Helmholtz equation in Laplace equation of cylinder I have a cylinder of height $H$ and radius $a$. I am about to find $u(\rho, \phi, z)$ that solves the equation $$\Delta u = 0 $$ with the given boundary conditions $$u(a, \phi, z) =0,$$ $$u(\rho, \phi, 3) =0,$$ $$u(\rho, \phi, 0) =f(\rho, \phi).$$ The problem here is not the process but instead the properties of the Helmholtz equation I get. Seperating the variables I get: $$\Delta G(z)-\lambda G(z)=0$$ $$\Delta w(\rho,\phi)+\lambda w(\rho,\phi)=0$$ The second equation is a Helmholtz equation. I know the properties of the Helmholtz equation. But can I safely from the second equation with the boundary condition $u(a, \phi, z) =0$ know that $\lambda > 0$? Because the function about the surface when $z=0$ is not zero, that makes me confused. Because in the case when I have disc (not a cylinder) with the same boundary condition at radius $a$ I know from the properties of Helmholtz equation that $\lambda >0$.

Yes, if $w$ is not-identically-zero function with zero boundary values in some domain $D$, and $\Delta w+\lambda w=0$ holds in $D$, then you can conclude $\lambda>0$. Proof: $$0 = \int_D w(\Delta w+\lambda w) = \int_D (-|\
abla w|^2 + \lambda w^2)$$ where the first term was integrated by parts. If $\lambda\le 0$, we have a contradiction: the integral on the right is negative.