# Can $e^{ax}$ be said to be the eigenfunction of the operator $\frac{d^{(n)}}{dx}$?

I'm gradually getting familiar with operators (as they are used in QM) and the terminology surrounding them, and I was wondering whether all the (to me) well-known operators have straight-forward, elementary functions, as seems to be the case with $\frac{d^{(n)}}{dx}$, because $$\frac{d^{(n)}}{dx}e^{ax}=a^n e^{ax}$$

and could one say that the spectrum of these eigenfunctions is degenerate, since $a$ can vary (I know "spectrum" is usually used for the set of eigenvalues, but it seems appropriate here)? Is this the correct interpretation? If so, what is the eigenfunctions and -values for the following differentialoperators (if you could point me in the direction of a resource that either collects them or - even better - shows how they are obtained, that would be very acceptable):

• $\int dx$
• $\nabla$ (grad)
• $\nabla \cdot$ (div)
• $\nabla^2$ (Laplace)
• If you have a cool one, please do just throw it in there!

Thanks!

For each value of $a$, $$\frac{d^n}{dx^n} e^{ax} = a^n e^{ax} ,$$ and so $e^{ax}$ is an eigenfunction of the operator linear differential operator $\frac{d^n}{dx^n}$ of eigenvalue $a^n$. Conversely, the $a^n$-eigenspace of this operator (regarded as a map, e.g., from $C^{\infty}(\Bbb R)$ to itself) has dimension $n$, and it's not too hard to write down an explicit basis thereof. (NB that for $a = 0$, the eigenspace is qualitatively different from the general case, as the solutions of $\frac{d^n}{dx^n} f = 0$ are just the polynomials of degree $< n$.) With a little more work (mostly involving passing to the complex setting), we can show that the spectrum of this operator is $\Bbb R$ itself.

For most of the other operators you mention, the input and output objects are of different types, so without more structure it doesn't make any sense to talk about eigenfunctions; for example, the gradient maps functions to vector fields. A little more subtly, the operator $f \mapsto \int f \,dx$ maps functions to equivalence classes of functions (where $f \sim \hat{f}$ iff $\hat{f} - f$ is a constant).

The exception to this is the Laplacian $\Delta := \nabla^2$; in this case, the eigenvalue equation, $$\Delta f = -\lambda f$$ is essentially the interesting and well-studied Helmholtz equation. The eigenvalues and eigenfunctions in this case depend on the (fixed, and often bounded) domain $\Omega$ of $f$, or, if you like, compact Riemannian manifold $(\Omega, g)$ with boundary. For general $\Omega$ the eigenfunctions are complicated, but when $\Omega$ is the $n$-sphere, the eigenvalues are the rather tractable spherical harmonics, which are important, for example, in the quantum mechanics of the hydrogen atom.

For any such domain $\Omega$, it's particularly interesting to restrict attention to eigenfunctions that vanish on the boundary (i.e., the functions $f$ that, in addition to the above equation, satisfy $f\vert_{\partial \Omega} = 0$). In this setting, we can ask whether the spectrum of eigenvalues for such eigenfunctions determine $\Omega$ (say, up to isometry), or a little more poetically, "Can One Hear the Shape of a Drum?" This question was posed by Mark Kac in a rightfully celebrated 1966 article by that title (though the origins of this circle of ideas are rather earlier), with emphasis on bounded domains $\Omega \subset \Bbb R^2$, and this question wasn't answered (in this case) until 1992, in the negative.

Another operator familiar from vector calculus that has the same domain and codomain is the curl operator, $\nabla \times\,\cdot\,$, on $\Bbb R^3$ (or any $3$-dimensional Riemannian manifold); its eigenvectors are the subject of this question.

• Note that the counter examples I'm aware of for the shape of a drum question are either high-dimensional or have boundary. As far as I know it's still open if there exists a pair of non-isometric, isospectral compact surfaces (without boundary) in $\mathbb{R}^3$. – user7530 Feb 19 '16 at 18:16
• Fascinating read, thanks! Could you elaborate on the part about the integration-operator? I don't really know about equivalence-classes, and don't understand the notation you used, but if you could throw me a wiki-page, I'd be grateful (I assume that there is no explicit function that serves as eigenvector?)! – Bobson Dugnutt Feb 19 '16 at 18:29
• You're welcome, I hope you found it useful! The point regarding the integration operator is that the output of the operator is not a function, but rather a family of functions, corresponding to the fact that any (nice) function $f(x)$ has infinitely many antiderivatives; at least provided that $f(x)$ is defined on an interval, if we call one antiderivative $F(x)$, then the others are the functions $F(x) + C$ (for more, just see en.wikipedia.org/wiki/Antiderivative ). In particular, since the domain and codomain of the integral operator are different, there's no notion of eigenvector. – Travis Willse Feb 19 '16 at 18:33
• You already mentioned it, but it might be worth stressing that, ususally, one does not look at the eigenvalues of the differential operator itself, but rather the operator endowed with suitable boundary conditions. Otherwise, the functions $f(x,y)=e^{ax}e^{by}$ are all eigenfunctions for the Laplacian in $\mathbb{R}^2$ for every value of $a$ and $b$. For instance, the spectrum of the Laplacian with homogeneous Neumann boundary conditions is different from the spectrum of the Laplacian with homogeneous Dirichlet boundary conditions (the former include zero, the latter doesn't). – bartgol Feb 19 '16 at 18:35
• That said, there are, of course, functions $f$ for which $\lambda f$ is an antiderivative for $f$. By definition, this means that $(\lambda f)' = \lambda f' = f$. For $\lambda = 0$, this forces $f = 0$. If $\lambda \neq 0$, we can rearrange the equation to read $f' = \lambda^{-1} f$, which is essentially the eigenvalue problem for $\frac{d}{dx}$. – Travis Willse Feb 19 '16 at 18:36

There is no "the" eigenfunction. The set of all eigenfunctions form a eigenspace. If $T:V \to V$ is a linear operator, the collection of all $v \in V$ satisfying $$Tv = \lambda v$$ for some scalar $\lambda$ is the eigenspace of $T$. It is not hard to see if $v$ is an eigenfunction of $T$, then so is $\mu v$ for any scalar $\mu$.

The operators gradient and divergence are not endomorphic. That is, they don't map spaces to themslves, so they don't have an eigenspace.

The eigenfunctions of $\nabla^2$ are given by the Helmholtz equation.

• Thanks! But does $\mu e^{ax}$ then constitute an eigenbasis for $\frac{d^{n}}{dx}$? I guess this would be the same as asking if $\mu e^{ax}$ is the general solution to that DE? – Bobson Dugnutt Feb 19 '16 at 18:13
• – Henricus V. Feb 19 '16 at 18:14

Yes, it's an eigenfunction but there are a lot. In particular, you need to attach some $n$th roots of unity since what you're really doing is solving an $n$th order linear ODE with constant coefficients, i.e., finding all the roots $r$ of $r^n - \lambda$ where $$\frac{d^{(n)}}{dx} y(x) = \lambda \cdot y(x).$$