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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 $0=\sin(\sqrt{-(\alpha+\lambda^2)}) \Leftrightarrow \sqrt{-(\alpha+\lambda^2)}=n\pi$

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

Thanks in advance

  • $\begingroup$ I think there's a mistake in the text. The equation follows that $ \lambda = i\sqrt{(n\pi)^2+\alpha} $. What is $A$? $\endgroup$ – Dylan Jan 14 at 15:04
  • $\begingroup$ $\phi=(\varphi,z)$ 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 $\endgroup$ – eraldcoil Jan 14 at 17:27

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