ERROR APPROXIMATION

1. NUMERICAL METHODS
V. MUNEESWARI
ASST PROF OF MATHEMATICS,
SAC WOMEN’S COLLEGE,
CUMBUM.

2. AN OVERVIEW
 The Method of Finite Differences
 Error Approximations and Dangers
 Approxmations to Diffusions
 Crank Nicholson Scheme
 Stability Criterion

3. Finite Differences
 Approximating the derivative with a
difference quotient from the Taylor series
 Function of One Variable
 Choose mesh size Δx
 Then uj ~ u(jΔx)

4. First Derivative Approximations
 Backward difference: (uj – uj-1) / Δx
 Forward difference: (uj+1 – uj) / Δx
 Centered difference: (uj+1 – uj-1) / 2Δx

5. TAYLOR EXPANSION
 u(x + Δx) = u(x) + u΄(x)Δx + 1/2 u˝(x)(Δx)
+ 1/6 u˝΄(x)(Δx) + O(Δx)
 u(x – Δx) = u(x) – u΄(x)Δx + 1/2 u˝(x)(Δx)
- 1/6 u˝΄(x)(Δx) + O(Δx)
2
3
4
4
2
3

6. TAYLOR EXPANSION
u΄(x) = u(x) – u(x – Δx) + O(Δx)
Δx
u΄(x) = u(x + Δx) – u(x) + O(Δx)
Δx
u΄(x) = u(x + Δx) – u(x – Δx) + O(Δx)
2Δx
2

7. SECOND DERIVATIVE APPROXIMATION
 Centered difference: (uj+1 – 2uj + uj-1) / (Δx)
 Taylor Expansion
u˝(x) = u(x + Δx) – 2u(x) + u(x – Δx) + O(Δx)
(Δx)
2
2
2

8. FUNCTION OF TWO VARIABLES
u(jΔx, nΔt) ~ uj
 Backward difference for t and x
(jΔx, nΔt) ~ (uj – uj ) / Δt
(jΔx, nΔt) ~ (uj – uj ) / Δx
n
n n-1
n-1
n
∂u
∂t
∂u
∂x

9. FUNCTION OF TWO VARIABLES
 Forward difference for t and x
(jΔx, nΔt) ~ (uj – uj ) / Δt
(jΔx, nΔt) ~ (uj – uj ) / Δx
n+1
n+1 n
n
∂u
∂t
∂u
∂x

10. FUNCTION OF TWO VARIABLES
 Centered difference for t and x
(jΔx, nΔt) ~ (uj – uj ) / (2Δt)
(jΔx, nΔt) ~ (uj – uj ) / (2Δx)
n+1
n+1 n-1
n-1
∂u
∂t
∂u
∂x

11. FINITE DIFFERENCE EQN.
 uj
n+1 - uj
n = un
j+1 - 2uj
n + un
j-1
t x
( ) 2
Error: The local truncation error is O( t)
from the left hand side and is O( x)2 from
the right hand side.

12. Assumptions
 Assume that we choose a small change in
x, and that the denominator on both sides
of the equation are equal.
 We are now left with the scheme:
uj
n+1 = un
j+1 - un
j + un
j-1
 Solving u with this scheme is now easy to
do once we have the initial data.

13. Initial Data
 Let u(x,0) = h(x) = a step function with
the following properties:
h(x) = 0 for all j except for j = 5, so
hj = 0 0 0 0 1 0 0 0 0 0 0 ….
 Initially, only a certain section, which is
at j = 5 is equal to the value of 1.
 “j” serves as the counter for the x
values.

14. How to solve?
 We know u0
j = 1 at j = 5 and 0 at all other j
initially (given by superscript 0).
 We can plug into our scheme to solve for u1
j at
all j’s.
 u1
j = u0
j-1 - u0
j + u0
j+1
 u1
5 = -1; u1
4 = 1; u1
6 = 1
 Now we can continue to increase the # of
iterations, n, and create a table…

15. Solution for 4 iterations
4 1 -4 10 -16 19 -16 10 -4 1 0
3 0 1 -3 6 -7 6 -3 1 0 0
2 0 0 1 -2 3 -2 1 0 0 0
1 0 0 0 1 -1 1 0 0 0 0
0 0 0 0 0 1 0 0 0 0 0
1 2 3 4 5 6 7 8 9 10
j values
n
-
v
a
l
u
e
s

16. Analysis of Solution
 Is this solution viable?
 Maximum principle states that the solution must
be between 0 and 1 given our initial data
 At n = 4, our solution has already ballooned to u
= 19!
 Clearly, there are cases when the finite
difference method can pose serious problems.

17. Charting the Error
 Assume the solution is constant and equal to 0.5 (halfway between
the possible 0 and 1)

18. CONCLUSION
 While the finite difference method is easy
and convenient to use in many cases,
there are some dangers associated with
the method.
 We will investigate why the assumption
that allowed us to simplify the scheme
could have been a major contributor to
the large error.