Solutions of Sample Problems for Third In-Class Exam Math 246, Spring 2011, Professor David Levermore

Transcription

1 Soluions of Sample Problems for Third In-Class Exam Mah 6, Spring, Professor David Levermore Compue he Laplace ransform of f e from is definiion Soluion The definiion of he Laplace ransform gives L[f]s lim T T e s e d lim T T This limi diverges o + for s because in ha case T e s d T d T, e s d which clearly diverges o + as T For s > an inegraion by pars shows ha T e s d e s T T e s s s d e s s e s T s T e st s e st + s s Hence, for s > one has ha L[f]s lim T [T e st s e st s s + lim T s ] + s T e st s e st s Find he Laplace ransform Y s of he soluion y of he iniial-value problem where d y d + dy d + y f, y, y, f { cos for < π, π for π You may refer o he able on he las page DO NOT ake he inverse Laplace ransform o find y, jus solve for Y s! Soluion The Laplace ransform of he iniial-value problem is L[y ]s + L[y ]s + L[y]s L[f]s,

2 where L[y]s Y s, L[y ]s sy s y sy s, L[y ]s s Y s sy y s Y s s To compue L[f]s, firs wrie f as f u π cos + u π π cos u π cos + u π π cos u π cos π + u π π Referring o he able on he las page, iem 6 wih c π and j cos, iem 6 wih c π and j, iem wih a and b, and iem wih n and a hen show ha L[f]s L[cos]s L[u π cos π]s + L[u π π]s L[cos]s e πs L[cos]s + e πs L[]s e πs s s + + e πs s The Laplace ransform of he iniial-value problem hen becomes s Y s s + sy s + Y s e πs s s + + e πs s, which becomes s + s + Y s s 6 e πs s s + + e πs s Hence, Y s is given by Y s s + s + s e πs s s + + e πs s Find he inverse Laplace ransforms of he following funcions You may refer o he able on he las page a Fs s + 5, Soluion Referring o he able on he las page, iem wih n and a 5 gives L[ e 5 ]s s + 5 Muliplying his by yields L[ e 5 ]s s + 5 You herefore conclude ha L [ s + 5 ] e 5

3 s b Fs s s 6, Soluion The denominaor facors as s s +, so he parial fracion decomposiion is s s s s s s + 5 s + 5 s + Referring o he able on he las page, iem wih n and a, and wih n and a gives L[e ]s s, L[e ]s s +, whereby s s s L[e ]s L[e ]s L [ 9 5 e + 6e ] s 5 You herefore conclude ha [ ] L s 9 s 5 s 6 e e c Fs s e s s s + 5 Soluion Complee he square in he denominaor o ge s + Referring o he able on he las page, iem wih a and b gives L[e s cos]s s + Iem 6 wih c and j e cos hen gives L[u e cos ]s e s s s + You herefore conclude ha [ ] L e s s u e cos s s + 5 Consider he marices A i + i, B + i Compue he marices a A T, Soluion The ranspose of A is i + i A T + i

4 b A, Soluion The conjugae of A is i i A i c A, Soluion The Hermiian ranspose of A is i i A i d 5A B, Soluion The difference of 5A and B is given by i 5 + i i + i5 5A B + i i5 7 e AB, Soluion The produc of A and B is given by i + i 7 6 AB + i 8 7 i i 8 i i 7 + i i i6 7 i5 8 + i7 6 + i6 f B Soluion Observe ha i is clear ha B has an inverse because 7 6 deb de The inverse of B is given by B deb

5 5 Consider he marix A a Find all he eigenvalues of A Soluion The characerisic polynomial of A is given by pz z raz + dea z z 5 z 6 The eigenvalues of A are he roos of his polynomial, which are ±, or simply and 5 b For each eigenvalue of A find all of is eigenvecors Soluion using he Cayley-Hamilon mehod from noes One has 6 A + I, A 5I 6 Every nonzero column of A 5I has he form α for some α These are all he eigenvecors associaed wih Similarly, every nonzero column of A + I has he form α for some α These are all he eigenvecors associaed wih 5 c Diagonalize A Soluion If you use he eigenpairs,, hen se V, D 5,, 5 Because dev + 6 8, you see ha V 8 You conclude ha A has he diagonalizaion A VDV 5 8 You do no have o muliply hese marices ou Had you sared wih differen eigenpairs, he seps would be he same as above bu you would obain a differen diagonalizaion 5

