Math 405 ALL Announcements Spring
2004
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January |
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Syllabus,
HomeWork,
Text Web Site,
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- Last
3/11/2004
5pm This Week. Mar 9 & 11 .
"If you do not know where
you are going, it does not matter which way you go. "
- Lewis Carroll.
1/13/04
Tuesday Welcome to Math 405 Spring 2004.
- Organize study groups (one student should send me a
message with copies to each student),
- each student should send me a resume with contact
info.
- email to me regarding this class must have a subject in
the form: 405: Topic, sender(s).
- Example1: 405: Resume, Vollmar
- Example2: 405: Minutes, Anderson, Riley,
Vollmar. N.B.: Minutes of Study groups must be cc'ed to all
members
- Study groups
- Before Class:: Anderson, Murphy, Pierson
- Before Class:: Arnett, Kuster, Linder, Weisner
- After Class:::::: Boyd, Feese, Johnson
- After Class:::::: Campbell, Lorenz, Venard
- Tues 2pm :::::: Bullock, Walhite, Walker
- Thur 1pm::::::::: Harcourt, Passamaneck, Raper
- TR 2-4pm :::::: Akers, Riley, Vollmar
- Fri 11am ::::::::: Blakemore, Blom, Fink, Rechtin
- Maple intro
- Issues, resolution, communication
- Chapter 1
1/15 2.1, 2.2 Thursday
- Homework Due Tuesday 1/27
- 1.1 1, 5, 11, 15, Maple 19
- 1.2 1a, 5-13
- 1.3 alt odd, 19
- 2.1 1, 13, 21, 23, 27, 29, 35, 37
- 2.2 1, 9, 21, 27, 31
- 2.3 3, 7, 14, 19, 24
- 2.4 1, 7, 13, 17, 23-26, 27, 28, 33
- More Chapter 2 Homework.
1/20/2004
Tuesday
- Extended discussion of 1.2 #7 and 2.1 #29
Projects:
brief preliminary presentation(1/22 2-3minutes),
presentation of results(1/29, 5min), written report (2/5)
- I recommend that the projects be worked and written using
MAPLE, but you may use any tools of your choice.
- The written projects should be professional technical
reports. The oral presentations and written reprorts should focus
on statement of the problem(s), issues, methods, models, results,
interpretations, conclusions. Technical details associated with
model solution and analysis should receive brief mention in the oral
presentation and be relegated to a detailed technical appendix in the
written report.
- Project assignment by Study group
- Before Class:: Anderson, Murphy, Pierson...........
1-5
Mixture
- Before Class:: Arnett, Linder, Weisner.................
6, 7, 8 Investment
- After Class:::::: Boyd, Feese,
Johnson................... 9,10,11,12 Loan Payoffs
- After Class:::::: Campbell, Lorenz,
Venard............ 8, 12, 13 Retirement
- Tues 2pm :::::: Bullock, Walhite,
Walker.............. 15,16,17 Population
- Thur 1pm::::::::: Harcourt, Passamaneck,
Raper..... 26,29,30 Vertical motion
- TR 2-4pm :::::: Akers, Riley,
Vollmar....................
31,32 Baseballs
- ????
::::::::::::::::::
Missing......................................... 21
-23 More Balls in motion
- ..
- Unassigned but available for
credit....................... 27,
28 Millikan's oil drops
- Unassigned but available for
credit.......................
24
Skydive, also check the following link: http://aci.mta.ca/TheUmbrella/Physics/P3401/Investigations/FreefallJH.html
- Others unassigned problems for individual credit: 14,
18, 19, 20, 33
1/22/2004
Thursday
- 2.3 Models: Mixture/Diffusion Models
1/27/2004 Tuesday
- 2.4 Existence and Uniqueness in Linear vs. Nonlinear
equations,
1/29/2004
Thursday
- 2.5 Autonomous Equations
- graphing and interpretation of y ' = f(y)
- roots of f(y)=0 correspond to constant (equilibrium
solutions
- solution may be stable or unstable
- sketch motion of y on y axis as indicated by sign of y'
- convacity
- Population Models, discussion of rational for variable
birth rates due to
2/3/2004 Tuesday
- Homework Assignment Previous
Assignment for Chapter 2,
- 2.5 odd 3-17
- 2.6 alt odd
- 2.7 Use Maple dsolve({IVP},numeric): alt
odd
- 2.8 odd 1-11, any of 13- 19
you can do
- Chapter
3.Homework
- Solution of logistic equation by partial
fractions
- 2.6 exact equations (subscriptsbelow
indicate partial derivatives)
- What is an exact equation: M dx + N
dy = 0, looks like
- chain rule applied to H(x,y) = k
where M = Hx and N = Hy
- How to test for exactness: Nx
- My = 0
- How to find H(x,y):
- ad hoc: integrate M wrt x, N wrt y,
Match terms
- systematic H = Int(M,x) +
c(y), diff the previous result wrt y, set equal to N, solve for
c', then c by solving the subsequent simple DE.
