We consider systems of differential equations of the form. x 1 = ax 1 + bx 2,

Transcription

1 Chapter 5 Systems We consider systems of differential equations of the fm (5) a + b, c + d, where a, b, c, and d are real numbers At first glance the system may seem to be first-der; however, it is coupled, and this two-dimensional system is me closely related to a second-der differential equation Befe finding the solution, some notation is introduced to simplify the fm of the system The new notation implicitly removes most of the and + operations The fm of the system suggests arranging the unknown functions, vertically and enclosed in brackets (ie matri fm) (5) This new object is similar to a number function Indeed, operations may be perfmed on (5) Let a R We define u u u u au : v v, a a :, v v av and a + b c + d a c b + d The new object can be part of mathematical statements We say a c b d if and only if a c and b d Meover, we set a b a + b : c d c + d 5

2 54 5 Systems Then (5) is equivalent to a c b d, often written me simply as a c b d 5 Real Eigenvalues In the calculations and epressions that follow both the matri fm (above) and the component fm will be given and separated by They are equivalent Based on our discussion in class, we epect a solution of (5) to have the fm ξ (5) e rt ξ e rt ξ e rt ξ where as of now, ξ, ξ, and r are unknown To be me concrete, we consider a specific system Eample 5 Consider the system of differential equations (54) Our guess, (5), is substituted in (54), and we find re rt ξ ξ e rt ξ 4 ξ Rearranging gives r ξ (55) 4 r ξ rξ ξ + ξ rξ 4ξ + ξ ( r)ξ + ξ 4ξ + ( r)ξ Geometrically the last system of equations describe two lines passing through the igin We do not want a unique solution (zero in this case), f that would say ξ ξ, and our guess would not produce anything interesting We want the lines in (55) to be the same, so we require the slopes to be equal That is, r r The roots, called eigenvalues, are r, Left to find are ξ and ξ Inserting r back in (55), we find ξ + ξ 4ξ ξ As epected this system has infinitely many solutions, one of them is ξ ξ ξ, and, a solution is e t e t e t

3 5 Real Eigenvalues 55 A similar calculation f r shows ξ and, a solution is e t ξ ξ, e t e t The roots are called eigenvalues, and the vects ξ are called eigenvects You will learn me about these when you take Linear Algebra The general solution is therefe, c e t + c e t Eample 5 Find the general solution of the ODE using the approach in Eample y y y, y(), y () c e t + c e t c e t c e t Solution We may turn the ODE into a system of ODEs by setting y and y Then the second-der ODE is equivalent to, () A comparison of (54) and (55) in Eample shows that we can immediately find the equations f ξ in our guess Here r ξ rξ + ξ r ξ + ( r)ξ ξ The polynomial f r is just the determinate the upper-left diagonal element times the lower-right minus the product of the remaining two diagonal elements Here, we find r( r) r r Note that this is the characteristic polynomial f the iginal second-der ODE! The roots (eigenvalues) are r, Inserting r in the equations f ξ, we find ξ + ξ ξ ξ As epected this system has infinitely many solutions, one of them is The other root, r gives and ξ ξ ξ + ξ ξ + ξ, ξ ξ, ξ ξ

4 56 5 Systems The general solution is therefe c e t + c e t c e t + c e t c e t c e t Since we set y in the iginal transfmation, the top line is the general solution to the second-der ODE, and the bottom line is the derivative of the top ( y ) To find c and c we use the initial data - c + c Solving, we find c c, and the solution is e t + e t y e t + e t y e t e t Drawing The Phase Ptrait We need a way of representing solutions graphically To do this we will plot trajecties in the, plane That is, we think of any point in the, plane as initial data, and we plot the trajecty from that point F eample, suppose we choose () and () in the previous eample Then we would find c and c (check this! Also you should have been able to guess this by simple inspection) Thus the solution in this case is (t) e t and (t) e t We see that / which has a graph represented by a straight line (y ) Similarly, if we choose () and (), we would find c and c, and the same line would result If we start any where on the line,, we will find c and c 6 Thus the solution will be in the fm (t) ce t, (t) ce t The ratio is always - the solution is always on the line, and the trajecty moves toward the igin The left diagram in Figure below graphs this trajecty along with the trajecty starting at () and () The right diagram shows the Phase Ptrait f the system The arrows indicate the directions of the trajecties The figure is called a Saddle Homewk 5 (Real Eigenvalues) In Problems -5 find the general solution, and sketch a phase ptrait, ()

