Chaper Three Sysems of Linear Differenial Equaions In his chaper we are going o consier sysems of firs orer orinary ifferenial equaions. These are sysems of he form x a x a x a n x n x a x a x a n x n x n a n x a x a nn x n. Here x,,x n are all unknown funcions of an he coefficiens a ij are all consan. We can express his sysem in marix noaion as x x n or X AX In aiion, here may be an iniial coniion o saisfy, x x n i.e., a a n X a n a nn n x x n Solving hese sysems will require many of he noions evelope in he previous chaper. 3. Soluion Mehos for Linear Sysems Consier he iniial value problem for he linear sysem X AX X. Suppose he n by n marix, A, has real eigenvalues,..., n wih corresponing linearly inepenen eigenvecors E,..., E n. Then he general soluion o he sysem of equaions is given by X C e E...C n e n E n. where C,..., C n enoe arbirary consans. We will show in a momen how his soluion was foun bu firs, le us check ha i oes solve he equaion. For each j, j n, noe ha, e j j e j an AE j j E j. Then
an X C e E...C n e n n E n AX A C e E...C n e n E n C e AE...C n e n AE n C e E...C n e n n E n Clearly hen, he soluion. saisfies equaion.. We will now show how his soluion is foun. Theorem. Suppose he n by n marix A has real eigenvalues,..., n wih corresponing linearly inepenen eigenvecors E,..., E n. Then he general soluion of. is given by. Proof- Since he eigenvecors are linearly inepenen, he n by n marix P E,..., E n whose columns are he eigenvecors, has rank equal o n. Then by heorem..3 here exiss a marix P such ha PP P P I. In aiion, AP AE,..., AE n E,..., n E n PD where D iag,..., n enoes he iagonal marix whose iagonal enries are he eigenvalues. Now wrie equaion as X APP X. Then Now P an P are consan marices so P X P APP X. P X P X an if we le Y P X, hen our equaion reuces o Y P APY. Bu AP PD an so P AP P PD D an our equaion reuces furher o Y DY. Tha is, y y n y n y n This sysem is uncouple since each equaion in he sysem conains a single unknown funcion. The uncouple sysem is easily solve o fin
y C e Y y n. C n e n Then since Y P X, X PY which is wha we were suppose o prove. y E...y n E n C e E...C n e n E n Corollary.- The unique soluion o he iniial value problem is foun by choosing he arbirary consans C i o be he unique soluion of he sysem of algebraic equaions PC, where C C,..., C n. If he eigenvecors are muually orhogonal hen he arbirary consans are given by C i E i E i E i i,..., n Proof- I follows from ha X PY if C E...C n E n. Bu recalling secion..4, we see PC C E...C n E n. If he eigenvecors are muually orhogonal hen he arbirary consans can be foun more easily by noing ha if C E...C n E n hen for any i i n, Bu E i E j if i j so E i C E...C n E n E i C E E i...c n E n E i C E E i...c n E n E i C i E i E i Since he E i s are eigenvecors E i E i an he resul is prove. For each i, i n, e i E i is a vecor value funcion of ha saisfies he sysem. These n funcions are sai o be linearly epenen if here exis consans C,..., C n, no all zero, such ha C e E...C n e n E n for all. If he n funcions are no linearly epenen, hey are sai o be linearly inepenen. Since he vecors E,..., E n are linearly inepenen vecors, i is clear ha e E,..., e n E n are linearly inepenen as funcions. Thus he sysem has n linearly inepenen soluions. Examples. To illusrae he use of heorem., we are going o consier he iffusion of a conaminan in a -imensional meium. Assume he meium is a long channel in which he conaminan can ravel from lef o righ or righ o lef bu here is no iffusion in he he ransverse irecion. We will hink of he channel as being ivie ino a number of cells 3
