MATH HOMEWORK #6 PART A SOLUTIONS Problem 7..5. Transform the given initial value problem into an initial value problem for two first order equations. u + 4 u + 4u cost, u0, u 0 Solution. Let x u and x u ; then u x. In terms of the new variables, we have x + 4 x + 4x cost with initial conditions x 0 and x 0. Along with these initial conditions, we havhe system x x x 4 x 4x + cost. Problem 7..7. Systems of first order equations can sometimes bransformed into a single equation of higher order. Consider the system x x + x, x x x. a Solvhe first equation for x and substitute into the second equation, thereby obtaining a second order equation for x. Solvhis equation for x and then determine x also. b Find the solution of the given system that also satisfies the initial conditions x 0, x 0. c Sketch the curve, for t 0, given parametrically by the expressions for x and x obtained in part b. Solution. a Solving the first equation for x, we have x x + x. Substituting into the second, equation, we find x + x x x + x x + 4x + x 0 which has characteristic equation 0 r + 4r + r + r +. Thus the general solution is Then x t c e t + c e t. x x + x c e t c e t + c e t + c e t c e t c e t. b Imposing the initial conditions, we obtain c + c, c c
so c 5/ and c /. Thus x t 5 e t e t, x t 5 e t + e t. c Problem 7... Consider two interconnected tanks similar to those from the previous problem. Initially, Tank contains 60 gal of water and Q 0 oz of salt, and Tank contains 00 gal of water and Q 0 oz of salt. Water containing q oz/gal of salt flows into Tank at a rate of gal/min. The mixture in Tank flows out at a rate of 4 gal/min, of which half flows into Tank, whilhe remainder leaves the system. Water containing q oz/gal of salt also flows into Tank from the outside at a rate of gal/min. The mixture in TAnk leaves it at a rate of gal/min, of which some flows back into Tank at a rate of gal/min, whilhe rest leaves the system. a Draw a diagram that depicts the flow process described above. Let Q t and Q t, respectively, bhe amount of salt in each tank at tim. Write down differential equations and initial conditions for Q and Q that model the flow process. b Find the equilibrium values Q E and QE in terms of the concentrations q and q. c Is it possible by adjusting q and q to obtain Q E 60 and QE 50 as an equilibrium state? d Describe which equilibrium states are possible for this system for various values of q and q. Solution. a Let Q t, Q t bhe amount of salt in the respectivanks at tim. Nothat the volume of each tank remains constant. Based on conservation of mass, the rate of increase of salt, in any given tank, is given by rate of increase rate in rate out.
The rate of salt flowing into Tank is [ ] [ oz r in q gal ] + gal min The rate out is so Similarly for Tank, r out Thus we havhe system [ Q 60 dq dq [ Q oz 00 gal ] [ gal ] q + Q min 00 ] [ oz 4 gal ] Q gal min 5 q + Q 00 Q 5. q + Q 0 Q 00. Q t q + Q 00 Q 5 Q t q + Q 0 Q 00 with initial conditions Q 0 Q 0 and Q 0 Q 0. b We find the equilibrium values by setting 0 Q t q + Q 00 Q 5 0 Q t q + Q 0 Q 00 oz min oz min. and solving this system yields Q E 54q + 6q and Q E 60q + 40q. c From the previous part, we sehat this amounts to solving 54q + 6q 60 60q + 40q 50. Upon solving this system, we find q 7/6 and q /, but this latter equality is not physically possible. d Solving the system 0 q + QE 00 QE 5 for q and q, we find 0 q + QE 0 QE 00 q 45 QE 00 QE q 0 QE + 00 QE.