- If not exact, check for integrating
factor I which fits (I N)x - (I M)y
= Ix N + I Nx - Iy M - I My
= 0
- If I is a function of
x, Iy = 0 and Ix / I = (
Nx
- My ) / N ,
solve for I if RHS is a function of x
- If I is a function of y, Ix = 0 and Iy
/ I = ( Nx
- My ) / -M , solve
for I if RHS is a function of y.
- Separable equations are exact and linear equations have
the integrating factor discussed above.
2/5/2004 Thursday
- Two student presentations
- Brief intro to Maple Basics
- Check out the Maple New User's Tour in
the Maple Help Menu.
2/10/2004 Tuesday
- Please Read Minutes,
which discusses both study groups and minutes.
- 2.8 Existence and Uniqueness Theorem for First order IVP's
2/12/2004 Thursday
- Test 1 over chapters 1 and 2 is postponed to next Thursday.
- Homework from Chapters 1& 2 due before the test next
Thursday
- Homework for chapter 3 Prior Chapter 2
assigment,
- 3.1: All odd 1, 3, 5, ...
- 3.2: Alt odd 1,5,9,....
- 3.3: Alt odd 1,5,9, ....
- 3.4: Alt odd 1-19, all odd 21-41
- 3.5: Alt odd 1-13, all odd 15-41
- More Chapter 3.
- Lecture Topics for today from Sections 3.1, 3.2, 3.4, 3.5.
Read all
these section and start on the homework
- Linear DE's can be written as: y" + B(t) y' + C(t)
y = g(t) .
- The DE is homogeneous (on an interval I) if g(t) = 0
(for all t in I).
- Solutions exists on an interval I for which B(t),
C(t) and g(t) are continuous.
- A unique solution exists on the interval when the
values of the function and its derivative are specified at a
single point in the interval: y(t0)
= y0 and y'(t0)
= y'0 are specified.
- Characteristic equation for 2HOLDECC's
(2nd
order Homongeneous
Ordinary
Linear
Differential
Equations
with Constant
Coefficients).
- The first order HOLDECC equation y' + r y = 0 has y(t)
= e(-r t) as a solution, try this in a 2HOLDECC
y" + b y + c y = 0 to produce ( r2
+ b r + c) e(-r t) = 0;
setting the first factor equal to zero is the characteristic equation.
- Read 3.1 for solution technique and possible solution
behavior when there are two different real solutions to the
characteristic equation. Example. ( r2
+ 4 r + 3) = 0......
- Read 3.4 for solution technique and possible solution
behavior when there are two real different complex solutions to the
characteristic equation. ( r2 + 4 r
+ 5) = 0 ......
- Read 3.5 for solution technique and possible solution
behavior when there is one repeated real solution to the characteristic
equation. ( r2 + 4 r + 4) = 0 ...
- If your group has only two members, you can merge with
another 2 or 3 member
group if you wish. It is not necessary to merge if your two
member group work
well. However you cannot meet as a one member group if the other
members do not show up. It is permissible to meet simultaneously
with other
groups without merging, and if you do so and report it as such in both
sets of minutes you can then get credit as having met. Groups meeting
simultaneously should be reported in the minutes as an action
item, and the other group members should be listed as visitors.
Both group minutes must reflect this simultaneous meeting. I
encourage you to arrange a simultaneous meeting with other groups for
practice.
- Study Group meetings must meet for at least 30 minutes and
you must do something
2/17/2004 Tuesday
- General advice: Life is short, if the rest
of your group is not serious and you are, join another group. I will
request an evaluation (confidential), by each member of the
contribution of other group members at the end of the semester.
This should not effect your study group grade, unless the majority
reports that you were seldom there and did little. You do have a
responsibility to your group.
