Download Homogeneous Equations

Homogeneous Differential Equations:
Definition:
Homogeneous Equations:
A differential equation is said to be homogeneous if there is no isolated
constant term in the equation, e.g., each term in a differential equation for y has y or
some derivative of y in each term.
Ordinary differential equation
A differential equation is an equation which contains the derivatives of a variable, such as
the equation
here x is the variable and the derivatives are with respect to a second variable t. The
letters a, b, c and d are taken to be constants here. This equation would be described as a
second order, linear differential equation with constant coefficients. It is second order
because of the highest order derivative present, linear because none of the derivatives are
raised to a power, and the multipliers of the derivatives are constant. If x were the
position of an object and t the time, then the first derivative is the velocity, the second the
acceleration, and this would be an equation describing the motion of the object. As
shown, this is also said to be a non-homogeneous equation, and in solving physical
problems, one must also consider the homogeneous equation.
Let y be an unknown function
in x with y(n) the nth derivative of y, then an equation of the form
is called an ordinary differential equation (ODE) of order n; for vector valued
functions,
,
it is called a system of ordinary differential equations of dimension m.
When a differential equation of order n has the form
it is called an implicit differential equation whereas the form
is called an explicit differential equation.
A differential equation not depending on x is called autonomous.
A differential equation is said to be linear if F can be written as a linear combination of
the derivatives of y
with ai(x) and r(x) continuous functions in x. The function r(x) is called the source term;
if r(x)=0 then the linear differential equation is called homogeneous, otherwise it is
called non-homogeneous or inhomogeneous.
Explanation:
First Order Homogeneous DE
A first order homogeneous differential equation involves only the first derivative of a
function and the function itself, with constants only as multipliers. The equation is of the
form
and can be solved by the substitution
The solution which fits a specific physical situation is obtained by substituting the
solution into the equation and evaluating the various constants by forcing the solution to
fit the physical boundary conditions of the problem at hand. Substituting gives
Applications:
Homogeneous Linear Equations with Constant Coefficients
A second order homogeneous equation with constant coefficients is written as
Where a, b and c are constant. This type of equation is very useful in many applied
problems (physics, electrical engineering, etc..). Let us summarize the steps to follow in
order to find the general solution:
(1)
Write down the characteristic equation
This is a quadratic equation. Let
and
be its roots we have
;
(2)
If and are distinct real numbers (this happens if
general solution is
), then the
(3)
If
(which happens if
), then the general solution is
(4)
If and are complex numbers (which happens if
general solution is
Where
), then the
,
That is,
Example: Find the solution to the IVP
Solution: Let us follow the steps:
1
Characteristic equation and its roots
Since 4-8 = -4<0, we have complex roots
. Therefore,
and
;
2
General solution
;
3
In order to find the particular solution we use the initial conditions to determine
and . First, we have
.
Since
we get
,
From these two equations we get
,
Which implies?
Homogeneous equations with constant coefficients
The first method of solving linear ordinary differential equations with constant
coefficients is due to Euler, who realized that solutions have the form ezx, for possiblycomplex values of z. The exponential function is one of the few functions that keep its
shape even after differentiation. In order for the sum of multiple derivatives of a function
to sum up to zero, the derivatives must cancel each other out and the only way for them
to do so is for the derivatives to have the same form as the initial function. Thus, to solve
we set y = ezx, leading to
Division by e zx gives the nth-order polynomial
This equation F(z) = 0, is the "characteristic" equation considered later by Monge and
Cauchy.
Formally, the terms
of the original differential equation are replaced by zk. Solving the polynomial gives n
values of z, z1, ..., zn. Substitution of any of those values for z into e zx gives a solution e zix.
Since homogeneous linear differential equations obey the superposition principle, any
linear combination of these functions also satisfies the differential equation.
When these roots are all distinct, we have n distinct solutions to the differential equation.
It can be shown that these are linearly independent, by applying the Vandermonde
determinant, and together they form a basis of the space of all solutions of the differential
equation.
Examples
has the characteristic equation
This has zeroes, i, −i, and 1 (multiplicity 2). The
solution basis is then
This corresponds to the real-valued solution basis
The preceding gave a solution for the case when all zeros are distinct, that is, each has
multiplicity 1. For the general case, if z is a (possibly complex) zero (or root) of F(z)
having multiplicity m, then, for
,
is a solution of
the ODE. Applying this to all roots gives a collection of n distinct and linearly
independent functions, where n is the degree of F(z). As before, these functions make up
a basis of the solution space.
