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Computerised Mathematical Methods in Numerical Differentiation
Forward Difference Approximation
Engineering (HG3MCE)
Nonlinear functions
𝑓 ′ (𝑥) ≈
𝑓(𝑥+∆𝑥)−𝑓(𝑥)
∆𝑥
∆𝑥→0
Bisection method
Half the interval between 2 points successively to find
root.
𝑐=
𝑎+𝑏
2
Centred Difference Approximation
, if f(c)>0, a=c; if f(c)<0, b=c; iterate.
𝑥 = √𝑎 ⟹ 𝑓(𝑥) = 𝑥 2 − 𝑎
𝑓(𝑥𝑛 )
𝑓′ (𝑥𝑛 )
𝑥=
1
𝑎
⟹ 𝑓(𝑥) = 𝑎 −
𝑥𝑛 = 𝑥𝑛−1 −
h = Δx
h
h/2
h/4
…
𝑏−𝑎
3
ℎ
1
2
2𝑛−1 −1
𝑁𝑛 (ℎ) = 𝑁𝑛−1 ( ) +
Δx=h
h/2n
h/2n
…
f’n(x)
a
b
…
N1(h)=f’(x)
#
#
#
…
N2(h)
#
#
…
…
#
…
ℎ
[𝑁𝑛−1 ( ) − 𝑁𝑛−1 (ℎ)]
2
N1(h) = f’(x) (using a difference approximation)
Numerical Integration (Quadrature)
Uses parabolas (quadratic
functions) using 3 points that use
the x-axis intercept as the next
approximation
𝑏±√𝑏2 −4𝑎𝑐
ℎ2
Richardson’s Extrapolation
𝑥𝑛−1 −𝑥𝑛−2
2𝑐
𝑓(𝑥+ℎ)−2𝑓(𝑥)+𝑓(𝑥−ℎ)
𝑓∗ = 𝑏 +
𝑓(𝑥𝑛−1 )−𝑓(𝑥𝑛−2 )
Muller’s method
𝑓 ′ (𝑥) ≈
n
0
1
…
2ℎ
1
𝑥
Secant method
Uses straight lines dictated by 2 points on the graph that
close in on the root
𝑥𝑛 = 𝑥𝑛−1 − 𝑓(𝑥𝑛−1 )
𝑓(𝑥+ℎ)−𝑓(𝑥−ℎ)
Extrapolation Techniques σ(h2)
Find better estimate by using 2 approximations (a & b)
and cancel errors.
Newton Raphson method
Differentiate function, 𝑥𝑛+1 = 𝑥𝑛 −
𝑓 ′ (𝑥) ≈
𝑏
𝐼 = ∫𝑎 𝑓(𝑥) 𝑑𝑥
Fixed Point Theorem
Recast function into fixed point form, g(x):
Trapezoidal Rule (1 subinterval)
f(x)=0  x=…=g(x)  |g’(x)|max <1  only one root
Composite Trapezoid Rule
Divide h into n smaller subintervals of width h/n:
𝑏
𝐼 ≈ ∫𝑎 𝑃1 (𝑥) 𝑑𝑥 =
If xn+1 iteration converges, |g’(root)|<1
𝑓(𝑎)+𝑓(𝑏)
2
ℎ where 𝒉 =
𝒃−𝒂
𝑵
ℎ
𝐼 ≈ [𝑓(𝑎0 ) + 2𝑓(𝑎1 ) + 2𝑓(𝑎3 ) + ⋯ + 2𝑓(𝑎𝑛−1 ) + 𝑓(𝑏)]
Interpolation
2
Straight lines: f(x0) = mx0 + d, 𝑚 =
𝑓(𝑥1 )−𝑓(𝑥0 )
Simpsons Rule (2 subintervals)
Interpolate with a quadratic polynomial:
𝑥1 −𝑥0
