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Let $f(z)=f(x+iy)=u(x,y)+iv(x,y)$. Consider the following ODE:

$\frac{df}{dz}-\lambda f(z)=0$ in $\Omega$

where $\Omega\subset\mathbb{C}$ is a connected open subset of the complex plane and $\lambda\in \mathbb{C}$.

Now, clearly if we begin with the assumption that $f$ is holomorphic in $\Omega$ ($u$ and $v$ are harmonic), this equation can be solved using classical ODE methods and one would easily find that $f(z)=Ce^{\lambda z}$ where $C\in\mathbb{C}$. However, when I look at the definition of the differentiation operator in this case:

$\frac{d}{dz}=\frac{\partial_x-i\partial_y}{2}$

It seems like we can try to solve this deceptively simple equation under a much weaker assumption; namely, the assumption that $u,v\in C^1(\Omega)$ (Here $\Omega$ is understood to be embedded in $\mathbb{R}^2$ instead of being in $\mathbb{C}$). Now one can see things get a lot trickier:

$$\frac{df}{dz}-\lambda f(z)=\frac{1}{2}(\partial_x-i\partial y)(u+iv)-\lambda(u+iv)=0$$ $$\implies \Big[\frac{1}{2}(u_x+v_y-\lambda u\Big)+i(v_x-u_y-\lambda v)=0$$ $$\implies \frac{1}{2}(u_x+v_y)-\lambda u = 0$$ $$\frac{1}{2}(v_y-u_x)-\lambda v=0$$

Doesn't look so simple now, does it? Under the assumption that $f$ is harmonic, we could have used the Cauchy-Riemann equations to decouple these two PDEs and arrived at the expected solution, but now under our weaker assumption we're forced to deal with a system of coupled first-order PDEs - not the nicest situation.

The most I've been able to do with this is write the system in a matrix-vector form which seems to suggest a kind of spectral theory problem:

$$\begin{pmatrix} \frac{1}{2}\partial_x - \lambda & \frac{1}{2}\partial_y \\ -\frac{1}{2}\partial_y & \frac{1}{2}\partial_x-\lambda \end{pmatrix} \begin{pmatrix} u \\ v\end{pmatrix} = \begin{pmatrix} 0 \\ 0\end{pmatrix}$$

but I certainly have no clue how to solve this. So, what do you guys think? How does one solve this ODE-turned-PDE, given these difficulties? Or, is there something about the problem statement itself (even with the weaker assumptions) that forces $f$ to be holomorphic?

Remark: I didn't find this problem in any book or paper, I just thought of it while studying complex analysis. I haven't been able to find any literature that provides any way of tackling a problem like this.

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    $\begingroup$ Pick any entire function $g$ and then $f(z)=g(\bar z)e^{\lambda z}$ satisfies the equation; conversely if $f$ is real analytic (so it is represented by Taylor series $\sum c_{n,m}z^n\bar z^m$), identifying coefficients shows that $f$ is as above; not sure if the equation here forces $f$ to be real analytic as I am not that well versed in general PDE's $\endgroup$
    – Conrad
    Commented Nov 29, 2022 at 2:48
  • $\begingroup$ Very nice observation there! That would, indeed satisfy the equation - and $g(\overline{z})$ would certainly be a non-holomorphic candidate that fits the weak assumption. I didn't think of that! However, I can't help but wonder if we can generalize even further, as if $g$ is entire, that would still imply that the real and imaginary components would be $C^\infty$, correct? $\endgroup$ Commented Nov 29, 2022 at 19:04
  • $\begingroup$ This also seems to heavily imply that, under this weak assumption, it becomes very important to consider the equation with some kind of boundary condition (as in the theory of PDEs) in order to get any meaningful solution. $\endgroup$ Commented Nov 29, 2022 at 19:08

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