- Lecture topic: Linear Differential
operators:
- An operator L is a function which acts on
other functions and returns a function: L: domain --> range. A
prime example of an operator is the derivative operator D(f)(t) = f
'(t). (Note: f is a function, D(f) is a function, D(f)(t) is the
value of D(f) at t. This is precisely the syntax required by
MAPLE.) Another example is the Equality or Identity operator E(f)
= f
- The derivative operator is linear in the
sense that D(a f + b g) = a D(f) + b D(g) for constants a and b.
In general, any operator L is linear if L(a
f + b g) = a L(f) + b L(g) .
- For operators L and M, we can define the sum, L + M, and
composition, L@M, of operators by the usual rules for
forming sums and composition of functions (remember, operators
themselves are functions):
- (L + M)(f) = L(f) + M(f) and
- (L@M)(f) = L(M(f)).
- The notation L@M is expressive but clumsy and usually
the notation LM is used. L@L is written as L2.
Using this notation, the second derivative of y is (D2)(y)
or just
- We can also form new operators by multiplying by
constants , i.e., (kL)(f) = k L(f), or even by multiplying by a
function, (B*L)(f)(t) = B(t)*L(f)(t). This can be very tricky
since L M , kL and B*L and meaning often must be
inferred from context.
- This can be cleared up (at some notational expense) by
defining a (linear) multiplication operator Mg(f)(t) =
g(t)*f(t). If we regard a constant k as a (constant) function,
then (Mk@L)(f) = Mk((L)(f)) = k L(f) and (MB@L)(f)
= MB(L(f) = B*L(f). We can the see that all the various
multiplications can be viewed as the same type of operator
compositions. However, we will avoid this type of pedantic formalism in
favor of context. It is also just not the best way to think about it.
- It is easy to show that if L and M are linear operators,
then kL, aL+bM, L@M, B*L are all linear operators.
- Linear differential operators can be used to
represent Linear differential equations:
- Example: y ' + 4 y = 0
can be written L(y) = 0, with L = (D + 4 E)
where D and E are the derivative and Equality operators: L(y) =
(D + 4 E)(y) = D(y) + 4 E(y) = y ' + 4 y = 0
- If Q = ( D2 + 5 D + 4 E ), Q(y) = 0 is the
differential equation y" + 5 y ' + 4 y = 0. The characteristic
equation approach to solution can be written as
- Q( ert ) = ( r2 + 5 r + 4 ) ert
= 0 , requiring r = -4 or r = -1,
- hence Q(e-4t ) = and Q( e-t
) = 0, and Q is linear,
- which implies Q( a e-4t + b e-t
) = a Q(e-4t ) + b Q( e-t ) = 0
- The operator Q can be "factored", like a polynomial,
into (D + 4 E)@(D + E)
or (D + E)@(D + 4 E). Equality of the three forms is checked by
verifying that both factorizations produe the same result when applied
to y. For instance
- (D + 4 E)@(D + E)(y) = (D + 4 E)(y' + y) = y" + y' +
4 y' + 4 y = y" + 5 y' + 4 y = Q(y)
2/18/2004 Wednesday
- Some new Maple worksheets have been posted
on the web site.
- 405_0201X_Exercises.mws,
local, URL
Updated 2/18/04. Exercise 21 (direction fields, initial condition
analysis). The URL points to the most recent version of this file.
- I am trying to get Maple at a reduced
price for sudents in this course. Savings
of up to $50. Do not buy it now. When I
tell you to, you will be able to purchase online and download it.
2/19/2004 Thursday
2/24/2004
Tuesday
- Please be prepared to return to results presentations
related to the paper assignments on Thursday.
- All groups which will not or do not make preentations on
Thursday should schedule an out of class time (during your regular
group meeting times is my suggestion) to present results to me.
All result presentations must be completed by next Thurday, March 4,
and papers written and submitted by Thursday March 11.
