baylor campus in april 2020 welcome to flight 3326

MTH 3326 Baylor Mathematics

baylor campus in april 2020

Welcome to Flight 3326https://www.baylor.edu/canvas/

qin sheng (tim) Stay Safe for All in this Class fall 2020 1 / 25


Stay home if you have symptoms or are asymptomatic

1 Please stay home if you have Covid-19 alikesymptoms, no matter how mild they are.

2 Please stay home until you recover if you areasymptomatic.

3 All lecture notes will be online, quizzes and testscan be done online.

4 All office hours and Math Lab tutorials are virtual.5 Using face masks/coverings properly are essential

in our classes.



basic info

MTH 3326: Partial Differential EquationsTentative Syllabus, Fall 2020

# Sec. 504: TTh 09:30–10:45 am; Sec. 505: TTh 02:00–03:15 pm @ SR 203/Hybrid

Instructor: Professor Qin “Tim” ShengOffice Location: Sid Richardson 302.FOffice Phone: 254-710-1241E-Mail: Qin [email protected]: http://sites.baylor.edu/qin sheng/Office Hours: TTh: 11:00 am–1:00 pm virtually via Zoom or Skype

University Mask Requirement: Students and instructors must properly wear facecoverings over their mouths and noses while in the classroom. In addition, social distanc-ing must be maintained within all classroom environments. Any student who does notbring a face covering or refuses to wear one will be dismissed from class and be consideredabsent. If a student who is not wearing face covering refuses to leave class, the instructormay dismiss in-person class for that day. Such a student will be subject to disciplinaryaction through the Student Code of Conduct.

TEXT: Applied Partial Differential Equations by John Davis, 2010 (my lecture notes will also beused)

COURSE COVERAGE & SUGGESTED EXERCISES: This course will cover the follow-ing sections of the textbook (tentatively): 1.1-4, 2.1-6, 3.1-6, 5.3-6, 6.1-2, 7.1-4. We mayuse computer software packages, such as Matlab and Mathematica, for illustrations of solutionproperties and project.

HOMEWORK: As we go along section by section for the covered materials in the textbook, youare required to complete all suggested exercises in time. You are required to keep a notebook solelyfor this homework. If you have questions please see me either during my office hours or by makingan appointment. You may also consult with our TA and tutors in the Math Lab.

EXAMS, QUIZZES, PROJECTS AND GRADING POLICIES: There will be three in-semester tests, one middle semester project, eight quizzes, and one final exam (or final project).

See a tentative test and quiz schedule on our course calendar.

Quiz 1 08/27 Quiz 4 09/24 Quiz 7 10/22 Quiz 10 11/19Quiz 2 09/03 Quiz 5 10/01 Quiz 8 10/29Quiz 3 09/10 Quiz 6 10/08 Quiz 9 11/05Test I 09/17 Test II 10/15 Test III 11/12 Final Ex. 12/4&8

↪→ Official Final Exam Information:504: Friday, Dec 4, 2020, 9:00–11:0 am; 505: Tuesday, Dec 8, 9:00-11:00 am

↪→ Website: http://www.baylor.edu/registrar/doc.php/358880.pdf

↪→ Method of Evaluation:

3 tests, each one counts 13% toward your final (no makeup tests without justified reasons)10 quizzes (or online homework assignments), each counts 4% toward your final1 final comprehensive exam (or comprehensive project), 21%

