Find all analytic functions $f:\mathbb{C} \longrightarrow \mathbb{C}$ such that
$$f'(z)=-2f(z),~z \in \mathbb{C}$$ and
$$f(0)+f'(0)=1$$
the only thing that majorly concerns me is making sure my answer is unique. as i'm sure it's possible that there may be a class of functions anyway
the preferred method i believe is to expand these as power series such as
$$\sum_{n=0}^{\infty}na_nz^{n-1} = -2\sum_{n=0}^{\infty}a_nz^{n}$$
do a bit of rearranging and then use the fact that co-effecient of power series are uniquely defined.
i'm sure theres a bit of rearranging magic to be done as considering the co-effecients at the moment gives
$$na_n = -2 a_n$$ which of course only happens at $n=-2$ apparently, but we have it that $n\geq0$ so that doesn't make sense. i can either presume there to be no solution or try the aforementioned rearrangement magic. by all means if you have a hint for how to solve this using the power series i'm all ears.
anyway, the method i used.
we have
$$f'(z)=-2f(z)$$ and these are meant to be analytic, so they're differentiable and integratable. so considering it to be a first order homogeneous ODE we solve using integrating factor
$$f'(z)+2f(z)=0 \implies$$
$$e^{2z}f'(z)+2e^{2z} = 0 \implies$$
$$\frac{d}{dz}\left(e^{2z}f(z)\right) = 0 \implies$$
$$e^{2z}f(z) = C \implies$$
$$f(z) = Ce^{-2z}$$
Differentiating and using the initial conditions given above we have
$$f'(z) = -2Ce^{-2z}$$ and
$$f'(0)+f(0)=1 \implies -2Ce^{0}+Ce^{0} = 1 \implies$$
$$-C=1 \implies C = -1$$
so
$$f(z) = -e^{-2z}$$ which indeed satisfies the requirements of both being analytic and the initial conditions.
now, with ODE's we use Initial conditions in order to guarantee uniqueness of solutions but assuming that $$f(z) = -e^{-2z} + D$$ is also a solution, plugging this into the above we obviously get $D = 0$
but…. is this function unique? and is this a valid arguement
Cheers for the help.
Incidently i've just figured out how to do it via power series. Consider a generating function
$$f(z) \leftrightarrow a_0,a_1,a_2,a_3,…a_n$$
then
$$2f(z) \leftrightarrow 2a_0,2a_1,2a_3,…,2a_n$$
and $$f'(z) \leftrightarrow a_1,2a_2,3a_3,…,n a_n$$
then for the equation
$$f'(z)+2f(z) = 0$$ we collate co-effecients and get
$$a_1 + 2a_0 = 0$$
$$2a_1+2a_2 = 0$$
etc…
Rearranging we get
$$a_1 = -2a_0$$
$$a_2 = -a_1$$
$$a_3 = -\frac{2}{3}a_2$$
… etc
back substituting we get
$$f(z) \leftrightarrow a_0,-2a_0,2a_0,-\frac{4}{3}a_0,\frac{4}{6}a_0,…,\frac{(-2)^{n}}{n!}$$
this implies then that
$$f(z) = a_0\sum_{n=0}^{\infty}\frac{(-2)^{n}}{n!}z^n$$
which is agrees with the above, ie$ f(z) = a_0 e^{-2z} $
using initial conditions again gives us $a_0$ = -1 and so $$f(z) = -e^{-2z}$$ so i was correct i believe. if someone could confirm, i'd greatly appreciate it.
Best Answer
You do not need to use theroems about unicity of solutions of ODEs in order to prove that your solution is the only one. If $f(z)$ is such that $f'(z)=-2f(z)$, let $g(z)=f(z)e^{2z}$. Then $g'(z)=f'(z)e^{2z}+2f(z)e^{2z}=0$. So, $g$ is constant. In other words, the only functions $f$ such that $f'(z)=-2f(z)$ are those of the form $f(z)=Ke^{2z}$. And the only $K$ for which you have $f(0)+f'(0)=1$ too is $K=-1$ (as you know).