Continuous and Discrete Inverse of the 1D Laplacian

Jos van Kan

March 26, 2007

Abstract We are interested in constructing a continuous and discrete inverse to

the 1D Laplacian on the entire real axis.

1 Formulation of the problem

Consider the ODE

u − τ^{2}u′′= f  (1)

on the entire real axis. (τ > 0) Boundary conditions only require that the solution remains bounded in ±∞.

1.1 Continous solution

We rewrite (1)as follows

u − τu′+ τu′− τ^{2}u′′= f  (2)

and multiply on both sides of the equation by the integrating factor e^{−x/τ} . We integrate from x to ∞ and obtain

τe^{−x/τ} (u + τu′) = ∫_{x}^{∞} e^{−ξ/τ} f(ξ) dξ  (3)

as is easily verified by differentiating. We bring the factor −τe^{−x/τ} to the other side and redefine the integration variable to obtain

u + τu′= \frac{1}{τ} ∫_{0}^{∞} e^{−ξ/τ} f(x + ξ) dξ  (4)

We now repeat the procedure, this time with integrating factor e^{x/τ}. We integrate from −∞ to x to obtain:

τe^{x/τ} u = \frac{1}{τ} ∫_{−∞}^{x} e^{η/τ} ∫_{0}^{∞} e^{−ξ/τ} f(η + ξ) dξ dη  (5)

or equivalently, redefining η′= x − η and dropping the prime

u = \frac{1}{τ^{2}} ∫_{0}^{∞} ∫_{0}^{∞} e^{−(ξ+η)/τ} f(x + ξ − η) dξ dη  (6)

Introducing new variables t = ξ + η and s = ξ − η this transforms to

u = \frac{1}{2τ^{2}} ∫_{−∞}^{∞} f(x + s) ∫_{|s|}^{∞} e^{−t/τ} dt ds  (7)
= \frac{1}{2τ} ∫_{−∞}^{∞} e^{−|s|/τ} f(x + s) ds  (8)
= \frac{1}{2τ} ∫_{0}^{∞} e^{−s/τ} (f(x + s) + f(x − s)) ds  (9)

Remark, that this result makes sense for distributions f that increase exponentially slower than e^{x/τ}. If f is bounded over and has a limit for x to ±∞ so has u and in fact

lim_{x→±∞} u(x) = lim_{x→±∞} f(x)  (10)

as is easily proved.