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pdes for electrostatics & heat flow 473This equation represents an implicit scheme for the temperature T i,j , where theword “implicit” means that we must solve simultaneous equations to obtain the fullsolution for all space. In contrast, an explicit scheme requires iteration to arrive at thesolution. It is possible to solve (17.82) simultaneously for all unknown temperatures(1 ≤ i ≤ N) at times j and j +1. We start with the initial temperature distributionthroughout all of space, the boundary conditions at the ends of the bar for all times,and the approximate values from the first derivative:T i, 0 , known, T 0,j , known, T N, j , known,T 0,j+1 = T 0,j =0, T N, j+1 =0, T N, j =0.We rearrange (17.82) so that we can use these known values of T to step thej =0solution forward in time by expressing (17.82) as a set of simultaneous linearequations (in matrix form):⎛( 2η +2) ⎞−1⎛ ⎞T 1,j+1( −1 2η +2) −1T ( −1 2η +2) 2,j+1−1T 3,j+1). .. . .. . ..( −1 2η +2) .−1⎜T n−2,j+1 ⎟⎜(⎝−1 2η +2) ⎟ ⎝ ⎠⎠T n−1,j+1⎛T 0,j+1 + T 0,j + ( 2η − 2) ⎞T 1,j + T 2,jT 1,j + ( 2η − 2) T 2,j + T 3,jT 2,j + ( 2η − 2) T 3,j + T 4,j=. (17.83).⎜ T n−3,j + ( 2η − 2) T n−2,j + T n−1,j⎟⎝T n−2,j + ( ⎠2η − 2) T n−1,j + T n,j + T n,j+1Observe that the T ’s on the RHS are all at the present time j for various positions,and at future time j +1 for the two ends (whose T s are known for all times viathe boundary conditions). We start the algorithm with the T i,j=0 values of theinitial conditions, then solve a matrix equation to obtain T i,j=1 . With that we know−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 473