Polynomials: f(xn) = axi3 + bxi2 + cxi + d
Lagrange Polynomials
𝑃(𝑥) = 𝑃0 (𝑥) + 𝑃1 (𝑥) + ⋯ + 𝑃𝑛 (𝑥)
i
xi
f(xi)
x0
0
1
2nd order, 𝑃0 (𝑥) = (𝑥
nth
0 −𝑥1 ) (𝑥0 −𝑥2 )
f(x0)
(𝑥−𝑥𝑛−1 )
𝑛 −𝑥0 ) (𝑥𝑛 −𝑥1
𝑛 −𝑥𝑛−1
order, 𝑃𝑛 (𝑥) = (𝑥
… (𝑥
)
f(x1)
ℎ𝑛
3
𝑓(𝑥𝑛 )
)
𝑎1 = 𝑓[𝑥0 , 𝑥1 ] =
0
a0
a1
a2
…
f[xi, xi+1,…, xn]
an
𝑎2 = 𝑓[𝑥0 , 𝑥1 , 𝑥2 ] =
𝑓[𝑥1 ,𝑥2 ]−𝑎1
𝑎𝑛 = 𝑓[𝑥𝑖 , … , 𝑥𝑛 ] =
𝑓[𝑥𝑖+1 ,…,𝑥𝑛 ]−𝑓[𝑥1 ,…,𝑥𝑛−1 ]
Pn(xi) = f(xi)
𝑥2 −𝑥0
𝑥𝑛 −𝑥𝑖
[𝑓(𝑎0 ) + 4𝑓(𝑎1 ) +
[(𝑓0 + 𝑓2𝑛 ) + 4𝑜𝑑𝑑𝑠 + 2𝑒𝑣𝑒𝑛𝑠]
𝐼≈
i
xi
f[xi]
f[xi, xi+1]
f[xi, xi+1, xi+2]
…
𝑥1 −𝑥0
3
3/8 Simpsons Rule using Newton-Cotes formula for N=3 (3
subintervals)
For nth order polynomial: 𝑃𝑛 (𝑥) = 𝑎0 + 𝑎1 (𝑥 − 𝑥0 ) +
𝑎2 (𝑥 − 𝑥0 )(𝑥 − 𝑥1 ) + ⋯ + [𝒂𝒏 (𝒙 − 𝒙𝟎 ) … (𝒙 − 𝒙𝒏−𝟏 )]
𝑓(𝑥1 )−𝑓(𝑥0 )
ℎ𝑛
2𝑓(𝑎2 ) + 4𝑓(𝑎3 ) + 2𝑓(𝑎4 ) + ⋯ + 4𝑓(𝑎2𝑛−1 ) + 𝑓(𝑏)] =
Divided Difference method
nth order polynomial from (n+1) points.
𝑎0 = 𝑓[𝑥0 ] = 𝑓(𝑥0 )
3
Divide b-a into 2n subintervals: 𝐼 ≈
𝑓(𝑥0 )
(𝑥−𝑥0 ) (𝑥−𝑥1 )
ℎ
Composite Simpsons Rule
x1
(𝑥−𝑥1 ) (𝑥−𝑥2 )
𝑏
𝐼 ≈ ∫𝑎 𝑃2 (𝑥) 𝑑𝑥 = [𝑓(𝑎) + 4𝑓(𝑎1 ) + 𝑓(𝑏)]
1
-
3ℎ
8
[𝑓(𝑎) + 3𝑓(𝑎1 ) + 3𝑓(𝑎2 ) + 𝑓(𝑏)]
f(x)
P(x)
Extrapolation for σ(h4)
𝐼 ∗ = 𝐼2𝑛 + 𝜀 = 𝐼2𝑛 +
2
-
…
…
…
…
…
n
-
𝐼2𝑛 −𝐼𝑛
15
h
a0=a
Richardson’s Extrapolation
ℎ
1
2
4 𝑛−1 −1
𝑁𝑛 (ℎ) = 𝑁𝑛−1 ( ) +
N=2
subintervals
h
a1
an=b
ℎ
[𝑁𝑛−1 ( ) − 𝑁𝑛−1 (ℎ)]
2
N1(h) corresponds to composite trapezoid rule, N2(h)
corresponds to Simpsons Rule, N2(h/2) corresponds to
composite Simpsons Rule
N
2
4
8
…
N1(h)
#
#
#
…
N2(h)
#
#
…
…
#
…
Linear Algebra
Numerical methods for ODEs
Gaussian Elimination for solving a linear system of equations
Ax=b where “A” is a nxn matrix of the left hand side of
equations, “b” is a vector of the right hand side, and the
𝑥
vector “x”=(𝑦). Place “A” and “b” alongside each other in
𝑧
a table and manipulate each row until “A” becomes an
upper triangle matrix.
