CHAPTER 1. Vector Valued Functions of One-Variable (Time)

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1 A SERIES OF CLASS NOTES FOR TO INTRODUCE LINEAR AND NONLINEAR PROBLEMS TO ENGINEERS, SCIENTISTS, AND APPLIED MATHEMATICIANS DE CLASS NOTES 3 A COLLECTION OF HANDOUTS ON SYSTEMS OF ORDINARY DIFFERENTIAL EQUATIONS (ODE's) CHAPTER 1 Vector Valued Fuctios of Oe-Variable (Time) 1. Vector Valued Fuctios i Oe Variable (Time) 2. Liear Idepedece of Vector Value Fuctios of Oe Variable (Time) Ch. 1 Pg. 1

2 R 3 Hadout #1 VECTOR VALUED FUNCTIONS OF ONE VARIABLE (TIME) Prof. Moseley Physical quatities such as velocity ad force are cosidered to be vectors sice they have magitude ad directio i three dimesioal physical space. Hece it is stadard to model 3 them as elemets i R. However, they ofte vary with time (ad space). This leads to cosideratio of time varyig vector-valued fuctios of the form where I =(a,b). (We use the traspose otatio to save space.) More geerally, if we cosider a system with state variables (e.g., several particles, cocetratios of several chemicals i a chemical reactor, or several species i a eco-system), we must cosider vector -valued fuctios of time i the form (here the word vector refers to the fact that we are cosiderig colum vectors or -tuples of fuctios rather tha that they are elemets i a abstract vector space): R (I) = (I,R )={ :I R } where I =(a,b) as well as matrix-valued fuctios of time the form: R m (I) = ( sice our system or model may also vary with time. Eve more geerally, recall the example of time varyig vectors. Suppose V is a real vector space (which we thik of as a state space). Now let V(I) = {x(t):i V}=F (I,V) where I = (a,b) R. That is, V is the set of all vector valued fuctios o the ope iterval I. (Thus we allow the state of our system to vary with time.) To make V(I) ito a vector space, we must Ch. 1 Pg. 2

3 equip it with a set of scalars, vector additio, ad scalar multiplicatio. The set of scalars for V(I) is the same as the scalars for V (i.e.,r). Vector additio ad scalar multiplicatio are simply fuctio additio ad scalar multiplicatio of a fuctio. To avoid itroducig to much otatio, the egieerig covetio of usig the same symbol for the fuctio ad the depedet variable will be used (i.e., istead of y=f(x), we use y=y(x) ). Hece istead of, for a fuctio i V(I), we use = (t). The cotext will explai whether is a vector i V or a fuctio i V(I). 1) If, V(I), the we defie + poitwise as, ( + )(t) = (t) + (t). 2) If V(I) ad á is a scalar, the we defie (á )(t) V(I) poitwise as (á )(t) = á (t). The proof that V(I) is a vector space is left to the exercises. We use the otatio V(t) istead of V(I), whe, for a math model, the iterval of validity is ukow ad hece part of the problem. Sice V is a real vector space, so is V(t). V(t) ca the be embedded i a complex vector space as described above. To defie limits ad hece derivatives i a abstract time varyig vector space, we eed more structure. Suppose V is a ier product space. The it will have a iduced orm which iduces a metric ad hece a topology. Sice the real umbers are a field with absolute value, the defiitio of the limit of a "vector" valued fuctio of t,, as t approaches t,, ca be defied The the derivative as the limit of the differece quotiet ca be 0 defied,. For R we have a ier product. Also, there are several way m m defie a orm o R ad R. Rather tha carry out this log process, for R ad R it is much simpler to just defie the derivative compoetwise. For time varyig vectors i R this leads to the subspaces A (R (I)) = A (I,R )={ = [x,...,x ] :I R x is aalytic} C(R (I)) = C(I,R ) ={ = [x,...,x ] :I R x is cotiuous} R (I) = (I,R )={ :I R } T i T 3 i m We leave it to future study to show that the usual ier product o R ad ay orm o R will i fact result i compoetwise differetiatio. DEFINITION #1. If ad A are as give i (2) ad (3), the. That is, for vectors ad matrices we compute derivatives (ad itegrals) compoetwise. We state oe theorem o the properties of derivatives of vector ad matrix valued fuctios. Ch. 1 Pg. 3

4 m THEOREM #1. Let A,B R (I),, R (I), ad c R. Assumig all derivatives exist,,,,, ad. Ch. 1 Pg. 4

