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Unformatted text preview: APMA 2101 Supplementary Handout Introduction to Applied Mathematics Inhomogeneous ODE BVPs: Greens Functions and Eigenfunction Expansions 1 Greens Function for the Laplacian The Laplacian operator (often written in coordinate-invariant and dimension-invariant form as 2 ) appears as a term in the partial differential equations describing most of the conservation laws of continuum me- chanics, electrodynamics, and in many other PDE systems. The potential equation with sources is 2 u = f . This inhomogeneous problem is also known as Poissons equation, whereas the homogeneous problem is also known as Laplaces equation. Here f can be gravitational mass density, electrical charge density, or other types of sources in other areas of physics and engineering. N.B.: As it turns out, the Laplacian with a + sign has all negative eigenvalues (think of taking two derivatives of the sine function to land back at the negative of the original). In order to be able to deal with an operator with all positive eigenvalues, we will include a - sign in our definition, referring to- 2 as the Laplacian throughout the balance of these notes. Observe that the Greens function for- 2 is just the negative of the Greens function for 2 , by linearity. In one dimension, the Poisson problem reduces simply to- d 2 dx 2 u ( x ) = f ( x ) . (1) We examine this equation as a boundary value problem on the interval 0 x 1, with homogeneous boundary conditions on the value of u , namely u (0) = 0 and u (1) = 0. We note that the fundamental solutions of the Laplacian operator in one dimension are u 1 ( x ) = 1 and u 2 ( x ) = x , and we construct the general solution to the inhomogeneous problem by the method of variation of parameters (a simple exercise, since the Wronskian of the fundamental set is equal to 1) as u ( x ) = c 1 + c 2 x + Z x f ( ) ( - x ) d. If we evaluate the boundary conditions at x = 0 and x = 1, we see that u (0) = c 1 and u (1) = c 1 + c 2 1 + Z 1 f ( )( - 1) d, which means that c 1 = 0 and c 2 =- R 1 f ( )( - 1) d. (Note that c 2 is independent of x , but c 2 balances the forcing f across the entire interval.) Upon plugging c 1 and c 2 into the general solution for u ( x ), splitting the range of integration into the two subranges (0 , x ) and ( x, 1) and regrouping terms, we derive u ( x ) = Z x f ( ) (1- x ) d + Z 1 x f ( ) x (1- ) d. This can be rewritten in the compact and insight-producing form: u ( x ) = Z 1 f ( ) G ( x, ) d, (2) where we have introduced the Greens function for the one-dimensional Laplacian operator,- d 2 dx 2 : G ( x, ) x (1- ) , x (1- x ) , x . (3) 1 Observe that we have, in effect, inverted the differential operator- d 2 dx 2 ! We have written the solution u ( x ) as a linear functional of the forcing term f ( x ). The inverse operator to the differential operator in (1) is the integral operator in (2). The range of integration spans the domain of definition of the boundary valuethe integral operator in (2)....
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This note was uploaded on 11/02/2009 for the course APMA 2102 taught by Professor Keyes during the Spring '08 term at Columbia.
- Spring '08