How to find all analytic functions such that $f(z)=-f''(z)$

Question

I'm trying to find all analytic functions such that $f(z)=-f''(z)$

I keep on trying but the solution seems a bit elusive...

I'm thinking of $\{a\cdot e^{iz},a\cdot e^{-iz}|a\in \mathbb{C}\}$ but cannot prove it...

any advice? :)

Answer 1

It the function is entire, it has a power series $f(z) = \sum_{n=0} {f^{(n)}(0) \over n!} z^n$, and hence we obtain $f(z) = \sum_{n=0} {f^{(2n)}(0) \over (2n)!} z^{2n} + \sum_{n=0} {f^{(2n+1)}(0) \over (2n+1)!} z^{2n+1}$, from which we get $f(z) = f(0)\sum_{n=0} {(-1)^n \over (2n)!} z^{2n} + f^{(1)}(0)\sum_{n=0} {(-1)^n \over (2n+1)!} z^{2n+1} = f(0) \cos z + f^{(1)}(0) \sin z$.

Answer 2

Let $x_0 \in \Bbb{C}$. Suppose $f(x_0) = A, f'(x_0) = B$. For $n > 1$, $f^{(n)}(x_0) = -f^{(n-2)}(x_0)$. It follows that $f^{(2n)} = (-1)^nA, f^{(2n+1)}(x_0) = (-1)^nB$. Since $f$ is analytic, there exists some neighborhood $D$ of $x_0$ such that for all $x\in D$ $$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(x_0)(x-x_0)^n}{n!} = $$ $$\sum_{n=0}^\infty \frac{(-1)^nA(x-x_0)^{2n}}{(2n)!} + \sum_{n=0}^\infty \frac{(-1)^nB(x-x_0)^{2n+1}}{(2n+1)!} = $$ $$A\cos(x-x_0) + B\sin(x-x_0)$$ Due to analytic extension, this equality holds on $\Bbb{C}$.

Answer 3

If you want to avoid power series you can use that your equation is linear, hence the space of solution is a vector space. Rewriting it as first order ODE and using the Cauchy-Lipschitz theorem we get that for all initial values $(x_0, v_0)\in \mathbb{C}^2$ exists a unique global solution. Thus the vector space of solutions is two dimensional. One easily verifies that $e^{iz}, e^{-iz}$ are linearly independend solutions and form therefore a basis of the solution space.

Answer 4

I guess another way would be to look at the equation $$f''(z)+f(z)=0,z \in \mathbb{C} $$

We know that the solution to this equation has the form :$$f(x)=f(0)\cos x+ f'(0) \sin x, x \in \mathbb{R}$$

Then by Identity Theorem, $$f(z)=f(0)\cos z+ f'(0) \sin z$$