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Ordinary Differential Equations – Linear differential equations…

Ordinary Differential Equations

5. Examples of linear differential equations and their applications We consider some examples of systems of linear differential equations with constant coefficients

1 11 1 1

2 21 1 2

1 1

......

...

n n

n n

n n nn

y a y a yy a y a y

y a y a yn

′ = + +′ = + +

′ = + +

(5.1)

and the linear n-th-order differential equation (n > 1) with constant coefficients (5.2) ( ) ( 1)

1 1 0... 0n nnx a x a x a x−− ′+ + + + =

We discuss suitable methods to compute solutions and the fundamental matrix. In particular, we compute the fundamental matrix for all planar systems (n = 2) and classify these systems according to their properties in phase space. We also introduce the notion of stability for linear differential equations.

5.1. Planar systems 1. Equation of small oscillations. We study the linear second-order differential equation 1 0 0x a x a x+ + = (5.3) for real coefficients a0, a1. This equation arises from problems in mechanics and electricity. Oscillations of a spring. A mass m coupled to a spring is submitted to a linear force. Newton’s equation of motion reads mx Dx= − where D is a positive constant. It is of the form (5.3) with a0 = D/m and a1= 0. If friction is taken into account then a1 is a positive constant. Small oscillations of the mathematical pendulum. We consider a weightless rod of length l attached at one end and bearing a point mass m at the other end. If x denotes the angle by which the rod deviates from the vertical then . For small deviations sin(x) ≈ x and we get an equation of the form (5.3) with a

( )2 sinml x mgl x= −

0 = g/l and a1= 0.

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Ordinary Differential Equations – Linear differential equations…

Again, if we take into account a force due to friction that is proportional to the velocity then a0 = a1 is a positive constant. Inverted pendulum. If we turn the pendulum upside down, then the sign of the force due to gravitation changes sign and therefore a0 = - g/l. Electrical circuits. The charge y of a condenser of capacity C in a closed electrical circuit with inductivity L and resistance R satisfies the second order differential equation

which is of the form (5.3) with a1 0Ly Ry C y−+ + = 0 = 1/LC and a1= R/L. 1.1. Transformation to a first-order system. The second-order differential equation (5.3) is equivalent to a two-component linear first-order system with matrix A given by

(5.4) 0 1

0 1A

a a

= − −

It is convenient to rewrite A using the trace and the determinant or its eigenvalues. Then

( ) ( ) 1 2 1 2

0 1 0 1det

AA tr A λ λ λ λ

= = − − +

(5.5)

where

( ) ( ) ( )

( ) ( ) ( )

2

1

2

2

4det2

4det2

tr A tr A A

tr A tr A A

λ

λ

− −=

+ −=

(5.6)

are the (possibly complex) eigenvalues of A. If there is no friction (no resistance), then tr(A) = 0 which means that the system is area preserving in phase space. 1.2. Equation in phase space. The equation (5.3) describes a motion in the two-dimensional phase space with coordinates ,x y x=

)

. The matrix A defines the direction field, which is a composition of a rotation and a dilation of the plane. The orbit of a solution is the set of points {( }( ), ( ) ,x t y t t∈ or if we consider y as a function of x the set of points {x, y(x)} where y is (at least formally) a solution of the non-linear first-order differential equation

( ) ( )0 1( ) det

( ) ( )dy x x xa a A tr

dx y x y x= − − = − + A

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Ordinary Differential Equations – Linear differential equations…

which is of homogeneous type and can be solved (at least in principle) by elementary integration methods (see ch.2.3). Another important quantity defined on the phase space is the energy of the system. The energy functional associated to (5.3) is defined by ( ) 2

01 12 2,E x x x a x= + 2 (5.7)

If x(t) is a solution of (5.3) then

( ) ( ) ( )20 1,d E x x x x a x a x tr A x

dt= + = − = 2 (5.8)

