Solutions of the Sample Problems for the Third In-Class Exam Math 246, Fall 2017, Professor David Levermore

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1 Solutions of the Sample Problems for the Third In-Class Exam Math 6 Fall 07 Professor David Levermore Compute the Laplace transform of ft t e t ut from its definition Solution The definition of the Laplace transform gives Lf]s lim T T 0 e st t e t ut dt lim T T t e s t dt This limit diverges to + for s because in that case for every T > we have T t e s t dt which clearly diverges to + as T T t dt T For s > an integration by parts shows that T t e s t dt t e s t T T e s t s + s dt t e s t s e s t T s T e s T s e s T + s Hence for s > we have that Lf]s lim T e s T T s e s T e s s + e s s lim T e s s + e s s + s T e s T s e s s + e s s e s + e s s + e s T s s ] Consider the following MATLAB commands >> syms t s Y; f heavisidet*tˆ + heavisidet **t tˆ ]; >> diffeqn sym DDyt 6*Dyt + 0*yt f; >> eqntrans laplacediffeqn t s; >> algeqn subseqntrans { laplaceyttsts y0 Dy0 } {Y }; >> ytrans simplifysolvealgeqn Y; >> y ilaplaceytrans s t a Give the initial-value problem for yt that is being solved b Find the Laplace transform Y s of the solution yt

2 You may refer to the table on the last page transform to find yt just solve for Y s! DO NOT take the inverse Laplace Solution a The initial-value problem for yt that is being solved is y 6y + 0y ft y0 y 0 where the forcing ft can be expressed either as { t for 0 t < ft t for t or in terms of the unit step function as ft t + ut t t Solution b The Laplace transform of the initial-value problem is where Ly ]s 6Ly ]s + 0Ly]s Lf]s Ly]s Y s Ly ]s s Ly]s y0 s Y s Ly ]s s Ly ]s y 0 s Y s s To compute Lf]s we first write ft as ft t + ut t t t + ut jt where by setting jt t t we see that jt t + t + t + 9 t 6t 9 t t Referring to the table on the last page item with a 0 and n and with a 0 and n shows that Lt ]s s Lt]s s whereby item 6 with c and jt t t shows that L ut jt ] s e s Lj]s e s L t + t ] s e s s + s Therefore Lf]s L t + ut jt ] s s e s s + s The Laplace transform of the initial-value problem then becomes s Y s s 6 sy s + 0Y s s e s s + s which becomes s 6s + 0Y s s + 9 s e s s + s Therefore Y s is given by Y s s 9 + s s s 6s + 0 e s + s

3 Find the Laplace transform Y s of the solution yt of the initial-value problem where y + y + y ft y0 y 0 ft { cost for 0 t < π t π for t π You may refer to the table on the last page transform to find yt just solve for Y s! Solution The Laplace transform of the initial-value problem is where Ly ]s + Ly ]s + Ly]s Lf]s Ly]s Y s Ly ]s s Ly]s y0 s Y s Ly ]s s Ly ]s y 0 s Y s s To compute Lf]s first write f as ft ut π cost + ut πt π cost + ut π t π cost cost + ut πjt π where by setting jt π t π cost we see that jt t + π π cost + π t cost DO NOT take the inverse Laplace Here we have used the fact that cost is π-periodic Referring to the table on the last page item 6 with c π shows that Lf]s Lcost]s + Lut πjt π]s Lcost]s + e πs Ljt]s Lcost]s + e πs Lt cost]s Then item with a 0 and b and item with n and a imply that Lf]s s s + + e πs s s s + The Laplace transform of the initial-value problem then 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

4 Find the inverse Laplace transforms of the following functions You may refer to the table on the last page a F s s + 5 s b F s s s 6 s e s c F s s s + 5 Solution a Referring to the table on the last page item with n and a 5 gives Lt e 5t ]s s + 5 Therefore we conclude that L ]t L s + 5 s + 5 ] t t e 5t Solution b Because the denominator factors as s s+ we have the partial fraction identity s s s s s s + 5 s + 5 s + Referring to the table on the last page item with n 0 and a and with n 0 and a gives Le t ]s s Le t ]s s + Therefore we conclude that ] L s 9 t L 5 s s 6 6 s + 5 s L s 9 5 et e t ] t ] t L s + Solution c Complete the square in the denominator to get s + Referring to the table on the last page item with a and b gives Le t s cost]s s + Item 6 with c and jt e t cost then gives Lut e t cost ]s e s s s + Therefore we conclude that ] L e s s t ut e t cost s s + 5 ] t