- Lecture Topic Differential equations as
linear operators (References Sections 3.2-3.5, 4.1, 4.2)
- Linear differential are operators of the
form
- L(y) = An(t) y(n)
+ An-1(t) y(n-1) + .. + A2(t) y'' + A1(t)
y' + A0(t) y
- The Existence and Uniqueness Theorem
(EUT) says there is a solution of L(y) = g(t) on any interval I
for which all the coefficient functions Ak(t) and g(t) are
continuous and the lead coefficient An(t) is nonzero. Usually, the emphasis on the continuity of the coeficients
and the nonzero behavior of An is enforced by dividing
by An(t)
and requiring the resulting coefficient functions to be continuous. The solution is uniquely determined by the "state" of the
system at any point t0 in I. The state of the system is the value of
the solution function and its derivatives at t0. If
y(t) is the solution of an nth order differential equation, the values
of y, y', y'', .. y(n-1) are sufficient to determine
the system state,
since the values of the higher order derivatives can be determined from
the DE itself..
- Linear differential operators look like polynomials in D
with possibly variable coeffiecients.
- If the coefficients are constant, the linear operators
act even more like polynomials, factoring (under functional
composition) into lower order operators that commute ( F1@F2
=
F2@F1);
- example L = D2 + 3D +2E = ( D +
2E )@( D + E ) = ( D + E )@( D + 2E )
- This commutative property reveals the VERY IMPORTANT FACT that all the
derivatives of solutions of HOLDECC are also solutions of the
same
HOLDECC. Let L be a LDOCCs, and L(y) = 0, then
- L(y(k)) = (L@Dk)(y) = (Dk@L)(y)
= Dk(L(y) = Dk(0) = 0:
- The structure of the null space:
The set of solutions of L(y) = 0
- The linearity of L implies that
the null space of L, N(L) =
{y(t): L(y) = 0 }, is a linear subspace of C(n)(I),
the set of functions on the interval I with continuous n-th derivatives.
- The EUT implies that L(y) = 0 has n
solutions functions Bk(t), k=0..n-1 which satisfy, at a
given t0 in I,
- Bk( j )(t0)
= 1 if j = k, = 0 otherwise
- These function span the null space:
given any set of possible state values, the linearity of L
implies
- y(t) = s0 B0(t)
+ s1 B1(t) + ... + sn-1 Bn-1(t)
is a solution, and
- y( j )(t0)
= s0 B0( j )(t0) + s1 B1( j (t0) + ... sj
Bj( j )(t0) + ... + sn-1
Bn-1(
j )(t0) = sj
- so this y is the (unique) solution
corresponding to the state vector at t0.
- These functions are linearly
independent since
- y(t) = s0 B0(t)
+ s1 B1(t) + ... + sn-1 Bn-1(t)
= 0 for all t in I, means that the value of y and all of its
derivatives are zero at t0 and these values correspond to
the values of the sj's, hence all sj = 0.
- Since the Bk(t) span N(L) and they are independent they
form a basis for N(L) , which
is an n dimmensional linear subspace of C(2)(I).
- Solution of linear nonhomogenous
equations:
- If v and w are both solutions to L(y) =
g (corresponding to different ICs), then z = w - v is a solution to
L(y)
= 0:
- L( z ) = L( w - v ) = L( w ) - L( v ) = g -g = 0
- Another SUPERPOSITION rule: if L(y1)
= g1 and L(y2) = g2 then
- L( y1 + y2 ) = L( y1 ) + L( y2 ) = g1
+ g2
This allows a DE like y" + 4 y = 5 + t2 + t3e(-3t)
to be solved independently for groups of terms on the RHS.
- Solutions of repeated factor linear
operators:
- For ANY linear differential operator
(specifically including variable coefficients)
- L( t y ) = t L( y ) + LD(
y ) where LD is a new operator made by formal
differentiation of L by D.
- If L is an LDOCC and L(y) = 0, then LD(y)
is a linear combination of y and its derivatives (see 3.2.1 above), all
of which
satisfy L(y) = 0. Consequently, LD(y) is in N(L), the null space of L, and
L( LD(y) ) = 0.
- Direct calculation then shows
that if L(y) = 0 then L2( t y ) = 0:
- L( L( t y ) ) = L( t L( y ) + LD(
y ) ) = L( t 0 + LD(y) ) = L( LD(y)) = 0
2/26/2004 Thursday
- Students in this class may now purchase MAPLE directly for
$75. The UofL COPYIT centers were selling it (with a
printed Manual) for $125. If you have purchased Version 9
recently for the full price, you may be able to send the Manual back
(if it is in mint condition) for a $35 refund. Contact me for the
special URL to the Maple sale site.