If the coefficients Ai of the differential equation are real, then real-valued solutions are
generally preferable. Since non-real roots z then come in conjugate pairs, so do their
corresponding basis functions xkezx, and the desired result is obtained by replacing each
pair with their real-valued linear combinations Re(y) and Im(y), where y is one of the
pair.
A case that involves complex roots can be solved with the aid of Euler's formula.
Examples
Given
. The characteristic equation is
which
has zeroes 2+i and 2−i. Thus the solution basis {y1,y2} is
a solution if and only if
for
Because the coefficients are real,


we are likely not interested in the complex solutions
our basis elements are mutual conjugates
. Now y is
.
The linear combinations
and
will give us a real basis in {u1,u2}.
Application:
Simple harmonic oscillator
The second order differential equation
D2y = − k2y,
which represents a simple harmonic oscillator, can be restated as
(D2 + k2)y = 0.
The expression in parenthesis can be factored out, yielding
(D + ik)(D − ik)y = 0,
which has a pair of linearly independent solutions, one for
(D − ik)y = 0
and another for
(D + ik)y = 0.
The solutions are, respectively,
y0 = A0eikx
and
y1 = A1e − ikx.
These solutions provide a basis for the two-dimensional "solution space" of the second
order differential equation: meaning that linear combinations of these solutions will also
be solutions. In particular, the following solutions can be constructed
and
These last two trigonometric solutions are linearly independent, so they can serve as
another basis for the solution space, yielding the following general solution:
yH = A0cos(kx) + A1sin(kx).
Damped harmonic oscillator
Given the equation for the damped harmonic oscillator:
the expression in parentheses can be factored out: first obtain the characteristic equation
by replacing D with λ. This equation must be satisfied for all y, thus:
Solve using the quadratic formula:
Use these data to factor out the original differential equation:
This implies a pair of solutions, one corresponding to
and another to
The solutions are, respectively,
and
where ω = b / 2m. From this linearly independent pair of solutions can be constructed
another linearly independent pair which thus serve as a basis for the two-dimensional
solution space:
However, if |ω| < |ω0| then it is preferable to get rid of the consequential imaginaries,
expressing the general solution as
This latter solution corresponds to the underdamped case, whereas the former one
corresponds to the overdamped case: the solutions for the underdamped case oscillate
whereas the solutions for the overdamped case do not.
Fredholm Theory:
Homogeneous equations
Much of Fredholm theory concerns itself with finding solutions for the integral equation
This equation arises naturally in many problems in physics and mathematics, as the
inverse of a differential equation. That is, one is asked to solve the differential equation
Lg(x) = f(x)
where the function f is given and g is unknown. Here, L stands for a linear differential
operator. For example, one might take L to be an elliptic operator, such as
in which case the equation to be solved become the Poisson equation. A general method
of solving such equations is by means of Green's functions, namely, rather than a direct
attack, one instead attempts to solve the equation
LK(x,y) = δ(x − y)
where δ(x) is the Dirac delta function. The desired solution to the differential equation is
then written as
This integral is written in the form of a Fredholm integral equation. The function K(x,y) is
variously known as a Green's function, or the kernel of an integral. It is sometimes called
the nucleus of the integral, whence the term nuclear operator arises.
In the general theory, x and y may be points on any manifold; the real number line or mdimensional Euclidean space in the simplest cases. The general theory also often requires
that the functions belong to some given function space: often, the space of squareintegrable functions is studied, and Sobolev spaces appear often.
The actual function space used is often determined by the solutions of the eigenvalue
problem of the differential operator; that is, by the solutions to
Lψn(x) = ωnψn(x)
where the ωn are the eigenvalues, and the ψn(x) are the eigenvectors. The set of
eigenvectors form a Banach space, and, when there is a natural inner product, then the
eigenvectors form a Hilbert space, at which point the Riesz representation theorem is
applied. Examples of such spaces are the orthogonal polynomials, which occur as the
solutions to a class of second-order ordinary differential equations.
Given a Hilbert space as above, the kernel may be written in the form
where
is the dual to ψn. In this form, the object K(x,y) is often called the Fredholm
operator or the Fredholm kernel. That this is the same kernel as before follows from the
completeness of the basis of the Hilbert space, namely, that one has
Since the ωn are generally increasing, the resulting eigenvalues of the operator K(x,y) are
thus seen to be decreasing towards zero.
Non Homogeneous differential equations:
Definition:
An equation which is not homogeneous is called non homogeneous.
