Mathe IIILecture 7Mathe IIILecture 7

2

Second Order Differential Equations

x = F t, x, x

2

2

d x d dxx = =

dt dt dt

The simplest possible equation of this type is:

x = k

by integrating : x = kt + A

integrating (2) : 2k x = t + At + B

2 for arbitrary A,B

3

consider the equation : x = F t, x F t, x, x

change the variable : y = x

y = F y , t,

This is a first order equation.

if is its general solutiony = g t

then by integrating we find

the general solution of

x = g t

4

Example:

x = x + t

y = y + t x y =

,-t -t tye ' = te y = x = Ae - t - 1

t 21x = Ae - t - t + B

2

5

If is continuous in the three variables ,

then for each three given numbers the

equation has a solution satisfying

0 0

0 0 0 0

F(t, x, x) t, x, x

t , x ,a

x = f t

dfx = f t ,a = f t = t

dt

x = F t, x, x

6

If are two solutions of the

equation then so is

for arbitr

homo

ary

geneou

s1 2

1 2

u (t),u (t)

Au (t)+ Bu (t) A,B

x + a t x + b t x = f t

Linear Equations

The corresponding homogeneous equation

x + a t x + b t x = 0

7

1 2 1 2

1 2

Au (t)+ Bu (t) + a(t) Au (t)+ Bu (t) +

+b(t) Au (t)+ Bu (t) =

Linear Homogeneous Equation

The set of solutions of the

equation is a linear space

x + a t x + b t x = 0

1 1 1

2 2 2

Au (t)+ a(t)Au (t)+ b(t)Au (t) +

+ Bu (t)+ a(t)Bu (t)+ b(t)Bu (t) =

0

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Linear Homogeneous Equation

if the two solution are proportional

i.e.

then is the

of the equation.

1 2

1 2

1 2

u (t),u (t)

u (t) αu (t),

Au (t)+ Bu (t

t

)

no

general solution

x + a t x + b t x = 0

2 1u (t) αu (t)

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Let be two solutions of the

equation

then is a solution of the

equation

1 2

1 2

w (t),w (t)

nonhomogeneous

w (t) - w (t)

homogeneous

x + a t x + b t x = f(t)

1 1 1

2 2 2

w (t)+ a(t)w (t)+ b(t)w (t)

w (t)+ a(t)w (t)+ b(t)w (t) =

= f(t)

= f(t) - f(t) = 0

is a solution of the

equation1 2

ho

w (t) - w (t)

mogeneous

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.1 2 1 2w (t) - w (t) = Au (t)+ Bu (t)

x + a t x + b t x = f(t)

where are two independent

solution of the homogeneous equation.1 2u (t),u (t)

.1 2 1 2w (t) = w (t)+ Au (t)+ Bu (t)

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x + a t x + b t x = f(t)

The general solution of the nonhomogeneous equation

is

where are two independent

solution of the homogeneous equation

and is a solution of the nonhomogeneous equation

1 2

1 2

Au (t)+ Bu (t)+ w*

u ,u

w*

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Linear Equations with constant coefficients

x + ax + bx = 0

The homogeneous equation:

search a solution of the form:

rte2 rt rt rtr e + are + be = 0

, rt rt rt 2 rte = re e = r e rt 2e r + ar + b = 0

2r + ar + b = 0

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2r + ar + b = 0

,2 21 2

1 1 1 1r = - a + a - b r = - a - a - b

2 4 2 4

The solutions:

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The general solution of the equation

is 1 2r t r tAe + Be

If :2a - 4b > 0x + ax + bx = 0

If :2a - 4b = 0

The general solution of the equation

is rtA+ Bt e

21,2

1 1r = - a a - b

2 4

1r = - a

2

If :2a - 4b < 0

cos

The general solution of the equation

is rtAe βt + B 21 1r = - a, β = b - a

2 4

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Example:

x - 3x = 02r - 3 = 0 r = 3

t 3 -t 3x = Ae + Be

x - 4x + 4x = 02r - 4r + 4 = 0 r = 2

2tx = A+ Bt e

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The nonhomogeneous equation

x + ax + bx = f(t) b 0

We need to find a particular solution

If : f(t) k

search for a constant solution

k

x(t)b

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The nonhomogeneous equation

x + ax + bx = f(t) b 0

If : is a polynomial f(t)

search for a polynomial solution

Example : 2 x - 4x + 4x = t + 2

try : 2 x(t) = At + Bt + C

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2

x + ax + bx =

2A - 4(2At + B)+ 4(At + Bt + C)

4A = 1

2= 4At (4B - 8A)t + (2A - 4B + 4C)2= 1t + 0t + 22= t + 2

4B - 8A = 02A - 4B + 4C = 2

1 1 7A = ,B = ,C =

4 2 8

2t 21 1 7x = (A+ Bt)e + t + t +

4 2 82t 21 1 7

x = + t +(A t +4 2

+ Bt)e8

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The nonhomogeneous equation

x + ax + bx = f(t) b 0

If : qt f(t) = pe

Search for a solution of the form : qtAe

, qt qt qt 2 qt(Ae ) = Aqe (Ae ) = Aq e

qt 2x + ax + bx = Ae (q + aq + b) qt= pe

2

pA =

q + aq + bif 2 q + aq + b 0

if 2 q + aq + b = 0

20

Look for a solution of th

e

form :

or qt 2 qtBte Ct e

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Stability:

The general solution to:

x + ax + bx = f(t)

1 2Au (t)+ Bu (t)+ w* (t)

is Globally Asymptotically Stable (Stable) if:

lim lim 1 2[Au (t)+ Bu (t)+ w* (t)] = w* (t)t ∞ t ∞

: limor 1 2[Au (t)+ Bu (t)] = 0t ∞

for all values of A,B

22

For second order linear equations with constant coefficients:

two independent solutions for the homogeneous equation are:

, orrt rte te

where is a solution of 2 r r + ar + b = 0

lim lim when

rt rt

t te te = 0 r < 0

1 2 1 2 r + r = -a, r r = b

a > 0 b > 0

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Constrained Maximization

Lagrange Multipliers

max f(x, y) s.t. g(x, y) = c

(x, y, ) = f(x, y) - g(x, y) - c L

At a maximum point of the original problem

the derivatives of the Lagrangian vanish (w.r.t. all variables).