In order for the result to be physically meaningful, we must have q q 0. Then 0 q 45 QE 00 QE 00 QE 45 QE QE 0 QE 0 q 0 QE + 00 QE 0 QE 0 00 QE 9 QE QE so we must have 0 9 QE QE 0 QE. Problem 7..0. Find all eigenvalues and eigenvectors of the given matrix. 0 and Solution. r detri A det r r 4 r r + r + eigenvalues, For r, I A For r, 0 0 x + x 0 x x x x x x I A x 0 0 x x x x x x Thus we have eigenvalues and with corresponding eigenvectors,. Problem 7... Find all eigenvalues and eigenvectors of the given matrix. 4 4 4
Solution. r detri A det r 4 r 6r + r 6 r r r 4 r + so we have eigenvalues r,,. For r, I A 0 4 0 4 4 0 0 0 0 0 0 0 0 0 0 0 0 0 x, x 0 x x x 0 0 For r, I A 0 0 0 0 0 x x, 0 4 0 0 0 0 0 x x x x x 0 0 For r, I A 0 4 4 x 0, x 0 4 4 x 0 x 0 0 0 Thus we have eigenvalues r,, with corresponding eigenvectors 0,, 0. 0 0 0 0 0 0 0 Problem 7... Show that if λ and λ are eigenvalues of a Hermitian matrix A, and if λ λ, then the corresponding eigenvectors x and x are orthogonal. Solution. Since λ, λ are eigenvalues, then Ax i λ i x i for i,. Since A is Hermitian, then A A and its eigenvalues λ, λ are real. Putting these facts together, then λ x, x λ x, x Ax, x x, A x x, Ax x, λ x λ x, x λ x, x. 5
Subtracting, we have 0 λ x, x λ x, x λ λ x, x. Since λ λ, then λ λ 0. Thus we must have x, x 0, so x and x are orthogonal. Problem 7.4.. In this problem we otline a proof of Theorem 7.4. in the case n. Let x and x be solutions of Eq. for α < t < β, and let W bhe Wronskian of x and x. a Show that dw b Using Eq., show that dx dx x x + x x dx dx dw p + p W. c Find Wt by solving the differential equation obtained in part b. Ushis expression to obtain the conclusion stated in Theorem 7.4.. Solution. a This is an immediate consequence of the multilinearity of the determinant, but we provide a computational proof anyway. Observhat dx dx x x + x x dx dx Since then dw x x tx t x W[x, x ]t x tx t + x tx. tx t + x tx t x tx t tx t x tx t t x x tx t x which is the same as, establishing the desired equality. b By Eq., x i Ptx i for i,, so x p p x p p x x x p p p p tx t + x tx t tx t + x tx t x tx t x x x 6 p x + p x p x + p x p x + p x p x + p x
dw By part a and the multilinearity of the determinant, then dx dx x x x x + dx dx p x + p x p x + p x x x + p x x x x p + p x x p x + p x p x + 0 x p x x x + x 0 p x x x + p x p + p W x x x x x x x + p x where we are ablo cancel the second and third terms becaushey have repeated rows. c Using our expression from part c and separating variables, then dw ln W W p t + p t W ce p t+p t. If c 0, then W is identically 0 on the interval. Otherwise, if c 0, sinche exponential function is always strictly positive, then W never vanishes on the interval. Note: There is a part d for this problem, but it was not assigned. Problem 7.4.6. Consider the vectors x t t and x t t. t a Computhe Wronskian of x and x. b In what intervals are x and x linearly independent? c What conclusion can be drawn about the coefficients in the system of homogeneous differential equations satisfied by x and x? d Find this system of equations and verify the conclusions of part c. Solution. a W[x, x t t ]t det t t t t b The Wronskian is nonzero for t 0, so the functions are linearly independent on, 0 0,. c By Theorem 7.4., at least one of the coefficient functions the entries of Pt must be discontinuous at t 0. d Letting t t c t + c x c + c t t t c t c + c t t c 7
then x t c. 0 c We now try to determinhe matrix Pt. Observhat p p Px t t c p p t c Equating x and Px, we find t p p t t. 0 p p t }{{} A We compute A /t /t /t and multiplying both sides on the right by A yields p p P 0 p p /t. /t Thus x and x satisfy the system x 0 /t /t and indeed, we observhat the coefficients of P are discontinuous at t 0. Problem 7.4.7. Consider the vectors x t t t and answer the same questions as in Problem 6. Solution. a W [ x, x ] t det t x x e t t, t t tt. b The Wronskian vanishes at t 0 0 and t 0. Thus the vectors are linearly independent on, 0 0,,. c By Theorem 7.4., we must havhat one or more of the coefficients of the ODE is discontinuous at t 0 0 and t 0 ; otherwishe Wronskian would not vanish. d Letting then x c t t e t c t + c + c t c t + c t x t e t c. c 8 c c
We now try to determinhe matrix Pt. Observhat p p Px t c p p t c Equating x and Px, we find t e t p p t p p t. }{{} A We compute A tt tt e t te t t t and multiplying both sides on the right by A yields P p p p p Thus x and x satisfy the system x t tt 0 t t tt tt 0 t tt and indeed, we observhat two of the entries of P are discontinuous at t 0,. x. 9