1

Hybrid MTH 3326 (fall) 2020

August 24 – November 25, 2020 - Page I

Monday

Tuesday

Wednesday

Thursday

Friday

25 Aug.® Sec. 1.1

27 Sec. 1.2 & 1.3 Quiz 1

1 Sept.®

Sec. 1.3 & 1.4

3 Sec. 1.4 Quiz 2

8 Sec. 2.1

10 Sec. 2.2 Quiz 3

15 Sec. 2.3

17 Test 1

22 Sec. 2.4

24 Sec. 2.5 Quiz 4

29 Sec. 2.6 & 3.1

1 Oct.® Sec. 3.2 Quiz 5

6 Sec. 3.3

8 Sec. 3.4 Quiz 6

QS

Hybrid MTH 3326 (fall) 2020

August 24 – November 25, 2020 - Page II

Monday

Tuesday

Wednesday

Thursday

Friday

13 Oct.® Sec. 3.5

15 Test 2

20

Sec. 3.6

22 Sec. 5.3 & 5.4 Quiz 7

27 Sec. 5.5 & 5.6

29 Sec. 6.1 Quiz 8

3 Nov.® Sec. 6.2

5 Sec. 7.1 Quiz 9

10 Sec. 7.2

12 Test 3

17 Sec. 7.3

19 Sec. 7.4 Quiz 10

24 Review

QS

MA3326 Tentative list of suggested homework problems Fall 2020 Sec 1.1: 3, 5, 7 Sec 1.2: 3, 9 Sec 1.3: 3, 4, 5 Sec 1.4: 1—5, 8, 9, 11, 13 Sec 2.1: 2—5, 7 Sec 2.2: 2, 3, 5 Sec 2.3: 1—3, 5, 9 Sec 2.4: 1, 2, 5, 7 Sec 2.5: 1, 2 Sec 2.6: 1 Sec 3.1: 1—3, 5, 7 Sec 3.2: 3 Sec 3.3: 1, 2, 5 Sec 3.4: 2, 3 Sec 3.5: 1—4, 6 Sec 3.6: 1 Sec 5.3: 3 Sec 5.4: 1 Sec 5.5: 1, 3, 5 Sec 5.6: 1, 2 Sec 6.1: 1—3 Sec 6.2: 1 Sec 7.1: 1—4, 6, 9 Sec 7.2: 1, 2 Sec 7.3: 5—7 Sec 7.4: 1, 3

Virtual Office Hours: 11:00 am – 1:00 pm. Use Email,Skype (ID: Baylor.Sheng), Zoom, Phone

STAY SAFE!



basic info

Monday : 3 :00-5 :00 pm * 6 :00-8 :00 pm

Tuesday : 3 :00-5 :00 pm * 6 :00-8 :00 pm

Wednesday : 3 :00-5 :00 pm * 6 :00-8 :00 pm

Thursday : * 6 :00pm-8 :00 pm

Friday : 3 :00pm-5 :00 pm

MATH TUTORING LABMATH TUTORING LABMATH TUTORING LAB

Days and Times

Access the lab on Zoom

*6:00-8:00 pm CST MTWR:

August – October (8/24 – 10/30):

https : //baylor .zoom .us/j/97706577727?

pwd=dTBwZ0J4OWxONTUxc2FiRFViRDlQdz09

November – December (11/02 – 12/01):


pwd=ZkJETWtmUzBpYnpGdFFadVFMdm90Zz09

Zoom Links:

August– October (8/24 – 10/29):


pwd=Q051VUczQ25VUEs5T29NS01nWUVVQT09

November – December (11/02 – 12/01):


pwd=Zkt5ZGpwSm45VUtIcEErQkl5S0lpQT09

3:00-5:00 pm CST MTW&F:

Math Tutoring Lab ⇒ to access the lab on ZoomMatlab Ref ⇒ to find a copy of the book

STAY SAFE!



basic info

For most of us, Partial Differential Equations (PDEs) is a difficult, butrewarding, piece of knowledge to learn. The importance andpervasiveness of PDEs in our increasingly complex society is evidentand well-documented.

Very few people can learn PDEs by observation only. It invariably takesa great deal of time, patience, practice, more practice, and even morepractice to learn the necessary skills and techniques to succeed.

It is an absolute prerequisite to master college algebra, calculus andordinary differential equations (ODEs) in order to succeed in PDEs.




A Brief Introductionhttps://sites.baylor.edu/qin sheng/

introduction & section 1.1


MTH 3326 Sec. 1.1

Stay safe for all of us!

Sec. 1.1. Objectives1 To distinguish between ordinary and partial derivatives.

f ′(x) =dfdx

(x) ⇒ fx (x , t) =∂f∂x

(x , t), fxx (x , t) =∂2f∂x2 (x , t)

2 To distinguish between ODEs and PDEs.

y ′(t) + 3y(t) = g(t) ⇒ ut = a2uxx , uxx + uyy = 0

3 To verify solutions of PDEs by using chain rules or implicitdifferentiations.

4 To recall initial and boundary conditions.


MTH 3326 Sec. 1.1

derivatives

Notations of ordinary derivatives.

y ′, f ′,dfdx,

dgdt, y ′′, y (5),

d5gdx5 (x),

dnφ

dξn (ξ), . . .

Chain rule. Example 1: Given f (x) = sin(6x + π). We wantdfdx.

dfdx

(x) =ddx

sin(6x + π) =d sin(u)

du× du

dx

= cos(u)

(d(6x + π)

dx

)= 6 cos(6x + π).

Chain rule. Example 2: Given f (x) = φ(6x + π). We wantdfdx.

dfdx

(x) =dφ(u)

du× du

dx= φ′(u)× 6 = 6φ′(6x + π).


