Download Basic Theory of Linear ODE`s

Basic Theory of Linear O.D.E.’s
Definition: A linear O.D.E. of order n, in the independent variable x and the dependent
variable y is an equation that can be expressed in the form:
a 0 (x)y( n ) + a1 (x)y( n!1) + ..... + a n!1 (x)y" + a n (x)y = b(x)
We will assume that a0(x), a1(x), …. , an(x) and b(x) are continuous real-valued functions
on the interval [a, b] and a0(x) ≠ 0 for at least one x in [a, b].
Remark: A linear ODE satisfies the following conditions:
1- The dependent variable y and its derivatives occur to the first degree only.
2- No products of y and/or any of its derivatives appear in the equation.
3- No transcendental functions of y and/or its derivatives occur.
Definition: Equations that not satisfy the above conditions are called nonlinear.
Remark:
1) The functions a0(x), a1(x), …. , an(x) are called the coefficients of the equation.
2) The right-hand member b(x) is called the non-homogeneous term.
3) If b(x) = 0, then the equation reduces to
a 0 (x)y( n ) + a1 (x)y( n!1) + ..... + a n!1 (x)y" + a n (x)y = 0
and it is called a homogeneous linear differential equation (H.L.D.E.).
Example:
d 3y
d2y
dy
!
cos(x)
+ 3x + x 3y = e x is a linear differential equation of order 3,
1)
3
2
dx
dx
dx
where a0(x) = 1, a2(x) = - cos(x), a3(x) = 3x, a4(x) = x3 and b(x) = ex, they are all
continuous function on ! .
Definition: If a0, a1, …. , an are all constants, then we have a linear differential equation
with constant coefficients, otherwise we have an equation with variable coefficients.
Example:
1) 3y’’’ + 2y’’ – 4y’ – 7y = sin x is an equation with constant coefficients.
2) 2y’’ + xy’ – ex y = 0 is an equation with variable coefficients.
Theorem: Let
a 0 (x)y( n ) + a1 (x)y( n!1) + ..... + a n!1 (x)y" + a n (x)y = b(x)
with a0(x), a1(x), …. , an(x) and b(x) are continuous real-valued functions on the interval
[a, b] and a0(x) ≠ 0 for at least one x in [a, b].
Let x0 be a point on the interval [a, b] and let c0, c1, cn-1 be n arbitrary constants.
Conclusion:
There exists a unique function f(x) defined on [a, b] such that f(x) is a solution of the
linear differential equation and satisfies
f(x0) = c0, f’(x0) = c1, f’’(x0) = c2, … , f(n-1)(x0) = cn-1.
Remark: The theorem extends the I.V.P. discussed for first order differential equations
to linear differential equations of order n.
Example:
Consider the I.V.P.
" d2y
dy
3
x
$ 2 + 3x + x y = e
$ dx
dx
#
$ y(1) = 2
$%y '(1) = !5
a0(x) = 1, a1(x) = 3x, a3(x) = x3, and b(x) = ex, they are continuous functions everywhere.
We have a linear differential equation order 2. The real numbers c0 = 2 and c1 = -5.
Then, according to the theorem there exists an unique function f(x) continuous
everywhere that is the solution of the equation and f(1) = 2 and f(x) = -5.
Corollary: Given the H.L.D.E.
a 0 (x)y( n ) + a1 (x)y( n!1) + ..... + a n!1 (x)y" + a n (x)y = 0
with a0(x), a1(x), …. , an(x) are continuous real-valued functions on the interval [a, b] and
a0(x) ≠ 0 for at least one x in [a, b].
Let f(x) be a solution of the H.L.D.E. such that f(x0) = 0, f’(x0) = 0, f’’(x0) = 0, … ,
f(n-1)(x0) = 0 where x0 is in [a, b],
Conclusion:
f(x) = 0 for all x in [a, b].
Example:
y’’’ + 2y’’ + 4xy’ + x2y = 0
y(2) = y’(2) = y’’(2) = 0
The unique solution that satisfies the I.V.P. is f(x) = 0
Definition: Given k functions defined on an interval [a, b], f1(x), f2(x), … , fk(x), and k
constants, c1, c2, … , ck. The expression
c1f1(x) + c2f2(x) + … + ckfk(x)
is called a linear combination of f1(x), f2(x), … , fk(x).
Basic Theorem on Homogeneous Linear Differential Equations
Given the n-th order H.L.D.E.
a 0 (x)y( n ) + a1 (x)y( n!1) + ..... + a n!1 (x)y" + a n (x)y = 0
let f1(x), f2(x), … , fk(x) be k solutions of the H.L.D.E. on the interval [a, b].
Conclusion:
The linear combination c1f1(x) + c2f2(x) + … + ckfk(x) is a solution of H.L.D.E. on
[a, b] where c1, c2, … , ck are k arbitrary constants.
Example:
d2y
+y = 0,
dx 2
The functions f1(x) = sin x and f2(x) = cos x are solutions of the equation, since taking
the second derivative of the function f1'' (x) = ! sin!x and substituting into the equation,
we get: –sin x + sin x = 0. Similarly for the other function.
