# ordinary differential equation order 2-limit problem

We consider the problem $$\begin{cases} y''+2y'+\lambda y=0\\ y'(0)=y(1)=0 \end{cases}$$

1. Prove that there is no eigenvalue $\lambda <1$ for this problem.

2. Prove that the eigenvalues are $\lambda_n=n^2\pi^2+1$ and the eigenfunctions associated are $y_n(x)=e^{-x} \sin(n\pi x)$.

For 1. I try the following:

We put $\lambda <1$ is equivalent to say that $\lambda-1 <0$, so we can put $$\lambda-1=-\alpha^2<0, \ \alpha \in \mathbb{R}^\star.$$ Then the equation is $$y''+2y'+(1-\alpha^2)y=0.$$ We plug $y(x)=e^{rx}$ in the equation, then we obtain the characteristic equation $$r^2+2r+(1-\alpha^2)=0$$ it admits two distinct solutions: $r_1=-1-|\alpha|$ and $r_2=-1+|\alpha|$ so there is equivalent to say that the two distinct solutions are $r_1=-1-\alpha$ and $r_2=-1+\alpha$.

Then, the general solution of the equation is $$y(x)=C_1e^{(-1-\alpha)x}+C_2 e^{(-1+\alpha)x}$$ We have

• $y'(0)=0$ implies $C_1(1-\alpha)+ C_2(-1+\alpha)=0$ and
• $y(1)=0$ implies $C_1e^{-1-\alpha} +C_2e^{-1+\alpha}=0$.

To find $C_1$ and $C_2$, we resolve the system $$\begin{cases} (-1-\alpha)C_1 +(-1+\alpha)C_2=0\\ e^{-1-\alpha}C_1 +e^{-1+\alpha}C_2=0 \end{cases}$$ We have $$\det = (-1-\alpha)e^{-1+\alpha} +(-1+\alpha)e^{-1-\alpha}$$ My question is: why $\det=0$?

For question 2. I try the following: in the case $\lambda >1$ we put $\lambda-1=\alpha^2>0$. Then, the equation is $y''+2y'+(1+\alpha^2)y=0$. We plug $y(x)=e^{rx}$ and we obtain the caracterisctic equation $r^2+2r+(1+\alpha^2)=0$. Then the general solution of the equation is $$y(x)= e^{-x} (C_1 \cos(\alpha x)+C_2 \sin(\alpha x)]$$ We have $$y'(x)= -e^{-x}(C_1 \cos(\alpha x)+C_2 \sin(\alpha x)) +e^{-x}(-C_1 \sin(\alpha x) +C_2 \cos(\alpha x))$$ Then $$y'(0) => C_1=C_2$$ and $$y(1)=0 => C_1 (\cos(\alpha)+\sin(\alpha))=0$$

My question is: how we can find $\alpha \in \mathbb{R}^\star_+$ then $\cos(\alpha)+ \sin(\alpha)=0$?

Thank's for the help

We may assume $\alpha>0$. $$\text{det}=e^{-1}\bigl(-\alpha(e^\alpha-e^{-\alpha})-e^\alpha-e^{-\alpha}\bigr)<0.$$
• Why we have $det <0$? Please. We know that $e^\alpha +e^{-\alpha}>0$, but what the sign of $-\alpha(e^{\alpha}-e^{-\alpha}) - (e^{\alpha}+e^{-\alpha})$? and why? Please – user415040 May 12 '17 at 11:02
• $\alpha>0\implies e^\alpha>e^{-\alpha}$ – Julián Aguirre May 12 '17 at 11:09
• and if $\alpha <0$ then $e^{-\alpha} > e^{\alpha}$? But how we explain it? – user415040 May 12 '17 at 11:12
• Read the first line. Since you only care about $\alpha^2$, you may assume from the beginning that $\alpha>0$. – Julián Aguirre May 12 '17 at 11:18