Answer Key to Short Answer Questions for
Fatigued: A Case on Blood
1. The values collected from a CBC can reveal a great deal of information about a
patients health. This information can be broken down into three broad categories,
which are listed below.
Solution Assignment 3
1
Exercise 1
The Reynolds number is computed simply by inserting the appropriate values into the given equation: a) 1.29 0.5 0.003 = 113.29 Re = 1.708 10-5 b) 1000 10 0.1 Re = = 570.451 1.753 10-3 c) 1.29 10 2.0 107 = 1.511 1013 Re =
Solution Assignment 4
1
Exercise 1
t 2 j-1 (un - 2uj + uj+1 ) n n 2 h
The Euler forward method is given by: u j = uj + n+1 n
Here we put the dirichlet boundary conditions into a part of the function f . Hence un+1 = un + rAun + fB with r = (t 2 )/h2 and f
Solution Assignment 5
1
Exercise 1
Prior to inserting the ansatz into the numerical scheme it is advantegous to simplify the equation: 1 1 un+1 = un + tAun + tA(un + tAun ) 2 2 First introducing variables u we can rewrite it for a typical point as: ~ 1 un
Solution Assignment 6
1
Exercise 1
The consitency order of the Richardson method is analysed by inserting the TaylorExpansions around the point considered u(x, t): u t + t u u(x, t - t) = u(x, t) - t + t u(x, t + t) = u(x, t) + 1 2u 2 1 3u 3 1 4u 4 t + t
Solution Assignment 7
1
Exercise 1
The consitency order of the Richardson method is analysed by inserting the TaylorExpansions around the point considered u(x, t) = un,j : un+1,j = un,j + un-1,j = un,j un,j-1 = un,j un,j+1 = un,j u 1 2u 2 1 3u 3 1 4u 4 t
Solution Assignment 8
1
Exercise 1
Prior to applying the Euler-Forward method, the stationary heat equation must be transformed into the instationary form: u 2 u 2 u - 2 - 2 = 0. t x y After spatial discretisation the equation is: 1 uj,l = 2 (4uj,l - uj-1
Solution Assignment 9
1
Exercise 1
To find the entries of the matrix K the following integral has to be computed for every Kij :
1
Kij =
0
i j (x) (x) dx x x
Hence for the monomials we have the following derivatives:
1 x 2 x 3 x 4 x
= 1 = 2x = 3x2 = 4x3 1
Solution Assignment 10
1
Exercise 1
The following script shows one possible way to solve the first three points of the exercise. function F = ass10_1 % This function solves the poisson equation with some right hand side and % computes the error in the l2n
Solution Assignment 11
1
Exercise 1
Only some functions must be added to the Matlab script provided in the internet. The collection of scripts becomes: File:fem2d.m function F = fem2d % Little examples, which implements a basic FEM program. This % script
Solution Assignment 12
1
Exercise 1
Ivn+1 - Ivn 1 1 = Iun + Iun+1 t 2 2 Iun+1 - Iun 1 1 = Kun + Kun+1 t 2 2
With the trapezoidal rule its possible to get the following system of coupled equations:
Sorting these terms gives the following global system of l
Institute of Scientific Computing Technical University Braunschweig Prof. Hermann G. Matthies, Ph. D. Dipl.-Inform. Oliver Kayser-Herold
Winter Term 2002/2003
Introduction to PDE and numerical methods: Test 1 (Time: 45 min.)
Problem 1: Consider the 2D sta
Institute of Scientific Computing Technical University Braunschweig Prof. Hermann G. Matthies, Ph. D. Dipl.-Inform. Oliver Kayser-Herold
Winter Term 2002/2003
Introduction to PDE and numerical methods: Test 2 (Time: 45 min.)
Problem 1: Consider the transp
Why Is this true?
An individual excitatory graded potential generated by an impulse from Neuron F will cause less than
1mV depolarization at the axon hillock of Neuron G.
For the following: neuron F releases neurotransmitter onto the dendrite of Neuron G.