MTH 3326 Sec. 1.1

derivatives

Implicit differentiations. Example: Given3xy + sin(y2)

exy = 1. We

want y ′. We rewrite the equation to 3xy + sin(y2) = exy . Thus,

(3xy + sin(y2))′ = (exy )′ ⇒ (3xy)′ + (sin(y2))′ = (exy )′.

⇒ 3y + 3xy ′ + sin(y2)2yy ′ = exy (y + xy ′).⇒ 3y + [3x + 2y sin(y2)]y ′ = yexy + xexyy ′.⇒ [3x + 2y sin(y2)− xexy ]y ′ = yexy − 3y .

⇒ y ′ =yexy − 3y

3x + 2y sin(y2)− xexy .

Chain rules are used in calculations.


MTH 3326 Sec. 1.1

derivatives

Notations of partial derivatives.

yx , fy ,∂f∂x,∂g∂t, yxx , yt5 ,

∂5g∂x5 (x , y , t),

∂n+2φ

∂ξn∂t2 (ξ, t), . . .

Example. Given u(x , t) = f (2x − 3t). We want∂u∂x

and∂u∂t.

∂u∂x

=∂f (2x − 3t)

∂x=

df (w)

dw∂w∂x

= f ′(w)(2) = 2f ′(2x − 3t),

∂u∂t

=∂f (2x − 3t)

∂t=

df (w)

dw∂w∂t

= f ′(w)(−3) = −3f ′(2x − 3t).

Apparently, chain rules are used here.


MTH 3326 Sec. 1.1

ODEs and PDEs

If ordinary derivative(s) of the unknown function is/are used in anequation, that is an ODE.If partial derivative(s) of the unknown function is/are used in anequation, that is a PDE.Ordinary and partial derivatives are not used in the sameequation.Tips for homework problems: how to choose a proper domain forplotting a PDE solution, such as that for #4 in Sec. 1.1?

1 For time variable t ≥ 0, we only need to determine how much timewill be meaningful. We often choose t ∈ [0,10] if no references aregiven. Of course, we may readjust the interval later.

2 For spatial variables such as x , y , z we tend to see how we caneffectively display the solution feature. We may consider−10 ≤ x , y , z ≤ 10 if no physical limitations are available.


MTH 3326 Sec. 1.1

computer simulations

Movie Time!

sample prog page001 movie cos.m


MTH 3326 Sec. 1.2


Sec. 1.2. Objectives1 To derive a conservation law.

∂u∂t

(x , t) =∂φ

∂x= f (x , t)

2 To obtain typical conservation laws.3 To derive a linear wave equation

utt = c2uxx ,

where c is the speed of the wave.


MTH 3326 Sec. 1.2

derivation of conservation laws

We have flows around the world. Imagine that we have an arbitrary smallslice of 3-dimensional domain in the ocean, from the left to right as shown inFig. 1.4. The cross-section area is A (we don’t care about the shape of sucha cross-section). If u(x , t) is the density of the fluid, then the total mass of thefluid is

M(t) =

∫ b

au(x , t)Adx = A

∫ b

au(x , t)dx

between location indicators x = a and x = b at time t .


MTH 3326 Sec. 1.2

derivation of conservation laws

Therefore the following conservation is true:

dMdt

= Addt

∫ b

au(x , t)dx = Aφ(a, t)− Aφ(b, t) +

∫ b

af (x , t)Adx ,

where φ(x , t) is the physical flux and f (x , t) is the possible internal fluidgenerator (due to chemical or biological reactions, say). Thus,

ddt

∫ b

au(x , t)dx =

∫ b

aut (x , t)dx = −

∫ b

aφx (x , t)dx +

∫ b

af (x , t)dx ,

due to the Fundamental Theorem. Since the above is true for all b ≥ a, weconclude that

ut (x , t) = −φx (x , t) + f (x , t),

which is the conservation law. We note that φ may depend on u.


MTH 3326 Sec. 1.2

birth of the wave equation

The Great Wave off Kanagawa, 1829-1833


MTH 3326 Sec. 1.2


the hypotenuse ∆T =√

(∆x)2 + (∆u)2,

tan θ(x , t) =∆u∆x

,

sin θ(x , t) =∆u∆T

=∆u√

(∆x)2 + (∆u)2=

∆u/∆x√1 + (∆u/∆x)2

,

cos θ(x , t) =∆x∆T

=∆x√

(∆x)2 + (∆u)2=

1√1 + (∆u/∆x)2

.


MTH 3326 Sec. 1.2


Here is Newton’s Second Law: M−→a = M∂−→v∂t

= M∂2−→u∂t2 =

−→F .

But, what is our mass M?

M = ρ(x)|∆T | = ρ(x)√

(∆x)2 + (∆u)2.

