TITLE: Spectral theory - continuous spectrum
QUESTION [7 upvotes]: imagine that I have some differential operator $D$ that is defined on an interval $[a,b]$.
Now, assume that we take the boundary conditions in such a way that this operator is self-adjoint. Then, I have a few questions about this:
1.) Is it true that this operator cannot have a residual spectrum due to its self-adjointness.
2.) Where is the difference between continuous spectrum and a classical solution?
Probably, I need to specify question (ii). The classical theory of ODEs tells me that for any eigenvalue $\lambda $ I will find two independet solutions that solve $D f = \lambda f$. The thing is of course, that these two solutions do not satisfy the boundary conditions of the operator. 
But what happens when I have a continuous spectrum? Will I find eigenvalues ( I heard that one gets a whole interval of eigenvalues, is this true?) ? Will I get solutions to my eigenvalue that satisfy the boundary conditions? I think I still did not understand what I get in the case of a continuous spectrum.
Unfortunately, I am not a mathematician, so I am not so good at teaching this to myself from books in funtional analysis, so I would very much appreciate any assistance.

REPLY [7 votes]: This is a continuation of what I posted earlier. What follows are two examples of how the theory is applied to the trigonometric functions where $V=0$. The equation is in the limit point case on $[0,\infty)$ because $e^{i\sqrt{\lambda}x}\in L^{2}[0,\infty)$ while $e^{-i\sqrt{\lambda}x}$ is not, where $\sqrt{\lambda}$ is the branch whose branch cut is along the positive $x$ axis. So there is only one condition at $x=0$ of the form $Af(0)+Bf'(0)=0$ which can be imposed in order to arrive at a selfadjoint problem. No condition is needed at $\infty$, and none can be imposed--that's the consequence of the limit point case.
The most general domain for $Lf = -\frac{d^{2}}{dx^{2}}f$ consists of all twice continuously differentiable functions $f$ on $(0,\infty)$ such that $f$, $f'$ are continuous at $0$, and $f,Lf \in L^{2}[0,\infty)$. Then $L$ becomes selfadjoint when imposing one condition at $x=0$ on the domain of the form $Af(0)+Bf'(0)=0$ for real constants $A$, $B$ with $A^{2}+B^{2}\ne 0$. I'll look at a couple of cases because one gives only continuous spectrum, while the other gives rise to continuous spectrum and an eigenvalue, too.
Example 1 (Continuous Spectrum Only): In this example, consider the equation
$$
              -f''-\lambda f = g,\\
               f(0)=0.
$$
Start by letting $\phi_{\lambda}$, $\psi_{\lambda}$ be the unique classical solutions of $Lf=\lambda f$ with
$$
       \begin{array}{cc}
              \phi_{\lambda}(0) = 0 &  \psi_{\lambda}(0)=1 \\
              \phi_{\lambda}'(0) = 1 & \psi_{\lambda}'(0)=0.
       \end{array}
$$
These solutions are
$$
            \phi_{\lambda}(x) = \frac{\sin{\lambda x}}{\sqrt{\lambda}},\;\;
            \psi_{\lambda}(x) = \cos(\sqrt{\lambda}x).
$$
The Wronskian $W(\phi_{\lambda},\psi_{\lambda})=\phi_{\lambda}\psi_{\lambda}'-\phi_{\lambda}'\psi_{\lambda}\equiv -1$. For the limit point case, the trick to this problem is choosing the unique $m(\lambda)$ such that
$$
      \psi_{\lambda}+m(\lambda)\phi_{\lambda} \in L^{2}[0,\infty),\;\;\; \lambda\notin\mathbb{R}.
$$
The above condition requires the coefficient of $e^{-i\sqrt{\lambda}x}$ to be 0. So
$$
    \frac{1}{2}-m(\lambda)\frac{1}{2i\sqrt{\lambda}}=0 \implies m(\lambda)=i\sqrt{\lambda}.
