Solving the IVP $f'(t) = \frac{i}{f(t)}$ for $f:\mathbb{R}\to\mathbb{C},\mathrm{Re}(f(t))>0, f(0) = c^2,c>0$

Question

Let $f:\mathbb{R}\to\mathbb{C}$ be such that $\mathrm{Re}(f(t))>0$, $f'(t) = \frac{i}{f(t)}$ for the imaginary unit $i$, and for a fixed $c > 0$, $f(0) = c^2$. How could I then solve for $f$? We can certainly write $f(t) = c^2 + \int_0^t\frac{i}{f(s)}ds$. But I don't really know how you can move forward from this.

Answer 1

$f'(t) f(t) = i$, so that $$ f(t)^2 = f(0)^2 + \int_0^t 2 f'(s) f(s) \, ds = c^4 + 2it \, . $$ The right-hand side is a complex number in the right halfplane and has two square roots, one of which, the “principal value”, has a positive real part. So the solution is $$ f(t) = \sqrt{c^4 + 2it} $$ with the principal value of the square root.

Answer 2

$f=u+iv, f'=u'+iv'$, so the differential equation becomes:

$$uu'-vv'+(u'v+uv')i=i\\ \Rightarrow uu'-vv'=0, ~~u'v+uv'=1\\ \Rightarrow u^2=v^2+c_1,~~ uv=t+c_2 $$

Initial condition: $f(0)=c^2=u(0)+iv(0)\Rightarrow u(0)=c^2, v(0)=0$

$$\Rightarrow c_1=c^4,~~ c_2=0 $$

$$\Rightarrow u^2=v^2+c^4,~~ uv=t $$

Next solve the quadratic equation for $u$ and $v$ and you get the function $f$.