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we consider the problem $$ \begin{cases} y''+2y'+\lambda y=0\\ y'(0)=y(1)=0 \end{cases} $$ The question is to prouve that this problem admits eigenvalues $\lambda >1$.

I try to do the following: we put $\lambda-1=\alpha^2$ where $\alpha \in \mathbb{R}^\star_+$. Then, the general solution of the differential equation is $$ y(x)= e^{-x}[C_1 \cos(\alpha x) + C_2 \sin(\alpha x)] $$ Then using the limis conditions, we have: $$ y(1)=0=> C_1 \cos(\alpha) + C_2 \sin(\alpha)=0 $$ and $$ y'(0)=0 => C_1 = \alpha C_2 $$ Then combined the two, we obtaine $$ C_2 (\alpha \cos(\alpha)+ \sin(\alpha))=0 $$ Is it true? Please. And finally who's the eigenvalues of this problem? Please

  • Any solutions of $α=-\tan(α)$, see https://math.stackexchange.com/questions/917846/fixed-point-iteration-for-x-tanx, https://math.stackexchange.com/questions/305538/numerical-solution-to-x-tan-x – Lutz Lehmann May 14 '17 at 18:06
  • But why we can divide on $\cos(\alpha)$? What about the case$ \cos(\alpha)=0$? –  May 14 '17 at 18:14
  • Then $α\ne 0$ and $\sin α=\pm 1$, so the original term is not zero. – Lutz Lehmann May 14 '17 at 18:18
  • Ok , i understand. So my solution is absolutly correct? Please –  May 14 '17 at 18:20
  • Yes, there is a solution between any two poles of the tangent function. More precisely, set $α=a+u$, $a=(k-\frac12)\pi$, $k=1,2,3,…$ then $(a+u)\tan(u)=1$ has a solution close to $(a+\frac4{3a})^{-1}$ for larger $k$. – Lutz Lehmann May 14 '17 at 19:40

1 Answers1

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Everything looks good so far. Your question is whether it's possible that $\lambda>1$ is a solution, equivalently $\alpha^2>0$. The nontrivial solution necessarily needs $C_1,C_2\neq 0$ (if one is 0 so is the other from your results). Thus $\alpha\cos(\alpha)+\sin(\alpha)=0$. At $\alpha=\pi/2$, the l.h.s. is $1$. At $\alpha=\pi$, the l.h.s. is $-\pi$, so there is a solution $\alpha \in(\pi/2,\pi)$.

Alex R.
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