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A basic fact about the Fourier transform regarding derivatives gives (having the Spectral Theorem of self-adjoint operators in mind) that $$ -\Delta = \mathcal{F}^{-1} M_{\xi^2} \mathcal{F},$$ where $\Delta$ is the Laplacian on $L^2(\mathbb{R})$, $\mathcal{F}$ the $L^2$ Fourier transform and $M_{\xi^2}$ the operator of multiplication by $\xi^2$.

I suppose similar diagonalization processes are possible for Laplacians on $L^2([0, \infty))$ and on $L^2([0, 1])$, where the role of the Fourier transform is replaced by the Laplace transform or the Fourier series. Unfortunately I cannot find any good informations on these things. In particular I would be interested in the role of the boundary conditions of self-adjoint realizations of the Laplacian when diagonalizing the operator. Any explanations or references would be appreciated.

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The way to recover (or discover) a spectral decomposition for a selfadjoint operator is through the singularities of the resolvent operation. The operator $L = -\frac{d^2}{dx^2}$ on $L^2[0,\infty)$ is not selfajdoint on its natural domain $\mathcal{D}(L)$ consisting of all twice absolutely continuous $f \in L^2$ for which $f''\in L^2$. A selfadjoint operator requires an additional endpoint condition at $0$ of the form $$ \cos\alpha f(0)+\sin\alpha f'(0)=0, $$ where $0 \le \alpha < \pi$ can be assumed. The resolvent operator $R(\lambda)=(L-\lambda I)^{-1}$ can then be constructed explicitly as follows. Let $$ \varphi_{\lambda}(x) = \sin\alpha\cos(\sqrt{\lambda}x)-\frac{1}{\sqrt{\lambda}}\cos\alpha\sin(\sqrt{\lambda}x) $$ This is the only classical solution (up to a scalar multiple) of the classical eigenvalue equation $$ -\varphi_{\lambda}''(x)-\lambda\varphi_{\lambda}(x) = 0 \\ \cos\alpha \varphi_{\lambda}(0)+\sin\alpha\varphi_{\lambda}'(0)=0. $$ The only classical eigenfunction $\psi_{\lambda}$ that is in $L^2[0,\infty)$ for $\lambda\notin\mathbb{R}$ is a constant multiple of $$ \psi_{\lambda}(x) = e^{i\sqrt{\lambda}x}. $$ These two functions can be used to construct the general resolvent operator because one satisfies the required condition at $0$, and the other satisfies the only requirement at $\infty$, which is that the function be square integrable.

Inverse of $L-\lambda I$: In what follows $\sqrt{\lambda}$ is taken to be the principle branch of the square root, with branch cut along the positive axis. Using the two functions $\psi_{\lambda}$ and $\varphi_{\lambda}$, the resolvent $(L-\lambda I)^{-1}$ becomes $$ R(\lambda)f=\frac{1}{w(\lambda)}\left[ \varphi_{\lambda}(x)\int_{x}^{\infty}f(t)\psi_{\lambda}(t)dt+\psi_{\lambda}(x)\int_{0}^{x}f(t)\varphi_{\lambda}(t)dt\right] $$ where $w$ is the Wronskian, which is independent of $x$: $$ w(\lambda) = W(\psi_{\lambda},\varphi_{\lambda})=\psi_{\lambda}\varphi_{\lambda}'-\psi_{\lambda}'\varphi_{\lambda} = -\cos\alpha-i\sqrt{\lambda}\sin\alpha $$ To verify that $R(\lambda)$ is the correct resolvent for $\lambda\notin\mathbb{R}$, consider \begin{align} \frac{d}{dx}R(\lambda)f & = \frac{1}{w(\lambda)}\left[\varphi_{\lambda}'(x)\int_{x}^{\infty}f(t)\psi_{\lambda}(t)dt-\varphi_{\lambda}(x)f(x)\psi_{\lambda}(x)\right] \\ & +\frac{1}{w(\lambda)}\left[\psi_{\lambda}'(x)\int_{0}^{x}f(t)\varphi_{\lambda}(t)dt+\psi_{\lambda}(x)f(x)\varphi_{\lambda}(x)\right] \\ &= \frac{1}{w(\lambda)}\left[\varphi_{\lambda}'(x)\int_{x}^{\infty}f(t)\psi_{\lambda}(t)dt+\psi_{\lambda}'(x)\int_{0}^{x}f(t)\psi_{\lambda}(t)dt\right] \\ \frac{d^2}{dx^2}R(\lambda)f & = \frac{1}{w(\lambda)}\left[\varphi_{\lambda}''(x)\int_{x}^{\infty}f(t)\psi_{\lambda}(t)+\psi_{\lambda}''(x)\int_{0}^{x}f(t)\varphi_{\lambda}(t)dt\right] \\ & + \frac{1}{w(\lambda)}\left[-\varphi_{\lambda}'(x)f(x)\psi_{\lambda}(x)+\psi_{\lambda}'(x)f(x)\varphi_{\lambda}(x)\right] \\ & = -\lambda R(\lambda)f - f \end{align} Therefore, $$ \left(-\frac{d^2}{dx^2}-\lambda\right)R(\lambda)f = f. $$ Using the above expressions, it is also easy to check that $$ \cos\alpha( R(\lambda)f)(0) +\sin\alpha (R(\lambda)f)'(0) = 0. $$ So $R(\lambda)f \in \mathcal{D}(L)$ and $(L-\lambda I)R(\lambda)f=f$ for $f\in L^2$. It is also true that $R(\lambda)(L-\lambda I)f=f$ for all $f\in\mathcal{D}(L)$. So $R(\lambda)f=(L-\lambda I)^{-1}f$, at least for all $\lambda\notin\mathbb{R}$.