wih he cells being separae from one anoher by faces. The cells are numbere, k n, wih cell k o he righ of cell k an o he lef of cell k. The faces bouning cell k are numbere as face k on he lef of cell k an face k on he righ sie of cell k. If we le u k enoe he concenraion level in cell k a ime, hen conservaion of conaminan is expresse as follows, u k flow in flow ou where flow in an flow ou are he raes of mass flow hrough face k an face k, respecively. If we assume ha F k, he mass flow rae across face k is proporional o he concenraion graien across face k, hen we can wrie F k D k u k u k where D k enoes he maerial epenen proporionaliy consan known as he coefficien of iffusion. Then we have he following equaion for he concenraion in cell k u k D k u k u k D k u k u k D k u k D k D k u k D k u k If he channel is comprise of n cells hen here are n ODE s for he n unknown concenraions u n. For simpliciy of he example, le us suppose n 3 an ha he ens of he channel are seale so he conaminan can move wihin he channel bu i can no escape a he ens. This amouns o assuming ha D D 4. Then he equaions we mus solve are as follows u D u D u u D u D D 3 u D 3 u 3 u 3 D 3 u D 3 u 3 If we make he aiional assumpion ha D D 3, hen he sysem of equaions in marix noaion, becomes u u u 3 u u u 3 Finally, le us assume ha in he iniial sae, he concenraion in cell 3 is (saurae) an he concenraion in cells an is zero (no conaminan). Then u u u 3 In orer o fin he general soluion of his sysem, we firs fin he eigenvalues an eigenvecors for he coefficien marix. Proceeing as we i in he previous chaper, we fin.. 4
E ; E ; 3 3 E 3 Since he coefficien marix is symmeric, he eigenvalues are all real an he eigenvecors are muually orhogonal. Now heorem. assers ha he general soluion o his sysem is u C e E C e E C 3 e 3 E 3 C C e C 3 e 3 We mus now fin values for C, C an C 3 so ha he iniial coniion is saisfie. Tha is, C C C 3. In general we woul have o solve his sysem of algebraic equaions in orer o eermine he C i s bu since he eigenvecors are muually orhogonal, we have C or 3C Similarly, we fin C an 6C 3 Then he unique soluion of he iniial value problem is u 3 e 6 e3 3 e 6 e3 3 3 e3 3 e 6 e3 If we plo he concenraions in he hree cells versus for 5, we see ha he concenraions in cells an are nearly inisinguishable from each oher an ha he concenraion in cell 3 rops from o /3 while he concenraions in he oher wo cells increase o /3. Tha is, he concenraion becomes he same in all hree cells, which is 5
wha we woul expec o happen if he ens of he channel are seale. u()..8.6.4.. 3 4 5 Channel wih seale ens ime If we assume D k for all five faces an he concenraion o he lef of cell an o he righ of cell 3 are equal o zero, his amouns o supposing ha he conaminan can escape from he channel an ha boh ens of he channel are conaminan free. In his case he coefficien marix changes o D D D D D D 3 D 3 D 3 D 3 D 4 an now he eigenvalues an eigenvecors are E E 3 E 3 The general soluion of he sysem is now equal o 6
u C e C e C 3 e The unique soluion o he iniial value problem now becomes u 4 e e e 4 e e 4 4 e 4 e e 4 e e 4 4 e We can see from he plo ha now he concenraion in all hree cells ens o zero as ens o infiniy, which means ha all he conaminan evenually leaves he channel. Noe ha as u 3 ecreases, u a firs increases, followe by u. This is because he conaminan flowing ou of cell 3 flows firs ino cell an hen from cell ino cell. Evenually, all he conaminan flows ou of he channel. u()..8.6.4.. 3 4 5 Channel wih open ens ime Noice ha in boh of hese wo examples he coefficien marices were symmeric. This was no a coincience. When he ifferenial equaions in he moel are erive from a conservaion principle (in his case i is he conservaion of conaminan) he coefficien marix will be symmeric. Since many of he mahemaical moels leaing o sysems of linear ODE s are erive from conservaion laws, i is no unusual o have symmeric coefficien marices an corresponingly muually orhogonal eigenvecors.. Le us now consier an example of an iniial value problem where he coefficien marix is no symmeric; e.g. 7