5 5 5 Compute the Green function gt for the following differential operators a L D b L D + 8D 9 Solution a The characteristic polynomial of L D is ps s Therefore its Green function gt is given by ] ] gt L t L t ps s Referring to the table on the last page item with a and n gives ] gt L s t e t Solution b The characteristic polynomial of L D +8D 9 is ps s +8s 9 Therefore its Green function gt is given by ] ] gt L t L t ps s + 8s 9 Because ps factors as ps s s + 9 we have the partial fraction identity s + 8s 9 s s s + 0 s + 9 Because s factors as s s s + we have the partial fraction identity s s s + s + s + By combining the above partial fraction identities we obtain s + 8s 9 0 s 0 s + 0 s + 9 Referring to the table on the last page item with a and n 0 and with a and n 0 gives ] ] L t e t L t e t s s + while item with a 0 and b gives ] L t sint s + 9 Therefore the Green function gt is given by ] gt L t s + 8s 9 ] ] 0 L t 0 s L t s + 0 et 0 e t 0 sint 0 L ] t s + 9

6 6 6 Transform the equation u +t u u sinht into a first-order system of ordinary differential equations Solution Because the equation is third order the first-order system must have dimension three The simplest such first-order system is d x x x x where x x u u dt x sinht + x t x x u 7 Consider two interconnected tanks filled with brine salt water The first tank contains 00 liters and the second contains 50 liters Brine flows with a concentration of grams of salt per liter flows into the first tank at a rate of liters per hour Well stirred brine flows from the first tank to the second at a rate of 5 liters per hour from the second to the first at a rate of liters per hour and from the second into a drain at a rate of liters per hour At t 0 there are 5 grams of salt in the first tank and 0 grams in the second Give an initial-value problem that governs the amount of salt in each tank as a function of time Solution The rates work out so there will always be 00 liters of brine in the first tank and 50 liters in the second Let S t be the grams of salt in the first tank and S t be the grams of salt in the second tank These are governed by the initial-value problem ds dt + S 50 S 00 5 S 0 5 ds dt S 00 5 S 50 S 50 S 0 0 You could leave the answer in the form given above However it can be simplified to ds dt 6 + S 5 S 0 S 0 5 ds dt S 0 S 0 S Consider the matrices Compute the matrices a A T b A c A H d 5A B e AB f B A i + i B + i

7 7 Solution a The transpose of A is i + i A T + i Solution b The conjugate of A is i i A i Solution c The Hermitian transpose of A is i i A H i Solution d The difference of 5A and B is given by i0 5 + i i0 + i5 5A B 0 + i i5 7 Solution e The product 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 Solution f Observe that it is clear that B has an inverse because 7 6 detb det Then the inverse of B is given by B detb Consider the vector-valued functions x t t + t x t a Compute the Wronskian Wrx x ]t b Find At such that x x is a fundamental set of solutions to dx dt Atx wherever Wrx x ]t 0 c Give a fundamental matrix Ψt for the system found in part b d For the system found in part b solve the initial-value problem dx dt Atx x 0 t

8 8 Solution a t Wrx x ]t det + t t t + 9 t t + 9 t Solution b Let Ψt + t t Because dψt dt At dψt t Ψt t t + t dt t 0 t t t + 9 t t t 0 t t + t + 9 AtΨt we have 8t 6t t 5 t t Solution c Because x t x t is a fundamental set of solutions to the system found in part b a fundamental matrix for the system found in part b is simply given by Ψt x t x t t + t t Solution d Because a fundamental matrix Ψt for the system found in part b was given in part c the solution of the initial-value problem is t xt ΨtΨ x + t t 0 t + t t 0 0 t + t 0 t t + 9 t 0 6t 6 Alternative Solution d Because x t x t is a fundamental set of solutions to the system found in part b a general solution is given by t xt c x t + c x t c + t t + c The initial condition then implies that x c + c c + c c + c 0 from which we see that c and c 0 The solution of the initial-value problem 5 is thereby t xt + t 0 t 0 t 5 t t 5

9 9 0 Given that is an eigenvalue of the matrix A 0 find all the eigenvectors of A associated with Solution The eigenvectors of A associated with are all nonzero vectors v that satisfy Av v Equivalently they are all nonzero vectors v that satisfy A Iv 0 which is 0 v v 0 0 v The entries of v thereby satisfy the homogeneous linear algebraic system v v + v 0 v v 0 v + v 0 We may solve this system either by elimination or by row reduction method we find that its general solution is v α v α v α for any constant α The eigenvectors of A associated with thereby have the form Consider the matrix α for any nonzero constant α A a Find all the eigenvalues of A b For each eigenvalue of A find all of its eigenvectors c Diagonalize A Solution a The characteristic polynomial of A is given by pz z traz + deta z z 5 z 6 By either The eigenvalues of A are the roots of this polynomial which are ± or simply and 5 Solution b using the Cayley-Hamilton method We have 6 A + I A 5I 6 Every nonzero column of A 5I has the form α for some α 0