Assume that there is no internal force, for a vertical movement, we have

ρ(x)√

(∆x)2 + (∆u)2utt = T (x + ∆x , t) sin θ(x + ∆x , t)− T (x , t) sin θ(x , t).

Or,

ρ(x)

√1 +

(∆u∆x

)2

utt =T (x + ∆x , t) sin θ(x + ∆x , t)− T (x , t) sin θ(x , t)

∆x.


MTH 3326 Sec. 1.2


Recall definitions of derivatives. We have lim∆x→0

∆u∆x

= ux . Thus,

utilizing a chain rule, we obtain

ρ(x)

√1 + (ux )2utt = (T (x , t) sin θ(x , t))x

= Tx (x , t) sin θ(x , t) + T (x , t) cos θ(x , t)θx (x , t).(1)

For the same reason, as ∆x → 0,

tan θ(x , t) = ux ,

sin θ(x , t) =ux√

1 + (ux )2,

cos θ(x , t) =1√

1 + (ux )2.


MTH 3326 Sec. 1.2


Further,

|ux | = | tan θ(x , t)| � 1,√

1 + (ux )2 ≈ 1, sin θ(x , t) ≈ ux (x , t)

cos θ(x , t) ≈ 1, tan θ(x , t) ≈ θ(x , t), θx (x , t) ≈ uxx (x , t).

Substitute the above into (1). We find that

ρ(x)

√1 + (ux )2utt (x , t) = Tx (x , t)ux (x , t) + T (x , t)uxx (x , t).

Hence,ρ(x)utt (x , t) = T (x , t)uxx (x , t).

⇒ utt (x , t) =T (x , t)ρ(x)

uxx (x , t) = c2uxx (x , t).


MTH 3326 Sec. 1.3


Sec. 1.3. Objectives1 To have Dirichlet boundary conditions (or first boundary

conditions).u(a, t) = φ1(t), u(b, t) = φ2(t)

2 To have Neumann boundary conditions (or second boundaryconditions).

ux (a, t) = ψ1(t), ux (b, t) = ψ2(t)

3 To have Robin boundary conditions (or third boundary conditions).

α1u(a, t) + β1ux (a, t) = θ1(t), α2u(b, t) + β2ux (b, t) = θ2(t)

4 To have periodic boundary conditions.

u(0, t) = u(`, t)


MTH 3326 Sec. 1.4

No Covid-19 Patient is Allowed in Class

Sec. 1.4. Objectives1 To review several most popular methods for solving ODEs.2 To recall differential operator notations.3 To solve first order linear ODEs.

y ′(t) + p(t)y(t) = 0

4 To use initial conditions.5 To use boundary conditions.6 To solve Cauchy-Euler equations such as

ax2y ′′(x) + bxy ′(x) + cy(x) = 0, a,b, c ∈ R.

7 To recall hyperbolic trig functions.


MTH 3326 Sec. 1.4

sample solutions

Let’s take a look at homework #6:

y ′ + 2ty = t . (∗)

This is a first order nonhomogeneous DE. Recall equation (1.13). Wehave an integrating factor

m(t) = e∫

p(t)dt = e∫

2tdt = et2.

Now,

y(t) =

∫et2

tdt + Cet2 =

12

∫et2

d(t2) + Cet2 =

12

∫eutdu + C

eu

=12eu + C

eu = Ce−u +12

= Ce−t2+

12.


MTH 3326 Sec. 1.4

sample solutions

To check the solution, we may differentiate y to yield

y ′(t) = −2Cte−t2.

Substitute y(t), y ′(t) to the DE (*), we may find that the generalsolution is indeed correct!Let’s take a look at homework #10:

y ′′ − 4y ′ + 5y = 0. (∗∗)

This is a second order linear homogeneous DE. Recall (1.14). Wesubstitute y = ert into DE (**) to obtain following characteristic equation

r2 − 4r + 5 = 0.


MTH 3326 Sec. 1.4

sample solutions

The characteristic equation has two complex roots:

r1,2 =−b ±

√b2 − 4ac

2a=

4±√

16− 202

=4±√−4

2= 2± i ,

where i =√−1. Therefore our final answer is

y(t) = C1e(2+i)t + C2e(2−i)t = C1e2teit + C2e2te−it

= C1e2t (cos(t) + i sin(t)) + C2e2t (cos(t)− i sin(t))

= e2t ((C1 + C2) cos(t) + (C1 − C2)i sin(t))

= e2t (C cos(t) + D sin(t)),

where C,D are free constants. Remember to check if y is correct!


baylor campus in april 2020 welcome to flight 3326

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