If there is no friction ( a1 = -tr(A) = 0) the energy of a solution remains constant and the solution follows the lines of constant energy in phase space. In this case the system is called conservative. Since the equation is linear the energy defines a quadratic form and the orbits are either ellipses (det(A) = a0 > 0) , hyperbolas (det(A) = a0 < 0), or straight lines (det(A) = a0 = 0). In the presence of friction in a mechanical system or electrical resistance in an electrical circuit ( a1 = -tr(A) > 0) the energy is decreasing. 1.3. Explicit solutions of the second order equation. According to the results of the previous chapter (theorem 4.7) the general complex valued solution is of the form

( ) ( ) ( )( ) ( ) ( )

1 1 2 2 1 2

1 2 1

exp exp ,

exp exp ,

x t c t c t

x t c t c t t

λ λ λ

2

λ

λ λ λ λ

= + ≠

= + = λ≡ (5.9)

The corresponding real solutions are listed in the following table. λ1, λ2 real , tr(A)2 ≥ 4det(A)

λ1 ≠ λ2 tr(A)2 > 4det(A)

λ1 = λ2 = λ, tr(A)2 = 4det(A)

( ) ( ) ( )1 1 2exp exp 2x t c t c tλ λ= + ( ) ( ) (1 2exp exp )x t c t c t tλ λ= + λ1, λ2 complex conjugate, tr(A)2 < 4det(A)

λ1 = µ – iω, λ2 = µ + iω

λ1 = – iω, λ2 = iω, tr(A) = 0

( ) ( ) ( ) ( )( )1 2exp sin cosx t t c t c tµ ω ω= + or ( ) ( ) (1 2sin cos )x t c t c tω ω= + or

( ) ( ) ( )exp sinx t C t tµ ω δ= + ( ) ( )sinx t C tω δ= + The inverted pendulum belongs to the first class since det(A) = a0 < 0. In all other models det(A) = a0 > 0 and damping -a1 = tr(A) ≤ 0 For weak damping i.e. small tr(A) ≤ 0 there are two complex conjugate, critical damping corresponds to a double real eigenvalue If both eigenvalues are real and different the system is called overdamped.

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Ordinary Differential Equations – Linear differential equations…

1.4. Computing exp(tA). A complex valued fundamental matrix of the associated first order system is easily computed if the eigenvalues are different. The matrix A defined in (5.4) resp. (5.5) has eigenvectors

1 21 2

1 1,v v

λ λ

= =

Therefore a fundamental matrix Y(t) is given by

( ) ( ) ( )( ) ( ) ( )( ) ( )

11 1 2 2

1 1 2 2

exp expexp exp

exp expt t

Y t v t v tt t

λ λλ λ

λ λ λ λ

= =

2 (5.10)

Since

( ) ( )2 1 1 2 2 1 2 1 2 1

1 0,

0 1v v v vλ λ λ λ λ λ

− = − − = −

we get

( ) ( )1 2 2 1

1 2 2 1

2 1

2 1 2 1 2 1

e e e e1expe e e e

t t t t

t t t ttAλ λ λ λ

λ λ λ λ

λ λ

λ λ λ λ λ λ

− −= − − −

(5.11)

If λ1 = µ – iω, λ2 = µ + iω (5.11) reads

( )( ) ( ) ( )( )( ) ( ) ( )2 2

cos sin sinexp

sin cos sin

t t t tetAt t

µ ω ω µ ω ω

ω ω µ ω ω ω µ ω

− = + + t

(5.12)

To compute exp(tA) in case λ1 = λ2 = λ we let λ1 → λ2 = λ . We obtain

( ) 2

1exp e

1t t t

tAt t

λ λλ λ−

= − + (5.13)

The direct computation goes as follows: By theorem 4.7. the functions exp(λt) and t⋅exp(λt) are solutions of the second order equation. The corresponding solution vectors are

1 2

1e , e

1t t t

v vt

λ λ

λ λ

= = +

from which we easily construct a fundamental matrix.

