Math 3120
Differential Equations
with
Boundary Value Problems
Chapter 4:
Higher-Order Differential Equations
Section 4-1: Preliminary Theory-Linear Equations
Initial-Value and Boundary-Value Problems
• Theorem 4.1.1: An initial value problem for a nth order Linear
DE has the general form
dny
d n 1 y
dy
P0 (t ) n  P1 (t ) n1    Pn1 (t )  Pn (t ) y  Gt 
dt
dt
dt
(1)
subject to the initial conditions
y t0   y0 , yt0   y0 , , y ( n 1) t0   y0( n 1)
• If the functions P0,…, Pn, and G are continuous on an open
interval I = (,  ), and that P0 is nowhere zero on I. If t = t0 is any
point in I, then there exists exactly one solution y = (t) that
satisfies the initial value problem. This solution exists
throughout the interval I .
Example 1
• Verify that
y (t )  2et  eist a unique solution to the IVP
y  y  0, y0  3, y0  1
Boundary-Value Problems
• Is a linear DE of order two or greater in which the dependent
variable y or its derivative is specified at different points. For
example,
y   p ( x) y   q( x) y  g ( x), y ( )  0, y (  )  0
where   
• The values
(2)
are called boundary conditions.
y ( )  0, y (  )  0
• A solution to Eq. (1) is a function y = (x) that satisfies the
differential equation on the interval  < x <  and that takes on
the specified values y0 and y1 at the endpoints.
Example 2: BVP has a unique solution
• Consider the boundary value problem
y  2 y  0, y (0)  1, y ( )  0
• The general solution of the differential equation is
y  c1 cos 2 x  c2 sin 2 x
• The first boundary condition requires that c1 = 1.
• From the second boundary condition, we have
c1 cos 2  c2 sin 2  0  c2   cot 2  0.2762
• Thus the solution to the boundary value problem is
y  cos 2 x  cot 2 sin 2 x
• This is an example of a nonhomogeneous boundary value
problem with a unique solution.
Example 3: BVP has No solution
• Consider the boundary value problem
y   y  0, y (0)  1, y ( )  6
• The general solution of the differential equation is
y  c1 cos x  c2 sin x
• The first boundary condition requires that c1 = 1, while the
second requires c1 = - a. Thus there is no solution.
Example 4: BVP has Infinite solution
• Consider the boundary value problem
y   y  0, y (0)  1, y ( )  1
• The general solution of the differential equation is
y  c1 cos x  c2 sin x
• However, if a = -1, then there are infinitely many solutions:
y  cos x  c2 sin x, c2 arbitrary
• This example illustrates that a non homogeneous boundary
value problem may have no solution, and also that under
special circumstances it may have infinitely many solutions.
Homogeneous Equations
• An nth order homogenous Linear DE is of the form
dny
d n 1 y
dy
P0 (t ) n  P1 (t ) n 1    Pn1 (t )  Pn (t ) y  0
dt
dt
dt
(3)
• An nth order nonhomogenous Linear DE is of the form
dny
d n 1 y
dy
P0 (t ) n  P1 (t ) n 1    Pn 1 (t )  Pn (t ) y  Gt 
dt
dt
dt
(4)
• When we talk about Linear equations of form (1), we will
assume that the coefficient functions Pi(t), i = 0,1,…n and G(t)
are continuous and P0(t) ≠ 0 for every t in the interval.
Differential Operator
• We can denote differentiation by the Capital letter D, i.e.,
dy
 Dy
dt
• This symbol D is called the Differential operator. Higher-Order
derivatives can also be expressed in terms of D. For example,
d  dy  d 2 y
2
   2  D( D) y  D y
dt  dt  dt
• An nth order differential operator or polynomial operator is
defined as
L  p n (t ) D n  p n 1 (t ) D n 1  ......  p1 (t ) D  p 0 (t )
• L is a linear operator, i.e., L{αf(x)+βg(x)}= αL(f(x))+βL(g(x))
Theorem 4.1.2: Superposition Principle
• If y1,…, yn are solutions of the homogeneous nth order linear
differential equation on an interval I:
dny
d n 1 y
dy
P0 (t ) n  P1 (t ) n 1    Pn1 (t )  Pn (t ) y  0
dt
dt
dt
(3)
• Then the linear combination y(t )  c y (t )  c y (t )    c y (t )
1 1
2 2
n n
where the ci , i = 0,1,…,n are arbitrary constants is also a
solution on the interval I.
Linear Independence and Linear Dependence
• A set of functions f1(x), f2(x),…., fn(x) is said to be linearly
dependent on an interval I if there exist constants c1, c2,…,cn,
not all zero, such that
c1 f1 ( x)  c2 f 2 ( x)    cn f n ( x)  0
for every x in the interval.
• If the set of functions is not linearly dependent on the
interval I, then it is said to be linearly independent.
Wronskian
• Suppose each of the functions f1(x), f2(x),…., fn(x) possesses at
least n-1 derivatives. The determinant
W  f 1 , f 2 , , f n  
f1
f1'
f2
f 2