Can write nth order ODE as n first order ODEs
Forward Euler Method
Initial condition: y(x0) = y0
yn+1 = yn + hf(xn, yn)
xn = x0 + nh
n
0
1
2
Modified Euler Method
1
𝑦𝑛+1 = 𝑦𝑛 + (𝑘1 + 𝑘2 )
xn
x0
x0+h
x0+2h
yn
y0
y1
y2
k1
-
k2
-
2
This method won’t work when the |pivot element| <<
|any other element of that row|.
𝑘1 = ℎ𝑓(𝑥𝑛 , 𝑦𝑛 )
With Scaled Partial Pivoting
Matrix, 𝐴 = 𝑎[𝑟𝑜𝑤 𝑛𝑢𝑚𝑏𝑒𝑟],[𝑐𝑜𝑙𝑢𝑚𝑛 𝑛𝑢𝑚𝑏𝑒𝑟] = 𝑎𝑖,𝑗
𝑘2 = ℎ𝑓(𝑥𝑛+1 , 𝑦𝑛+1 ) = ℎ𝑓(𝑥𝑛+1 , 𝑦𝑛 + 𝑘1 )
Index vector, k = (1, 2, 3, 4)
2nd order
𝑦𝑛+1 = 𝑦𝑛 + 𝑎𝑘1 + 𝑏𝑘2
Runge-Kutta Methods
Largest elements of each row, s = (-, -, -, -)
𝑘1 = ℎ𝑓(𝑥𝑛 , 𝑦𝑛 )
Index for iterations, m = 1 (for first step)
𝑘2 = ℎ𝑓(𝑥𝑛 + 𝛼ℎ, 𝑦𝑛 + 𝛽𝑘1 )
Pivot row,
𝑗=
Mod Euler when a=b=0.5, α=β=1; Midpoint method
when a=0, b=1, α=β=0.5
|𝑎𝑘𝑖 , 𝑚 |
∶ 𝑖 = 1, 2, 3, 4
𝑠𝑘𝑖
Pick kj = first index j with largest number and then swap
this with km in index vector. The number in km = pivoting
row (for all manipulations to be based upon in this turn).
k (=k1+k2) is a weighted sum of 1 (a+b=1).
4th order
1
𝑦𝑛+1 = 𝑦𝑛 + (𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4 )
6
Matrix Inversion
1 0
Inverted matrix, A-1  𝐴𝐴−1 = 𝐼 = [0 1
0 0
0
0]
1
𝑘1 = ℎ𝑓(𝑥𝑛 , 𝑦𝑛 )
Determinants
The det of an upper triangular matrix is simply the
product of the diagonal elements.
Iterative Methods
Matrix needs to be diagonally dominant for methods to
converge i.e. |𝑎𝑖,𝑗 | > ∑𝑗:𝑗≠𝑖|𝑎𝑖,𝑗 |
Gauss-Jacobi Method
Ax=b  (D+R)x=b  Dx=b-Rx where D is the diagonal
of A and R is the remainder.
=
D-1(b
–
Rx(n))
n
x1(n+1)
(n+1)
x2
x3(n+1)
…
-1 (start value)
-
Gauss-Seidel Method
Improves the Gauss-Jacobi by using the already
computed values of x(n+1) for the (n+1) stage.
