1

Reminder

• Second Hour Exam• Date: July 9 ( Monday).• Room: LC-C3, Time: 10 -11 am• Covers: Sections 5.1, 5.3 and 7.1 -

7.9 (inclusive).

2

Dirac Delta Function • Paul A. M. Dirac, one of the great physicists

from England invented the following function:• Definition: A function (t) having the following

properties:

• is called the Dirac delta function. It follows from (2) that for any function f(t) continuous in an open interval containing t = 0, we have

- 1, dt (t) (2)

and , 0 tallfor , 0(t) )1(

. )()( )( afdtattf

3

Remarks on Theory of Distribution.

• Symbolic function, generalized function, and distribution function.

4

Heuristic argument on the existence of -function.

• When a hammer strikes an object, it transfer momentum to the object. If the striking force is F(t) over a short time interval [t0, t1], then the total impulse due to F is the integral

momentum".in Change Impulse " that means This

motion, of law 2nd sNewton'By .dt F(t)Impulse

1

0

1

0

1

0

01

t

t

t

t

t

t) . )-mv(t mv(t dt

dtdvm F(t) dt

5

This heuristic leads to conditions 1 and 2.

6

What is the Laplace Transform of -function?

• By definition, we have

).(') :F.T.C.) (by theget we(*), of sidesboth ateDifferenti

(*)

thatfollowsIt .1)( ,

and , 0)( ,

that whenNote

. )( )( ))}(( {

t

0

atuaδ (t

a). u(ta)dxδ (x

dxaxat

dxax a t

edtatedtatesatL

t

-

-

t

asstst

7

Application:• Consider the symbolic Initial Value Problem:

x +9x=3δ \( t - π \) ; ital x \( 0 \) = 1 ital , x ' \( 0 \) = 0 . } {} # {} # size 12{ Thisrepresents a mass attached ¿a spring is released¿

rest } {} # size 12{1 meter below the equilibrium position for the mass−spring } {} # size 12{ system . begins ¿vibrate , but π seconds later , the mass is struck } {} # size 12{ by a hammer exerting an impulse on the mass . , where x \( t \) is the} {} # size 12{ displacement ¿ the equilibrium position at time t . } {} # size 12{We shall solveby the method of ℒ Transform . } {} } } {

8

Linear Systems can be solved by Laplace Transform.(7.9)

• For two equations in two unknowns, steps are:• 1. Take the Laplace Transform of both equations

in x(t) and y(t),• 2. Solve for X(s) and Y(s), then• 3. Take the inverse Laplace Transform of X(s)

and Y(s), respectively.• 4. Work out some examples.

9

Examples• Solve the following IVP:

• #21, P.438

0.y(0) ;0',0)0( );2(1'

yxxtuyx

10

Qualitative analysis of Differential Equations

• Consider three type of differential equations:

equations.nonlinear solvingfor proceduressolution general no are There

comments). (some equations,linear -Non are These 012042 (3)

ts.coefficien leith variabequation w linear a is This t.t variableindependen of functions

are and where0 (2)

ts.coefficienconstant ith equation wlinear a is this constants, are where0 (1)

22 .y(y'), or y"yy'y"-

c(t) a(t), b(t),c(t)yb(t)y'a(t)y"

a, b, ccyby'ay"

11

Reminder

• Second Hour Exam• Date: July 9 ( Monday).• Room: LC-C3, Time: 10 -11 am• Covers: Sections 5.1, 5.3 and 7.1 -

7.9 (inclusive).

12

Reminder

• Final Exam• Date: July 19 ( Thursday).• Room: LC-C3, Time: 10:30 am• Covers: All

13

Generally, we can write a 2nd order D.E. in the form:

namely equation, of typethisaboutn informatio equalitativ some have do We

(*).for proceduressolution general no are thereAlthough . (or) and offunction nonlinear

a is if nonlinear, isequation This).',,())('),(,(" (*)

y'y f(t,y,y')

yytftytytfy

14

The Energy Integral Lemma

.0 dtd i.e. , timeallfor constant a is

21 )(

quantity Then the i.e. , of tiveantiderivaan be Let .explicitly

t variableindependen or the on dependnot doesthat function continuous a is where (*)equation aldifferenti theosolution t a be Let Lemma.

