why am i facing problem while using two loops in summation?

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Question:why am i facing problem while using two loops in summation?

Posted: Khair Muhammad Saraz 35 Product: Maple

August 11 2024

i am writing code for an iterative process at the end i want to evaluate the summation expression with two loops but it is not evaluating kindly help me out here automatic_differentiation.mw

for i from 0 to N do x[i] := i*h end do

(1)

initial_condition := []; for i to N do initial_condition := [op(initial_condition), evalf(2*v*Pi*sin(Pi*x[i])/(a+cos(Pi*x[i])))] end do

u := proc (i) local u_x, u_xx, expr, j; u_x := (1/2)*(u[i+1]-u[i-1])/h; u_xx := (u[i-1]-2*u[i]+u[i+1])/h^2; expr := -alpha*u[i]*u_x+v*u_xx; expr end proc

odes := [seq(u(i, [seq(u[j], j = 1 .. N-1)]), i = 1 .. N-1)]

for i to N-1 do assign(o[i] = odes[i]) end do

>	$for i to N-1 do printf("u_%d = %s\n", i, convert(u(i), string)) end do$

u_1 = -alpha*u[1]*(20*u[2]-20*u[0])+1600*u[0]-3200*u[1]+1600*u[2]
u_2 = -alpha*u[2]*(20*u[3]-20*u[1])+1600*u[1]-3200*u[2]+1600*u[3]
u_3 = -alpha*u[3]*(20*u[4]-20*u[2])+1600*u[2]-3200*u[3]+1600*u[4]
u_4 = -alpha*u[4]*(20*u[5]-20*u[3])+1600*u[3]-3200*u[4]+1600*u[5]
u_5 = -alpha*u[5]*(20*u[6]-20*u[4])+1600*u[4]-3200*u[5]+1600*u[6]
u_6 = -alpha*u[6]*(20*u[7]-20*u[5])+1600*u[5]-3200*u[6]+1600*u[7]
u_7 = -alpha*u[7]*(20*u[8]-20*u[6])+1600*u[6]-3200*u[7]+1600*u[8]
u_8 = -alpha*u[8]*(20*u[9]-20*u[7])+1600*u[7]-3200*u[8]+1600*u[9]
u_9 = -alpha*u[9]*(20*u[10]-20*u[8])+1600*u[8]-3200*u[9]+1600*u[10]
u_10 = -alpha*u[10]*(20*u[11]-20*u[9])+1600*u[9]-3200*u[10]+1600*u[11]
u_11 = -alpha*u[11]*(20*u[12]-20*u[10])+1600*u[10]-3200*u[11]+1600*u[12]
u_12 = -alpha*u[12]*(20*u[13]-20*u[11])+1600*u[11]-3200*u[12]+1600*u[13]
u_13 = -alpha*u[13]*(20*u[14]-20*u[12])+1600*u[12]-3200*u[13]+1600*u[14]
u_14 = -alpha*u[14]*(20*u[15]-20*u[13])+1600*u[13]-3200*u[14]+1600*u[15]
u_15 = -alpha*u[15]*(20*u[16]-20*u[14])+1600*u[14]-3200*u[15]+1600*u[16]
u_16 = -alpha*u[16]*(20*u[17]-20*u[15])+1600*u[15]-3200*u[16]+1600*u[17]
u_17 = -alpha*u[17]*(20*u[18]-20*u[16])+1600*u[16]-3200*u[17]+1600*u[18]
u_18 = -alpha*u[18]*(20*u[19]-20*u[17])+1600*u[17]-3200*u[18]+1600*u[19]
u_19 = -alpha*u[19]*(20*u[20]-20*u[18])+1600*u[18]-3200*u[19]+1600*u[20]
u_20 = -alpha*u[20]*(20*u[21]-20*u[19])+1600*u[19]-3200*u[20]+1600*u[21]
u_21 = -alpha*u[21]*(20*u[22]-20*u[20])+1600*u[20]-3200*u[21]+1600*u[22]
u_22 = -alpha*u[22]*(20*u[23]-20*u[21])+1600*u[21]-3200*u[22]+1600*u[23]
u_23 = -alpha*u[23]*(20*u[24]-20*u[22])+1600*u[22]-3200*u[23]+1600*u[24]
u_24 = -alpha*u[24]*(20*u[25]-20*u[23])+1600*u[23]-3200*u[24]+1600*u[25]
u_25 = -alpha*u[25]*(20*u[26]-20*u[24])+1600*u[24]-3200*u[25]+1600*u[26]
u_26 = -alpha*u[26]*(20*u[27]-20*u[25])+1600*u[25]-3200*u[26]+1600*u[27]
u_27 = -alpha*u[27]*(20*u[28]-20*u[26])+1600*u[26]-3200*u[27]+1600*u[28]
u_28 = -alpha*u[28]*(20*u[29]-20*u[27])+1600*u[27]-3200*u[28]+1600*u[29]
u_29 = -alpha*u[29]*(20*u[30]-20*u[28])+1600*u[28]-3200*u[29]+1600*u[30]
u_30 = -alpha*u[30]*(20*u[31]-20*u[29])+1600*u[29]-3200*u[30]+1600*u[31]
u_31 = -alpha*u[31]*(20*u[32]-20*u[30])+1600*u[30]-3200*u[31]+1600*u[32]
u_32 = -alpha*u[32]*(20*u[33]-20*u[31])+1600*u[31]-3200*u[32]+1600*u[33]
u_33 = -alpha*u[33]*(20*u[34]-20*u[32])+1600*u[32]-3200*u[33]+1600*u[34]
u_34 = -alpha*u[34]*(20*u[35]-20*u[33])+1600*u[33]-3200*u[34]+1600*u[35]
u_35 = -alpha*u[35]*(20*u[36]-20*u[34])+1600*u[34]-3200*u[35]+1600*u[36]
u_36 = -alpha*u[36]*(20*u[37]-20*u[35])+1600*u[35]-3200*u[36]+1600*u[37]
u_37 = -alpha*u[37]*(20*u[38]-20*u[36])+1600*u[36]-3200*u[37]+1600*u[38]
u_38 = -alpha*u[38]*(20*u[39]-20*u[37])+1600*u[37]-3200*u[38]+1600*u[39]
u_39 = -alpha*u[39]*(20*u[40]-20*u[38])+1600*u[38]-3200*u[39]+1600*u[40]

(2)

for i from 2 to N-1 do T1[i] := (u[i+1]-u[i-1])/(2*h); T2[i] := u[i]*T1[i]; T3[i] := (u[i-1]-2*u[i]+u[i+1])/h^2; uN[i][1] := v*T3[i]-T2[i] end do

(3)

for i from 3 to N do for k to 4 do T1[i][k] := (uN[i+1][k]-uN[i-1][k])/(2*h); T2[i][k] := 0; for j from 0 to k do T2[i][k] := T1[i][k-j]*uN[i][j]+T2[i][k] end do; T3[i][k] := (uN[i-1][k]-2*uN[i][k]+uN[i+1][k])/h^2; uN[i][k+1] := (-T2[i][k]+T3[i][k])/(k+1) end do end do