# Does the operator $\mathrm{id}-t\Delta$ or its Green's function have a name?

Consider a Riemannian manifold and let

• $\mathrm{id}$ be the identity operator, let
• $\Delta$ be the scalar, negative-semidefinite Laplace-Beltrami operator, and let
• $t > 0$ be a parameter.

Does the operator $$A := \mathrm{id}-t\Delta$$ have an established name? Or alternatively, is there an established name for the corresponding Green's functions? Where do these objects show up? I've been working with this beast a lot lately and it would be very useful to at least have a Googleable name.

The operator shows up in a couple places, for instance if you take the heat equation $$\dot{u} = \Delta u$$ and discretize it using backward Euler then you get $$\frac{u-u_0}{t}=\Delta u$$ or $(\mathrm{id}-t\Delta)u=u_0$. So the Green's function can be viewed as a 1-step approximation to the heat kernel. (The operator cannot, however, be viewed as the first part of the Taylor series for the solution operator $e^{t\Delta}$ -- that would be $\mathrm{id} + t\Delta$.)

If you replace $\mathrm{id}$ with the identity matrix $I$ and $\Delta$ with the graph Laplacian $L$, then on a finite graph you can pick $t$ small enough that $I-tL$ can be viewed as the limit of the Neumann series $$\sum_{k=0}^\infty t^k L^k.$$

However, I suspect there are much better interpretations of the original operator $A$, which is why I'm asking the question!

Thanks.

• Generally, this is related to resolvent, which is very important in operator theory. In this specific situation with the Laplacian, I have seen it called the Helmholtz operator, Bessel operator (typically with t=1) etc. – timur Aug 7 '12 at 14:35
• Thanks. Though when $\Delta$ is negative semi-definite the Helmholtz operator looks like $k^2 + \Delta$, which doesn't behave like $\mathrm{id} - \Delta$ for any $k$. In particular, Green's functions of the Helmholtz operator look sinusoidal, whereas Green's functions of the operator in question look like exponential decay (roughly like $e^{-x/\sqrt{t}}$). – TerronaBell Aug 7 '12 at 15:31
• Yes, I know but people sometimes abuse (or abstract) the name to include also the case with a negative constant in front of the Laplacian. Perhaps "reaction-diffusion operator" would have been better. – timur Aug 7 '12 at 15:46
• In the pseudo-Riemannian case the physicists would probably refer to it as the (massive) Klein--Gordon operator. – mathphysicist Aug 17 '12 at 19:13

This is in fact just the resolvent of the Laplacian. The Helmholtz operator is $\Delta + k^2$ but $k$ is allowed to be any complex number so, assuming the convention that the Laplacian has nonnegative spectrum, one gets exponential decay for solutions when $k$ is a (nonnegative) real number while solutions have oscillatory slow decay (described by the classical far-field expansion) when $k$ is purely imaginary. The resolvent family $(\Delta + k^2)^{-1}$ (which differs from your operator by scalar multiplication only) is a holomorphic family of bounded operators on $L^2$ when the imaginary part of $k$ is strictly negative (for example). Part of scattering theory studies what happens when $k$ approaches the imaginary axis (limiting absorption) or even in some cases crosses this line (continuation of the resolvent). You can find out much more about this, e.g. in Vol. IV of Reed-Simon, Methods of Mathematical Physics, see also Melrose, Geometric Scattering Theory, and Vols. I and II of Partial Differential Equations by Michael Taylor.