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Ordinary Differential Equations – Linear differential equations…

2. General first order systems and their classification. Let A be an arbitrary real 2×2 matrix. Then A is either similar to the matrix given in (5.5) or A is a multiple of the unit matrix E. In the first case all results are similar to the results for the second order equation. To classify the systems it is sufficient to consider their normal forms, which are one of the following real matrices:

1

2

0 1, ,

0 0λ λ µ ω

λ λ ω µ −

U

U

(5.14)

where we used the notations introduced above. Diagonal case λ1 < λ2 < 0 λ1 < 0 < λ2 0 < λ1 < λ2

Stable node Saddle point nstable node

λ1 = λ2 < 0 λ1 < 0 = λ2 0 < λ1 = λ2

Stable star Marginally stable nstable star

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Ordinary Differential Equations – Linear differential equations…

Jordan case 0 1 = λ2 = λ = 0 = λ2 = λ = 0

Stable node Marginally unstable Unstable node

Complex case = 0 , ω > 0 ω > 0

Stable focus Center Unstable focus

λ1 = λ2 = λ < λ λ1

µ < 0 , ω > 0 µ µ > 0 ,

.1. Stability. We precise the meaning of stability and instability of linear equations. We

ble if ble

part,

ich

efinition: The stationary solution y(t) = 0 is called stable or Lyapunov stable if for every ε > 0 there is a δ > 0 such that for every y0 with | y0| < δ the solution y(t) with

2note that y(t) = 0 is always a stationary solution of a linear homogeneous system. We define stability and instability as a property of solutions with initial conditions close tothe zero solution. We say that y(t) = 0 is stable if all solutions remain bounded, asymptotically stable if all solutions tend to zero as t tends to infinity, and unstathere is an unbounded solution of the linear equation. We have seen that y(t) = 0 is staif there is no eigenvalue with positive real part ( with exception of the marginally unstable Jordan block), asymptotically stable if all eigenvalues have negative real and unstable if there is an eigenvalue with positive real part or a marginally unstable Jordan block. A more appropriate formulation of stability of a stationary solution, whalso extends to nonlinear systems is the concept of Lyapunov stability. D

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Ordinary Differential Equations – Linear differential equations…

initial condition y(t) = y0 exists for all t > 0 and satisfies | y(t) | < ε for all t > 0. The stationary solution is called asymptotically stable if in addition all solutions satisfy lim ( ) 0t

y t→∞

= .

As an application of the linear result we present the following result for nonlinear planar utonomous systems.

and

a Theorem 1: Let A M∈ ( )2 2,× ( )22 2: , ( )g g y O→ = y a differentiable

r the differentia ation function. Conside l equ y Ay′ = ( )g y+

I the stationary solution y = 0 of (5.15)

(5.15)

f all eigenvalues of A have negative real parts then asymptotically stable.

tability result will be a beautiful application of Gronwall-s equality. If y(t) is a solution of (5.15) then it satisfies the integral equation

(5.16)

Let |.| be a norm on R2. Since all eigenvalues of A have negative real parts there exist

positive constants a

is Proof: The proof of the sin

t

( ) ( ) ( ) ( )00

exp exp ( ) ( )y t At y A t s g y s ds= + −∫

, C such that ( )exp atAt x Ce x−≤ for any x ∈ R2.

n on g there is a positive constant K such that By the assumptio ( ) 2g y K y≤ if y r≤ .

Therefore, if y r≤ , then

( )g y Kr y≤

L

ater we shall choose r sufficiently small such that the constant Kr will be small. Taking e norm on both sides of (5.16) and using these estimates we obtain the following th

integral inequality which holds for all t such that y r≤ in [0,t]:

t

( ) ( )0

0

( )a t saty t Ce y CKr e y s ds− −−≤ + ∫

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Ordinary Differential Equations – Linear differential equations…

We set ( ) ( ) atz t y t e=

0

By Gronwall’s inequality ( ) CKrtz t C y e≤ .