10 0 These are all the eigenvectors associated with Similarly every nonzero column of A + I has the form α for some α 0 These are all the eigenvectors associated with 5 Solution c If we use the eigenpairs 5 then set V D Because detv we see that V 8 We conclude that A has the diagonalization 0 A VDV You do not have to multiply these matrices out Had we started with different eigenpairs the steps would be the same as above but we would obtain a different diagonalization A matrix A has the eigenpairs 0 5 a Give an invertible matrix V and a diagonal matrix D such that e ta Ve td V You do not have to compute either V or e ta! b Give a fundamental matrix for the system x Ax Solution a One choice for V and D is V D Solution b Use the given eigenpairs to construct the special solutions x t e t x t e t x t e 5t 0

11 Then a fundamental matrix for the system is Ψt x t x t x t e t e t e 5t e t e t e 5t 0 e t e 5t Alternative Solution b Given the V and D from part a a fundamental matrix for the system is Ψt Ve td e t e t 0 e t e t e 5t e t e t e 5t e 5t 0 e t e 5t What answer will be produced by the following Matlab command? >> A ; ]; vect val] eigsyma You do not have to give the answer in Matlab format Solution The Matlab command will produce the eigenpairs of A characteristic polynomial of A is pz z traz + deta z z 0 z 5z + so its eigenvalues are 5 and Because A 5I A + I we can read off that the eigenpairs are 5 Compute e ta for the following matrices a A 6 b A Solution a The characteristic polynomial of A is given by pz z traz + deta z z z The The eigenvalues of A are the roots of this polynomial which are ± Hence e ta e t coshti + sinht ] A I 0 e cosht t + sinht ] cosht sinht e t sinht cosht

12 Natural Fundamental Set Method Solution a The characteristic polynomial of A is pz z traz + deta z z z + z The associated second-order general initial-value problem is y y y 0 y0 y 0 y 0 y This has the general solution yt c e t + c e t Because y t c e t c e t the general initial conditions yield y 0 y0 c + c y y 0 c c This system can be solved to obtain c y 0 + y c y 0 y The solution of the general initial-value problem is thereby yt y 0 + y e t + y 0 y e t et + e t y 0 + et e t Therefore the associated natural fundamental set of solutions is N 0 t et + e t N t et e t whereby e ta N 0 ti + N ta et + e t 0 + et e t 0 e t + e t e t e t e t e t e t + e t y Eigen Method Solution a The characteristic polynomial of A is pz z traz + deta z z z + z The eigenvalues of A are the roots of this polynomial which are and Because A + I A I we can read off that A has the eigenpairs Set V D 0 0

13 Because detv we see that e e ta Ve td V t 0 0 e t e t 0 0 e t e t + e t e t e t e t e t e t + e t e t e t e t e t Solution b The characteristic polynomial of A is given by pz z traz + deta z 8z + 6 z The eigenvalues of A are the roots of this polynomial which is a double root Hence e ta e t I + t A I ] ] 0 e t + t 0 + t t e t t t Natural Fundamental Set Method Solution b The characteristic polynomial of A is pz z traz + deta z 8z + 6 z The associated second-order general initial-value problem is y 8y + 6y 0 y0 y 0 y 0 y This has the general solution yt c e t + c te t Because y t c e t + c te t + c e t the general initial conditions yield y 0 y0 c y y 0 c + c This system can be solved to obtain c y 0 and c y y 0 The solution of the general initial-value problem is thereby yt y 0 e t + y y 0 t e t te t y 0 + t e t y Therefore the associated natural fundamental set of solutions is whereby N 0 t te t N t t e t 0 e ta N 0 ti + N ta te t 0 + t t e t t t 6 + t e t

14 5 The characteristic polynomial of A 0 is pz z + 9z and the 0 natural fundamental set of solutions associated with t 0 for the operator D + 9D is N 0 t N t sint and N t 9 cost Compute e ta Solution The natural fundamental set method says that e ta N 0 ti + N ta + N ta Because N 0 t N t sint N t 9 cost and A we see that e ta sint cost cost sint sint + cost sint + cost 9 9 sint + 8 cost sint + cost sint cost 8 + cost + sint Solve each of the following initial-value problems a d x x x0 dt y 5 y y0 b d x x x0 dt y y y0 Solution a The characteristic polynomial of A is given by 5 pz z traz + deta z z z + z The eigenvalues of A are the roots of this polynomial which are and These have the form ± 7 whereby e ta e t cosh 7 t I + sinh 7 t 7 A I] e cosh t 7 t 0 + sinh 7 t ] e cosh 7 t t + sinh 7 t sinh 7 t sinh 7 t cosh 7 t sinh 7 t 7 7

