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.
1 Answer
$\begingroup$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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