. Then z(t) satisfies the integral inequality

0( ) ( )t

z t C y CKr z s ds≤ + ∫

W0

C

e choose r sufficiently small such that CKr = a/2. We get

20( ) aty t y e−≤

we choose y0 such that If 0y r≤ C , then the above inequality shows in particular that

( )y t r≤

Remark: It is clear that th also holds for higher dimensional autonomous

for all t > 0 concluding the proof.

eorem 1 systems.

tically et(A) = a0 > 0 and -a1 = tr(A) < 0. In this case the energy functional defined in

.7) is positive definite and strictly decreasing for nontrivial solutions. Functionals on hase space having this property will be called Lyapunov functionals. The existence of an

on

escribes a basic model for studying the dynamics of a drug in the body where A1 and A2 present the drug concentration in tissue and blood, re ectively. The rate k0 is due to

removal from the body by the kidneys. The corresponding system of differential quations is then

2y

For the second order differential equation (5.3) the trivial solution x = 0 is asymptostable if d(5pappropriate Lyapunov functional will imply the stability result. 3. Applications 3. 1. An example from pharmacokinetics. The reaction equati 21 0

121 2 0k k

kA A→ →←

re sp

d

e

1 12 2 21 1y k y k y= −= − + −

2 12 2 21 1 0y k y k y k

(5.17)

Hence

(5.18)

21 12

21 12 0

k kA

k k k−

= − −

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Ordinary Differential Equations – Linear differential equations…

and tr(A) < 0, det(A) = k21k0 > 0. The trivial solution is asymptotically stable: the drug ill completely disappear as one expects.

. 2. The mathematical pendulum with friction. The equation for the mathematical

w

3pendulum with friction is given by sinx x glγ+ = −

( )1 x−

(5.19)

y theorem 1 the stationary solution x = 0 is asymptotically stable.

ficients

e consider the differential equation

B

5.2. How to solve linear equations with constant coef W y Ay′ = for a real n×n matrix A. While it is straightforward to find all solutions if all eigenvalues f A have algebraic multiplicity one (see theorem putation of solutions in the ase of multiple eigenvalues is a little bit more involved.

step 1) Compute the characteristic polynomial P (σ) and compute its zeros which are the

) For simple eigenvalues λ find the corresponding eigenvectors w. construct the

f complex conjugate eigenvalues

o 4.5) the comc 1. A solution algorithm. We describe a routine to compute a fundamental system of any linear differential equations with constant coefficients. ( Aeigenvalues of A. (step 2solution as in theorem 4.5, that is set ( ) ty t e wλ= and take real and imaginary part in case o ,λ λ . (step 3)Let λ be an eigenvalue with m k > 1. First of all, solve for correspondingeigenvectors, i.e. solve

ultiplicity

( ) 0A E wλ− =

(oT

The number of linear independent solutions n1 is equal to the dimension of the nullspace

r kernel) of A – λE . he solutions of the differential equation are constructed as in (step 2) yielding n1 linearly dependent solutions for the eigenvalue λ.

you have found all solutions for this eigenvalue. If n1 < k, compute the nullspace of (A – λE)2, i.e. solve

in (step 4) If n1 = k, then

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Ordinary Differential Equations – Linear differential equations…

( )2 0A E wλ− = The number of linear independent solutions n2 is equal to the dimension of the nullspace

r kernel) of (A – λE)2. Take the n2 – n1 linear independent solutions, which do not belong to the nullspace of A – λE , i.e. (o

( ) 0A E wλ− ≠ . The corresponding solution y(t) of the differential equation is given by y(t) = exp(At)w. It can be explicitly computed as follows:

( ) ( )( )exp exptA w t E A E wλ λ= + −

( ) ( )( )

( )( exp exp

t

t E t A E w

e E tλ

λ λ= −

= +

( )

( )

2212 ...

t t

A E t A E w

e w te A E wλ λ

λ λ

λ

− + − +

= + −

T

)

he last equality is a consequence of ( ) 0nA E wλ− =

lex conjugate eigenvalues

for all n > 1. Again take real and

imaginary part in case of comp ,λ λ

If n2 < k, then compute the nullspace o proceed as before. (step 6) Repeat the procedure for the following powers of λE

.