$$
Then $\eta_{\lambda}=\psi_{\lambda}+i\sqrt{\lambda}\phi_{\lambda} \in L^{2}[0,\infty)$, and $W(\phi_{\lambda},\eta_{\lambda})=W(\phi_{\lambda},\psi_{\lambda})\equiv -1$. The resolvent operator associated with $L$ is $f=(\lambda I-L)^{-1}g$ or, equivalently,
$$
              f''+\lambda f = g,\\
               f(0)= 0.    
$$
The solution of this equation is
$$
           f=(\lambda I-L)^{-1}g = -\eta_{\lambda}(x)\int_{0}^{x}g(t)\phi_{\lambda}(t)\,dt
  -\phi_{\lambda}(x)\int_{x}^{\infty}g(t)\eta_{\lambda}(t)\,dt.
$$
Choosing $\eta_{\lambda}$ the way we did is important in order to guarantee that the integral on the right exists for $g \in L^{2}[0,\infty)$. As I mentioned before, finding the spectral measure of the interval $[\alpha,\beta]$--provided neither $\alpha$ nor $\beta$ are eigenvalues--is obtained by finding the generalized residue of the resolvent associated with the singularities in $[\alpha,\beta]\subset \mathbb{R}$:
$$
       E[\alpha,\beta]g=\lim_{\epsilon\downarrow 0}\frac{1}{2\pi i}\int_{\alpha}^{\beta}((u-i\epsilon)I-A)^{-1}g-((u+i\epsilon)I-A)^{-1}g \,du
$$
All of the singularities of the resolvent are found in the Titchmarsh m-function $m$. To see this, assume for the moment that $f$ is $0$ outside some interval $[0,R]$. Then the resolvent expression can be split into entire functions of $\lambda$ and one singular component. This is because $\phi_{\lambda}(x)$, $\psi_{\lambda}(x)$ are holomorphic in $\lambda$ for fixed $x$, and uniformly differentiable in $\lambda$ for all $x$ in such an interval $[0,R]$. The resolvent splits into
$$
    (\lambda I-L)^{-1}g = -\psi_{\lambda}\int_{0}^{x}g\phi_{\lambda}\,dt
            - \phi_{\lambda}\int_{x}^{\infty}g\psi_{\lambda}\,dt+
            -m(\lambda)\phi_{\lambda}\int_{0}^{\infty}g\phi_{\lambda}\,dt
$$
So, integrating the resolvent in $\lambda$ gives $0$ for all but the last term, and the last term is only non-zero because of the branch cut for $m(\lambda)$. So the spectrum of $L$ is $[0,\infty)$ and it's all continuous spectrum. Integrating over everything gives back $g$. That is $g=\lim_{\alpha\downarrow -\infty,\beta\uparrow\infty}E[\alpha,\beta]g$, and the convergence is guaranteed in $L^{2}[0,\infty)$. This gives the Fourier transform and its inverse:
$$
\begin{align}
       g(x) & =\frac{1}{2\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}g(t)\phi_{u}(t)\,dt\right)\phi_{u}(x)\sqrt{u}\,du \\
            & = \frac{1}{2\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}f(t)\frac{\sin(\sqrt{u}t)}{\sqrt{u}}\,dt\right)\frac{\sin(\sqrt{u}x)}{\sqrt{u}}\sqrt{u}\,du.
