Linear Systems with Constant Coefficients

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1 Liner Systems with Constnt Coefficients Here is system of n differentil equtions in n unknowns: x x + + n x n, x x + + n x n, x n n x + + nn x n This is constnt coefficient liner homogeneous system Thus, the coefficients ij re constnt, nd you cn see tht the equtions re liner in the vriles x,, x n nd their derivtives The reson for the term homogeneous will e cler when I ve written the system in mtrix form The primes on x,, x n denote differentition with respect to n independent vrile t The prolem is to solve for x,, x n in terms of t Write the system in mtrix form s Equivlently, x x n n n n nn x A x (A nonhomogeneous system would look like x A x+ ) It s possile to solve such system if you know the eigenvlues (nd possily the eigenvectors) for the coefficient mtrix n n n nn First, I ll do n exmple which shows tht you cn solve smll liner systems y rute force x x n Exmple Consider the system of differentil equtions dx dt x 9x 48x, dx dt x 6x +7x The ide is to solve for x nd x in terms of t One pproch is to use rute force Solve the first eqution for x, then differentite to find x : Plug these into second eqution: x 48 ( x 9x ), x 48 ( x 9x ) 48 ( x 9x ) 6x ( x 9x ), x +x 5x 0

2 This is constnt coefficient liner homogeneous eqution in x The chrcteristic eqution is m m 5 0 The roots re m 5 nd m 3 Therefore, Plug ck in to find x : x c e 5t +c e 3t x 48 ( x 9x ) ( ( 5c e 5t +3c e 3t ) 9(c e 5t +c e 3t ) ) 48 c e 5t 3 c e 3t The procedure works, ut it s cler tht the computtions would e pretty horrile for lrger systems To descrie etter pproch, look t the coefficient mtrix: A Find the eigenvlues: A λi 9 λ λ (λ+9)(λ 7)+6 48 λ +λ 5 This is the sme polynomil tht ppered in the exmple Since λ + λ 5 (λ+5)(λ 3), the eigenvlues re λ 5 nd λ 3 Thus, you don t need to go through the process of eliminting x nd isolting x You know tht x c e 5t +c e 3t once you know the eigenvlues of the coefficient mtrix You cn now finish the prolem s ove y plugging x ck in to solve for x This is etter thn rute force, ut it s still cumersome if the system hs more thn two vriles I cn improve things further y mking use of eigenvectors s well s eigenvlues Consider the system x A x Suppose λ is n eigenvlue of A with eigenvector v This mens tht Av λv I clim tht x ce λt v is solution to the eqution, where c is constnt To see this, plug it in: x cλe λt v ce λt (λv) ce λt (Av) A(ce λt v) A x To otin the generl solution to x A x, you should hve one ritrry constnt for ech differentition In this cse, you d expect n ritrry constnts This discussion should mke the following result plusile Suppose the mtrix A hs n independent eigenvectors v,, v n with corresponding eigenvlues λ,, λ n Then the generl solution to x A x is x c e λ t v + c n e λ nt v n

3 Exmple Solve dx dt dx dt x 9x 48x, x 6x +7x The mtrix form is x x The mtrix A x hs eigenvlues λ 5 nd λ 3 I need to find the eigenvectors Consider λ 5: 4 48 A+5I 6 3 The lst mtrix sys + 0, or Therefore, x Tke The eigenvector is (,) Now consider λ 3: A 3I The lst mtrix sys + 3 0, or 3 Therefore, 3 3 Tke The eigenvector is ( 3,) You cn check tht the vectors (,), ( 3,), re independent Hence, the solution is x x x c e 5t +c e 3t 3 Exmple Find the generl solution (x(t), y(t)) to the liner system The mtrix form is x Let y dx dt x+y dy dt 6x+y A x y

4 det(a xi) x 6 x x 3x 4 (x 4)(x+) The eigenvlues re x 4 nd x For x 4, I hve If (,) is n eigenvector, then So (, 3) is n eigenvector For x, I hve If (,) is n eigenvector, then So (, ) is n eigenvector The solution is A 4I A+I , , x c y e 4t +c 3 e t Exmple (Complex roots) Solve The chrcteristic polynomil is x 5x +5x, x 4x 3x 5 λ λ λ λ+5 The eigenvlues re λ ±i You cn check tht the eigenvectors re: +i λ i : i λ +i : Oserve tht the eigenvectors re conjugtes of one nother This is lwys true when you hve complex eigenvlue The eigenvector method gives the following complex solution: x x c e ( i)t +i +c e (+i)t i 4

