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458 chapter 1717.13.2.1 SOLUTION VIA LINEAR EQUATIONSBecause the basis functions φ i in (17.42) are known, solving for U(x) involvesdetermining the coefficients α j , which are just the unknown values of U(x) on thenodes. We determine those values by substituting the expansions for U(x) andΦ(x) into the weak form of the PDE (17.41) and thereby convert them to a set ofsimultaneous linear equations (in the standard matrix form):Ay = b. (17.46)We substitute the expansion U(x) ≃ ∑ N−1j=0 α jφ j (x) into the weak form (17.41):∫ badx ddx⎛⎞N−1∑⎝ α j φ j (x) ⎠ dΦ ∫ bdx = dx4πρ(x)Φ(x).j=0aBy successively selecting Φ(x)=φ 0 ,φ 1 ,...,φ N−1 , we obtain N simultaneous linearequations for the unknown α j ’s:∫ badx ddx⎛⎞N−1∑⎝ α j φ j (x) ⎠ dφ ∫ bidx = dx 4πρ(x)φ i (x), i=0,...,N− 1. (17.47)j=0aWe factor out the unknown α j ’s and write out the equations explicitly:∫ b∫ b∫ bα 0 φ ′ 0φ ′ 0 dx + α 1 φ ′ 0φ ′ 1 dx + ···+ α N−1 φ ′ 0φ ′ N−1 dx =aaa∫ b∫ b∫ bα 0 φ ′ 1φ ′ 0 dx + α 1 φ ′ 1φ ′ 1 dx + ···+ α N−1 φ ′ 1φ ′ N−1 dx =∫ ba∫ baaaa. ..∫ b∫∫ bα 0 φ ′ N−1φ ′ 0 dx + α 1 ···+ α N−1 φ ′ N−1φ ′ N−1 dx =aaa∫ b4πρφ 0 dx,4πρφ 1 dx,4πρφ N−1 dx.Because we have chosen the φ i ’s to be the simple hat functions, the derivatives areeasy to evaluate analytically (otherwise they can be done numerically):dφ i,i+1dx=⎧0, x