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Unformatted text preview: Linear ODEs Second order linear equations Many traditional problems involving ordinary equations arise as second order linear equations au 00 + bu + cu = f , more briefly as Lu = f . The problem is, given f , find u ; often we will want to find u that satisfies some auxiliary initial or boundary conditions. Here we have used the notation Lu : = a ( x ) u 00 + b ( x ) u + c ( x ) u , (1) so L takes a function u and gives a new function Lu . This operator L is a linear map because it has the two properties L ( u ) = Lu and L ( u + v ) = Lu + Lv , where is any constant and u and v are functions. One consequence is that if Lu = 0 and Lv = 0, then L ( Au + Bv ) = 0 for any constants A and B . The solutiond of Lu = 0 are often called the nullspace or kernel of L . These properties show that the nullspace of a linear map is a linear space . For instance, in the special case where Lu = u 00 + u we know that L cos x = 0 and L sin x = 0. Thus L ( A cos x + B sin x ) = 0 for any constants A and B . EXAMPLE Well show that any solution of Lu : = u 00 + u = 0 has the form u ( x ) = A cos x + B sin x . This will show that the nullspace of Lu : = u 00 + u = 0 has dimension two . First we must pick the constants A and B . Letting x = 0 we see that (if this is to work) A = u ( ) . Similarly, taking the derivative we get B = u ( ) . Let v ( x ) : = u ( ) cos x + u ( ) sin x . Our task is to show that u ( x ) = v ( x ) . Equivalently, if we let w ( x ) : = u ( x ) v ( x ) , we must show that w ( x ) 0. A key observation motivating us is that by linearity, w 00 + w = 0, and w ( ) = 0, w ( ) = 0. Introduce the function E ( x ) = 1 2 [ w 2 + w 2 ] . Then since w 00 = w , E ( x ) = w w 00 + ww = w ( w )+ ww = , so E ( x ) = constant (in many physical examples, this is conservation of energy ). But from w ( ) = and w ( ) = 0 we find E ( ) = 0. Since E ( x ) is a sum of squares, the only possibility is w ( x ) 0, as claimed. This Example generalizes. Assuming a ( x ) 6 = 0, the nullspace of (1) always has dimension 2. Let ( x ) and ( x ) be the solutions of the homogeneous equation Lu = 0 with ( ) = 1, ( ) = 0, and ( ) = 0, ( ) = 1, then every solution of the homogeneous equation Lu = 0 has the form u ( x ) = A ( x )+ B ( x ) for some constants A and B . The proof, which we dont give, has two parts....
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This note was uploaded on 03/06/2012 for the course MATH 260 taught by Professor Staff during the Spring '12 term at UPenn.
 Spring '12
 STAFF
 Linear Equations, Equations

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