5 e t ( (c +c )+i(c c ))cost+((c +c )+i(c c ))sint (c +c )cost i(c c )sint Note tht the constnts occur in the comintions c +c nd i(c c ) Something like this will lwys hppen in the complex cse Set d c +c nd d i(c c ) The solution is x e t ( d +d )cost+(d +d )sint d cost d sint x In fct, if you re given initil conditions for x nd x, the new constnts d nd d will turn out to e rel numers You cn get picture of the solution curves for system x f( x) even if you cn t solve it y sketching the direction field Suppose you hve two-vrile liner system x x y c d y This is equivlent to the equtions Then dx dt x+y nd dy dt cx+dy dy dx dy dt dx dt cx+dy x+y Tht is, the expression on the right gives the slope of the solution curve t the point (x,y) To sketch the direction field, pick set of smple points for exmple, the points on grid At ech point (x,y), drw the vector (x+y,cx+dy) strting t the point (x,y) The collection of vectors is the direction field You cn pproximte the solution curves y sketching in curves which re tngent to the direction field Exmple Sketch the direction field for x x y, y x+y I ve computed the vectors for 9 points: x y x y x+y vector 0 (0, ) 0 (, ) 0 (, 0) 0 (, ) (0, 0) 0 (, ) 0 (, 0) 0 (, ) 0 (0, ) 5

6 Thus, from the second line of the tle, I d drw the vector (, ) strting t the point (,0) Here s sketch of the vectors: y x While it s possile to plot fields this wy, it s very tedious You cn use softwre to plot fields quickly Here is the sme field s plotted y Mthemtic: The first picture shows the field s it would e if you plotted it y hnd As you cn see, the vectors overlp ech other, mking the picture it ugly The second picture is the wy Mthemtic drws the field y defult: The vectors lengths re scled so tht the vectors don t overlp In susequent exmples, I ll dopt the second lterntive when I disply direction field picture The rrows in the pictures show the direction of incresing t on the solution curves You cn see from these pictures tht the solution curves for this system pper to spirl out from the origin Exmple (A comprtment model) Two tnks hold 50 gllons of liquid ech The first tnk strts with 5 pounds of dissolved slt, while the second strts with pure wter Pure wter flows into the first tnk t 3 gllons per minute; the well-stirred micture flows into tnk t 4 gllons per minute The mixture in tnk is pumped ck into tnk t gllon per minute, nd lso drins out t 3 gllons per minute Find the mount of slt in ech tnk fter t minutes Let x e the numer of pounds of slt dissolved in the first tnk t time t nd let y e the numer of pounds of slt dissolved in the second tnk t time t The rte equtions re Simplify: ( dx dt 3 gl )( 0 ls ) ( + gl )( ) ( yls 4 gl )( ) xls, min gl min 50gl min 50gl ( dy dt 4 gl )( ) ( xls gl )( ) ( yls 3 gl )( ) yls min 50gl min 50gl min 50gl x 008x+00y, y 008x 008y Next, find the chrcteristic polynomil: 008 λ λ λ +06λ+0048 (λ+004)(λ+0) 6

7 The eigenvlues re λ 004, λ 0 Consider λ 004: A+004I This sys 0, so Therefore, Set The eigenvector is (,) Now consider λ 0: A+0I This sys + 0, so Therefore, Set The eigenvector is (,) The solution is x c e 004t When t 0, x 5 nd y 0 Plug in: 5 c 0 +c +c e 0t c Solving for the constnts, I otin c 5, c 5 Thus, x 5e 004t +5e 0t 5e 004t +5e 0t 5e 004t 5e 0t The direction field for the system is shown in the first picture In the second picture, I ve sketched in some solution curves c 7

8 The solution curve picture is referred to s the phse portrit The eigenvectors (,) nd (,) hve slopes nd, respectively These pper s the two lines (liner solutions) Consider the liner system x Ax Suppose it hs hs conjugte complex eigenvlues λ, λ with eigenvectors v, v, respectively This yields solutions e λt v, e λ t v If +i is complex numer, re(+i) ((+i)+( i)) ((+i)+(+i) ), im(+i) ((+i) ( i)) ((+i) (+i) ) I ll pply this to e λt v, using the fct tht ( e λt v ) e λ t v Then re ( e λt v ) im ( e λt v ) ( e λt v +e λ t v ), ( e λt v e λ t v ) The point is tht since the terms on the right re independent solutions, so re the terms on the left The terms on the left, however, re rel solutions Here is wht this mens If liner system hs pir of complex conjugte eigenvlues, find the eigenvector solution for one of them (the e λt v ove) Then tke the rel nd imginry prts to otin two independent rel solutions Exmple Solve the system x x y, y x+y Set A The eigenvlues re λ ±i Consider λ +i: A (+i)i i i i The lst mtrix sys i 0, so i The eigenvectors re i 8 i

