# Boundary Conditions at Infinity

## Abstract

Solutions of problems with boundary conditions involving a limit at infinity can be difficult. To build some intuition, we will begin with a simple example modelled by a linear differential equation. Consider the function u = e^{−x} = Σ _{m=0} ^{∞} (−x)^{m}/m!. We obviously have u(0) = 1 and u(∞) = 0. By the latter, we clearly mean that . This function satisfies the differential equation d^{2}u/dx^{2} − u = 0. Letting L denote d^{2}/dx^{2}, we have Lu−u = 0 which is in our standard format Lu + Ru = 0 with R =− 1. With L^{1} defined as a two-fold indefinite integration operator, we have L^{−1}Lu = L^{−1}u so that u = C_{0} + C_{1} x + L^{−1}u. (Since we are dealing with a linear ordinary differential equation, double decomposition is unnecessary.)

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## Suggested Reading

- 1.R. E. Meyer,
*Introduction to Mathematical Fluid Dynamics*, Wiley-Interscience (1971).Google Scholar