Eigen values of a boundary value problem .

Question

Consider the differential equation $$y’’+2y’+a(x)y=\lambda y, y’(0)=0=y’(1)$$ where $a(x):[0,1]\to (1,\infty)$ is a continuous function. Let $y$ be a positive eigen function corresponding to eigen value $\lambda$. Then which of the following statements are possibly true ?

$1$ $\lambda >0$.

$2.$ $ \lambda <0$.

$3.$ $\lambda =0$.

$4.$ $\int_0^1 (y’)^2dx=2\int_0^1 yy’dx+\int_0^1(a(x)-\lambda)y^2dx$.

If $\lambda =0$ then differential equation becomes $y’’+2y’+a(x)y=0$. I can solve it only by taking particular $a(x)$. I tried by choosing $a(x)=0$ which gives trivial solution. Then I tried by taking $a(x)=1$ again gives trivial solution. Given problem is not a Sturm Liouville problem so that I can say eigen value will be positive. I think I am not in right direction. Please help me to solve this problem or please some detailed solution. Thank you in advanced.

Answer 1

The answer is (4). In fact, multiplying both sides by $y$ and integrating, one has $$ \int_0^1y''ydx+2\int_0^1yy'dx+\int_0^1(a(x)-\lambda)y^2dx=0. \tag1$$ By Integration-By-Parts, one gets $$ \int_0^1y''ydx=\int_0^1ydy'=yy'\bigg|_0^1-\int_0^1(y')^2dx=-\int_0^1(y')^2dx . \tag2$$ Putting (2) into (1), one obtains (4).

Answer 2

Reject option $(3)$ by choosing $a(x)=2$ and $\lambda=0$. The solution in this case is $y(x)=0$ which can never be an eigenfunction.

For option $(1)$ if you choose $\lambda=2(>0)$ with $a(x)=2$, you get the eigenfunction $y=c, c\in\mathbb R-\{0\}$. For positive eigenfunction you need $c\in\mathbb R^+$, so option $(1)$ is possibly true.

Option $(2)$ can be discarded as well for any $\lambda<0$ where a zero solution is obtained on applying conditions. Option $(4)$ is true as proved by xpaul.