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Second order linear equations

Samy Tindel

Purdue University

Differential equations - MA 266

Taken fromElementary differential equations

by Boyce and DiPrima Samy T.Second order equationsDifferential equations 1 / 115

Outline

1Homogeneous equations with constant coefficients

2Homogeneous equations and Wronskian

3Complex roots of the characteristic equation

4Repeated roots, reduction of order

5Nonhomogeneous equations

6Variation of parameters

7Mechanical vibrations

8Forced vibrations

Samy T.Second order equationsDifferential equations 2 / 115

Second order differential equations

General form of equation:

d 2ydt 2=f? t,y,dydt Importance of second order equations:1Instructive methods of resolution

2Crucial for modeling in physics:

I

Fluid mechanics

IHeat transfer

IWave motion

IElectromagnetismSamy T.Second order equationsDifferential equations 3 / 115

Outline

1Homogeneous equations with constant coefficients

2Homogeneous equations and Wronskian

3Complex roots of the characteristic equation

4Repeated roots, reduction of order

5Nonhomogeneous equations

6Variation of parameters

7Mechanical vibrations

8Forced vibrations

Samy T.Second order equationsDifferential equations 4 / 115

General form of 2nd order linear equation

General form 1:

y ??+p(t)y?+q(t)y=g(t)

General form 2:

P(t)y??+Q(t)y?+R(t)y=G(t)

Remark:

2 forms are equivalent ifP(t)?=0

Initial condition:Given byy(t0) =y0andy?(t0) =y?0Two conditions necessary because two integrations performed

Samy T.Second order equationsDifferential equations 5 / 115

Homogeneous linear equations

Homogeneous equations:

When g≡0, that is

y ??+p(t)y?+q(t)y=0

Remark:

Nonhomogeneous solutions can be deduced from homogeneous ones

Homogeneous equations with constant coefficients:

ay ??+by?+cy=0, fora,b,c?R.Samy T.Second order equationsDifferential equations 6 / 115

Simple example

Equation:

y ??-y=0.(1)

Initial condition:

y(0) =2,andy?(0) =-1.

Two simple functions satisfying (1):

y=exp(t),andy=exp(-t).

Using linear form of (1):

fo rc1,c2?R, y=c1exp(t) +c2exp(-t). is solution to the equation. Samy T.Second order equationsDifferential equations 7 / 115

Simple example (2)

First conclusion:

We obtain an infinite family of solutions indexed byc1,c2.

Initial value problem:

with y(0) =2 andy?(0) =-1 we find ?c

1+c2=2

c

1-c2=-1

Solution:c1=12

andc2=32

Solution to initial value problem:

y=12 exp(t) +32 exp(-t).Samy T.Second order equationsDifferential equations 8 / 115

Generalization

Equation considered:

fo ra,b,c?R, ay ??+by?+cy=0.(2)

Characteristic equation:

ar

2+br+c=0.

Hypothesis:

Characteristic equation has 2 distinct real rootsr1,r2.

Conclusion:

general solution to (2) given b y: y=c1exp(r1t) +c2exp(r2t).(3)Proposition 1. Samy T.Second order equationsDifferential equations 9 / 115

Generalization: initial value

Initial value problem:

under assumptions of Prop osition1, ay ??+by?+cy=0,y(t0) =y0,y?(t0) =y?0.(4)

Solution to (4):

given b y y=c1exp(r1t) +c2exp(r2t), with c

1=y?0-y0r2r

1-r2exp(-r1t0),c2=y?0-y0r1r

2-r1exp(-r2t0)Samy T.Second order equationsDifferential equations 10 / 115

Example 1

Equation considered:

y ??+5y?+6y=0y(0) =2,y?(0) =3.(5)

Solution:

given b y y=9exp(-2t)-7exp(-3t)

Graph of solution:

Samy T.Second order equationsDifferential equations 11 / 115

Example 2

Equation considered:

4y??-8y?+3y=0y(0) =2,y?(0) =12

Solution:

given b y y=-12 exp?3t2 +52
exp?t2

Graph of solution:

Samy T.Second order equationsDifferential equations 12 / 115

Asymptotic behavior of solutions

3 cases:

under assumptions of Prop osition1, 1If bothr1,r2<0, then limt→∞y(t) =02Ifr1>0 orr2>0, exponential growth fory3Ifr1<0 andr2=0, then limt→∞y(t) =??RSamy T.Second order equationsDifferential equations 13 / 115

Outline

1Homogeneous equations with constant coefficients

2Homogeneous equations and Wronskian

3Complex roots of the characteristic equation

4Repeated roots, reduction of order

5Nonhomogeneous equations

6Variation of parameters

7Mechanical vibrations

8Forced vibrations

Samy T.Second order equationsDifferential equations 14 / 115

Definition of an operator

Let

I= (α,β), that is

We defineL[φ] :I→Rby:

L[φ] =φ??+pφ?+qφDefinition 2.