- Lecture Topic See Tuesday
- Continued from Tuesday
- If a 2nd order linear operator behaves like an
irreducible quadratics polynomial and does not permit linear
factorization over the real numbers it can be analyzed by completing
the square:
- If Q = (D2 + 2b D + b2 E + w2
E) = ( D + b E )2 + w2 E and Q( y ) = 0,
then
- D2( e(bt) y ) = e(bt)
( y" + 2b y' + b2) = e(bt) (-w2 y ) =
- w2 e(bt) y
- The operator equation D2 = -w2 E
or D2 + w2 E = 0 is easy solved by observing that
we know the two functions cos(wt) and sin(wt) both satisfy y" =
-y. In fact cos(wt) is one of the basic functions at t0
= 0 cited in 4.2 above having value = 1 at 0 and derivative = 0.
The other basic function is sin(wt)/w which has value = 0 and
derivative = 1.
- If Q(y) = 0 then z = e(bt) y is a
solution of z" + w2 z = 0, so z = e(bt)
y = c1 cos(wt) + c2 sin(wt)
and y = e(-bt) (c1 cos(wt) + c2 sin(wt)) which provides two
solutions of Q(y) = 0. That these solutions are sufficient will
be established later.
- The trick with e(bt) y can be used to
solve all three possible differentiable forms.
2/27/2004 Friday
- Monday 3/1 is the last day to
drop. If you wish to discuss your
current status see me Monday.
- Check your email often. I am
havig difficulty posting to the web site.
3/02/2004Tuesday
- Lecture Topic
3/04/2004 Thursday
- Homework Note prior homework for chapter 3 on 2/12/04.
- 3.5 38, 39, 40
- 3.6 alt odd 1- 29, 30, 31, 33-36
- 3.7 alt odd 1- 17, all 21-28, all 29-32
- 3.8 alt odd 1-29, do as many of the rest as you can
(Group project)
- 3.9 alt odd 1-25, do as many of the rest as you can
(Group Project)
- 4.1 alt odd 1-17, all the rest
- 4.2 alt odd 1-37, 39
- 4.3 alt odd 1- 17, 19, all 20-22
- 4.4 alt odd 1-17
- Lecture Topic
- Exercises with Emphasis on Fundamental
solutions,
- operator techniques
- Characteristic function techniques
- testing for independence with the
Wronskian determinant,
- If the Wronskian of a set of
functions is nonzero at any point, functions are independent.
- Using the wronskian matrix to determine
constants given initial conditionsWronskians
- Analysis of solutions after the constants
have been determined
- Essential Property of the Wronskian of
solutions of L(y)=0 : W ' = -An-1(t) W
- For any n solutions of an n-th order
LDE, the Wronskian determinant is either never zero (for a fundamental
set) or always zero (a dependent set of solutions)
- Derivatives of Determinants
- Properties of determinants used for
evaluation (easily verified for 2x2 matrices)
- a common factor in a row can be
factored out
- Value does not change if a multiple
of a row is added/subtracted to/from another row
- identical rows (or columns) force
determinant to be zero
- Annihilators of common functions. Given y(t) find a
Differential Operator which L(y)=0. Many different operator may
work. Most common function can be, and in our case are, annihilated by
LDOCCs - linear differential operators with constant coefficients. Our
interest is in least order LDOCCs.
- Constants: y=A, L = D
- Linear polynomials y = A + B t, L = D2
- n-th degree polynomial are annihilated by D(n+1)
.
- Exponentials ert by L = ( D - r E )
- t ert (and (A + B t )ert
by L = ( D - r E )2.
- If P(t) is any nth deg poly, P(t) ert
is annihilated by L = ( D - r E )n+1.
- cos(wt) and sin(wt) are annihilated by L = D2
+ w2 E .
- ert cos(wt) and ertsin(wt) are
annihilated by L = D2 - 2r D + ( r2 + w2
) E .
- If P(t) is any nth deg poly, then P(t) ert
cos(wt) and P(t) ertsin(wt) are
annihilated by L = ( D2 - 2r D + ( r2 + w2
) E )n+1 .
3/09/2004Tuesday
- Lecture Topic; Undetermined coefficients
(3.6, 4.3)
3/11/2004 Thursday
- Lecture Topic Reduction
of order (3.5) and Variation of parameters(3.7,
4.4)
Spring Break 3/14 - 3/21
3/23/2004Tuesday
- Lecture Topic
3/25/2004 Thursday
Test 3
Test 4
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