Then g(x) = c1 sin x + c2 cos x is a solution of the equation where c1 and c2 are any
arbitrary constant: g’’(x) = - c1 sinx – c2 cos x, substituting into the equation we get
- c1 sinx – c2 cos x + c1 sinx + c2 cos x = 0.
Given the H.L.D.E.
Definition: n functions defined on an interval [a, b], f1(x), f2(x), … , fn(x) are called
linearly dependent on [a, b] if there exist constants c1, c2, … , cn, not all of them zero,
such that the linear combination
c1f1(x) + c2f2(x) + … + cnfn(x) = 0
for all x in [a, b].
Example:
1) The functions f1(x) = x2 and f2(x) = 4x2 are linearly dependent on [-1, 1] since
there exist constants c1 = - 4 and c2 = 1 such that c1f1(x) + c2f2(x) = -4x2 + 4x2 = 0 for all
x in [-1, 1].
2) The functions f1(x) = ex , f2(x) = - ex and f3(x) = 8ex are linearly dependent on [-10, 10]
since there exist constants c1 = 5, c2 = 1 and c3 = 1 such that
c1f1(x) + c2f2(x) + c3f3(x) = 5ex – ex + 4ex = 0 for all x in [-10, 10].
Definition: n functions defined on an interval [a, b], f1(x), f2(x), … , fn(x) are called
linearly independent on [a, b] if they are NOT linear dependent on [a, b].
This means that f1(x), f2(x), … , fn(x) are linearly independent on [a, b] if the relation
c1f1(x) + c2f2(x) + … + cnfn(x) = 0 for all x in [a, b]
implies that c1 = c2 = … = cn = 0.
Example:
The functions f1(x) = x2 , f2(x) = x3 and f3(x) = x5 are linearly independent on [-5, 5] since
the relation
c1f1(x) + c2f2(x) + c3f3(x) = c1x2 + c2x3 + c3x5 = 0 for all x in [-5, 5]
if and only if c1 = c2 = c3 = 0.
Theorem: The n-th order H.L.D.E.
a 0 (x)y( n ) + a1 (x)y( n!1) + ..... + a n!1 (x)y" + a n (x)y = 0
always possesses n solutions f1(x), f2(x), … , fn(x) that are linearly independent.
Furthermore, if f1(x), f2(x), … , fn(x) are n linearly independent solutions of the equation,
then every solution f(x) of the equation can be expressed as a linear combination
f(x) = c1f1(x) + c2f2(x) + … + cnfn(x)
of those n solutions by a proper choice of constants c1, c2, … , cn.
Example:
Given the 2nd order H.L.D.E. y’’ + y = 0,
the functions f1(x) = sin x and f2(x) = cos x are two linearly independent solutions of the
equation on (-∞, ∞).
If f(x) is any other solution of the equation then f(x) = c1 sin x + c2 cos x, for particular
values of the constants c1 and c2.
For example f(x) = cos(x – π/3) is a solution of the equation,
Since cos(x – π/3) = cos x cos(π/3) + sin x sin(π/3) = ½ cos x + 3 2 sin x, then
f(x) =
3
2
sin x + ½ cos x =
3
2
f1(x) + ½ f2(x)
Definition: If f1(x), f2(x), … , fn(x) are n linearly independent solutions of the H.L.D.E.
a 0 (x)y( n ) + a1 (x)y( n!1) + ..... + a n!1 (x)y" + a n (x)y = 0
on [a, b], then the set { f1(x), f2(x), … , fn(x)} is called the fundamental set of solutions
of the equation.
The function f(x) = c1f1(x) + c2f2(x) + … + cnfn(x) where c1, c2, … , cn are arbitrary
constants is called a general solution of the equation.
Example:
Given the 3rd order H.L.D.E. y’’’ - 2y’’ – y’ + 2y = 0,
the functions f1(x) = ex, f2(x) = e-x, and f3(x) = e2x are three linearly independent solutions
of the equation on (-∞, ∞).
The set {ex, e-x, e2x} is the fundamental set of solutions.
Then f(x) = c1 ex+ c2 e-x + c3 e2x, where c1, c2 and c3 are arbitrary constants is a general
solution of the equation.
Definition: Let f1(x), f2(x), … , fn(x) be n real-valued functions defined on [a, b] such that
the (n -1)-st derivative of each function exists in [a, b].
The determinant
f1 (x)
f2 (x) !
fn (x)
W ( f1, f2 ,..., fn ) (x) =
f1' (x)
!
f2' (x)
!
!
!
fn' (x)
!
f1(n!1) (x) f2(n!1) (x) ! fn(n!1) (x)
is called the Wronskian of f1(x), f2(x), … , fn(x).
Remark: The Wronskian is also a function of x.