2

E(t) t

F(y(t))(y'(t)) E(t)

f(y).F'(y)fF(y)ty'

f(y)f(y),y"y(t)

15

Sketch of the proof.

.y"-f(y)(y')

f(y)y'(y')(y")F'(y)y'(t)(y')(y")E'(t)

F(y)(y')E(t)

0

221

that followsit , 21 Since 2

• Clearly we see that

.implicitly

solved becan this2in

orderfirst ofequation separable a is equation The .

Thus . )}("{}0)('{

F(y)][k dtdyy ;

kE(t)f(y)} {y"k} {E(t)

yfytE

16

Examples• 1. Consider the nonlinear equation:

n

. 636 that see we,simplicityfor

0k By taking .184324

131

24

thatNote .]2 have welemma,

by theThen . 24Let . 24

3/23/4

34

341

31

3131

yydtdy

yyyF(y)

F(y)[kdtdy

dyy F(y)yy" //

17

There are many other examples in your book. Select some of them depends on your field of interest and go over by yourself.

18

Series Solutions of Differential Equations.

• Recall Taylor polynomials of degree n, at x = a.• For example solution of the I.V.P.

Form. Lagrange theasknown 1

)

:RemainderTaylor On :

50100 , '3"

11

37

,a)(x)!(n

(f

(x)E(x)pf(x)(x)RNote

.), y'() y(yxyy

)(n)(n

n,an,an,f,a

19

Cauchy-Euler Equation Revisited.

.0at point singular a have

and tscoefficien the

thatsee we, 0 (**)

:Form Standard in theit rewrite weIf constants. real are and , 0 Where

00 (*)

:equationEuler -Cauchy theRecall

2

22

2

xaxcq(x)

axbp(x)

yaxcy'

axbxy"

b, ca., xcybxy'y" ax

20

Solutions are of the form xr

• Consider three possible cases, distinct, double and complex roots.

• Let w(r,x)=xr , and L[w](x)=0, we obtain ar2 + (b-a) r + c = 0.

• Note that L commute with w/ r , • (i.e. Dr L= L Dr) Thus w/ r is also a

solution.

21

Partial Differential Equations• Definition• One of the classical partial differential

equation of mathematical physics is the equation describing the conduction of heat in a solid body (Originated in the 18th century). And a modern one is the space vehicle reentry problem: Analysis of transfer and dissipation of heat generated by the friction with earth’s atmosphere.

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For example:• Consider a straight bar with uniform cross-

section and homogeneous material. We wish to develop a model for heat flow through the bar.

• Let u(x,t) be the temperature on a cross section located at x and at time t. We shall follow some basic principles of physics:

• A. The amount of heat per unit time flowing through a unit of cross-sectional area is proportional to with constant of proportionality k(x) called the thermal conductivity of the material.

xu /

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• B. Heat flow is always from points of higher temperature to points of lower temperature.

• C. The amount of heat necessary to raise the temperature of an object of mass “m” by an amount u is a “c(x) m u”, where c(x) is known as the specific heat capacity of the material.

• Thus to study the amount of heat H(x) flowing from left to right through a surface A of a cross section during the time interval t can then be given by the formula:

),(A) of )(()( txxutareaxkxH

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Likewise, at the point x + x, we have

• Heat flowing from left to right across the plane B during an time interval t is:

• If on the interval [x, x+x], during time t , additional heat sources were generated by, say, chemical reactions, heater, or electric currents, with energy density Q(x,t), then the total change in the heat E is given by the formula:

).,(utB) of )(()( txxt

areaxxkxxH

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E = Heat entering A - Heat leaving B + Heat generated .