𝑘1
2
2
ℎ
𝑘2
2
2
𝑘3 = ℎ𝑓 (𝑥𝑛 + , 𝑦𝑛 +
Can only use: New row = Old row + Multiple of another
row
x(n+1)
ℎ
𝑘2 = ℎ𝑓 (𝑥𝑛 + , 𝑦𝑛 +
0
-
1
-
Ill-conditioned Matrices

When |largest element in A| x |largest element in
A-1| >> size of matrix, n.

Eigenvalues differ by 2 orders of magnitude or
more.
)
)
𝑘4 = ℎ𝑓(𝑥𝑛 + ℎ, 𝑦𝑛 + 𝑘3 )
k (inside brackets) is a weighted sum of 6.
Multistep Methods
Use (k+1) past values of y to interpolate a polynomial,
and then evaluate yn+1 from integration. y’(x)=f(x,y) 
𝑥
𝑥
y=f(x,y)dx 𝑦𝑛+1 = ∫𝑥 𝑛+1 𝑦𝑛 = ∫𝑥 𝑛+1 𝑓(𝑥, 𝑦)𝑑𝑥
𝑛
Adams-Bashford (predictor)
3rd point 2nd order polynomial:
𝑦
̅̅̅̅̅̅
𝑛+1 = 𝑦𝑛 +
ℎ
12
𝑛
x
f(x)
(23𝑓𝑛 − 16𝑓𝑛−1 + 5𝑓𝑛−2 )
̅̅̅̅̅̅
𝑓𝑛+1 = 𝑓(𝑥𝑛+1 , ̅̅̅̅̅̅)
𝑦𝑛+1
Adams-Moulton (corrector)
ℎ
̅̅̅̅̅̅
𝑦𝑛+1 = 𝑦𝑛 + (9𝑓
𝑛+1 + 19𝑓𝑛 − 5𝑓𝑛−1 + 𝑓𝑛−2 )
24
-2h
f-2
-h
f-1
0
f0
PDEs
For unknown function, c(x, y), 𝐴
𝑓(
𝜕𝑐 𝜕𝑐
,
𝜕𝑥 𝜕𝑦
𝜕2 𝑐
𝜕𝑥 2
+𝐵
𝜕2 𝑐
𝜕𝑥𝜕𝑦
+𝐶
𝜕2 𝑐
𝜕𝑦 2
+
No-flux boundary condition:
, 𝑥, 𝑦, 𝑐) = 0
𝜕𝑐
B2-4AC <0 : elliptic
=0 : parabolic
>0 : hyperbolic
Ellipse
Parabola Hyperbola

Dirichlet boundary condition: φ(x,y) on S

Neumann boundary condition:

Mixed boundary condition: ϕ + ∇ϕ = 0
𝜕𝜙
𝜕𝑥
𝜕𝑥 2
+
𝜕2 𝜙
𝜕𝑦 2
=0
Compare with general formula, A=1, B=0, C=1 so
satisfies B2-4AC<0
Let φi,j = φ(xi, yj) where xi=hi and yj=kj (spacing of a 2D
grid) then use centred difference approximation to get:
𝜙𝑖,𝑗 =
𝑘2
2ℎ2 +2𝑘 2



(𝜙𝑖−1,𝑗 + 𝜙𝑖+1,𝑗 ) +
ℎ2
2ℎ2 +2𝑘 2
(𝜙𝑖,𝑗−1 + 𝜙𝑖,𝑗+1 )
Boundary conditions give the values of φ(x,0)
φ(0,y) φ(x,M) φ(N,y) so only internal points
need to be solved
This problem is symmetrical around y=x, so
φi,j=φj,i (i.e. only 1 half of the grid needs to be
solved)
Linear system of equations for unknown points
can be solved using Gaussian elimination