15 5 Therefore the solution of the initial-value problem is xt x0 e yt ta e y0 ta e cosh 7 t t + sinh 7 t sinh 7 t t sinh 7 t cosh 7 t sinh e cosh 7 t t sinh 7 t 7 cosh 7 t + sinh 7 t 7 Solution b The characteristic polynomial of A is given by pz z traz + deta z z + 5 z + The eigenvalues of A are the roots of this polynomial which are ± i Hence e ta e t costi + sint ] A I 0 e cost t + sint ] cost e t sint sint cost Therefore the solution of the initial-value problem is xt x0 e yt ta e y0 ta cost e t sint cost + e sint cost t sint sint + cost Remark We could have used other methods to compute e ta in each part of the above problem Alternatively we could have constructed a fundamental matrix Ψt and then determined c so that Ψtc satisfies the initial conditions 7 Find a general solution for each of the following systems a d x dt y b d x dt y c d x dt y x y 5 x y 5 x 5 y

16 6 Solution a The characteristic polynomial of A is given by pz z traz + deta z z + z The eigenvalues of A are the roots of this polynomial which is a double root Hence e ta e t I + t A I ] ] 0 e t + t 0 + t t e t t t Therefore a general solution is xt e yt ta c + t t e c t c t t c + t t c e t + c t e t t Solution b The characteristic polynomial of A pz z traz + deta z is given by The eigenvalues of A are the roots of this polynomial which are ±i Hence e ta costi + sint ] 0 A cost + sint ] 5 0 cost + sint 5 sint sint cost sint Therefore a general solution is xt e yt ta c cost + sint 5 sint c c sint cost sint c cost + c sint + c 5 sint sint cost sint Eigen Method Solution b The characteristic polynomial of A pz z traz + deta z is The eigenvalues of A are the roots of this polynomial which are ±i Because i 5 + i 5 A ii A + ii i + i we can read off that A has the eigenpairs + i i i i

17 Therefore the system has the complex-valued solution + i e it cost + i sint + i cost sint + i cost + i sint cost + i sint By taking real and imaginary parts we obtain the two real solutions + i cost sint x t Re e it cost + i cost + sint x t Im e it sint Therefore a general solution is xt cost sint c yt cost Solution c The characteristic polynomial of A cost + sint + c sint 5 is given by 5 pz z traz + deta z 6z + 5 z + 6 The eigenvalues of A are the roots of this polynomial which are ± i Hence e ta e t costi + sint ] A I 0 e cost t + sint ] 0 5 cost + e t sint sint 5 sint cost sint Therefore a general solution is xt e yt ta c cost + e c t sint sint c 5 sint cost sint c cost + c e t sint 5 sint + c e t sint cost sint Eigen Method Solution c The characteristic polynomial of A pz z traz + deta z 6z + 5 z is 5 The eigenvalues of A are the roots of this polynomial which are ± i Because i + i A + ii A ii 5 i 5 + i we can read off that A has the eigenpairs + i i i + i 7

18 8 Therefore the system has the complex-valued solution e +it e i t cost + i sint i e t cost i sint cost + sint + i sint i cost By taking real and imaginary parts we obtain the two real solutions x t Re e +it e i t cost cost + sint x t Im e +it e + i t sint sint cost Therefore a general solution is xt c yt e t cost cost + sint + c e t sint sint cost 8 Sketch the phase-plane portrait for each of the systems in the previous problem Indicate typical orbits For each portrait identify its type and give a reason why the origin is either attracting stable unstable or repelling Solution a Because the characteristic polynomial of A is pz z we see that µ and δ 0 Because A I we see that the eigenvectors associated with have the form α for some α 0 Because µ > 0 δ 0 and a > 0 the phase portrait is a counterclockwise twist source The origin is thereby repelling The phase portrait should show one orbit that emerges from the origin on each side of the line y x/ Every other orbit emerges from the origin with a counterclockwise twist Solution b Because the characteristic polynomial of A is pz z + 6 we see that µ 0 and δ 6 There are no real eigenpairs Because µ 0 δ 6 < 0 and a > 0 the phase portrait is a counterclockwise center The origin is thereby stable The phase portrait should indicate a family of counterclockwise elliptical orbits that go around the origin Solution c Because the characteristic polynomial of A is pz z + 6 we see that µ and δ 6 There are no real eigenpairs Because µ δ 6 < 0 and a < 0 the phase portrait is a clockwise spiral source The origin is thereby repelling The phase portrait should indicate a family of clockwise spiral orbits that emerge from the origin

19 9 A Short Table of Laplace Transforms Lt n e at n! ]s s a n+ for s > a Le at s a cosbt]s s a + b for s > a Le at sinbt]s b s a + b for s > a Lt n jt]s n J n s where Js Ljt]s Le at jt]s Js a where Js Ljt]s Lut cjt c]s e cs Js where Js Ljt]s and u is the unit step function