fn )
fn '



f1( n 1)
f 2( n 1)


f n( n 1)
where the primes denote derivatives, is called the Wronskian
of the functions.
Homogeneous Equations & Wronskian
• If y1,…, yn are solutions of the homogeneous nth order linear differential equation
on an interval I , then the set of solutions is Linearly independent iff Wronskian, is
nonzero at t0 in the interval I.
W  y1 , y 2 , , y n t 0  
y1 (t 0 )
y1 (t 0 )
y 2 (t 0 )
y 2 (t 0 )


y1( n 1) (t 0 )


y n (t 0 )
y n (t 0 )


y 2( n 1) (t 0 )  y n( n 1) (t 0 )
0
• Since t0 can be any point in the interval I, the Wronskian determinant needs to be
nonzero at every point in I.
• As before, it turns out that the Wronskian is either zero for every point in I, or it is
never zero on I.
Fundamental Solutions & Linear Independence
• Consider the nth order homogenous LDE:
dny
d n 1 y
dy
P0 (t ) n  P1 (t ) n 1    Pn1 (t )  Pn (t ) y  0
dt
dt
dt
(3)
• A set {y1,…, yn} of solutions of Eq. (3) with W(y1,…, yn)  0 on I is
called a fundamental set of solutions.
• If y1,…, yn are fundamental solutions, then W(y1,…, yn)  0 on I. It
can be shown that this is equivalent to saying that y1,…, yn are
linearly independent:
c1 y1 (t )  c2 y2 (t )    cn yn (t )  0 iff c1  c2    cn  0
General Solution of Homogenous Equations
• Since all solutions can be expressed as a linear
combination of the fundamental set of solutions, the
general solution of the equation on the interval is
y(t )  c1 y1 (t )  c2 y2 (t )    cn yn (t )
where ci’s are arbitrary constants
Example 5
• Verify that the given functions are solutions of the differential
equation, and determine their Wronskian.
 t 2 y  ty  0; 1, t , t 3
Nonhomogeneous Equations
• Consider the non homogeneous equation:
dny
d n 1 y
dy
P0 (t ) n  P1 (t ) n 1    Pn 1 (t )  Pn (t ) y  Gt 
dt
dt
dt
(4)
• Any function yp that has no arbitrary parameters and satisfies
the Eq. (4) is said to be a particular solution of the non
homogeneous equation.
• The general solution to the non homogeneous equation (4) is
y (t )  c1 y1 (t )  c 2 y 2 (t )    c n y n (t )  y p (t )
where yp is any particular solution and y1,…, yn are the
fundamental solutions of the associated homogeneous
differential equation (3) on the interval I and where ci’s are
arbitrary constants.
Theorem 4.1.7: Superposition Principle
Non Homogenous Equations
• If yp1,…, ypk be k particular solutions of the non homogeneous
nth order linear differential equation (4) on an interval I
corresponding, in turn, to k distinct functions g1,…, gk, i.e.,
a n (t ) y ( n )  a n 1 (t ) y ( n 1)  ......  a1 (t ) y '  a0 (t ) y  g i (t)
where i=1,….,k. Then
y p (t )  y p1 (t )  y p 2 (t )    y p k (t )
is a particular solution of (5).
(5)