u u u 3 3 4 u u u 3, u u u 3 Proceeing as in he previous chaper, we fin, E ; 3, E ; 3 4, E 3 an since he marix is no symmeric bu he eigenvalues are isinc, he eigenvecors are linearly inepenen bu no muually orhogonal. The general soluion o he sysem is u C e E C e E C 3 e 3 E 3 C e C e 3 C 3 e 4 If we assume he same iniial coniion as in he previous examples, hen in orer o fin he values for C i, we have o solve he following sysem of algebraic equaions C C C 3 Then C C 3 an C. The unique soluion o he iniial value problem is hen u e e3 e4 e e3 e4 e3 e3 e4 Exercises 8
Le A 3, A, A 3 3, A 4. Solve X A X X.Solve X A X X 3. Solve X A 3 X X 4. Solve X A 4 X X 3. Complex Eigenvalues In he previous secion, we have consiere firs orer sysems or linear ODE s where he coefficien marix ha real eigenvalues. In his secion we will see how o moify he approach slighly in orer o accommoae marices wih complex eigenvalues. We coul rea he case of complex eigenvalues an eigenvecors exacly as we hanle he real value case bu hen he soluions o he sysem of ODE s woul be complex value. In mos applicaions, we woul prefer o have real value soluions an in orer o achieve his we will procee as follows. If he marix A has a complex eigenvalue i, wih corresponing eigenvecor E u iv, ( E will hen have complex enries), hen he conjugae of will also be an eigenvalue for A an he corresponing eigenvecor will be he conjugae of E. The conjugae eigenvalue an eigenvecor will be enoe by i, an E u iv, respecively. Now e E an e E are wo linearly inepenen soluions o he original sysem. However, recall ha, e i cos i sin an hence, e E e i u iv e cos i sinu iv e cos u sin v ie cos v sin u U iv I is clear ha boh U an V are soluions o he original sysem since 9
e E U i V A e E AU iav so U i V AU iav Moreover, U an V are real value an linearly inepenen. Thus, if we prefer o have real value soluions o he sysem of ODE s, we can replace he complex value soluions e E an e E wih he real value soluions U an V. Examples. Consier he sysem x x x x The eigenvalues for A are i. To fin he eigenvecor E corresponing o i, we wrie A ii i which implies ha E e, e wih ie e. Tha is, e ie an E, i. Noe ha here was no nee o employ he secon equaion in A IE, since i mus conain he same informaion as he firs equaion. Now a complex value soluion o he sysem is given by X e i i If we prefer o have real value soluions we noe ha X e cos i sin i i e cos sin ie cos sin e cos cos sin ie sin cos sin Hence U e cos cos sin an U e sin cos sin are wo linearly inepenen real value soluions for he sysem. The general real value soluion is hen given by
X C e cos cos sin C e sin cos sin. Consier x x x 3 3 3 4 x x x 3 In one of he examples of he previous chaper, we foun he real eigenpair 4, E,, an complex eigenvalues 3i wih complex eigenvecors E,i,. Then X e 3i i is a complex value soluion for he sysem. A corresponing pair of real value soluions is foun from X e cos3 i sin 3 i e cos3 sin 3 ie sin 3 cos3 cos3 sin 3 e sin 3 ie cos3 Tha is, cos3 sin 3 U e sin 3 an U e cos3 are wo linearly inepenen real value soluions for he sysem. The general soluion is
X C e 4 C e cos3 sin 3 C 3 e sin 3 cos3 C e cos3 C 3 e sin 3 C e sin 3 C 3 e cos3 C e 4 Exercises In each problem, fin he real value general soluion an hen fin a paricular soluion which saisfies he iniial coniion.. X 4 X X 3. X 4 5 X X 3 3. X 4 X X 3 4. X 3 3 X X 5. X 3 3 X X 3.3 Exponenial of an n n marix A In he firs secion of he firs chaper, we learne ha he soluion of he scalar iniial value problem, x Ax, x x, where A enoes a real number, is x e A x for any consan x. We will learn in his secion ha his is also he soluion when x an x are vecors in R n an A is an n by n marix. Bu firs we mus explain how o inerpre e A when A is an n by n marix. For any real number x, he exponenial funcion e x can be efine as follows For an n n marix A, we can efine e x x! x 3! x3... 3.
