Advanced Calculus fi..

708 Advanced Calculus, Fifth EditionOne could then solve for F2 = 1 and all other F, = 0, obtaining go,2; in generalu, = g,,, is the solution when F, = 1 and Fa = 0 for a # p. The linear combinationis then the desired solution of (10.155); for substitution of u, in the left-hand sideof the first equation (10.155) gives F,, since g,l gives 1 and all other g's give 0; asimilar reasoning holds for the other equations. Thus the effect of all the Fa can bebuilt up by superposition of unit forces.By a limit process a similar result is obtained for the problem:One finds(10.158)where the g(x, s) are the solutions for a force F "concentrated at a point s." Thefunction g(x, s) is called the GreenSfinction for (10.157); it is 0 when x = 0 andx = L, has the value s(L - s)/L when x = s, and is linear in x between these values.Accordingly, g(x, s) has a "corner" at x = s, due to the concentrated force at thispoint, whereas aZg/ax2 = 0 otherwise.Similar results hold for quite general nonhomogeneous linear equations. Inparticular, a Green's function g(x, y; r, s) can be found for the problem (Poissonequation):t . - p(10.159)v2u = -F(x, y) inside R,u(x, y) = 0 onboundaryof R,for a general region R in the plane. The solutions of (10.159) are then given by aformula:U(X, y) = 11 g(x, y;r, s)F(r. s) dr ds. (10.160)RFor each (r, s) the function g satisfies the equation v2g = 0 except at x = s, y = r,where it has a discontinuity corresponding to a "point load." Also g(x, y; r, s) = 0when (x, y) is on the boundary of R; therefore u is given as a "linear combination"of functions all 0 on the boundary of R and is therefore itself 0 on the boundary.(See Section 5.17.)In order to solve the characteristic value problemv2u +Au = 0 in R,u = 0 on the boundary of R,' 'one can rewrite the problem in form (10.159): V2u = -hu; thereforeufx, y) = ASJg(x, y; r, s)u(r, s)dr ds.,- r-. 8(10.161)