Spectrum of $L$: There are two factors that influence the invertibility of $L-\lambda I$. First, for $\lambda$ real and positive, $R(\lambda)$ is not bounded, which means that $[0,\infty)$ is in the spectrum of $L$. Second, if $w(\lambda)=0$ and $\lambda < 0$, then $\lambda$ is an eigenvalue of $L$, and the resolvent has a first order pole at $\lambda$. The Wronskian vanishes if $\cos\alpha=0$ and $\lambda=0$, but this is not an eigenvalue. Otherwise, the Wronskian vanishes if $$ \sqrt{\lambda} = i\cot\alpha. $$ For real $\lambda < 0$, the value of $\sqrt{\lambda}$ lies on the positive imaginary axis, which means that $w(\lambda)=0$ can only occur if $\cot\alpha \ge 0$, which includes any $\alpha \in (0,\pi/2]$. For example, in the case where $\alpha=\pi/4$, the Wronskian vanishes at $\lambda=-1$, which corresponds to the fact that the condition $f(0)+f'(0)=0$ leads to a non-trivial eigenfunction $f(x)=e^{-x}$ of $L$ with eigenvalue $-1$, which is easily directly verified. $L$ has a negative eigenvalue $\lambda$ whenever $0 < \alpha < \pi/2$, but not for $\pi/2 \le \alpha < \pi$. So the spectrum of $L$ consists of continuous spectrum $[0,\infty)$, and one negative eigenvalue $\lambda = -\cot^2\alpha$ for $0 < \alpha \le \pi/2$.