6 6 6 Given ha is an eigenvalue of he marix A, find all he eigenvecors of A associaed wih Soluion The eigenvecors of A associaed wih are all nonzero vecors v ha saisfy Av v Equivalenly, hey are all nonzero vecors v ha saisfy A Iv, which is v v v The enries of v hereby saisfy he homogeneous linear algebraic sysem v v + v, v v, v + v You may solve his sysem eiher by eliminaion or by row reducion By eiher mehod you find ha is general soluion is v α, v α, v α, for any consan α The eigenvecors of A associaed wih herefore have he form α for any nonzero consan α 7 Transform he equaion d u d + du u sinh ino a firs-order sysem of d ordinary differenial equaions Soluion: Because he equaion is hird order, he firs order sysem mus have dimension hree The simples such firs order sysem is d x x x x, where x x u u d x sinh + x x u 8 Consider wo inerconneced anks filled wih brine sal waer The firs ank conains liers and he second conains 5 liers Brine flows wih a concenraion of grams of sal per lier flows ino he firs ank a a rae of liers per hour Well sirred brine flows from he firs ank o he second a a rae of 5 liers per hour, from he second o he firs a a rae of liers per hour, and from he second ino a drain a a rae of liers per hour A here are 5 grams of sal in he firs ank and grams in he second Give an iniial-value problem ha governs he amoun of sal in each ank as a funcion of ime x

7 Soluion: The raes work ou so here will always be liers of brine in he firs ank and 5 liers in he second Le S be he grams of sal in he firs ank and S be he grams of sal in he second ank These are governed by he iniial-value problem ds d + S 5 S 5, S, ds d S 5 S 5 S 5, S You could leave he answer in he above form I can however be simplified o ds d 6 + S 5 S, S, ds d S S, S 9 Consider he vecor-valued funcions x a Compue he Wronskian W[x,x ] +, x Soluion W[x,x ] de b Find A such ha x, x is a fundamenal se of soluions o dx d Ax wherever W[x,x ] Soluion Le Ψ + Because Ψ AΨ, one has d A Ψ d Ψ c Give a fundamenal marix Ψ for he sysem found in par b Soluion Because x, x is a fundamenal se of soluions o he sysem found in par b, a fundamenal marix for he sysem found in par b is simply given by Ψ x x + d For he sysem found in par b, solve he iniial-value problem dx d Ax, x

8 8 Soluion Because a fundamenal marix Ψ for he sysem found in par b was given in par c, he soluion of he iniial-value problem is x ΨΨ x Alernaive Soluion Because x, x is a fundamenal se of soluions o he sysem found in par b, a general soluion is given by x c x + c x c + + c The iniial condiion hen implies ha x c + c c + c, c + c from which we see ha c and c The soluion of he iniial-value 5 problem is hereby x + 5 Compue e A for he following marices a A Soluion The characerisic polynomial of A is given by pz z raz + dea z z z The eigenvalues of A are he roos of his polynomial, which are ± One hen has [ e A e coshi + sinh ] A I e [cosh + sinh ] cosh sinh e sinh cosh Alernaive Soluion The characerisic polynomial of A is pz z raz + dea z z z + z The associaed second-order general iniial-value problem is y y y, y y, y y

9 This has he general soluion y c e +c e Because y c e c e, he general iniial condiions yield y y c + c, y y c c This sysem can be solved o obain c y + y, c y y The soluion of he general iniial-value problem is hereby y y + y e + y y e e + e y + e e y The associaed naural fundamenal se of soluions is herefore N e + e, N e e, whereby e A N I + N A e + e + e e e + e e e e e e + e Second Alernaive Soluion The characerisic polynomial of A is pz z raz + dea z z z + z The eigenvalues of A are he roos of his polynomial, which are and Because A + I, A I, you can read off ha A has he eigenpairs,,, Se V, D Because dev, you see ha e e A Ve D V e e e e + e e e e e e + e e e e e 9

10 6 b A Soluion The characerisic polynomial of A is given by pz z raz + dea z 8z + 6 z The eigenvalues of A are he roos of his polynomial, which is, a double roo One hen has [ e A e I + A I ] [ e + e + ] Alernaive Soluion The characerisic polynomial of A is pz z raz + dea z 8z + 6 z The associaed second-order general iniial-value problem is y 8y + 6y, y y, y y This has he general soluion y c e + c e Because y c e + c e + c e, he general iniial condiions yield y y c, y y c + c This sysem can be solved o obain c y and c y y The soluion of he general iniial-value problem is hereby y y e + y y e e y + e y The associaed naural fundamenal se of soluions is herefore N e, N e, whereby e A N I + N A e + e 6 + e Solve each of he following iniial-value problems a d x d y x, 5 y x y 5 Soluion The characerisic polynomial of A is given by pz z raz + dea z z z + z