Steroid binding lipid soluble, steroid and th bind to receptos inside cell
One side binds hormone and one to dna
In absence of dna binds to chaperone that stabilizes proten
Reeptors not active until hormone binds to them
Receptors changes chape and cheprs
Repro 1- male sex
Resprodcutve organs- gonads
Produce gametes- repro cells
2- testes- secrete anrogens
Ovaries- esterogen and progesterone
Both found in both sexes
Adenal cortex- also makes sex hormones
Accessory -Ovary and penis ,
Gonads- lubrication and
Solution Assignment 2
1
Exercise 1
-x2 (x, t) -x2 = (4 2 t)(-1/2) e 42 t + (4 2 t)(-1/2) e 42 t t t t
To show that (x, t) is a solution the partial derivatives are computed: (Prod.Rule)
Using two times the chain rule we obtain:
-x2 -x2 (x, t) 1 x2 = 4 2 -
Solution Assignment 1
1
Exercise 1
= 0, t = 1, x 0 - 0 = 0, = 0, t = cos(x), x q.e.d 2 = - 2 sin(x) x2 2 =0 x2
a) With (x, t) = x we get:
and hence: b) We have (x, t) = sin(x). The derivatives are:
and hence: 0 + 2 sin(x) = (x, t) (x, t) = 2 sin(x) c) As
Solution Test 2
1
Problem 1
a) The weak form can be easily obtained by multiplying the PDE with the weighting function and integrating over the computational domain: u d - c t u d - 2 x 2u d = 0 x2
To avoid second derivatives the last integral should unde
2.8: From the expression for x(t), x(0) = 0, x(2.00 s) = 5.60 m and x(4.00 s) = 20.8 m. a) 5.60 m 0 .8 m 2.80 m s b) 204.00 s 0 5.2 m s c) 20.8 m 5.60 m 7.6 m s 2.00 s 2.00 s
2.43: a) Using the method of Example 2.8, the time the ring is in the air is
t
v0 y
2 v0 y
2 g ( y y0 ) g (5.00 m s) 2(9.80 m s 2 )( 12.0 m) (9.80 m s 2 )
2
(5.00 m s)
2.156 s, keeping an extra significant figure. The average velocity is then 1
2.70: The position of the cars as functions of time (taking x1 = 0 at t = 0) are
x1 1 2 at , 2 x2 D v 0 t.
The cars collide when x1 = x2; setting the expressions equal yields a quadratic in t,
1 2 at 2 v0 t D 0,
the solutions to which are
t
1 a
2.51: a) From Eqs. (2.17) and (2.18), with v0=0 and x0=0,
vx
x
t 0
t 0
( At Bt 2 ) dt
A 2 t 2 B 3 t 3
A 2 B 2 t t (0.75 m s 3 )t 3 (0.040 m s 4 )t 3 2 3 A 3 B 4 dt t t (0.25 m s 3 )t 3 (0.010 m s 4 )t 4 . 6 12
b) For the velocity to be a maximum,
Institute of Scientic Computing
Technical University Braunschweig
Prof. Hermann G. Matthies, Ph. D.
Dipl.-Inform. Oliver Kayser-Herold
Winter Term 2002/2003
Assignment 1
Due date: 31.10.2002
Introduction to PDE and numerical methods:
Heat equation/Dierent
Institute of Scientic Computing
Technical University Braunschweig
Prof. Hermann G. Matthies, Ph. D.
Dipl.-Inform. Oliver Kayser-Herold
Winter Term 2002/2003
Assignment 2
Due date: 7.11.2001
Introduction to PDE and numerical methods:
Partial Dierential Equ
Institute of Scientific Computing Technical University Braunschweig Prof. Hermann G. Matthies, Ph. D. Dipl.-Inform. Oliver Kayser-Herold
Winter Term 2000/2001 Assignment 3 Due date: 8.11.2001
Introduction to PDE and numerical methods: Dimensionless Forms
Institute of Scientific Computing Technical University Braunschweig Prof. Hermann G. Matthies, Ph. D. Dipl.-Inform. Oliver Kayser-Herold
Winter Term 2002/2003 Assignment 4 Due date: 21.11.2000
Introduction to PDE and numerical methods: Finite Difference m
Institute of Scientific Computing Technical University Braunschweig Prof. Hermann G. Matthies, Ph. D. Dipl.-Inform. Oliver Kayser-Herold
Winter Term 2002/2003 Assignment 5 Due date: 28.11.2002
Introduction to PDE and numerical methods: Von Neumann Stabili