Samy T.Second order equationsDifferential equations 15 / 115

Homogeneous equation in terms ofL

Equation considered:

Under conditions of Definition 2,

L[y] =0??y??+py?+qy=0

Initial conditions:

fo rt0?I, y(t0) =y0,andy?(t0) =y?0.Samy T.Second order equationsDifferential equations 16 / 115

Existence and uniqueness theorem

General linear equation:

y ??+p(t)y?+q(t)y=g(t),y(t0) =y0,y?(t0) =y?0.(6)

Hypothesis:t

0?I, whereI= (α,β).p,qandgcontinuous onI.

Conclusion:

There exists a unique functionysatisfying equation (6) onI.Theorem 3. Samy T.Second order equationsDifferential equations 17 / 115

Existence and uniqueness theorem (2)

Important conclusions of the theorem:1There exists a solution to (6).

2There is only one solution.

3The solutionyis defined and twice differentiable onI.

Back to equation (5):We had existence part.

Uniqueness is harder to see.

Major difference with first order equations:No general formula for solution to (6). Samy T.Second order equationsDifferential equations 18 / 115

Example of maximal interval

Equation considered:

?t2-3t?y??+t y?-(t+3)y=0,y(1) =2,y?(1) =1.

Equivalent form:

y ??+1t-3y?+t+3t(t-3)y=0,y(1) =2,y?(1) =1.

Application of Theorem 3:g(t) =0 continuous onRp(t) =1t-3continuous on(-∞,3)?(3,∞)q(t) =t+3t(t-3)continuous on(-∞,0)?(0,3)?(3,∞)1?(0,3)

We thus get unique solution on(0,3)Samy T.Second order equationsDifferential equations 19 / 115

A trivial example of equation

Equation considered:

y ??+p(t)y?+q(t)y=0,y(t0) =0,y?(t0) =0

Hypothesis:pandqcontinuous onIt

0?I

Application of Theorem 3:1y≡0 solves equation.2According to Theorem 3 it is the unique solution.

Samy T.Second order equationsDifferential equations 20 / 115

Principle of superposition

Equation considered:

y ??+p(t)y?+q(t)y=0.(7)

Hypothesis:y

1andy2are 2 solutions to equation (7).c

1andc2are 2 constants.

Conclusion:

y=c1y1+c2y2also solves (7).Theorem 4.

Additional question:

Are all the solutions of the formy=c1y1+c2y2?Samy T.Second order equationsDifferential equations 21 / 115

Proof of Theorem 4

Step 1:

p rovethat

L[c1y1+c2y2] =c1L[y1] +c2L[y2].

Step 2:

W eobtain

L[y1] =0,L[y2] =0=?L[c1y1+c2y2] =0.Samy T.Second order equationsDifferential equations 22 / 115

Wronskian

Consider:

Equationy??+p(t)y?+q(t)y=0.Two solutionsy1,y2on intervalI.t 0?I.

The Wronskian fory1,y2att0is:

W=W[y1,y2](t0) =?

????y

1(t0)y2(t0)

y ?1(t0)y?2(t0)? ????.Definition 5. Samy T.Second order equationsDifferential equations 23 / 115

Wronskian and determination of solutions

Equation:

back to (6) tha tis

L[y] =y??+p(t)y?+q(t)y=0.

Hypothesis:Existence of two solutionsy1,y2.Initial conditiony(t0) =y0andy?(t0) =y?0assigned.

Conclusion:

One can find c1,c2such that

y=c1y1+c2y2 satisfies (6) with initial condition iff

W[y1,y2](t0)?=0Theorem 6.

Samy T.Second order equationsDifferential equations 24 / 115

Complement to Theorem 6

Expression forc1,c2:Under assumptions of Theo rem6 w ehave c 1=? ????y

0y2(t0)

y ?0y?2(t0)? ????y

1(t0)y2(t0)

y ?1(t0)y?2(t0)? ????,andc2=? ????y

1(t0)y0

y ?1(t0)y?0? ????y

1(t0)y2(t0)

y ?1(t0)y?2(t0)?

Justification:c1,c2are solution to the system

?c

1y1(t0) +c2y2(t0) =y0

c

1y?1(t0) +c2y?2(t0) =y?0Samy T.Second order equationsDifferential equations 25 / 115

Example 1

Equation considered:

back to (5), that is y ??+5y?+6y=0.

2 solutions:

given b y y

1=exp(-2t),andy2=exp(-3t)

Expression of Wronskian:

fo rt?R,

W[y1,y2](t) =exp(-5t).

Solving the equation:W[y1,y2](t)?=0 for allt?R

=?initial value problem can be solved at anyt?R.Samy T.Second order equationsDifferential equations 26 / 115

Wronskian and uniqueness of solutions

Equation:

back to (6) tha tis

L[y] =y??+p(t)y?+q(t)y=0.

Hypothesis:Existence of two solutionsy1,y2.

Conclusion:

The general solution

y=c1y1+c2y2,withc1,c2?R includes all solutions to (6) iff: there existst0?Isuch thatW[y1,y2](t0)?=0.Theorem 7. Samy T.Second order equationsDifferential equations 27 / 115

Example: equations with constant coefficients

Equation considered:

fo ra,b,c?R, ay ??+by?+cy=0.(8)quotesdbs_dbs47.pdfusesText_47

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