Real part of the solution is not a solution?

Question

Take for example this very simple differential equation with complex coefficients:

\begin{equation}\tag{1} f'(t) = - i f(t) \end{equation}

where t is a real variable. The solution is $f(t)=c e^{-i t}$. Since in Physics only real quantities have sense, a physicist would take only the real part of this solution, i.e. $f(t)= \cos(t)$. However, it is easy to see that $f(t)=\cos(t)$ is not a solution of (1) anymore. So, the question is: Of which equation is $f(t)=\cos(t)$ the solution? Is there a connection with Eq.(1) ?

PS. If one looks for real $f$ solutions, then from Eq.(1) we must have $f'(t)=0$ and $-i f(t)=0$, i.e. $f(t)=0$, which is another different solution ! Which is the correct answer ?

Answer 1

Take $f(t)= \cos(t)$ as a solution for the given differential equation is indeed wrong.

The differential equation associated to this (kind of) solution is the harmonic oscillator equation

$$f''(t) = - f(t) $$

which is related to the the given differential equation by the fact that

$$f''(t)=(f'(t))'=(- i f(t) )'=-if'(t)=-i(-if(t))=-f(t)$$

Answer 2

This is a great question, and it illustrates why we sometimes need to be careful when extending things we do with real numbers to complex numbers.

It is true that, given a linear differential operator of the form

$$\mathscr L u=\sum_{k=0}^n c_k \mathrm D^k u$$

Where $u$ is possibly complex valued and the $c_k$s are real constants, that if $u$ satisfies $\mathscr L u=0$, then also $\mathscr L(\Re u)=0$ and $\mathscr L(\Im u)=0$. This is because both differentiation and real scalar multiplication commute with the real and imaginary part operators, e.g $\mathscr L(\Re u)=\Re (\mathscr L u)$.

However, when we allow the constants $c_k$ to be complex valued, this is no longer the case, because complex scalar multiplication does not commute with the real and imaginary part operators. Let $z\in \Bbb C$ and let $a,b\in\Bbb R$. Also let $x=\Re z$ and $y=\Im z$. Then:

$$\Re \big((a+\mathrm ib)z\big)=\Re\big((a+\mathrm ib)(x+\mathrm iy)\big)=\Re\big((ax-by)+\mathrm i(bx+ay)\big) \\ =ax-by$$ However, $$(a+\mathrm ib)\Re z=(a+\mathrm ib)x=ax+\mathrm ibx \neq \Re\big((a+\mathrm ib)z\big)$$

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For some geometric intuition, recall that multiplication by real numbers in the complex plane act like a scaling - they change only magnitude, not direction. However multiplication by imaginary numbers can change both magnitude and direction.

Answer 3

As a Physics enthusiast I would handle this ode as

$$ \left(\matrix{f'_r\\ f'_i}\right) = \left(\matrix{0& 1\\ -1& 0}\right)\left(\matrix{f_r\\ f_i}\right) $$

with $f = f_r + i f_i$