# How to find the eigenvalues of this operator?

The Question:

Consider the system

\begin{align} & Ly(x) \equiv y''(x)+y(x)=f(x),\qquad 0<x<1 \\ & y(0)=1, \qquad y(1)=0 \end{align}

(i) Find the eigenvalues and eigenfunctions of $L$

(ii) Find the eigenvalues of $M$, where

$$My(x) \equiv \int_0^1g(x,s)y(s)ds$$

and $g(x,s)$ is the associated Green's Function of the above system.

[Hint: There is no need to compute $g(x,s)$. Simply apply across the equation]

My Attempt:

(i) I found the eigenvalues $\lambda_n = 1-n^2 \pi ^2$ with corresponding eigenfunctions $y_n(x) = A_n \sin(n\pi x)$

(ii) I am not exactly sure what the hint means. I tried considering $LMy(x)$, which gives

\begin{align} LMy(x) & =L\biggl (\int_0^1g(x,s)y(s)ds \biggl) \\ & = \frac{d^2}{dx^2}\biggl (\int_0^1g(x,s)y(s)ds \biggl)+\biggl (\int_0^1g(x,s)y(s)ds \biggl) \\ & = \int_0^1\biggl( \frac{\partial^2}{\partial x^2}g(x,s)+g(x,s) \biggl)y(s)ds \end{align}

and I still don't know how to proceed.

Also, I am not sure how to use the fact that $g(x,s)$ is the associated Green's Function.

Any hints?

Here's a hint: The Green's function has the property

$$Lg(x; s) = \delta(x-s)$$

where $\delta(x)$ is the Dirac-delta function.

Hence,

$$L\big(My(x)\big) = \int_0^1 \delta(x-s) y(s) ds = y(x) \tag{1}$$

You'll also need to show that

$$L\big(My(x)\big) = M\big(Ly(x)\big) = \lambda_nMy(x) \tag{2}$$

This means proving

$$M\big(y''(x)\big) = \int_0^1 g(x;s)y''(s)ds = \int_0^1 \frac{\partial^2}{\partial s^2}g(x;s) y(s)ds$$

which can be easily done through integration by parts.

• But when you integrate twice by parts, don't you end up with $$\int_0^1 \frac{\partial^2}{\partial s^2}g(x,s)y(s)ds$$ instead? – glowstonetrees May 20 '18 at 16:28
• You do. Hence $M(Ly) = My''(x) + My = L(My)$ – Dylan May 20 '18 at 16:31
• I thought $$L(My) = \int_0^1 \biggl(\frac{\partial^2}{\partial x^2}g(x,s)+g(x,s) \biggl)y(s)ds$$ whereas $$M(Ly) = \cdots = \int_0^1 \biggl(\frac{\partial^2}{\partial s^2}g(x,s)+g(x,s) \biggl)y(s)ds$$ Or am I misunderstanding something? – glowstonetrees May 20 '18 at 16:33
• Yes, you are correct. These are equivalent because of symmetry, i.e. $g(x,s) = g(s,x)$ – Dylan May 20 '18 at 16:36
• Hmmmm, I know that $g(x,s)=g(s,x)$, but I am not too convinced that $$\int_0^1 \frac{\partial^2}{\partial x^2}g(x,s)ds = \int_0^1 \frac{\partial^2}{\partial s^2}g(x,s)ds$$ Take $g(x,s) = x^2s^2$ as an example? – glowstonetrees May 20 '18 at 16:43