11 The eigenvalues of A are he roos of his polynomial, which are and These have he form ± 7 One herefore has [ e A e cosh 7 I + sinh 7 7 A I] e [cosh 7 + sinh 7 ] 7 5 e cosh 7 + sinh 7 sinh sinh 7 cosh 7 sinh The soluion of he iniial-value problem is herefore x x e y A e y A e cosh 7 + sinh 7 sinh sinh 7 cosh 7 sinh e cosh 7 sinh 7 7 cosh 7 + sinh 7 7 b d x d y x, y x y Soluion The characerisic polynomial of A is given by pz z raz + dea z z + 5 z + The eigenvalues of A are he roos of his polynomial, which are ± i One herefore has [ e A e cosi + sin ] A I e [cos + sin ] cos e sin sin cos The soluion of he iniial-value problem is herefore x x e y A e y A cos e sin sin cos e cos + sin sin + cos Remark You could have used oher mehods o compue e A in each par of he above problem Alernaively, you could have consruced a fundamenal marix Ψ and hen deermined c so ha Ψc saisfies he iniial condiions

12 Find a general soluion for each of he following sysems a d x d y x y Soluion The characerisic polynomial of A is given by pz z raz + dea z z + z The eigenvalues of A are he roos of his polynomial, which is, a double roo One herefore has e A e [ I + A I ] [ ] e + + e A general soluion is herefore given by x e y A c + e c c c + c e + c e b d x d y 5 x y Soluion The characerisic polynomial of A pz z raz + dea z is given by The eigenvalues of A are he roos of his polynomial, which are ±i One herefore has [ e A cosi + sin ] [ A cos + sin ] 5 cos + sin 5 sin sin cos sin A general soluion is herefore given by x e y A c cos + sin 5 sin c c sin cos sin c cos + c sin + c 5 sin sin cos sin 5 Alernaive Soluion The characerisic polynomial of A is pz z raz + dea z + 6

13 The eigenvalues of A are he roos of his polynomial, which are ±i Because i 5 + i 5 A ii, A + ii, i + i you can read off ha A has he eigenpairs + i i,, i, i The sysem herefore has he complex-valued soluion + i e i cos + i sin + i cos sin + i cos + i sin cos + i sin By aking real and imaginary pars, you obain he wo real soluions + i cos sin x Re e i, cos + i cos + sin x Im e i sin A general soluion is herefore x cos sin c y cos c d x d y 5 x 5 y Soluion The characerisic polynomial of A cos + sin + c sin 5 is given by 5 pz z raz + dea z 6z + 5 z + 6 The eigenvalues of A are he roos of his polynomial, which are ± i One herefore has [ e A e cosi + sin ] A I e [cos + sin ] 5 cos + e sin sin 5 sin cos sin A general soluion is herefore given by x e y A c cos + e c sin sin c 5 sin cos sin c cos + c e sin 5 sin + c e sin cos sin

14 Alernaive Soluion The characerisic polynomial of A pz z raz + dea z 6z + 5 z is 5 The eigenvalues of A are he roos of his polynomial, which are ±i Because i + i A + ii, A ii, 5 i 5 + i you can read off ha A has he eigenpairs + i,, i i, + i The sysem herefore has he complex-valued soluion e +i e i cos + i sin i e cos i sin cos + sin + i sin i cos By aking real and imaginary pars, you obain he wo real soluions x Re e +i e i cos, cos + sin x Im e +i e + i sin sin cos A general soluion is herefore x c y e cos cos + sin + c e sin sin cos Skech he phase-plane porrai for each of he sysems in he previous problem Indicae ypical rajecories For each porrai idenify is ype and give a reason why he origin is eiher aracing, sable, unsable, or repelling a Soluion Because he characerisic polynomial of A is pz z, one sees ha µ and δ Because A I, we see ha he eigenvecors associaed wih have he form α for some α Because µ >, δ, and a > he phase porrai is a counerclockwise wis source The origin is hereby unsable The phase porrai should show here is one rajecory ha emerges from he origin on each side of he line y x/ Every oher rajecory emerges from he origin wih a counerclockwise wis

15 b Soluion Because he characerisic polynomial of A is pz z + 6, one sees ha µ and δ 6 There are no real eigenpairs Because µ, δ 6 <, and a > he phase porrai is a counerclockwise cener The origin is hereby sable The phase porrai should indicae a family of counerclockwise ellipical rajecories ha go around he origin c Soluion Because he characerisic polynomial of A is pz z + 6, one sees ha µ and δ 6 There are no real eigenpairs Because µ, δ 6 <, and a < he phase porrai is a clockwise spiral source The origin is hereby unsable The phase porrai should indicae a family of clockwise spiral rajecories ha emerge from he origin A Shor Table of Laplace Transforms L[ n e a n! ]s s a n+ for s > a L[e a s a cosb]s s a + b for s > a L[e a sinb]s L[ n j]s n J n s b s a + b for s > a where Js L[j]s 5 L[e a j]s Js a L[u cj c]s e cs Js where Js L[j]s where Js L[j]s and u is he uni sep funcion