# 2D Elliptic Eigenproblem

Consider the elliptic eigenproblem \begin{align} \nabla^2\phi&=0 \ \ \ \ \ \ \ \ \text{in \Omega}\\ \frac{\partial\phi}{\partial r}&=\lambda\phi \ \ \ \ \ \text{on \Gamma_1} \\ \phi&=0\ \ \ \ \ \ \ \ \text{on \Gamma_2\cup\Gamma_3}, \end{align} where $$\Omega$$ is the quarter circle $$0 and $$0<\Omega<\frac{\pi}{2}$$ and the boundary $$\partial\Omega=\Gamma_1\cup\Gamma_2\cup\Gamma_3$$ is defined as\begin{align} \Gamma_1&: r=1 \ \text{and} \ 0<\Omega<\frac{\pi}{2} \\ \Gamma_2&: 0

I am trying to find all non-trivial solutions $$H$$ using the method of separations of variables.

For $$H$$, I've determined that the only non-trivial case ($$p=$$ separation constant) is when $$\lambda=p^2>0.$$ When $$p^2>0$$, $$H''+pH=0\implies H(\theta)=A\cos(p\theta)+B\sin(p\theta).$$ Our initial conditions for $$H(0)=H\left(\frac{\pi}{2}\right)=0$$. This implies $$A=0$$ and a solution exists iff $$B\neq 0\implies \sin\left(\frac{p}{2}\pi\right)=0\implies \frac{p}{2}=k, \ \ \ k\in\mathbb{Z^+}.$$ So $$p=2k$$.

My question is, why does the solution I have state that the non-trivial solution is $$H(\theta)_k=\sin(2k\pi)$$ and not $$H(\theta)_k=\sin(k\pi)?$$ Shouldn't the latter be correct as $$p=2k$$ (i.e. how do we form our eigenfunction)?