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# C u0 t 0 u1 t u0 1 u0 2 u0 3

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Unformatted text preview: ) =....... = 0 I.C. u(x,0) = cos x ⇒ u(0, 0) = 0, u(1, 0) = 0.8090 = u(4, 0), u(2, 0) = 0.3090 = u(3, 0) The problem is symmetric w.r.t x = 0.5 (x, 0) = 0 ⟹ Now , = (2) , Put ting j = 0 in eqn. (1) , ⇒ , + , ( , For first time level , + , )= = ( ( + , , + , , )+ )+ , , + ( , + , )+ (3) , Putting i = 1, 2 in eqn. (3) ⇒ , ( , + , )= ( , + , )+ , , ( , + , )= ( , + , )+ , ⇒ , (0+ , ( & & ⇒ , = 0.5635 = 0.4027 ⇒ = 0.4098 & = , ( 0.8090+ 0.3090) + (0.3090) + , + )= , ⇒ , = 6.4432 , + 17 , = 0.5009 , , = = 0.4781 , = 0.4098, , + 0.3090) + (0.8090) , , , ⇒ ( , , )= , = , = 0.478, , =0 , For second time level Putting j = 1 in eqn. (1) ( + , , )+ , = ( , + , )+ , + ( , + , ) , putting i = 1, 2 ( + , )+ , = ( , + , )+ , + ( , + , ) , ( & , , + , )+ , = ( , + , )+ , + ( , + , ) , ⇒ , + & , + ⇒ , + & , + ⇒ , , , , = 0.5188, = 0.5188 = , For 3rd time level Putting j = 2 in eqn. (1) (0.3090) (0.4781+ 0.4098) + (0.4098) + = (0.8090) (0.8090 + 0.3090) = 0.0029 , (0.4098) + (0.4781) + = , = 0.5496 , , = 0.7283 , = 0.7283 = , , , =0= (0.3090) ( + , , )+ , = ( , + , )+ , + ( , + , ) putting i = 1, 2 ( + , )+ , = ( , + , )+ , + ( , + , ) , ( & , , + , )+ , = ( , + , )+ , + ( , + , ) , = 0.7722 = , , ⇒ , + & , + ⇒ , , = 0.7352 , , = 0.6650, = 0.6650 = , = 0.6583 , , = 0.7722 , , =0= ,...
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