10Ma2aAnNotes2 - 33 PART II: LINEAR EQUATIONS WITH CONSTANT...

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33 PART II: LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 15. Basic concepts 15.1. Linear equations. The standard form of a second order linear equation is L [ y ] y ±± + p ( t ) y ± + q ( t ) y = g ( t ) . The map y ±→ L [ y ]isa differential operation .I f g ( t ) 0, then the equation L [ y ]=0 is called homogenous . The operation L has constant coefficients if p ( t ) = const and q ( t ) = const. The standard form of n -th order equations is L [ y ]= g ( t ) with L [ y y ( n ) + p 1 ( t ) y ( n - 1) + ··· + p 0 ( t ) y. 15.2. Initial value problem. To solve the IVP( t 0 ,y 0 ± 0 ) for a second order equa- tion [not necessarily linear] is to Fnd a solution y ( t ) such that y ( t 0 )= y 0 ± ( t 0 y ± 0 . In the case of n -th order equations, the initial conditions have the form y ( t 0 y 0 ± ( t 0 y ± 0 , ..., y ( n - 1) ( t 0 y ( n - 1) 0 . Theorem. Consider a linear equation L [ y g . Suppose that g and the coefficients of L are continuous functions in some interval I . Then (i) all solutions extend to the whole interval, (ii) every IVP with t 0 I has a unique solution deFned on I . 15.3. Linear properties. Assume that the coefficients of a differential operation L are in C ( I ). Then we have a linear map (”operator”) L : C ( n ) ( I ) C ( I ) . Notation: ker L = { y : L [ y } . In other words, the kernel of L is the set of all solutions of the homogeneous equation. Lemma. ker L is a linear space of dimension n . Proof: (n=2) ±ix some t 0 , and consider the map R 2 ker L :( y 0 ± 0 ) ±→ y ( t ) , where y ( t ) is the solution of the corresponding IVP. Applying the main theorem we see that this map is linear, 1-to-1, and ”onto”. ± Recall that n elements (”vectors”) y 1 ,...,y n of a linear space are (linearly) inde- pendent if C 1 y 1 + + C n y n =0 C 1 = = C n .
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34 If the linear space has dimension n , then n independent vectors form a basis: every vector y has a unique representation as a linear combination of y j ’s. Any n +1 vectors of an n dimensional space are dependent. Let y 1 ,...,y n be solutions of an n -th order equation L [ y ] = 0. Then the collection y 1 n is called a fundamental system of solutions if the functions y j are inde- pendent as elements of ker L . It follows that every solution of L [ y ] = 0 has a unique representation as a linear combination y ( t )= C 1 y 1 ( t )+ ··· + C n y n ( t ) . Example. The functions 1 ,t,. ..,t n - 1 form a fundamental system of solutions of the equation y ( n ) = 0. The general solution ( the set of all solutions) is the set of polynomials of degree <n . Lemma. If y * is a solution of L [ y ]= g , then the general solution of L [ y g is y y * +ker L. [”particular integral” + ”complementary function”] 15.4. Reduction of order. The following substitution is quite useful. Theorem. If we know some solution y 1 of a homogeneous equation L [ y ]=0 of order n , then the substitution y = y 1 v , (where v = v ( t ) is a new unknown function) transforms the equation L [ y g into a linear equation of order n - 1 for v ± . Proof: Consider the case n =3, L [ y ] y ±±± + p 1 y ±± + p 2 y ± + p 3 y = g, [the coefficients are not necessarily constant]. We have L [ y y ±±± 1 v +3 y ±± 1 v ± y ± 1 v ±± + y 1 v ±±± + p 1 y 1 v + p 1 y ± 1 v ± + p 1 y 1 v ±± + p 2 y ± 1 v + p 2 y 1 v ± + p 3 y 1 v = g.
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This note was uploaded on 11/19/2010 for the course ANTH 122 taught by Professor 323 during the Spring '10 term at Centennial College.

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10Ma2aAnNotes2 - 33 PART II: LINEAR EQUATIONS WITH CONSTANT...

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