Theorem: The n solutions f1(x), f2(x), … , fn(x) of the n-th H.L.D.E.
a 0 (x)y( n ) + a1 (x)y( n!1) + ..... + a n!1 (x)y" + a n (x)y = 0
are linearly independent on [a, b] if and only if W(f1, f2, … , fn) (x) ≠ 0 for at least one x
on [a, b].
Example:
1) Show that the functions f1(x) = ex, f2(x) = e-x, and f3(x) = e2x are three linearly
independent solutions of the 3rd order H.L.D.E. y’’’ - 2y’’ – y’ + 2y = 0, on (-∞, ∞).
Let’s compute the Wronskian
e x e !x
e 2x
1 1 1
(
)
W e x , e !x , e 2x = e x
!e !x
ex
e !x
2e 2x = e x e !x e 2x 1 !1 2 = e 2x (!6) " 0 for all x.
1 1 4
4e 2x
2) Show that the functions f1(x) = x2, f2(x) = x3, and f3(x) = x -2 are three linearly
independent solutions of the 3rd order H.L.D.E. x3y’’’ – 6xy’ + 12y = 0, on (-∞, ∞).
Let’s compute the Wronskian
x2 x3
x !2
2
!3
2
!2x !3
2 3 !2
2
!3
2 3x
3 2x !2x
!2 2x 3x
W(x , x , x ) = 2x 3x !2x = x
!x
+x
!4
!4
2 6x
6x
6x
2
6x
2 6x 6x !4
(
)
(
)
(
)
= x 2 18x !2 + 12x !2 ! x 3 12x !3 + 4x !3 + x !2 12x 2 ! 6x 2 = 30 ! 16 + 6 = 20 " 0
The Non-Homogeneous Linear Differential Equation
Given the n-th order non-homogeneous linear differential equation
a 0 (x)y( n ) + a1 (x)y( n!1) + ..... + a n!1 (x)y" + a n (x)y = b(x)
then the n-th order H.L.D.E.
a 0 (x)y( n ) + a1 (x)y( n!1) + ..... + a n!1 (x)y" + a n (x)y = 0
is called the corresponding homogeneous equation.
Remark: Every non-homogeneous differential equation has a corresponding
homogeneous equation; we obtain it from the non-homogeneous replacing the nonhomogeneous term by zero.
Theorem: Let v(x) be a solution of the non-homogeneous equation
a 0 (x)y( n ) + a1 (x)y( n!1) + ..... + a n!1 (x)y" + a n (x)y = b(x)
Let u(x) be a solution of the corresponding homogeneous equation
a 0 (x)y( n ) + a1 (x)y( n!1) + ..... + a n!1 (x)y" + a n (x)y = 0 .
Conclusion:
Then u(x) + v(x) is also a solution of the non-homogeneous equation.
Example:
Given the non-homogeneous equation y’’ + y = x,
The function v(x) = x is a solution of the equation.
The corresponding homogeneous equation is y’’ + y = 0, and u(x) = sin x is a solution of
this equation.
Then v(x) + u(x) = x + sin x is also a solution of the non-homogeneous equation.
Theorem: Let yp(x) be a solution of the n-th order non-homogeneous equation
a 0 (x)y( n ) + a1 (x)y( n!1) + ..... + a n!1 (x)y" + a n (x)y = b(x)
that does not involve and arbitrary constant.
Let yc(x) = c1f1(x) + c2f2(x) + … + cnfn(x) where c1, c2, … , cn are arbitrary constants be
the general solution of the corresponding homogeneous equation
a 0 (x)y( n ) + a1 (x)y( n!1) + ..... + a n!1 (x)y" + a n (x)y = 0 .
Conclusion:
Every solution y(x) of the non-homogeneous equation can be expresses as
y(x) = yc(x) + yp(x) = c1f1(x) + c2f2(x) + … + cnfn(x) + yp(x).
Remark: yc(x) + yp(x) is called the general solution of the non-homogeneous linear
differential equation.
yc(x) is called the complementary solution of the non-homogeneous equation.
yp(x) is called a particular solution of the non-homogeneous equation, it does not
involve any arbitrary constant.
Example:
d2y
= 4,
1) Given the non-homogeneous equation
dx 2
The function yp(x) = 2x2 is a particular solution of the non-homogeneous equation.
d2y
The corresponding homogeneous equation is 2 = 0 , and f1(x) = 1 and f2(x) = x are two
dx
linearly independent solutions of the equation.
Then yc(x) = c1 +c2x is the complementary solution.
Therefore y(x) = yc(x) + yp(x) = c1 + c2x + x is the general solution of the nonhomogeneous equation.
d2y
!y=4,
dx 2
The function yp(x) = -4 is a particular solution of the non-homogeneous equation.
d2y
The corresponding homogeneous equation is 2 ! y = 0 , and f1(x) = ex and f2(x) = e-x
dx
are two linearly independent solutions of the equation.
Then yc(x) = c1 ex+c2e-x is the complementary solution.
Therefore y(x) = yc(x) + yp(x) = c1ex + c2e-x - 4 is the general solution of the nonhomogeneous equation.
2) Given the non-homogeneous equation