) If n2 = k, then you have found all solutions for this eigenvalue. f (A – λE)3 and

A – until the dimension of e nullspace and hence the number of independent solution of the differential equation

. Examples. We present a few explicit computations of solution in the case of multiple

the Cauchy problem

(5.20)

If A denotes the matrix in (5.20) then PA(λ) = (λ – 1)3 and therefore λ = 1 is a triple igenvalue of A. Let {e1, e2, e3} denote the standard orthono

compute

(step 5

thequals k. 2eigenvalues. 2.1. Solve

1 2 1 1− 0 1 1 , (0) 20 0 1 3

y y y ′ = =

e rmal basis of R3. We

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Ordinary Differential Equations – Linear differential equations…

0 2 10 0 1A E

− − =

We see that the nullspace of A – E is spanned by e1. Hence y1(t) = exp(t) e1 solves (5.20). e compute (A – E)2.

0 0 0

W

( ) 0 0 0A E 20 0 2

0 0 0

d e

− =

The nullspace of (A – E)2 is spanned by e1 and e2. Only e2 yields a new solution of the ifferential equation. According to the general procedur described above it is given by

( )2 2 2 2( ) 2t t ty t e e te A E e e e te e= + − = +

R

1t

3. The third linear independent solution is

We note that since (A – E)3 = 0 its nullspace is R3. Take e3 as the third vector spanning

( ) ( ) 3( ) 0 1 01 0

y t e e te A E e t e A E e e te t e= + − + − = + +

22 23 3 3

12

0 1 1t t t t t t

−

0

We have computed the fundamental matrix

( ) ( )2

1 2 3

1 2( ), ( ), ( ) 0 1

0 0 1

t

t t tY t y t y t y t e t

−

= =

Since Y(0) = E we have Y(t) = exp(tA). The solution of the Cauchy problem is therefore

+

+ (5.21)

2.2. Solve the Cauchy problem

2 21 2 1 1 3( ) 0 1 2 2 3t t

t t t t ty t e t e t

− + = =

0 0 1 3 3

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Ordinary Differential Equations – Linear differential equations…

1 1 1 11 1 1 , (0) 22 2 2 1

y y − − − −

(5.22)

If A denotes the matrix in (5.20) then PA(λ) = λ 3 and therefore λ = 0 is a triple eigenvalue of A. The nullspace of B is spanned by the vectors

y′ = =

1 2 00 1

1 11 ,w w− − = =

Hence y1(t) = w1 and y2(t) = w2 solve (5.22). We note that since A2 = 0 its nullspace is R3.

ake w3 = w2 = (0,1,0) as the third vector spanning R3. The third linear independent

e have computed the fundamental matrix

+

We easily check that

t

The solution of the Cauchy problem is therefore

1 2

t

T

solution is

0 1 t 3 2 2( ) 1 1 1

0 2 2y t e tAe t t

t

= + = + = + − −

W

( ) ( ( ), ( ),Y t y t y t y= )1 2 3

1 1( ) 1 0 1

0 1 2

tt t

t

− − = −

( ) (exp tA y= )3 1 3 2 1 3

1( ) ( ), ( ), ( ) ( ) ( ) 1

2 2 1 2

t t tt y t y t y t y t y t t t t

t t

+ − − + = + − − −

1 1

2 2 22 2 1 2 1 1 4

t t t tt

t t t

+ + = +

− − − − − −

( ) 1y t t t t= +

(5.23)

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Ordinary Differential Equations – Linear differential equations…

2.3. Consider the equations of motion for two coupled pendulums

( )( )

x ax k x yy ay k y x= − − −

= − − −

(5.24)

here a and k are positive constants. The matrix of the corresponding first-order system for the vector

w

( ), , , Tx x y y

0 1 0 0a k

A

=

is given by

0 00 0 0 1

0 0

k

k a k

− − −

p

−

it has degenerate purely imaginary eigenvalues , 2i a i a k± ± +

(

( )2 ( )

x y a x

x y a k x y

+ = − +

− = − + − (5.25)

and can be solved by the rocedure described above. In this case, however, it is more convenient to “decouple” the

equations of motion by considering the equations for the sum and the difference of x and y, which are given by

)y

which are easily solved.

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