Week - 2 - Explicit and Implicit Analysis

f(u)=u^3+9u^2+4u

k(u)=df/du=3u^2+18u+4

Explicit Analysis:∆F=k∆u, ∆F=1, 3 steps

step 1

u0=0⇒k(u0) = 4, ∆F = 1, ∆u1 = ∆F/k(u0) = 0.25, u1 = u0 + ∆u1 = 0.25

step 2

k(u1) = k(0.25)=8.69, ∆F = 1, ∆u2 = ∆F/k(u1) = 1/8.69, u2 = u1 + ∆u2 = 0.25+0.12=0.37

step 3

k(u2) = k(0.37)=11.07, ∆F = 1, ∆u3 = ∆F/k(u2) = 1/11.07, u3 = 0.37+0.09=0.46

Fext = ∆F + ∆F + ∆F = 3.0

fint = 0.46^3+9*0.46^2+4*0.46=3.84

step i ∆F ∆ui ui (Fext)i (fint)i fint − Fext = R

1 1 0.25 0.25 1 1.58 0.58

2 1 0.04 0.37 2 2.76 0.76

3 1 0.02 0.46 3 3.84 0.84

Implicit Analysis:

δuj+1 = −[k(uj)]−1R

step 1

u0=0⇒ k(u0) = 4,∆F = 1,∆u1 = ∆F/k(u0) = 0.25, u1 = u0 + ∆u1 = 0.25

check the residual, Fext=∆F = 1, fint=0.25^3+9*0.25^2+4*0.25=1.578125,R0=fint-Fext=0.578125

R0>10^-2⇒ Newton-Raphson iterations are necessary.

Calculate the correction to u1 = u1(0)

δu(1) = −[k(u(0)1)]−1R(0)=-(8.6875)^-1*0.578125=-0.06655

⇒ The updated value of u(1)1 = u1(0) + δu(1)=0.25-0.06655=0.18345

Check the residual again, Fext = 1, fint=0.18345^3+9*0.18345^2+4*0.18345=1.043

R(1) = fint − Fext=1.043-1=0.043

||R(1)||> 10^−2 ⇒ another Newton-Raphson iteration is necessary.

Calculate the new total correction to u1 = u1(0)

δu(2) = δu(1) − [k(u(1)1)]−1R(1)=-0.06655-(3*0.18345^2+18*0.18345+4)^-1*0.043

=-0.06655-(0.043/7.4031)=-0.06655-0.0058=-0.07235

The updated value of u(2)1 = u(0)1 + δu(2)=0.25-0.07235=0.17765

Check the residual, Fext = 1,fint=0.17765^3+9*0.17765^2+4*0.17765=1.000246

R(2) = fint − Fext=0.000246

||R(2)||< 10−2 ⇒ no further iterations are necessary.

Therefore, the final value is u1 = u(2)1 = 0.17765.

step 2

K(u1)=3*0.17765^2+18*0.17765+4=7.2924

∆F = 1, ∆u2 = ∆F/k(u1) =0.13713

u2 = u1 + ∆u2 =0.17765+0.13713=0.31478

Check the residual, Fext = ∆F = 1, fint=0.31478^3+9*0.31478^2+4*0.31478=2.1821

R(0) = fint − Fext =0.1821>10−2 ⇒ Newton-Raphson iterations are necessary.

Calculate the correction to u2 = u2(0)

δu(1) = −[k(u(0)2)]−1R(0)=−[k(0.31478)]−1R(0)=−0.1821/9.9633=-0.0183

⇒ The updated value of u2(1) = u(0)2 + δu(1) =0.31478-0.0183=0.29648

Check the residual again, Fext = 1,fint=0.29648^3+9*0.29648^2+4*0.29648=2.00312

R(2) = fint − Fext = 0.00312<10^-2⇒ no further iterations are necessary.

Therefore, the final value is u2 = u2(1) = 0.29648

step 3

k(u2)=3*0.29648^2+18*0.29648+4=9.6

∆F = 1, ∆u3 = ∆F/k(u2)=0.1042

u3 = u2 + ∆u3=0.29648+0.1042=0.4

Check the residual, Fext = ∆F = 1, fint=0.4^3+9*0.4^2+4*0.4=3.104

R(0) = fint − Fext =0.104>10−2 ⇒ Newton-Raphson iterations are necessary.

Calculate the correction to u3 = u3(0)

δu(1) = −[k(u(0)3)]−1R(0)=-[k(0.4)]−1R(0)=-0.104/11.68=-0.009

⇒ The updated value of u3(1) = u(0)3 + δu(1)=0.391

Check the residual again, Fext = 1,fint=0.391^3+9*0.391^2+4*0.391=2.99967

||R(3)|| = ||fint − Fext|| = 0.00033<10^-2⇒ no further iterations are necessary.

Therefore, the final value is u3 = u3(1) = 0.4

step i ∆F ui (Fext)i (fint)i fint − Fext = R

1 1 0.17765 1 1.000246 0.000246

2 1 0.296482 2 2.00312 0.00312

3 1 0.46 3 2.99967 0.00033

Aim of this challenge: write detailed calculation for this problem with Implicit and Explicit schemes, compare to each other, find advantages and disadvantages.

Conclusion: In this problem, it is evident from the plot that the explicit analysis drifts from the exact solution. Implicit method is more accurate compared to exact result. The reason is Newton-Raphson is used to make equilibrium between internal and external forces.

Plot: