0
$\begingroup$

I'm given a BVP, $y''+2y'=\lambda y$; $\ y(0)=0,\ y(1)=0.$

Is $\lambda=-1$ an eigenvalue? If not why? If it is, how do we find the corresponding eigenfunction? I'm also supposed to solve the above eigenvalue problem and find the corresponding eigenfunctions $y_n(x)$, but I really don't understand the question.

  • 0
    $\lambda=-1$ is an eigenvalue if the problem has non-trivial solutions, try solving the equation with $\lambda$ as a parameter first, then decide for which $\lambda$ it has non-trivial solutions. – 2017-02-14
  • 0
    By $\lambda = -1$ as a parameter, do you mean that we just plug it in so it becomes the ordinary differential equation $y'' + 2y' =-y$? – 2017-02-14
  • 0
    No, solve the linear equation $$y''+2y'-\lambda y=0$$ and then decide for which $\lambda$ it has non-trivial solutions (do $z=y'$ and solve the linear system). – 2017-02-14

1 Answers 1

0

The characteristic equation of $$y''+2y'=\lambda y\tag{1}$$ is $$ r^2+2r-\lambda =0 $$ whose solution is $r_{1,2}=-1\pm\sqrt{1+\lambda}$.

Case 1: If $\lambda=-1$, then $r_1=r_2=-1$ and hence $-1$ has the multiplicity 2. Therefore the general solution of (1) is $$ y=C_1e^{-x}+C_2e^{-x}.$$ Using the boundary values $y(0)=y(1)=0$ gives $C_1=C_2=0$ and hence $\lambda=-1$ is not an eigenvalue.

Case 2: If $\lambda>-1$, then the general solution of (1) is $$ y=C_1e^{r_1x}+C_2e^{r_2x}.$$ Using the boundary values $y(0)=y(1)=0$ gives $C_1=C_2=0$ and hence If $\lambda>-1$, then $\lambda$ is not an eigenvalue.

Case 3: If $\lambda<-1$, then $r_{1,2}=-1\pm i\sqrt{-1-\lambda}$ and hence the general solution of (1) is $$ y=C_1e^{-x}\cos(\sqrt{-1-\lambda}x)+C_2e^{-x}\sin(\sqrt{-1-\lambda}x).$$ Using the boundary value $y(0)=0$ gives $C_1=0$. Using $y(1)=0$ gives $C_2e^{-1}\sin(\sqrt{-1-\lambda})=0$. So if $\sqrt{-1-\lambda}=n\pi$ or $\lambda=-1-n^2\pi^2$ for $n\in N$, (1) has nonzero solutions $$ y=C_2\sin(n\pi x)$$ Hence $\lambda=-1-n^2\pi^2$ for $n\in N$ are all eigenvalues with the corresponding eigenfunctions $y=\sin(n\pi x)$.

  • 0
    I believe you need a parentheses like $-(1+n^2\pi^2)$ – 2017-02-14
  • 0
    @user409800, yes, you're right. – 2017-02-14