MATH 2280-002
Math 2280 Section 002 [SPRING 2013]
Lecture Notes: 01/28/2013
The set F of real-valued functions on
R is a real vector space.
Vector addition is just usual addition of functions
( f + g )( x ) = f ( x ) + g ( x ) and scalar multiplication is also what you’d expect
( cf )( x ) = c ( f ( x )) .
F is called the function space . Just like
R n
, we can talk about linear independence, spanning sets, bases, dimension, and subspaces for F . As you may remember from linear algebra, F is infinite dimensional, but it does have finite vector spaces. For example,
P n
= { polynomials of degree n or less } is an n -dimensional vector space.
The following is a generalization of a definition we saw earlier in the semester:
Definition.
An n th order linear DE is a differential equation of the form p
0
( x ) y
( n )
+ p
1
( x ) y
( n − 1)
+ ...
+ p n − 1
( x ) y
0
+ p n
( x ) y = f ( x ) .
We will usually assume p
0
( x ) , p
1
( x ) ,...
p n
( x ) , and f ( x ) are continuous functions on some open interval I . If f ( x ) = 0 , the differential equation is called a homogeneous linear equation. If f ( x ) = 0 , the DE is called nonhomogeneous . We’ve talked in some detail about first-order linear
DE’s and a way to solve them (integrating factors).
Theorem (Superposition) .
Let y
1
, y
2
, ..., y n be n solutions of the homogeneous linear equation y
( n )
+ p
1
( x ) y
( n − 1)
+ ...
+ p n − 1
( x ) y
0
+ p n
( x ) y = 0 on an open interval I . Then, for any n constants c
1
, c
2
,..., c n
, the linear combination c
1 y
1
+ c
2 y
2
+ ...
+ c n y n is also a solution to the DE on the inverval I .
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MATH 2280-002 Lecture Notes: 01/28/2013
In other words, the vector space spanned by the solutions y
1
, y
2
, ..., y n consists of solutions to our linear DE. This tells us that the set of all solutions to a linear DE is a subspace of F . This subspace is called the solution space for the given linear DE. The principle of superposition only deals with homogeneous linear DE, but in a few lectures, we’ll see a similar statement for nonhomogeneous linear DE with similar implications.
Excerise.
Verify that the principle of superposition holds.
Example.
Since 1 , x , x
2
, and x
3 are solutions to y
(4)
= 0 so is c
1 x
3
+ c
2 x
2
+ c
3 x + c
4
.
Actually, in this case, we know (by integrating 4 times) that c
1 x
3 solution to y
(4)
= 0 . That means 1 , x , x
2
, and x
3
+ c
2 x
2
+ c
3 x + c
4 is the general form a spanning set for the solution space, so the solution space is the 4-dimensional space P
4
.
The example above suggests that the solution space to a n th next theorem will show that this is indeed the case.
order linear DE is n -dimensional. The
Theorem (Existence and Uniqueness) .
Suppose the functions p
1
( x ) ,..., p n
( x ) , and f ( x ) are continuous on an open interval I containing the point a . Then the inital value problem y
( n )
+ p
1
( x ) y
( n − 1)
+ ...
+ p n − 1
( x ) y
0
+ p n
( x ) y = f ( x ) , y ( a ) = b
0
, y
0
( a ) = b
1
, ..., y
( n − 1)
( a ) = b n − 1 has a unique solution.
(Notice that there are no restrictions on the real numbers b
0
, ..., b n − 1
.) This existence and uniqueness theorem tells us that the solution space of an n th order linear DE must be n -dimensional because we need n conditions to obtain a unique solution. When we find n linearly independent solutions to a n th order linear DE, we can write down a general solution to the DE. This gives us a very good reason to care about the linear independence of solutions: When we find n linearly independent solutions to an n th order linear DE, we know that all solutions will just be linear combinations of these n solutions.
Question.
When are n solutions linearly independent?
Remember that y
1
, y
2
, ..., y n are linearly independent if c
1 y
1
+ c
2 y
2
+ ...
+ c n y n
= 0 only when c
1
= c
2
= ...
= c n
= 0 .
This definition may be hard to digest if you haven’t thought about it in a while, so let’s consider the easiest, nontrivial case.
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MATH 2280-002 Lecture Notes: 01/28/2013
Question.
When are two functions linearly independent?
By the definition above, y
1 and y
2 are linearly independent if c
1 y
1
+ c
2 y
2
= 0 only when c
1
= c
2
= 0 . If y
1 and y
2 are linearly dependent then either c
1
= 0 or c
2
= 0 and c
1 y
1
= − c
2 y
2
.
In other words, y
1 is a constant multiple of y
2
. The situation becomes more difficult for three functions y
1
, y
2
, y
3
. If y
1
, y
2
, and y
3 are linearly dependent, none of the functions may be a constant multiple of one of the other two functions; all you know, in this case, is that one function is a linear combination of the other two. The point is that, in general, it can be very difficult to check whether functions are linearly independent or not.
Warning.
Intervals matter when considering linear independence. Show that x and | x | are linearly independent on (-2,2) but not on (0,2).
Fortunately, we have the Wronskian for solutions to a DE!
Definition.
The Wronskian of n functions f
1
, f
2
,..., f n is
W ( f
1
, f
2
, ..., f n
) = f
1 f
0
1
.
..
f
1
( n − 1) f
2 f
2
· · · f
0
2
· · ·
.
..
( n − 1) f n f
0 n
.
..
· · · f
( n − 1) n
Theorem.
Suppose y
1
, y
2
,..., y n are n solutions to the homogeneous n th order linear equation y
( n )
+ p
1
( x ) y
( n − 1)
+ ...
+ p n − 1
( x ) y
0
+ p n
( x ) y = 0 on an open interval I on which p
1
( x ) , p
2
( x ) ,..., and p n
( x ) are continuous.
(a) If y
1
, y
2
,..., y n are linearly dependent, W ( y
1
, y
2
, ..., y n
) = 0 at every point I .
(b) If y
1
, y
2
,..., y n are linearly independent, W ( y
1
, y
2
, ..., y n
) = 0 at every point I .
There are only two possibilites either the Wronskian is identically 0 on I or it is never 0 on I . This can be very useful because you only need to check the Wronskian at a single point in the interval to check if the solutions are linearly independent or not.
Exercises.
• Show that the functions y
1
= 7 , y
2
= 2 + 4 x
2
, y
3
= − 1 + 12 x
2 are linearly dependent by computing the Wronskian. Find a nontrivial linear combination of these functions that vanishes identically.
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MATH 2280-002 Lecture Notes: 01/28/2013
• Verify that y
1
= sin( x ) , y
2
= cos( x ) , and y
3
= e
− x are solutions to the DE y
000
+ y
00
+ y
0
+ y = 0 .
Compute the Wronskian of y
1
, y
2
, y
3 to show that these three solutions are linearly independent.
What is the general solution to the above differential equation?
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