How to Show the Eigenvalues of a Sturm-Liouville Equation are Real Given No Boundary Condition

Question

How can I show that the eigenvalues of $\Psi_{xx}+u(x) \Psi = \lambda \Psi$ are real without a boundary condition?

Answer 1

Your equation can be written as $$\left(\frac{d^2}{dx^2}+u(x)\right)\Psi=\lambda\Psi.$$ If we define $A:=\frac{d^2}{dx^2}+u,$ then it will suffice to show that $A$ is formally self-adjoint. If we let $f,g\in C_0^\infty(\mathbb{R})$, then we can calculate \begin{align*}(Af,g)_{L^2}&=\int\limits_\mathbb{R} \left(\frac{d^2}{dx^2}+u(x)\right)f(x)g(x)\, dx\\ &=\int\limits_\mathbb{R}f(x)\left(\frac{d^2}{dx^2}+u(x)\right)g(x)\, dx\\ &=(f,Ag)_{L^2}, \end{align*} where I integrated by parts twice to move that two derivatives from $f$ to $g$. Now, it just follows from spectral theory.

Answer 2

Many thanks for the answer. But I still do not see how does that prove that all eigenvalues are real. I started a proof but I am stuck:

Suppose that $~\lambda \in \mathbb{C}~$ is an eigenvalue and $~\varphi~$ its corresponding eigen-function. Then $~H[\varphi] = \lambda \varphi~$ and taking complex conjugates we also obtain that $~\overline{H[\varphi]} = \overline{\lambda \varphi}~$. We can obtain the following two equations:

$$ H[\varphi]~ \overline{\varphi} = \lambda~ \varphi ~\overline{\varphi} $$ $$ \overline{H[\varphi]}~ {\varphi} = \overline{\lambda} ~\overline{\varphi}~ \varphi $$ And subtracting the second from the first we get:

$$ H[\varphi] ~~\overline{\varphi} - \overline{H[\varphi]} ~~{\varphi} + (\lambda - \overline{\lambda}) ~\varphi ~\overline{\varphi} = 0 $$

where $$ ~H[\varphi]~ \overline{\varphi} - \overline{H[\varphi]}~ {\varphi} = (- \varphi_{xx} + u~ \varphi)~\overline{\varphi} - (-\overline{\varphi}_{xx}~+~u ~\overline{\varphi}) ~\varphi = -~ \varphi_{xx}~ \overline{\varphi} - \overline{\varphi}_{xx} ~\varphi $$

So we can rewrite the equation as:

$$ \frac{d}{dx} (\varphi_x ~\overline{\varphi} - \varphi ~\overline{\varphi}_x) + (\lambda - \overline{\lambda})~ \varphi ~\overline{\varphi} = 0 $$

Answer 3

We have

$(\bar \Psi \Psi_x)_x = \bar \Psi_x \Psi_ x + \bar \Psi \Psi_{xx}; \tag 1$

thus,

$\bar \Psi \Psi_{xx} = (\bar \Psi \Psi_x)_x - \bar \Psi_x \Psi_ x; \tag 2$

given that

$\Psi_{xx} + u(x)\Psi = \lambda \Psi, \tag 3$

where $\lambda$ is an eigenvalue associated with eigenfunction $\Psi$, we have

$\bar \Psi\Psi_{xx} + u(x)\bar \Psi \Psi = \lambda \bar \Psi\Psi; \tag 4$

via (2),

$(\bar \Psi \Psi_x)_x - \bar \Psi_x \Psi_ x + u(x)\bar \Psi \Psi = \lambda \bar \Psi\Psi; \tag 5$

also

$ \bar \Psi_x \Psi_ x = \overline{\Psi_x} \Psi_x; \tag 6$

so (5) yields

$(\bar \Psi \Psi_x)_x - \overline{\Psi_x} \Psi_x + u(x)\bar \Psi \Psi = \lambda \bar \Psi\Psi; \tag 7$

$\displaystyle \int_{-L}^L (\bar \Psi \Psi_x)_x \;dx - \int_{-L}^L \overline{\Psi_x} \Psi_x \; dx + \int_{-L}^L u(x)\bar \Psi \Psi \; dx = \lambda \int_{-L}^L \bar \Psi\Psi \;dx; \tag 8$

$\displaystyle \int_{-L}^L (\bar \Psi \Psi_x)_x \;dx = \bar \Psi(L) \Psi_x(L) - \bar \Psi(-L) \Psi_x(-L) \to 0 \tag 9$

as

$L \to \infty; \tag{10}$

thus,

$-\displaystyle \int_{-\infty}^\infty \overline{\Psi_x} \Psi_x \; dx + \int_{-\infty}^\infty u(x)\bar \Psi \Psi \; dx = \lambda \int_{-\infty}^\infty \bar \Psi\Psi \;dx; \tag{11}$

both the functions $\bar \Psi \Psi$ and $\overline{\Psi_x} \Psi_x$ are real, being products complex conjugate pairs; hence all integrands and indeed all integrals appearing in (11) are real; furthermore,

$\displaystyle \int_{-\infty}^\infty \bar \Psi\Psi \;dx \ne 0, \tag{12}$

else the function $\Psi(x) \equiv 0$ identically, forbidden by definition for eigenfunctions; hence,

$\lambda \in \Bbb R \tag{12}$

as well.

The eigenvalues of Sturm-Liouville are real