Assignment 1.pdf - Assignment 1 AMME2000/BMET2960...

Assignment 1
AMME2000/BMET2960
Engineering Analysis/Biomedical Engineering 2
SID: 480225063
Semester 1, 2019
Tutorial Time: Friday 9am - 11am
Tutorial Room: Chem Eng Lecture Room 2
1

Contents
1 Taylor Series Approximations 3
1.1 First Derivative of f ( x ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Taylor Expansion of Difference Stencils . . . . . . . . . . . . . . . . . . . . . 3
1.3 Plotting Difference Stencils Compared to the Analytic
Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.4 Euclidean Norm of Errors in Respect to Δ x . . . . . . . . . . . . . . . . . . 8
2 Heat Equation using Fourier Series 10
2.1 Heat Equation as t → ∞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.2 Deriving the Temporal Solution and the Overall Solution to T ( x, t ) . . . . . 10
2.3 Formulating the Final Solution of T ( x, t ) . . . . . . . . . . . . . . . . . . . 11
2.4 Plotting T ( x, t ) with t = 0 and t = 0 . 1 and Finding Time Within 10% of
the Equilibrium at x = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.5 Plotting Different and Finding Optimal a . . . . . . . . . . . . . . . . . . . 14
3 Appendix 16
3.1 Appendix A: Matlab Code for Subsection 1.3 . . . . . . . . . . . . . . . . . . 17
3.2 Appendix B: Matlab Code for Subsection 1.4 - Plotting the log-log plot of
L
2
-norm against Δ x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.3 Appendix C: Matlab Code for Subsection 2.4 - Plotting T ( x, 0) and T ( x, 0 . 1) 22
3.4 Appendix D: Matlab Code for Subsection 2.4 - Finding t when T (0 , t ) =
110 % of the steady state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
3.5 Appendix E: Matlab Code for Subsection 2.5 - Plotting T(x,0.2) with varying a 24
3.6 Appendix F: Matlab Code for Subsection 2.5 - Finding optimal a such that
temperature reaches 2 o C before t = 0 . 1 s . . . . . . . . . . . . . . . . . . . 25
2

1 Taylor Series Approximations
1.1 First Derivative of f ( x )
Since
d
dx
cos(2 x ) = - 2 sin(2 x )
and
d
dx
sin( x ) -
1
2
x
2
= cos( x ) - x
f ( x ) = cos(2 x ) (sin( x ) -
1
2
x
2
)
= ⇒ f
0
( x ) =
d
dx
cos(2 x ) sin( x ) -
1
2
x
2
+ cos(2 x )
d
dx
sin( x ) -
1
2
x
2
= - 2 sin(2 x ) sin( x ) -
1
2
x
2
+ cos(2 x )(cos( x ) - x )
= cos(2 x )(cos( x ) - x ) - 2 sin(2 x ) sin( x ) -
1
2
x
2
1.2 Taylor Expansion of Difference Stencils
For simplicity, let’s allow the notation f ( x
i + n
) ≡ f
i + n
such that n ∈ Z
Using Taylor Series expansion of the basic functions centered at x = x
i
up to the 5th
derivative
f
i +1
= f
i
+
df
dx
x = x
i
Δ x +
1
2!
d 2 f
dx 2
x = x
i
(Δ x )
2
+
1
3!
d 3 f
dx 3
x = x
i
(Δ x )
3
+ · · · +
1
5!
d 5 f
dx 5
x = x
i
(Δ x )
5
∵ ( x
i +1
- x
i
= Δ x )
Similarly,
f
i - 1
= f
i
-
df
dx
x = x
i
Δ x +
1
2!
d 2 f
dx
2
x = x
i
(Δ x )
2
-
1
3!
d 3 f
dx
3
x = x
i
(Δ x )
3
+ · · · -
1
5!
d 5 f
dx
5
x = x
i
(Δ x )
5
∵ ( x
i - 1
- x
i
= - Δ x )
f
i +2
= f
i
+
df
dx
x = x
i
2Δ x +
1
2!
d 2 f
dx
2
x = x
i
(2Δ x )
2
+
1
3!
d 3 f
dx
3
x = x
i
(2Δ x )
3
+ · · · +
1
5!
d 5 f
dx
5
x = x
i
(2Δ x )
5
∵ ( x
i +2
- x
i
= 2Δ x )
3

∴ f
i +2
= f
i
+ 2
df
dx
x = x
i
Δ x +
4
2!
d
2
f
dx 2
x = x
i
(Δ x )
2
+
8
3!
d
3
f
dx 3
x = x
i
(Δ x )
3
+ · · · +
32
5!
d
5
f
dx 5
x = x
i
(Δ x )
5
Similarly,
f
i - 2
= f
i
+
df
dx
x = x
i
( - 2Δ x )+
1
2!
d 2 f
dx
2
x = x
i
(2Δ x )
2
+
1
3!
d 3 f
dx
3
x = x
i
( - 2Δ x )
3
+ · · · +
1
5!
d 5 f
dx
5
x = x
i
( - 2Δ x )
5
∵ ( x
i - 2
- x
i
= - 2Δ x )
∴ f
i - 2
= f
i
- 2
df
dx
x = x
i
Δ x +
4
2!
d 2 f
dx
2
x = x
i
(Δ x )
2
-
8
3!
d 3 f
dx
3
x = x
i
(Δ x )
3
+ · · · -
32
5!
d 5 f
dx
5
x = x
i
(Δ x )
5
Computing the four point forward difference stencil 1
f
+
x
( x
i
) ≈
- f
i +2
+ 6 f
i +1
- 3 f
i
- 2 f
i - 1
6Δ x
=
6
df
dx
x = x
i
Δ x -
12
4!
d 4 f
dx
4
x = x
i
(Δ x ) 4 -
24
5!
d 5 f
dx
5
x = x
i
(Δ x ) 5
6Δ x
=
df
dx
x = x
i
-
(Δ x ) 3
12
d 4 f
dx
4
x = x
i
-
(Δ x ) 4
30
d 5 f
dx
5
x = x
i
(1)
Hence the leading order error for this stencil is denoted as
O ((Δ x )
3
) = -
(Δ x ) 3
12
d 4 f
dx
4
x = x
i
Computing the four point backward difference stencil 1
f
-
x
( x
i
) ≈
2 f
i +1
+ 3 f
i
- 6 f
i - 1
+ f
i - 2
6Δ x
=
6
df
dx
x = x