\end{align}
$$
To be sure this is a non-standard form, but becomes recognizable after substituting $u=v^{2}$ so that $du=2v\,dv = 2\sqrt{u}\,dv$. Now the above becomes just about as classical as it gets:
$$
          g(x)=\frac{1}{\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}g(t)\sin(vt)\,dt\right)\sin(vx)\,dv.
$$
Example 2 (Mixed Point and Continuous Spectrum): This example is a good way to see that trigonometric expansions can be fairly complex. Here I'm using the same operator as before, but with a different endpoint condition
$$
                      -f''-\lambda f = g,\\
                       f(0)+f'(0) = 0.
$$
So the general unrestricted $L$ is as before, but this domain is further restricted to functions for which $f(0)+f'(0)=0$. In keeping with the previous ideas, choose solutions $\phi_{\lambda}$ and $\psi_{\lambda}$ such that
$$
            \begin{array}{cc}
  \phi_{\lambda}(0)+\phi_{\lambda}'(0)= 0, & \psi_{\lambda}(0)+\psi_{\lambda}'(0) = 2, \\
  \phi_{\lambda}(0)-\phi_{\lambda}'(0)= 2, & \psi_{\lambda}(0)-\psi_{\lambda}'(0) = 0.
            \end{array}
$$
You can check that the solutions are
$$
      \phi_{\lambda}(x) = \cos(\sqrt{\lambda}x)-\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}},\;\;\;
      \psi_{\lambda}(x) = \cos(\sqrt{\lambda}x)+\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}.
$$
The Wronskian is constant, $W(\phi_{\lambda},\psi_{\lambda})=\phi_{\lambda}\psi_{\lambda}'-\phi_{\lambda}'\psi_{\lambda}\equiv 2$. So the resolvent will have a factor of $1/2$ that didn't exist previously because we must divide by the Wronskian in forming the Green function solution. The function $m$ is again chosen so that $\psi_{\lambda}+m(\lambda)\phi_{\lambda} \in L^{2}[0,\infty)$ for $\lambda\notin\mathbb{R}$, which means that $m$ must be chosen so that the $e^{-i\sqrt{\lambda}t}$ term vanishes. So,
$$
      \left(\frac{1}{2}-\frac{1}{2i\sqrt{\lambda}}\right)+m(\lambda)\left(\frac{1}{2}+\frac{1}{2i\sqrt{\lambda}}\right) = 0\\
      m(\lambda) = \frac{i\sqrt{\lambda}-1}{i\sqrt{\lambda}+1} =
             \frac{\sqrt{\lambda}+i}{\sqrt{\lambda}-i}
        =\frac{(\sqrt{\lambda}+i)^{2}}{\lambda+1}
        =\frac{\lambda+2i\sqrt{\lambda}+1}{\lambda+1}.
$$
Now there are two non-holomorphic pieces: a pole at $\lambda=-1$ and the $\sqrt{\lambda}$ term on $[0,\infty)$. So there are two kinds of spectrum: point spectrum at $\lambda=-1$ and continuous spectrum on $[0,\infty)$. This time what we're integrating is
$$
    \frac{1}{2}m(\lambda)\phi_{\lambda}(x)\int_{0}^{\infty}g(t)\phi_{\lambda}(t)\,dt
$$
The $1/2$ comes because of having to divide by the Wronskian. The negative of the residue at $\lambda=-1$ is what we what (the integration is clockwise):
$$
         \phi_{-1}(x)\int_{0}^{\infty}g(t)\phi_{-1}(t)\,dt \\
$$
And,
$$
     \phi_{-1}(x)=\frac{e^{i^{2}x}+e^{-i^{2}x}}{2}-\frac{e^{i^{2}x}-e^{-i^{2}x}}{2i^{2}}=e^{i^{2}x}=e^{-x}
$$
Notice this comes out normalized already so that its $L^{2}$ norm is $1$. The expansion for this case is
$$
       g=e^{-x}\int_{0}^{\infty}g(t)e^{-t}\,dt+\frac{1}{\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}g(t)\phi_{u}(t)\,dt\right)\phi_{u}(x)\frac{\sqrt{u}}{u+1}\,du,
$$
where $\phi_{u}(x)=\cos(\sqrt{u}x)-\frac{\sin{\sqrt{u}x}}{\sqrt{u}}$. Warning: Check all the constants for yourself; I'm always getting those things wrong.