all non-trivial solutions are unbounded over $\Bbb R$.

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  • $\begingroup$ Why does $ie^{it}\bar{z}+e^{it}\bar{z}'=iz'+z$ hold? $\endgroup$
    – Rafaelo
    Nov 22, 2017 at 8:47
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    $\begingroup$ Because of the original equation, you get in reverse $e^{it}\bar z=z'$ and conjugated $\bar z'=e^{-it}z$. $\endgroup$ Nov 22, 2017 at 11:07
  • $\begingroup$ @LutzLehmann I am stuck understanding how you draw the conclusion after "This implies". Thanks in advance. $\endgroup$ May 30, 2020 at 21:28
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    $\begingroup$ @PabloJeken : Immediately you of course only get $(\sqrt3+i)c_1=2\bar c_1$. But then also $2c_1=(\sqrt3-i)\bar c_2$. Now combine these and apply "nice" factors to get this symmetric form, using $2=(\sqrt3+1)(\sqrt3-1)$. $\endgroup$ May 30, 2020 at 22:04

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