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pdes for electrostatics & heat flow 441The x =0boundary condition U(x =0,y)=0can be met only if B =0. The x = Lboundary condition U(x = L, y)=0can be met only forkL = nπ, n =1, 2,... . (17.11)Such being the case, for each value of n there is the solution( nπ)X n (x)=A n sinL x . (17.12)For each value of k n that satisfies the x boundary conditions, Y (y) must satisfy they boundary condition U(x, 0)=0, which requires D = −C:( nπ)Y n (y)=C(e kny − e −kny ) ≡ 2C sinhL y . (17.13)Because we are solving linear equations, the principle of linear superposition holds,which means that the most general solution is the sum of the products:U(x, y)=∞∑n=1( nπ) ( nπ)E n sinL x sinhL y . (17.14)The E n values are arbitrary constants and are fixed by requiring the solution tosatisfy the remaining boundary condition at y = L, U(x, y = L)=100 V:∞∑E n sin nπ x sinh nπ = 100 V. (17.15)Ln=1We determine the constants E n by projection: Multiply both sides of the equationby sin mπ/Lx, with m an integer, and integrate from 0 to L:∞∑E n sinh nπn∫ L0dx sin nπ ∫mπ Lx sinL L x = dx 100 sin mπ x. (17.16)LThe integral on the LHS is nonzero only for n = m, which yields0{0, for n even,E n =4(100)nπ sinh nπ , for n odd. (17.17)Finally, we obtain the potential at any point (x, y) asU(x, y)=∞∑n=1,3,5,...400( nπx) sinh(nπy/L)nπ sin . (17.18)L sinh(nπ)−101COPYRIGHT 2008, PRINCET O N UNIVE R S I T Y P R E S SEVALUATION COPY ONLY. NOT FOR USE IN COURSES.ALLpup_06.04 — 2008/2/15 — Page 441