Week - 2 - Explicit and Implicit Analysis
Explicit:
We lineralize y(t^(k+1))=y^(k+1) around previous time step t^(k)
`(y(`t^(k+1)`)`-y(`t^k`))/`Δt= f(`y^k`,`t^k`)+0Δt
Implicit :
Its backward Euler method in which it takes the function at t^(k+1) to get the value at t^k
`(y(`t^(k+1)`)`-y(`t^k`))/`Δt= f(`y^(k+1)`,`t^(k+1)`)+0Δt
F(u)= u3+9u2+4u
Logic: Assume `ΔF=1`
Step1 Uo=0
k(Uo)=4, ΔF= 1
ΔUo = ΔF/k(Uo)= 0.25
U1 = Uo +ΔU = 0+0.25= 0.25
U1 = 0.25 k(U1) = 8.6875
ΔU1 =0.115
U2= U1+ΔU1 = 0.365
|
Step i |
ΔF |
ΔUi |
Ui |
Fex |
Fint |
R= fex-fin |
|
1 |
1 |
0.25 |
0.25 |
1 |
1.578 |
0.578 |
|
2 |
1 |
0.115 |
0.365 |
2 |
2.707 |
0.707 |
Implicit
F(u)= u3+9u2+4u
Logic: Assume `ΔF=1`
Step1 Uo=0
k(Uo)=4, ΔF= 1
ΔU1 = ΔF/k(Uo)= 0.25
U1 = Uo +ΔU = 0+0.25= 0.25
Fex = ΔF =1 Fin= u3 + 9u2 +4u =1.578
R = Fex –Fin = 0.578
||R||> 10^-2, newton Raphson method
δu1= -[k(u1)]^-1* R
= -0.066-0.1348
= -0.0723
U1=0.1776
Fex= 1 fin=0.999 R= 0.001
||R||< 10^-2