Explanation:
First Order Non-homogeneous Differential Equation
An example of a first order linear non-homogeneous differential equation is
Having a non-zero value for the constant c is what makes this equation nonhomogeneous, and that adds a step to the process of solution. The path to a general
solution involves finding a solution to the homogeneous equation (i.e., drop off the
constant c), and then finding a particular solution to the non-homogeneous equation (i.e.,
find any solution with the constant c left in the equation). The solution to the
homogeneous equation is
By substitution you can verify that setting the function equal to the constant value -c/b
will satisfy the non-homogeneous equation.
It is the nature of differential equations that the sum of solutions is also a solution, so that
a general solution can be approached by taking the sum of the two solutions above. The
final requirement for the application of the solution to a physical problem is that the
solution fits the physical boundary conditions of the problem. The most common
situation in physical problems is that the boundary conditions are the values of the
function f(x) and its derivatives when x=0. Boundary conditions are often called "initial
conditions". For the first order equation, we need to specify one boundary condition. For
example:
Substituting at x=0 gives:
Inhomogenous equations
The inhomogenous Fredholm integral equation
may be written formally as
f = (K − ω)φ
which has the formal solution
A solution of this form is referred to as the resolvent formalism, where the resolvent is
defined as the operator
Given the collection of eigenvectors and eigenvalues of K, the resolvent may be given a
concrete form as
with the solution being
A necessary and sufficient condition for such a solution to exist is one of Fredholm's
theorems. The resolvent is commonly expanded in powers of λ = 1 / ω, in which case it is
known as the Liouville-Neumann series. In this case, the integral equation is written as
and the resolvent is written in the alternate form as
Applications:
Fredholm determinant
The Fredholm determinant is commonly defined as
where
and
and so on. The corresponding zeta function is
The zeta function can be thought of as the determinant of the resolvent.
The zeta function plays an important role in studying dynamical systems. Note that this is
the same general type of zeta function as the Riemann zeta function; however, in this
case, the corresponding kernel is not known. The existence of such a kernel is known as
the Hilbert-Polya conjecture.
Applications:
Charging a Capacitor
An application of non-homogeneous differential equations
A first order non-homogeneous differential equation
has a solution of the form :
.
For the process of charging a capacitor from zero charge with a battery, the equation is
.
Using the boundary condition Q=0 at t=0 and identifying the terms corresponding to the
general solution, the solutions for the charge on the capacitor and the current are:
.
In this example the constant B in the general solution had the value zero, but if the charge
on the capacitor had not been initially zero, the general solution would still give an
accurate description of the change of charge with time. The discharge of the capacitor is
an example of application of the homogeneous differential equation
Capacitor Discharge
An application of homogeneous differential equations
A first order homogeneous differential equation
Has a solution of the form:
.
For the process of discharging a capacitor which is initially charged to the voltage of a
battery, the equation is
.
Using the boundary condition and identifying the terms corresponding to the general
solution, the solutions for the charge on the capacitor and the current are:
.
Since the voltage on the capacitor during the discharge is strictly determined by the
charge on the capacitor, it follows the same pattern.
Nonhomogeneous equation with constant coefficients
To obtain the solution to the non-homogeneous equation (sometimes called
inhomogeneous equation), find a particular solution yP(x) by either the method of
undetermined coefficients or the method of variation of parameters; the general solution
to the linear differential equation is the sum of the general solution of the related
homogeneous equation and the particular solution.
Suppose we face
For later convenience, define the characteristic polynomial
We find the solution basis
as in the homogeneous (f=0) case. We now
seek a particular solution yp by the variation of parameters method. Let the
coefficients of the linear combination be functions of x:
Using the "operator" notation D = d / dx and a broad-minded use of notation, the ODE in
question is P(D)y = f; so
With the constraints
the parameters commute out, with a little "dirt":
But P(D)yj = 0, therefore
This, with the constraints, gives a linear system in the u'j. This much can always be
solved; in fact, combining Cramer's rule with the Wronskian,
The rest is a matter of integrating u'j.
The particular solution is not unique;
for any set of constants cj.
also satisfies the ODE
Example
Suppose y'' − 4y' + 5y = sin(kx). We take the solution basis found above {e(2 + i)x,e(2 − i)x}.
And so
(Notice that u1 and u2 had factors that canceled y1 and y2; that is typical.)
For interest's sake, this ODE has a physical interpretation as a driven damped harmonic
oscillator; yp represents the steady state, and c1y1 + c2y2 is the transient.
