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QUESTION SEVEN Approximate the solution to the following partial differential equations using the backward – difference algorithm. ∂u ∂t − ∂ 2 u ∂t 2 =0 , 0<x <2 , 0 <t u ( 0 ,t) =u ( 0 ,t) =0 , 0 <t u ( x, 0 )=sin ( π) 2 x, 0 ≤x≤ 2 usem=4 ,T=0.1∧N=2 ¿ compare your results ¿ the actual solution u ( x,t )=e −( π 2 /4 ) t sin π 2 x SOLUTION z, w ij − w ij−1 z − w i−1 ,j−2 w ij+¿ w ij−1 z 2 = 0 ¿ let z h 2 =α w i,j − w i,j−1 −α ( w i +1 ,j−2 w i,j + w i,j−1 , j=0 ) ( 1 +2 α ) w i,j −αw i +1, j−αw i−1, j− wi,j−1=0 −αw i−1 , j+ ( 1+2 α ) w i,j−α w i +1 ,j− w i, j−1=0 ……… .1 −αw i−1 , j+ ( 1+2 α ) w i,j −αw i + 1= w i ,j−1 i=1 equation … … .1 −αw 0 j +( 1+2 α ) w ij −αw 2 ,j =w 1 j−1 i=2 −αw 1 j + ( 1+ 2 α ) w 2 j −αw 3 j=w 3 j−1 ………b i=3 − w 2 j ( 1+ 2 α ) w 3 j−αw 4 j=w 3 j−1 ……………c j=1 equation ( 1)becomes

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QUESTION SEVENApproximate the solution to the following partial differential equations using the backward difference algorithm.

SOLUTION

becomes

becomes

Therefore from 1 substituting all these valueEquation1

Equation2

Equation3

Equation4

Equation5

Equation5

Equation6

Hence equation

So the backward substitution ijxiwwijU(xi,w)

110.50.050.632950.65204

211.00.050.895130.88394

311.50. 050.632950.62504

120.50.10.566570.55249

221.00.10.801260.78134

321.50.10.566570.55249


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