• And taking into simplification the principle C above, E = c(x) m u, where m = (x) V . After dividing by (x)(t), and taking the limits as x , and t 0, we get:

• If we assume k, c, are constants, then the eq. Becomes:

),()( )(),(),()( txtuxxctxQtx

xuxk

x

),( 2

22 txp

xu

tu

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Boundary and Initial conditions

• Remark on boundary conditions and initial condition on u(x,t).

• We thus obtain the mathematical model for the heat flow in a uniform rod without internal sources (p = 0) with homogeneous boundary conditions and initial temperature distribution f(x), the follolwing Initial Boundary Value Problem:

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.0 , )()0,(, 0 , 0),(),0(

, 0 ,0 , ),( ),( 2

2

LxxfxuttLutu

tLxtxxutx

tu

One Dimensional Heat Equation

28

• Introducing solution of the form • u(x,t) = X(x) T(t) .• Substituting into the I.V.P, we obtain:

The method of separation of variables

.0)()('' and 0 )( )('

have weThus Constants.)()(''

)( )('

eq. following the toleads this.00 , )()('' )(')(

xkXxXtkTtTxXxX

tTtT

L, t x tTxXtTxX

29

• Imply that we are looking for a non-trivial solution X(x), satisfying:

• We shall consider 3 cases:• k = 0, k > 0 and k < 0.

Boundary Conditions

0)()0(0)()(''

LXXxkXxX

30

• Case (i): k = 0. In this case we have• X(x) = 0, trivial solution• Case (ii): k > 0. Let k = 2, then the D.E gives

X - 2 X = 0. The fundamental solution set is: { e x, e -x }. A general solution is given by: X(x) = c1 e x + c2 e -x

• X(0) = 0 c1 + c2 = 0, and

• X(L) = 0 c1 e L + c2 e -L = 0 , hence

• c1 (e 2L -1) = 0 c1 = 0 and so is c2 = 0 .• Again we have trivial solution X(x) 0 .

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Finally Case (iii) when k < 0.• We again let k = - 2 , > 0. The D.E. becomes:• X (x) + 2 X(x) = 0, the auxiliary equation is • r2 + 2 = 0, or r = ± i . The general solution:• X(x) = c1 e ix + c2 e -ix or we prefer to write:

• X(x) = c1 cos x + c2 sin x. Now the boundary conditions X(0) = X(L) = 0 imply:

• c1 = 0 and c2 sin L= 0, for this to happen, we need L = n , i.e. = n /L or k = - (n /L ) 2.

• We set Xn(x) = an sin (n /L)x, n = 1, 2, 3, ...

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Finally for T (t) - kT(t) = 0, k = - 2 .

• We rewrite it as: T + 2 T = 0. Or T = - 2 T . We see the solutions are

: try we condition,initial e satisfy th To . conditions boundary the

and D.E the satisfies ) ( ) ( ) , (function the Thus

. .. 3, 2, 1, n , ) (2) / (

t T x X t x u

e b t T

n n n

t L nn n

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u(x,t) = un(x,t), over all n. • More precisely,

• This leads to the question of when it is possible to represent f(x) by the so called

• Fourier sine series ??

. )( sin)0,(

:havemust We

. sin),(

1

1

2

xfxL

ncxu

xL

nectxu

n

tL

n

n

34

Jean Baptiste Joseph Fourier (1768 - 1830)

• Developed the equation for heat transmission and obtained solution under various boundary conditions (1800 - 1811).

• Under Napoleon he went to Egypt as a soldier and worked with G. Monge as a cultural attache for the French army.

35

Example

• Solve the following heat flow problem

• Write 3 sin 2x - 6 sin 5x = cn sin (n/L)x, and comparing the coefficients, we see that c2 = 3 , c5 = -6, and cn = 0 for all other n. And we have u(x,t) = u2(x,t) + u5(x,t) .

.0 , 5sin62sin3)0,(00, (),0(

.0 , 0 , 7 2

2

π x xxxu , , t t) utu

txxu

tu

36

Wave Equation

• In the study of vibrating string such as piano wire or guitar string.

37