Lecture 2.2: Linear independence and the Wronskian Matthew Macauley - PowerPoint PPT Presentation

Lecture 2.2: Linear independence and the Wronskian Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4340, Advanced Engineering Mathematics M. Macauley (Clemson) Lecture 2.2: Linear independence and the Wronskian Advanced Engineering Mathematics 1 / 8

Motivation Question To solve an n th order linear homogeneous ODE, we need to (somehow) find n linearly independent solutions, i.e., a basis for the solution space. Given n functions y 1 , . . . , y n , how can we determine if they are linearly independent? The case n = 1 is easy: { y 1 } is independent iff y 1 �≡ 0. The case n = 2 isn’t difficult, but we have to be careful: { y 1 , y 2 } is independent iff y 1 � = cy 2 . Example Are the functions y 1 ( t ) = sin 2 t and y 2 ( t ) = sin t cos t linearly independent? Things can get much more complicated if n > 2. Example Are the functions y 1 ( t ) = e 2 t , y 2 ( t ) = e − 2 t , and y 3 ( t ) = cosh 2 t linearly independent? What about y 1 ( t ) = e 2 t − 5 e − 2 t , y 2 ( t ) = 3 e − 2 t − cosh 2 t , and y 3 ( t ) = e 2 t + e − 2 t + cosh 2 t ? M. Macauley (Clemson) Lecture 2.2: Linear independence and the Wronskian Advanced Engineering Mathematics 2 / 8

The Wronskian Definition Let f 1 , . . . , f n be functions defined on an interval [ α, β ]. Their Wronskian is defined by f 1 ( x ) f 2 ( x ) · · · f n ( x ) � � � � � � f ′ 1 ( x ) f ′ 2 ( x ) · · · f ′ n ( x ) � � � � W ( f 1 , . . . , f n )( x ) = � . . . � ... . . . � � . . . � � � � � f ( n − 1) f ( n − 1) f ( n − 1) � ( x ) ( x ) · · · ( x ) � � n 1 2 Example Compute the Wronskian of y 1 ( t ) = cos t and y 2 ( t ) = sin t . Theorem If W ( f 1 , . . . , f n ) �≡ 0, then { f 1 , . . . , f n } is linearly independent. Remark Unfortunately, W ( f 1 , . . . , f n ) ≡ 0 does not necessarily imply that { f 1 , . . . , f n } is linearly independent. Example: f 1 ( x ) = x 2 and f 2 ( x ) = | x | · x . M. Macauley (Clemson) Lecture 2.2: Linear independence and the Wronskian Advanced Engineering Mathematics 3 / 8

Key property of the Wronskian Abel’s theorem Let y 1 , y 2 be solutions to y ′′ + a ( t ) y ′ + b ( t ) y = 0, where a , b are continuous in [ α, β ]. Then the Wronskian is � a ( t ) dt , W ( y 1 , y 2 )( t ) = Ce − for some constant C. Moreover, W ( t ) is either identically 0, or never zero, on [ α, β ]. Proof Claim : The Wronskian W ( y 1 , y 2 ) is a solution to the first order ODE W ′ + a ( t ) W = 0. M. Macauley (Clemson) Lecture 2.2: Linear independence and the Wronskian Advanced Engineering Mathematics 4 / 8

Reduction of order Abel’s theorem Let y 1 , y 2 be solutions to y ′′ + a ( t ) y ′ + b ( t ) y = 0, where a , b are continuous in [ α, β ]. Then the Wronskian is � a ( t ) dt , W ( y 1 , y 2 )( t ) = Ce − for some constant C. Moreover, W ( t ) is either identically 0, or never zero, on [ α, β ]. Corollary If we only know one solution y 1 ( t ) of an 2nd order ODE, then we can solve for the other: 2 − y ′ 1 � a ( t ) dt / y 1 . y ′ y ′ y 2 = W ( y 1 , y 2 ) / y 1 = e − 2 + a ( t ) y 2 = g ( t ) : y 1 Proof M. Macauley (Clemson) Lecture 2.2: Linear independence and the Wronskian Advanced Engineering Mathematics 5 / 8

Reduction of order Example Solve y ′′ + 4 y ′ + 4 y = 0. M. Macauley (Clemson) Lecture 2.2: Linear independence and the Wronskian Advanced Engineering Mathematics 6 / 8

More applications of Abel’s theorem Abel’s theorem Let y 1 , y 2 be solutions to y ′′ + a ( t ) y ′ + b ( t ) y = 0, where a , b are continuous in [ α, β ]. Then the Wronskian is � a ( t ) dt , W ( y 1 , y 2 )( t ) = Ce − for some constant C. Moreover, W ( t ) is either identically 0, or never zero, on [ α, β ]. Example Compute the Wronskian of y 1 ( t ) = sin 2 t and y 2 ( t ) = sin t cos t . M. Macauley (Clemson) Lecture 2.2: Linear independence and the Wronskian Advanced Engineering Mathematics 7 / 8

More applications of Abel’s theorem Abel’s theorem Let y 1 , y 2 be solutions to y ′′ + a ( t ) y ′ + b ( t ) y = 0, where a , b are continuous in [ α, β ]. Then the Wronskian is � a ( t ) dt , W ( y 1 , y 2 )( t ) = Ce − for some constant C. Moreover, W ( t ) is either identically 0, or never zero, on [ α, β ]. Example There is no ODE of the form y ′′ + a ( t ) y ′ + b ( t ) y = 0 that has both y 1 ( t ) = t 2 and y 2 ( t ) = t + 1 as solutions. M. Macauley (Clemson) Lecture 2.2: Linear independence and the Wronskian Advanced Engineering Mathematics 8 / 8