5 5 Real Eigenvalues 57 Figure Upper: Invariant lines f Eample 5,, + Lower: Phase Ptrait (Saddle) c c 4 c 4 5 c e t + e t e t + c e t + c Answers e t e t e t + c e t + c e t e t

6 58 5 Systems 6 c 4 + c e t 5 Comple Eigenvalues In this section we mostly drop the component fm of the equations (get used to it) As in previous problems we use Euler s fmula to change eponents with comple numbers to oscillaty functions If z a + ib is a comple number, the comple conjugate is z a ib That is, i is replaced with i In addition, the real and imaginary parts are denoted Both are real numbers Re(z) a Im(z) b Eample 5 Solve the initial-value problem 8, () Solution As in the previous eample, the system f ξ is r 8 ξ r The nontrivial solution is obtained by requiring the determinate to be zero That is, ( r)( r) ( )(8) r + 4 The eigenvalues (roots) are r ±i The procedure is the same as in Eample Suppose r i The equations f ξ are ξ ( i)ξ + 8ξ ξ + ( i)ξ It is not obvious the two equations are multiples of one another However, you can verify that ( + i) times the second equation gives the first equation A non-trivial solution is required If we opptunistically set ξ in the second equation, then ξ i So + i ξ will wk, and one solution is () + i e i t Since the other root (eigenvalue) is just the comple conjugate of the first, a second solution is found simply by changing i in the first solution to i (its comple conjugate) So the general solutions is (56) C + i e i t + C i e i t

7 5 Comple Eigenvalues 59 The comple numbers are undesirable We hide them by using Euler s fmula: e iθ cos θ + i sin θ The eponential functions in (56) are replaced, and the result epressed as the real plus imaginary part - C + i (cos t + i sin t) + C i (cos t i sin t) Read across the first line and pick out all the terms without an i Then do the same f the bottom line (that is, find the real part) The result is cos t sin t Re() (C + C ) cos t The imaginary part consists of all the terms with an i in front In particular, cos t + sin t iim() i(c C ) sin t Finally, set c C + C and c i(c C ) Then is the general solution c cos t sin t cos t + c cos t + sin t sin t There is a slight sht-cut to this procedure Since there is so much symmetry in the solutions f the two roots (they differ by replacing i with -i), one might epect that all the necessary infmation is contained in one of the solutions This is the case The solution f r i (first part of Equation (56)) is () + i e i t + i (cos t + i sin t) To find the general solution, only the real and imaginary parts of this solution need to be found Again rewriting as real plus imaginary, () cos t sin t cos t + sin t + i cos t sin t The general solution is () c Re( () ) + c Im( () ) as befe c cos t sin t cos t + c cos t + sin t sin t To construct the phase ptrait f this problem, we first compute a trajecty starting at and y We find, c and c, and the solution is (t) cos t + sin t y(t) sin t We notice that 4 + y + y and recognize this as an ellipse In fact, any initial data produces an ellipse in the y plane Figure depicts the result It is called a Center Focus

8 6 5 Systems Figure Phase Ptrait f Eample 5 The figure is called a Center Homewk 5 (Comple) In Problems -5 find the general solution, and sketch a phase ptrait Answers c e t cos t + c cos t + sin t e t sin t cos t + sin t cos t sin t c e t + c sin t e t cos t 5 cos t 5 sin t c + c cos t + sin t cos t + sin t 4 c e t cos t + c cos t + sin t e t sin t cos t + sin t 5 c cos t cos t + sin t + c sin t cos t + sin t

