Boundary Conditions at Infinity
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 d2u/dx2 − u = 0. Letting L denote d2/dx2, we have Lu−u = 0 which is in our standard format Lu + Ru = 0 with R =− 1. With L1 defined as a two-fold indefinite integration operator, we have L−1Lu = L−1u so that u = C0 + C1 x + L−1u. (Since we are dealing with a linear ordinary differential equation, double decomposition is unnecessary.)
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- 1.R. E. Meyer, Introduction to Mathematical Fluid Dynamics, Wiley-Interscience (1971).Google Scholar