5 Hadout #2 LINEAR INDEPENDENCE OF VECTOR Prof. Moseley VALUED FUNCTIONS OF ONE VARIABLE (TIME) It is importat that you uderstad the defiitio of liear idepedece i a abstract vector space. DEFINITION #1. Let V be a vector space. A fiite set of vectors liearly idepedet (.i.) if the oly set of scalars c 1, c 2,..., c k which satisfy the (homogeeous) vector equatio V is is c 1 = c 2 = = c = 0; that is, (1) has oly the trivial solutio. If there is a set of scalars ot all zero satisfyig (1) the S is liearly depedet (.d.). DEFINITION #2. Let,..., (I,R ) where I = (a,b). Now let J = (c,d) (a,b) ad for i = 1,...,k, deote the restrictio of to J by the same symbol. The we say that S = {,..., } (J,R ) (I,R ) is liearly idepedet o J if S is liearly idepedet as a subset of (J,R ). Otherwise S is liearly depedet o J. Applyig Defiitios #1 ad 2 to a set of k fuctios i the fuctio space T C (I,R ) ={ = [x 1,...,x ] :I R exists ad is cotiuous} we obtai: THEOREM #1. The set S = {,..., } C (I,R ) where I = (a,b) is liearly idepedet o I if (ad oly if) the oly solutio to the equatio c 1 (t) + + c k (t) = 0 t I (1) is the trivial solutio c 1 = c 2 = = c k =0 (i.e., S is a liearly idepedet set i the vector space C (I,R ) ). If there exists c 1, c 2,,c R, ot all zero, such that (1) holds, (i.e, there exists a otrivial solutio) the S is liearly depedet o I (i.e., S is a liearly depedet set i the vector space C (I,R ) which is a subspace of (I,R ) ). Ofte people abuse the defiitio ad say the fuctios i S are liearly idepedet or liearly depedet o I rather tha the set S is liearly idepedet or depedet. Sice it is i geeral use, this abuse is permissible, but ot ecouraged as it ca be cofusig. Note that Eq. (1) is really a ifiite umber of equatios i the two ukows c 1 ad c 2, oe for each value of x i the iterval I. Four theorems are useful. THEOREM #2. If a fiite set S C (I,R ) where I = (a,b) cotais the zero fuctio, the S is Ch. 1 Pg. 5

6 liearly depedet o I. THEOREM #3. If f is ot the zero fuctio, the S = { } C (I,R ) is liearly idepedet. THEOREM #4. Let S = {, } C (I,R ) where I= (a,b). If either or is the zero vector i C (I,R ) (i.e., is zero o I), the S is liearly depedet o I. THEOREM #5. Let S = {, } C (I,R ) where I= (a,b) ad suppose either or } g is the zero fuctio. The S is liearly depedet if ad oly if oe fuctio is a scalar multiple of the other (o I). PROCEDURE. To show that S = {,..., } is liearly idepedet it is stadard to assume (1) ad try to show c 1 = c 2 = c 3 =... = c k = 0. If this ca ot be doe, to show that S is liearly depedet, it is madatory that a otrivial solutio to (1) be exhibited. t 2 T 3t 2 T EXAMPLE #1. Determie (usig DUD) if S = { [ e, si t, 3 t ], [ e, si t, 3 t ] } is liearly idepedet. ( We prefer colum vectors, but use the traspose otatio to save space.) Proof. (This is ot a yes-o questio). We assume t R. ad try to solve. Note that, i this cotext, the zero vector is the zero fuctio for all three compoets defied by t R. (3) The oe vector equatio (2) ca be writte as the three scalar equatios t 3t c 1 si t + c 2 si t = 0. t R (4) 2 2 c t + c 3 t = 0 Sice these equatios must hold t R, it is really a ifiite umber of algebraic equatios (there are a ifiite umber of values of t) i the two ukows c 1 ad c 2. Ituitively, uless we are very lucky, the two ukows, c ad c, ca ot satisfy a ifiite umber of equatios. To Ch. 1 Pg. 6

7 show this, we simply select two equatios (i.e. values of t) that are "idepedet". Choosig t = 0 ad t = 1 (so as to make the algebra easy) we obtai. 0 3(0) c 1 si 0 + c 2 si 0 = 0 (5) 2 2 c (0) + c 3 (0) = 0 1 3(1) c 1 si (1) + c 2 si (1) = 0 (6) 2 2 c (1) + c 3 (1) = 0 or if we simplify c 1 + c 2 = = 0 (7) = 0 3 c 1 si (1) + c 2 si (1) = 0 (8) c + c 3 = 0 Note that the secod ad third equatio i the first set yield 0 = 0. This is obviously true, but is ot helpful i showig c 1 = c 2 = 0. Also, we ca divide by e i the first equatio i the secod set ad by si(1) 0 i the secod equatio i the secod set. Igorig 0=0 we obtai: c 1 + c 2 = 0 2 c 1 + c 2 e = 0 (9) c 1 + c 2 = 0 c + c 3 = 0 Hece for S to be liearly idepedet, we eed oly show that these equatios imply c 1 = c 2 = 0. We could use Gauss elimiatio, but for relatively simple equatios examiatio may be faster. Note that the first ad third are the same ad yield c 2 = ) c 1. Substitutig ito the last equatio we obtai c ( ) c 1 ) = 0. Hece ) 2 c 1 = 0. Thus c 1 = 0 ad hece c 2 = c 1 = 0. Sice we have proved that the oly solutio to the "vector equatio (2) is the trivial solutio c = c = 0, the set S is liearly idepedet. Ch. 1 Pg. 7