# Eigenvalues of a partial differential equation

Why $$\lambda_n=sgn(n)\pi i \sqrt{n^2+\alpha}$$?

I have this:

$$\varphi_{xx}-(\alpha+\lambda^2)\varphi=0$$ and $$\varphi(0)=\varphi(1)=0$$ then $$\varphi(x)=c\sin(\sqrt{-(\alpha+\lambda^2)}x)+d\cos(\sqrt{-(\alpha+\lambda^2)}x)$$, and $$d=0$$ then $$\varphi(x)=c\sin(\sqrt{-(\alpha+\lambda^2)}x)$$ and $$0=\sin(\sqrt{-(\alpha+\lambda^2)}) \Leftrightarrow \sqrt{-(\alpha+\lambda^2)}=n\pi$$

$$\sqrt{-(\alpha+\lambda^2)}=n\pi\Rightarrow \lambda_n=sgn(n)\pi i \sqrt{n^2+\alpha}$$?

• I think there's a mistake in the text. The equation follows that $\lambda = i\sqrt{(n\pi)^2+\alpha}$. What is $A$? – Dylan Jan 14 at 15:04
• $\phi=(\varphi,z)$ with $z=\varphi'$ $\phi'+A\phi=0$ and $\phi(0)=\phi^0$ (abstract form Cauchy) $A(\varphi,z)=(-z,-{\partial}_{x}^{2}+\alpha \varphi)$ $A$ is differential operator – eraldcoil Jan 14 at 17:27