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 $0<r<1$ 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<r<1 \ \text{and} \ 0=0 \\ \Gamma_3&: 0<r<1 \ \text{and} \ 0=\frac{\pi}{2}. \end{align}

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)?


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