Equation with variable coefficients
A linear ODE of order n with variable coefficients has the general form
Examples
A particular simple example is the Cauchy-Euler equation often used in engineering
First order equation
Examples
with the initial condition
Using the general solution method:
The integration is done from 0 to x, giving:
Then we can reduce to:
Where κ is 2 from the initial condition.
A linear ODE of order 1 with variable coefficients has the general form
Dy(x) + f(x)y(x) = g(x).
Equations of this form can be solved by multiplying the integrating factor
throughout to obtain
which simplifies due to the product rule to
which, on integrating both sides, yields
In other words: The solution of a first-order linear ODE
y'(x) + f(x)y(x) = g(x),
with coefficients that may or may not vary with x, is:
where κ is the constant of integration, and
Examples
Consider a first order differential equation with constant coefficients:
This equation is particularly relevant to first order systems such as RC circuits and massdamper systems.
In this case, p(x) = b, r(x) = 1.
Hence its solution
Applications of homogeneous & Nonhmogeneous
equations to matrices:
You recall that a linear differential equation
was called homogeneous if
, and non-homogeneous or inhomogeneous
otherwise. We use the same terminology for systems of linear equations and for matrix
equations:
A matrix equation
is called homogeneous if is the zero vector (all entries are zero). A system of linear
equations is called homogeneous if the equivalent matrix equation is homogeneous.
Homogeneous matrix equations have some special properties:
1.
The matrix equation
Always has at least one solution, the zero solution
(Here 0 stands for a column vector all of whose entries are zero.)
2.
If the column vectors
and
are two solutions to the matrix equation
Then so is any linear combination of them,
.
3.
The complete solution to a matrix equation
Is always given in the form
where
,
, ...
are solutions and , , ...,
are parameters. The
number of parameters depends on the dimension of the ``solution space.''
You can see why property (1) holds; a system of linear equations like
Will always be satisfied by setting all the variables equal to zero. (This is the same reason
a homogeneous linear differential equation can always be satisfied by setting
Property (2) depends on the linearity of multiplication by
.)
. If
Then we have that
Property (3) also really comes from the linearity, since if we have
Then we have that
This is the same reason that the general solution to a homogeneous linear differential
equation is a linear combination of particular solutions, such as
In the case of differential equations, the number of different particular solutions, or the
number of constants in the general solution, depends on the order of the differential
equation; one solution for a first order equation, two different solutions for a second order
equation, etc. In the case of matrix equations, the number of particular solutions is the
number of paramters in the general or complete solution, the dimension of the solution
space.
We can also see property (3) in action by solving a matrix equation. Here's the equation:
The augmented matrix of this equation has the row echelon form
So we can write down the complete general solution
We can rewrite this as
The particular solutions from which we can put together this complete solution are
The really nice thing we get out of this is a method for finding solutions to nonhomogeneous systems of linear equations (or non-homogeneous matrix equations.) It
works exactly the same way as solutions for linear differential equations:
If the matrix equation
Have one particular solution
, and the associated homogeneous equation
Has the complete solution
homogeneous equation is
, then the complete solution to the original non-
Example:
Has the complete solution (which we computed earlier)
Which we can rewrite as
This is the sum of the solution to the associated homogeneous system, which we wrote
down in the previous example,
And a particular solution to this inhomogeneous system
Example: The homogeneous system of linear equations
Has the complete solution
The non-homogenous system
Has one particular solution
To get the complete solution to the non-homogeneous system
We add these together:
Exercise 1 Rewrite the systems of linear equations of Exercise 4 (in the last handout) as
matrix equations.
Exercise 2 Rewrite the following matrix equations as systems of linear equations.
Exercise 3 Carry out the following matrix multiplications, or explain why they cannot
be carried out.
Exercise 4 Solve the following matrix equations using row-reduction.
Exercise 5 What does it say about the set of solutions to a system of linear equations if
it’s augmented matrix, when put into row-reduced form:
(a.) Has a row whose leading entry is in the last column (the column corresponding to the
constant terms)?
(b.) has all zeroes in the last column?
(c.) Has a column, other than the last column, in which no row has a leading entry?
(d.) Has rows with leading entries in every column except the last one?
Exercise 6 Write down the associated homogeneous matrix equations for the matrix
equations in exercise 4. Now write down the complete solution to each of these
homogeneous matrix equations.
Exercise 7 The matrix equation
has one solution given by
Give the complete solution to this matrix equation.
From
Name
To
Roll.No.9022
Ahmad Sheeraz
Sir Atif Firdous