Solving Adams-Bashforth Integration for ODE

Running head: MEC3456 LAB 07
MEC3456 LAB 07
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MEC3456 LAB 072
Question 1:
The Adams-Bashforth integration for finding the solution of the ODE dy/dt = f(t,y) are given
by the following integral update formula
yn+ 1= yn + ∫
ti
ti+1
f ( t , y ) dt (1)
Where, yn+ 1 = y(tn+1 ¿
The method can be derived using polynomial interpretation as given below.
From equation (1) using the fundamental theorem of calculus we get
y(tn+1 ¿ = y(tn ¿ + ∫
tn
tn+1
y' ( t ) dt
Here, let A = ∫
tn
t n+1
y' ( t ) dt=∫
tn
t n+ 1
f (t , y ( t ) )dt
Now, for calculating the value of A the value of the interpolating polynomial P(t) as an
approximation f(t,y(t)) is used. Hence, the value of the interpolating polynomial in the
Lagrange form is
L(x) := ∑
j=0
k
yjlj(x ), where the expression of lj ( x )= ∏0 ≤m ≤ k∧m ≠ j
x −xm
xj−xm
Hence, the interpolation formula will be
P(t) = f ( tn , yn )∗t−tn−1
tn−tn−1
+f ( tn−1 , yn−1 )∗t−c
tn−1−tn
Hence, A = ∫
tn
tn+1
f ( t , y ( t ) ) dt ∫
tn
tn+1 f ( tn , yn ) ( t−tn−1 )
tn −tn−1
+ f ( tn−1 , yn−1 )∗t−tn
tn−1−tn
dt
Now, after integration and simplification gives,

MEC3456 LAB 073
(½)(tn+ tn +1 ¿ ( f ( tn , yn ) −f ( tn−1 , yn−1 ) )−tn−1 f ( tn , yn ) +tn f (tn−1 , yn−1)
= (½)(t n+ tn +1−2 tn−1 ¿ f ( tn , yn) + (½)(2t n−t n+1−t n ¿ f (tn−1 , yn−1 )
Now, tn−1 , c and tn+1 are equally spaced hence it can be considered that
tn−¿ tn−1 = tn+1−¿ tn = h. Hence, the value of A will be
A = (3/2)h*f(tn , yn ¿ – (½)h*f(tn−1 , yn−1 ¿
Hence, by Adams-Bashforth method the iteration equation will be this
y(tn+1 ¿ = y(tn ¿+ ( 3 h
2 )f ( tn , y n ) −( h
2 )f ( tn−1 , yn−1 )
Question 2:
2a:
The given second order differential equation is
( d
dx ) ( x3∗du
dx ) + λxu=0 (1)
The second order equation is defined inside the interval [1, 2] having boundary conditions
u(1) = 0 and u(2) = 0.
Hence, the above equation can be rewritten as
3 x2∗( du
dx )+ x3∗d2 u
d x2 + λx u= 0 (2)
Now, approximating (2) by finite difference method at an arbitrary point xi
3 xi
2∗ui+1 – ui
h + xi
3∗ui+1 – 2∗ui+ui−1
h2 + λ xi ui =0
Where, ui = u(xi) and holds true for all i.

MEC3456 LAB 074
Q2b:
Now, by relaxation method the equation is first converted to two unknowns of equation
du
dx = y
x^3* d2 u
d x2 = - λxy− λu −3 x2∗y
dy/dx ¿ ¿- y− λu −3 x2∗y)/ x^3
In this method at first discretization by the method of finite differences is performed with the
mesh spacing of h= 0.25 for approximating the eigenvalue problem. The given boundary
conditions are u(1) = 0, u(2) = 0. An initial guess of λ = 0 is assumed and the solution is
improved iteratively. The first order differential equations are approximated by trapezoidal
rule here.
Q2c:
Now, the second order differential equation is converted to two first order equations.
du
dx = y
x^3* d2 u
d x2 = - λxy− λu −3 x2∗y
d2 u
d x2 =¿- λxy− λu −3 x2∗y)/ x^3
Now, u(2) = 0 and u(1) = 0 can be initialized to solve the system numerically by evaluating
the next iteration u(3). Here, the step size need to be h= 1.