Ok. Suppose that I have a function $f(x,y)$ that have a odd laplacian in relation to $x$, so $\Delta f(-x,y) = \Delta f(x,y)$. Is my original function still odd?
I think it's true, because, if partial derivative $\frac{\partial f}{\partial x}(-x,y) = \frac{\partial f}{\partial x}(x,y) \implies \int \frac{\partial f}{\partial x}(-x,y)dx = \int \frac{\partial f}{\partial x}(x,y) dx \stackrel{-x = t}\implies \\ -\int \frac{\partial f}{\partial t}(t,y)dt = \int \frac{\partial f}{\partial x}(x,y) dx \implies -f(t,y) + c_1(y) = f(x,y) + c_2(y) \implies -f(-x,y) + c_1(y) = f(x,y)+c_2(y) \implies f$ is odd.
So, because laplacian is $\frac{\partial^2}{\partial x^2}$, we have that if laplacian is odd in relation to $x$, then $\frac{\partial f}{\partial x}$ is even in relation to $x$, and $f$ is odd in relation to $x$. It's true? Thanks!