Parabolic PDEs
Heat equation in 1 dimension:
𝜕𝑐
𝜕𝑡
= ∇2 c =
𝜕2 𝑐
𝜕𝑥 2
 A=1, B=C=0
Use forward difference approximation on time derivative,
centred difference on spatial derivative:
𝑐𝑖,𝑗+1 − 𝑐𝑖,𝑗 𝑐𝑖+1,𝑗 − 2𝑐𝑖,𝑗 + 𝑐𝑖−1,𝑗
=
𝑘
ℎ2
⟹ 𝑐𝑖,𝑗+1 = 𝑟𝑐𝑖+1,𝑗 + (1 − 2𝑟)𝑐𝑖,𝑗 + 𝑟𝑐𝑖−1,𝑗
Where ci,j = c(x,t) at spatial position xi and time tj (=jk)
and 𝑟 =
≈
𝑐𝑁+1,𝑗 −𝑐𝑁−1,𝑗
2ℎ
𝜕𝑐
𝜕𝑥
= 0 at x=hN
= 0 ⟹ 𝑐𝑁+1,𝑗 = 𝑐𝑁−1,𝑗
This means that all ci,j are determined by values inside
the boundaries
Initial condition [e.g. c(x,0)=sin(xπ)] must satisfy
boundary conditions [e.g. c(0,t)=0]
= ∇ϕ = 0
Elliptic PDE
Describes steady state behaviour, e.g.2D Laplace’s:
𝜕2 𝜙
|
𝜕𝑥 x=hN
Data on boundary, S = boundary condition
∇2 𝜙 =
This suggests the solution of next time point (j+1) can
be found given all values of i for time point j are known.
cj = Acj-1 where state vector at time jk,
cj=[c1,j, c2,j, …, cn,j]t and A is nxn matrix:
𝑨=[
1 − 2𝑟
⋮
𝑟
⋯
𝑟
⋱
⋮ ] but only valid for r≤0.5
⋯ 1 − 2𝑟
Crank-Nicholson Method
Acj+1 = Bcj + bj where bj = boundary data vector, 𝑨 =
2 + 2𝑟 ⋯
−𝑟
2 − 2𝑟 ⋯
𝑟
[ ⋮
⋱
⋮ ] and 𝑩 = [ ⋮
⋱
⋮ ]
−𝑟
⋯ 2 + 2𝑟
𝑟
⋯ 2 − 2𝑟
Hyperbolic PDEs
Wave equation:
𝜕2 ϕ
𝜕t2
− 𝑐2
𝜕2 ϕ
𝜕x2
= 0  A=1, B=0, C<0
φ=(φ,v) where u=x+ct and v=x-ct
Approximate φi,j using centred difference approximation
at time jk and position ih:
𝜙𝑖,𝑗+1 − 2𝜙𝑖,𝑗 + 𝜙𝑖,𝑗−1
𝜙𝑖+1,𝑗 − 2𝜙𝑖,𝑗 + 𝜙𝑖−1,𝑗
= 𝑐2
𝑘2
ℎ2
2
⟹ 𝜙𝑖,𝑗+1 = 2𝜙𝑖,𝑗 − 𝜙𝑖,𝑗−1 + 𝑟 (𝜙𝑖+1,𝑗 − 2𝜙𝑖,𝑗 + 𝜙𝑖−1,𝑗 )
where 𝑟 =
𝑘𝑐
ℎ
For t=k (j=1) use Taylor expansion:
𝜙𝑖,1 ≈ 𝜙(𝑖ℎ, 𝑘) = 𝜙(𝑖ℎ, 0) + 𝑘
𝜕𝜙(𝑖ℎ,0)
𝜕𝑡
+
k2 𝜕2 ϕ(ih,0)
2
𝜕t2
Where the first 2 terms are found from initial conditions
and 3rd term from wave eq
Valid for kc≤h
S
φ(x,M)
φN,M
𝑘
ℎ2
φi,j+1
φi-1,j
φ(0,y)
φi,j
h
k
φi+1,j
h
k
φi,j-1
φ0,0
φ(x,0)
φ(N,y)