2nd order diff equations

8/7/2019 2nd order diff equations

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SOLUTION OF LINEAR DIFFERENTIAL EQUATIONS

We want to solve differential equation of the form

(1)

with an initial condition

at time

.


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Theorem: If we can find any two linearly independent a

solutions

and

(2)

then the general solution of the system of equations (1)

starting from any given initial condition is

(3)

where the constants

and

are given by the initialcondition.

aTwo solutions are said to be linearly dependent if one is a constant multiple of the other,

i.e., if

!

"

and#

!

#

"

.


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Notice that in (1),

operates on the vector

to give the

vector

. Generally the derived vector is different from the

source vector, both in magnitude and direction.

Eigenvector: special directions in the state space such that if

the vector

is in that direction, the resultant vector

also

lies along the same direction. It only gets stretched or

squeezed.

Eigenvalue: the factor by which any eigenvector expands or

contracts when it is operated on by the matrix

.


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When the matrix

operates on the vector

, and if

happens to be an eigenvector, then we can write

where

is the eigenvalue. This yields

where

is the identity matrix of the same dimension as

.

This condition would be satisfied if the determinant

.


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Thus

This is called the characteristic equation, whose roots are the

eigenvalues. Thus, for a

matrix one gets a quadratic

equation which in general yields two eigenvalues.


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Finding Eigenvectors: By the definition of eigenvector,

This leads to the two equations

(4)

These two equations always turn out to be identical.


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SUMMARY

Eigenvalues are obtained by solving the equation

and the eigenvectors are obtained, for each real eigenvalue,

from the equation


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USING EIGENVECTORS TO SOLVE DIFF. EQNS.

The definition of eigenvector tells us that if an initial

condition is located on an eigenvector, then the

vector

remains along the same eigenvector and therefore the whole

solution also remains along that eigenvector.

The equation (4) may yield three different types of results:

1. eigenvalues real and distinct,

2. eigenvalues complex conjugate,

3. eigenvalue real and equal.


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Let

and

eigenvalues,

and

eigenvectors.

If we place an initial condition on

,

(dynamics is constrained along the eigendirection). This is

like an 1D differential equation

whose solution is

. Therefore the solution of the differentialequation

along the eigendirection is

Similarly, for any initial condition placed along

we have

another solution

"

Therefore the general solution can be constructed as

"

(5)


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EXAMPLE

Let the system of equations be given by

(6)

The matrix has eigenvalues and . For ,

the eigenvector is given by

. To choose any point on

this eigenvector, set

. This gives

. Thus


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For this initial condition the solution is

Similarly, for , the eigenvector is

. To

choose a point on this eigenvector, take

. This gives

. Thus the second eigenvector becomes

and the solution along this eigenvector becomes


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Hence the general solution of the system of differential

equations is

where the constants

and

are to be determined from the

initial condition.


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For example, if the initial condition is

at

, then

this equation gives

Solving, we get

and

. Thus the solution

of the differential equation with this initial condition is

or, in terms of the individual coordinates


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EIGENVALUES COMPLEX CONJUGATE

Complex eigenvalues always occur as complex conjugate

pairs. If

is an eigenvalue,

is also aneigenvalue. Let

be an eigenvector corresponding to the

eigenvalue

. This is a complex-valued vector. It is

easy to check that

, the conjugate of the vector

, isassociated with the eigenvalue

.

Though complex-valued eigenvectors cannot represent any

specific direction in the real-valued state-space, their physicalsignificance derives from the fact that the eigenvector

equation

holds.


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This allows us to obtain a solution as

(7)

Here the left hand side is a real-valued function and the right

hand side is complex valued, expressible in the form

(

) which is a linear combination of the functions

and .

Therefore the real part and the imaginary part must

individually be solutions of the differential equation. This

allows us to write two real-valued solutions from which the

general solution can be constructed.


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Let us illustrate this with the system of equations

(8)

for which the eigenvalues are

. For

the

eigenvector equation is

Thus the eigenvector direction is given by

. Tochoose a specific eigenvector, we take

, so that

is an eigenvector.


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Thus a complex-valued solution is

(9)

which is a linear combination of the real part and the

imaginary part.


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Hence the two linearly independent real-valued solutions are

and

Therefore a general solution is

(10)

The other eigenvalue supplies no new information, as it is thecomplex conjugate of the first one.


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As a final check, one can differentiate

and

from (10)to obtain back (8) which would imply that (10) is really a

solution for (8).

The case of imaginary roots is a special case of this solution,

where

. Thus the solutions for imaginary eigenvalues

are

(11)

Here also, the values of

and

have to be obtained from

the initial condition.


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To illustrate, let an initial condition be

. Substituting

this value of

at

in (11) we get

and

.

Thus the trajectory is given by

and

Thus each state variable follows a sinusoidal variation, while

lags behind

by

.

Signals and Networks


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EIGENVALUES ARE REAL AND EQUAL

We have only one real-valued eigenvector

associated with

the eigenvalue

. So we have only one solution

In this case the rule is to look for a second solution of the

form

(12)

so that the general solution is

(13)

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EXAMPLE

Let

(14)

Here both the eigenvalues are equal to 1. For this eigenvalue

the eigenvector equation is

, so that an eigenvector

. Thus one nontrivial solution is

We now seek another solution of the form (12) with

.

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Substituting this into (14) we get

This reduces to

Since these equations must hold independent of

, each termin the above equations must be zero.

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g

Hence

and

Solving these, we have

,

and

.

Since we can take any values of

and

satisfying these

equations, we take

and

. This gives another

linearly independent solution of (14) as

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g

Hence the general solution of (14) is

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Example 2:

L

R CE

i

v

KCL gives

or,

KVL gives

or,

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In matrix form,

The second order equation can be obtained in terms of

or

.

Differentiating the first equation,

or,

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We first take the homogenious part:

whose roots are

The critical value of the resistance is given by

or,

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The undamped (

) natural frequency of oscillation

Now we mould the equation in the form

which demands that

or,

Notice that in this case we have to define

as

.

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Now let us look at it from the point of view of the

homogenious first order equations

:

To calculate the eigenvalues,

.

or,

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This gives

which is the same equation as derived in the 2nd order case.

Therefore the eigenvalues are the same as the roots of the

characteristic equation.

In terms of

and

the eigenvalues are:

The eigenvalues will be real for

and complex conjugate

for

.

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Now let

H,

F, and

, for which the

eigenvalues are real:

and

.

Substituting the parameter values we have

as

First take

and calculate its corresponding

eigenvector.

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The two lines give

and

These are identical equations, giving the eigenvector

. Similarly, for the other eigenvalue

we get the eigendirection .

Now, we select specific (arbitrary) eigenvectors along the

eigendirections. Choose

A and

V as the first

eigenvector, and V and A along the second.

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v

i

1

2

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The solution is straightforward obtained as:

and

are to be obtained from the initial conditions.

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Now take

H,

F, and

, for which the

eigenvalues are complex conjugate:

.

Substituting the parameter values we have

as

First take

and calculate its corresponding

eigenvector.

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2nd order diff equations