e A I A! A 3! A3... 3. In he special case ha A is a iagonal marix hen 3. implies ha A iag,..., n n e A e iage,..., e n e n We mus collec a few facs abou he exponenial of a marix. For any real numbers a an b, e ab e a e b bu for n n marices A an B e AB e A e B if an only if AB BA For any n n marix A, e A is inverible an he inverse is given by e A e A I A! A 3! A3... (i.e., e A is inverible for any A ). If I enoes he n n ieniy marix, hen e I e I, an for any n n marix A, e A I. Now we can sae an prove, Theorem 3.- For any n n marix A,he unique soluion of he iniial value problem is given by X e A v. X AX, X v Proofe A v v Av! A v 3! 3 A3 v... ea v Av A v! A3 v... A v Av! A v 3! 3 A3 v... Ae A v Finally, X e A v Iv v. Noe he following consequences of 3. : 3
If Av hen e A v v If A v hen e A v v Av If A 3 v hen e A v v Av! A v Now suppose v is an eigenvecor for A corresponing o he eigenvalue ; i.e. A Iv. Then e A v e AI e I v e I e AI v e e AI v Bu A Iv an i follows ha e AI v v. Tha is If A Iv hen e A v e v. 3. 3 Nex, suppose v is an eigenvecor for A corresponing o he eigenvalue, an w saisfies A Iw v. Then A I w A Iv. In ha case, e A w e I e AI w Tha is, If e I e AI w e e AI w e w A Iw e w v A Iw v hen e A w e w v 3. 4 Finally, if v an w are as above, an u saisfies A Iu w hen A I 3 u A I w A Iv an in his case so e A u e I e AI u e u w v If A Iu w hen e A u e u w v 3. 5 The resuls 3. 3 o 3. 5 will now be use in orer o fin he soluion for a sysem of firs orer linear ODE s when he marix A oes no have a full se of eigenvecors. 3.4 Repeae Eigenvalues The marix A 3 is immeiaely seen o have eigenvalues,, 3. We say ha he eigenvalue has algebraic mulipliciy an 3 has algebraic mulipliciy. Ofen, bu no always, a repeae eigenvalue like can pose ifficulies is solving he sysem of linear ifferenial equaions having A as is coefficien marix. In his example, he eigenvecors associae wih are foun as follows: 4
A I This ells us ha x 3, while x an x are free so he eigenvecor associae wih is given by E x x x x Evienly, here are wo inepenen eigenvecors associae wih, E an E Then he geomeric mulipliciy of, he number of inepenen eigenvecors, is he same as he algebraic mulipliciy. For 3 we fin E 3 Then he general soluion of he linear sysem X A X is X C e E C e E C 3 e 3 E 3 Evienly, in his case ha he fac ha he eigenvalue is repeae oes no pose a problem for solving he associae linear sysem of ifferenial equaions. On he oher han, consier he marix A 3 whose eigenvalues are foun o be, bu since A I here is only he single eigenvecor E we see ha he algebraic mulipliciy is while he geomeric mulipliciy is jus. In orer o solve he associae sysem of linear ifferenial equaions, we nee anoher inepenen 5
vecor. For his purpose, we look for E saisfying A IE E Then, since E is an eigenvecor for, A IE an, A I E A IE Bu A I so any choice of E saisfies his equaion. A simple choice for E woul be, E Then, accoring o 3. 3 an 3. 4, wo inepenen soluions o X AX are given by an he general soluion o he sysem is X C X e A E e E X e A E e E E We refer o E as a generalize eigenvecor for A. Aiional Examples.Consier X A X where e e C e e A 3 Since A is riangular, he eigenvalues are seen o be 3 an 3. We fin he eigenvecor E,, corresponing o 3 an E,, corresponing o 3. An aiional inepenen vecor is neee in orer o form he general soluion o X A X. If we suppose E 3 saisfies hen Noe ha A IE 3 E A I E 3 A IE. 6