The expression for the resolvent always has the positive real axis as part of its set of singularities, along with a possible negative eigenvalue. So that's the spectrum of $L$, which depends on the condition at $0$ through the parameter $\alpha$. Integrating around the resolvent set on a negatively oriented contour $C$ is known to give $\frac{1}{2\pi i}\int_{C}R(\lambda)fd\lambda = f$. A negatively oriented contour is needed because $R(\lambda)=\frac{1}{L-\lambda I}$ instead of $\frac{1}{\lambda I - A}$. The function $\varphi_{\lambda}$ is an entire function of $\lambda$, while $\psi_{\lambda}$ has a branch cut along the positive real axis. The Wronskian also contributes a branch cut, as well as a pole at the negative eigenvalue of $L$, if there is one. The above equation then leads to a complete eigenfunction expansions of $f\in L^2$ in terms of eigenfunctions of $L$. This is obtained from a single residue at the eigenvalue, and from a continuous integral expansion coming from the branch cut. For $\mu > 0$, the contributing part of the integral with respect to the spectral parameter involves \begin{align} &\frac{1}{2\pi i}\{R(\mu+i0)f-R(\mu-i0)f\} \\ =&\frac{1}{2\pi i}\varphi_{\mu}(x) \int_{0}^{x}f(t)\left(\frac{\psi_{\mu+i0}(t)}{w(\mu+i0)}-\frac{\psi_{\mu-i0}(t)}{w(\mu-i0)}\right)dt \\ &+\frac{1}{2\pi i}\left(\frac{\psi_{\mu+i0}(x)}{w(\mu+i0)}-\frac{\psi_{\mu-i0}(x)}{w(\mu-i0)}\right)\int_{x}^{\infty}f(t)\varphi_{\mu}(t)dt \\ =&\frac{1}{2\pi i}\varphi_{\mu}(x)\int_{0}^{x}f(t)\left(\frac{e^{i\sqrt{\mu}t}}{-\cos\alpha-i\sqrt{\mu}\sin\alpha}-\frac{e^{-i\sqrt{\mu}t}}{-\cos\alpha+i\sqrt{\mu}\sin\alpha}\right)dt \\ &+\frac{1}{2\pi i}\left(\frac{e^{i\sqrt{\mu}x}}{-\cos\alpha-i\sqrt{\mu}\sin\alpha}-\frac{e^{-i\sqrt{\mu}x}}{-\cos\alpha+i\sqrt{\mu}\sin\alpha}\right)\int_{x}^{\infty}\varphi_{\mu}(t)f(t)dt \\ =&\frac{1}{\pi(\cos^2\alpha+\mu\sin^2\alpha)}\varphi_{\mu}(x)\int_{0}^{x}f(t)\{\sqrt{\mu}\sin\alpha\cos(\sqrt{\mu}t)-\cos\alpha\sin(\sqrt{\mu}t) \}dt \\ &\frac{1}{\pi(\cos^2\alpha+\mu\sin^2\alpha)}\{\sqrt{\mu}\sin\alpha\cos(\sqrt{\mu}x)-\cos\alpha\sin(\sqrt{\mu}x) \}\int_{x}^{\infty}f(t)\varphi_{\mu}(t)dt \\ =&\frac{\sqrt{\mu}}{\pi(\cos^2\alpha+\mu\sin^2\alpha)}\varphi_{\mu}(x)\int_{0}^{\infty}f(t)\varphi_{\mu}(t)dt \end{align} This leads to an $L^2[0,\infty)$ expansion for $f$ $$ f(x) = \frac{1}{\pi}\int_{0}^{\infty}\frac{\sqrt{\mu}}{\cos^2\alpha+\mu\sin^2\alpha}\varphi_{\mu}(x)\int_{0}^{\infty}f(t)\varphi_{\mu}(t)dtd\mu+\mbox{possible residue term} $$ And this also gives the $L^2$ Parseval identity $$ \int_{0}^{\infty}|f(x)|^2 = \frac{1}{\pi}\int_{0}^{\infty}\left|\int_{0}^{\infty}f(t)\varphi_{\mu}(t)dt\right|^2\frac{\sqrt{\mu}}{\cos^2\alpha+\mu\sin^2\alpha}d\mu+\mbox{(possible residue term)}|(f,\varphi_{\mu_e})_{L^2}|^2 $$ If $f\in\mathcal{D}(L)$ and if $\mu_e$ is the possible negative eigenvalue, then $$ Lf = \frac{1}{\pi}\int_{0}^{\infty}\frac{\mu\sqrt{\mu}}{\cos^2\alpha+\mu\sin^2\alpha}\varphi_{\mu}(x)\int_{0}^{\infty}f(t)\varphi_{\mu}(t)dtd\mu + \mu_{e}\cdot\mbox{ possible residue term} $$ For example, if $\alpha=0$, then everything reduces to the Fourier sine transform expansion after the spectral change of variable $\mu=s^2$: \begin{align} f & = \frac{1}{\pi}\int_{0}^{\infty}\sqrt{\mu}\varphi_{\mu}(x)\int_{0}^{\infty}f(t)\varphi_{\mu}(t)dtd\mu \\ & = \frac{1}{\pi}\int_{0}^{\infty}\sqrt{\mu}\frac{\sin(\sqrt{\mu}x)}{\sqrt{\mu}}\int_{0}^{\infty}f(t)\frac{\sin(\sqrt{\mu}t)}{\sqrt{\mu}}dt d\mu \\ & = \frac{2}{\pi}\int_{0}^{\infty}\sin(sx)\int_{0}^{\infty}f(t)\sin(st)dtds \\ Lf & = \frac{2}{\pi}\int_{0}^{\infty}s^2\sin(sx)\int_{0}^{\infty}f(t)\sin(st)dtds. \end{align} Similarly, if you let $\alpha=\frac{\pi}{2}$, you arrive at the Fourier cosine representation. But you get all the oddball representations as well for the other possible $\alpha$, including the ones where there is a negative eigenvalue. The added eigenvalue results in a one-dimensional projection onto the corresponding eigenvector. The normalized eigenvector projection $(f,\varphi_{\mu})\varphi_{\mu}$ is obtained by finding the residue of the resolvent at the eigenvalue.

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  • $\begingroup$ Thank you for your extensive answer! I'll read through it tomorrow and then accept it if no questions appear. $\endgroup$ – agb Aug 26 '16 at 20:41
  • $\begingroup$ @agb : I had time to come back and make sure the powers of $\mu$ and algebraic signs are right, and to add a bit more exposition. I think the solution is now nicely cleaned up. $\endgroup$ – DisintegratingByParts Aug 28 '16 at 4:12

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