9 5 Repeated Eigenvalues 6 5 Repeated Eigenvalues Of course the eigenvalues need not all be distinct Consider f eample (57) Proceeding as befe, the system f ξ is r ξ (58) r The nontrivial solution is obtained by requiring the determinate to be zero That is, ( r)( r) ( ) r 4r + 4 ξ The eigenvalues (roots) are r, The equations f ξ are A nontrivial solution is and one solution is ξ ξ ξ + ξ ξ (), e t In analogy with second-der, linear, constant coefficient, homogeneous ODEs, we might epect the second solution to be (59) () t () t e t Unftunately, this is NOT a solution What to do? Somehow a candidate f the second solution has to be constructed from the only solution we have Based on previous eperiences, (59) has to be close to the crect solution We alter it slightly and try again Set (5) () t () + ηe t t e t η + e t We have to see if a choice f η eists so that () solves (57) The left side of (57) (the time derivative of (5)) is (5) () () + t () + η The right side is (5) However, η e t η e t + t te t + te t t η η e t e t η + η e t e t

10 6 5 Systems Equating (5) and (5), canceling the common e t and the other common terms, we find η η + η η After rearranging (r here) r r η η η me generally the system f η is r η (5) r ξ Notice (5) is an iteration of (58) F the current eample, (5) in component fm is η η η + η As f the system f ξ, the two equations f η are multiples of one another Any solution will suffice The simplest way to find a solution is to set either η η to zero Hence, η and η will wk, and the second solution is () t e t + e t, and the general solution is c e t + c ξ te t + e t To construct a phase ptrait we could start on the line y We will discover that 4 c 6 and c That is, trajecties remain on the line and move out to infinity The remaining trajecties are not so easy to predict Figure depicts the situation Eample 54 Find the solution of the following system of differential equations 9, () 4 Solution The system f ξ is r 9 r ξ ξ The nontrivial solution is obtained by requiring the determinate to be zero That is, ( r)( r) ( )(9) r

11 5 Repeated Eigenvalues 6 Figure Phase Ptrait f Eample y, y + y The eigenvalues (roots) are r, The equations f ξ are ξ + 9ξ ξ + ξ A nontrivial solution is ξ, ξ, and () The system f η is η 9 η Any non-trivial solution is sufficient The choice η and η is a solution Therefe, the general solution is c + c t + To find the solution, the initial data must be applied At t () c 4 + c We find c 4 and c 4, and the solution is + 4t (t) 4 4t The phase ptrait in Eample 54 is a bit strange since all of the points on the line y / do not move In fact, all of the trajecties are parallel to this line Figure 4 depicts the situation

12 64 5 Systems Figure 4 Phase Ptrait f Eample 54 Homewk 5 (Repeated) In Problems -5 find the general solution, and the solution if initial data is provided Also sketch a phase ptrait Answers c e t + c te t + e t c + c t + / /4 c + c t + /4 4 c e t + c te t / t + 4t () e t

13 54 Me on Phase Ptraits Me on Phase Ptraits If we are given the matri a b A c d iginating from a system of ODEs, it is possible to sketch the phase ptrait by inspection Indeed, set Δ det(a) ad cb, T r(a) a + d, where det(a) is the determinant of the matri and T r(a) is called the trace Then the eigenvalues are λ τ τ ± 4Δ Thus we see Δ < implies one positive and one negative eigenvalue, and the phase ptrait is a saddle Negative τ and τ 4Δ < implies an inward sprial We summerize the results in Figure 5 Figure 5 summary of phase ptraits f given τ and Δ Below is a gallary of common phase ptraits

14 66 5 Systems Figure 6 Left: λ < λ < Right: λ < λ <, but ξ and ξ close Figure 7 Left: < λ λ Right: Comple roots λ ± i