A I 4 4 an i follows ha A I E 3 if x E 3 x for any choice of x an x as long as E 3 E. If we choose E 3,, hen X e A E 3 e E 3 E an X e 3, X e, X 3 e are hree linearly inepenen soluions o X A X..Consier X A X where A This marix has eigenvalue, wih algebraic mulipliciy 3. The geomeric mulipliciy is an he only eigenvecor is E,,. In orer o fin wo more generalize eigenvecors, we firs look for E saisfying Since A IE E so ha A I E. 4 A I we see ha E x, x, x 3 mus have x x 3 an x is arbirary. The simples choice for E is E,,. To fin anoher generalize eigenvecor, we look for E 3 saisfying A IE 3 E so ha A I 3 E 3 A I E. In his case, A I 3 is he zero marix (all zeroes) so any choice for E 3 will work as long as i is inepenen of E an E. The simples choice is E,,. Since we have chosen E an E 3 such ha 7
A I E an A I 3 E 3 we have hree inepenen soluions o X A X. They are X e A E e E X e A E e E E X 3 e A E e E 3 E E an he general soluion for he sysem is X C e C e C 3 e Exercises For each of he following marices, fin he eigenvalues an fin he algebraic an geomeric mulipliciy of each eigenvalue. Then fin he general soluion o X A X. A A 3 A 3 3 3 A 4 3 3 3 A 5 3 3 3 A 6 3 3 3 5. The Inhomogeneous Equaion The inhomogeneous version of he iniial value problem. is he following X AX F 5. X We alreay know he soluion o he homogeneous equaion, i is X H C e E...C n e n E n 5. where i, E i : i n enoe he eigenpairs for A. If we apply he meho of variaion of parameers ha was iscusse in chaper, hen we assume he paricular soluion for 5. is of he form If we subsiue 5. 3 ino 5. hen we fin X p C e E...C n e n E n 5. 3. C e E...C n e n E n F. 5. 4 This is he analogue for sysems of wha we encounere in chaper for a single equaion. 8
We can wrie 5. 4 in marix noaion as follows e E,, e n E n C C n F. Here Y e E,, e n E n is he marix whose j-h column is he j-h componen of he general homogeneous soluion, X j e E. We refer o Y as he funamenal marix for 5.. Since he columns are inepenen, he funamenal marix is inverible, hence C Y F an C n C C n Ys F ss Now 5. 3 implies X p YC Y Ys F ss 5. 5 We will illusrae wih an example. Consier he sysem X X The eigenpairs for his marix are E,, E, an Y e e e We recall now ha here is a simple formula for he inverse of a by marix; i.e., if A a c b hen A e A c b a where e A a bc. In his case, we have 9
Y e 3 e e an e e e e Then Ys F ss X p YC se s e s e s s e e e e e e e e e This is a paricular soluion for he inhomogeneous sysem. Noe ha X p so, if here were iniial coniions o be saisfie, hese coul be accommoae by choosing he consans he homogeneous soluion. I may be worh noing ha i is possible o avoi ealing wih he funamenal marix an is inverse. Consier he simpler case where he eigenvecors of A are muually orhogonal. Then i follows from 5. 4 ha for each j, C j j e j E j F where j E j E j. Then C j j e j s F s E js 5. 6 an finally, X p e E e s F s E s n e n E n e ns F s E ns We will illusrae wih an example. Consier he sysem X 3 3 X This coefficien marix has he following eigenpairs E,, E,, E 3, 3 5, Then 5. 6 implies,
an so he paricular soluion is C e s ss e e C e s s e C j e 5s ss 5 e5 5 e5 X p e E e E 5 e 5 5 E 3 Exercises Le A 3, A, A 3 3, A 4. Fin X p X A X. Fin X p X A X 3. Fin X p X A 3 X 4. Fin X p X A 4 X e