9 Tke This yields the eigenvector (i,) Write down the complex solution e (+i)t i e t e it i e t i (cost+isint) e t sint + icost cost + isint Tke the rel nd imginry prts: The generl solution is ree t sint + icost e t sint, cost + isint cost ime t sint + icost e t cost cost + isint sint x c e t sint +c cost e t cost sint The eigenvector method produces solution to (constnt coefficient homogeneous) liner system whenever there re enough eigenvectors There might not e enough eigenvectors if the chrcteristic polynomil hs repeted roots I ll consider the cse of repeted roots with multiplicity two or three (ie doule or triple roots) The generl cse cn e hndled y using the exponentil of mtrix Consider the following liner system: x A x Suppose λ is n eigenvlue of A of multiplicity, nd v is n eigenvector for λ e λt v is one solution; I wnt to find second independent solution Recll tht the constnt coefficient eqution (D 3) y 0 hd independent solutions e 3x nd xe 3x By nlogy, it s resonle to guess solution of the form Here w is constnt vector Plug the guess into x A x: x te λt w x te λt λ w +e λt w A(te λt w) Compre terms in te λt nd e λt on the left nd right: A w λ w nd w 0 While it s true tht te λt 0 0 is solution, it s not very useful solution I ll try gin, this time using x te λt w +e λt w Then Note tht Hence, x te λt λ w +e λt w +λe λt w A x te λt A w +e λt A w te λt λ w +e λt w +λe λt w te λt A w +e λt A w 9

10 Equte coefficients in e λt, te λt : A w λ w so (A λi) w 0, A w w +λ w so (A λi) w w In other words, w is n eigenvector, nd w is vector which is mpped y A λi to the eigenvector w is clled generlized eigenvector Exmple Solve x 3 8 x 5 3 λ 8 5 λ (λ+3)(λ 5)+6 λ λ+ (λ ) Therefore, λ is n eigenvlue of multiplicity Now 4 8 A I 4 The lst mtrix sys + 0, or Therefore, Tke The eigenvector is (,) This gives solution Next, I ll try to find vector w such tht Write w (c,d) The eqution ecomes e t (A I) w c d Row reduce: The lst mtrix sys tht c+d, so c d + In this sitution, I my tke d 0; doing so ( ) produces w,0 This work genertes the solution te t +e t 0 0

11 The generl solution is ( ) x c e t +c te t +e t 0 The first picture shows the direction field; the second shows the phse portrit, with some typicl solution curves This kind of phse portrit is clled n improper node Exmple Solve the system x 0 x The eigenvlues re λ nd λ (doule) I ll do λ first A I The lst mtrix implies tht 0 nd 0, so the eigenvectors re For λ, c c c A I The lst mtrix implies tht 0 nd c, so the eigenvectors re c 0 c c c 0 I ll use v (0,,) Next, I find generlized eigenvector w (,,c ) It must stisfy (A I) w v

12 Tht is, 0 0 c Solving this system yields, c + I cn tke c 0, so, nd w c 0 The solution is x c e t +c e t 0 +c te t 0 +e t 0 I ll give rief description of the sitution for n eigenvlue λ of multiplicity 3 First, if there re three independent eigenvectors u, v, w, the solution is x c e λt u+c e λt v +c e λt w Suppose there is one independent eigenvector, sy u One solution is Find generlized eigenvector v y solving A second solution is e λt u (A λi) v u te λt u+e λt v Next, otin nother generlized eigenvector w y solving A third independent solution is (A λi) w v t e λt u+te λt v +e λt w Finlly, comine the solutions to otin the generl solution The only other possiility is tht there re two independent eigenvectors u nd v These give solutions Find generlized eigenvector w y solving e λt u nd e λt v (A λi) w u+ v The constnts nd re chosen so tht the eqution is solvle w yields the solution te λt ( u+ v)+e λt w The est wy of explining why this works involves something clled the Jordn cnonicl form for mtrices It s lso possile to circumvent these techniclities y using the exponentil of mtrix c 05 y Bruce Ikeng

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