Systems. x 1 = ax 1 + bx 2, x 2 = cx 1 + dx 2,

Transcription

1 Systems We consider systems of differential equations of the fm x ax + bx, x cx + dx, 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 x, x vertically and enclosed in brackets ie matrix fm x x This new object is similar to a number function Indeed, operations may be perfmed on Let a IR We define u u u u au : v v a a : v v av 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 x c d x ax + bx : cx + dx Then is equivalent to x x a b c d x x, often written me simply as x a b c d x Real Eigenvalues In the calculations and expressions that follow both the matrix fm above and the component fm will be given and separated by They are equivalent Based on our discussion in class, we expect a solution of to have the fm x ξ ξ e rt x ξ e rt x ξ e rt where as of now, ξ, ξ, and r are unknown To be me concrete, we consider a specific system

2 Example Consider the system of differential equations x x x x + x 4 x 4x 4 + x Our guess,, is substituted in 4, and we find re rt ξ ξ 4 ξ ξ e rt rξ ξ + ξ rξ 4ξ + ξ Rearranging gives r 4 r ξ ξ rξ + ξ 4ξ + rξ 5 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 5 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 5, we find ξ + ξ 4ξ ξ As expected this system has infinitely many solutions, one of them is ξ ξ ξ, and, a solution is x A similar calculation f r shows ξ e t x e t x e t ξ ξ, and, a solution is The general solution is therefe, x c x e t + c e t e t x e t x e t x c e t + c e t x c e t c e t

3 Example Find the general solution of the ODE using the approach in Example x y y y, y, y Solution We may turn the ODE into a system of ODEs by setting x y and x y Then the second-der ODE is equivalent to x x, x ξ x A comparison of 4 and 5 in Example shows that we can immediately find the equations f ξ in our guess Here r ξ rξ + ξ r ξ + rξ 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 are r, Inserting r in the equations f ξ, we find ξ + ξ ξ ξ As expected this system has infinitely many solutions, one of them is ξ ξ ξ, The other root, r gives and The general solution is therefe x c ξ ξ + ξ ξ + ξ, e t + c e t ξ ξ x c e t + c e t x c e t c e t Since we set y x 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 x y x To find c and c we use the initial data - c Solving, we find c c, and the solution is x e t + e t + c y e t + e t y e t e t

4 Complex 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 exponents with complex numbers to oscillaty functions If z a + ib is a complex number, the complex 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 Rez a Imz b Example Solve the initial-value problem x 8 x, x Solution As in the previous example, the system f ξ is r 8 ξ r ξ 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 Example Suppose r i The equations f ξ are iξ + 8ξ ξ + iξ 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 x + i Since the other root eigenvalue is just the complex conjugate of the first, a second solution is found simply by changing i in the first solution to i its complex conjugate So the general solutions is + i x C e i t i + C e i t 6 The complex numbers are undesirable We hide them by using Euler s fmula: e iθ cos θ + isin θ The exponential functions in 6 are replaced, and the result expressed as the real plus imaginary part - + i i x C cos t + isin t + C cos t isin t 4 e i t

5 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 Rex C + C cos t sin t cos t The imaginary part consists of all the terms with an i in front In particular, cos t + sin t iimx ic C sin t Finally, set c C + C and c ic C Then x c cos t sin t cos t + c cos t + sin t sint is the general solution 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 expect 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 6 is x + i e i t + i cos t + isin 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, x cos t sin t cos t + sin t + i cos t sint The general solution is x c Rex + c Imx x c cos t sin t cos t + c cos t + sin t sint as befe Homewk Complex In Problems -5 find the general solution, and sketch a phase ptrait x x 4 x 4 x x 5 x 4 x 5 x 5 5 x x 5

6 x c e t cos t cos t + sin t x c e t cos t sin t x c 4 x c e t 5 x c 5cos t cos t + sin t cos t cos t + sin t cos t cos t + sin t + c e t + c e t sin t cos t + c + c e t + c Answers sint cos t + sint 5sin t cos t + sin t sint cos t + sin t sin t cos t + sin t Repeated Eigenvalues Of course the eigenvalues need not all be distinct Consider f example x x 7 Proceeding as befe, the system f ξ is r r ξ ξ 8 The nontrivial solution is obtained by requiring the determinate to be zero That is, r r r 4r + 4 The eigenvalues roots are r, The equations f ξ are A nontrivial solution is and one solution is ξ ξ ξ + ξ ξ x In analogy with second-der, linear, constant coefficient, homogeneous ODEs, we might expect the second solution to be x tx t e t 9 6, e t

7 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 experiences, 9 has to be close to the crect solution We alter it slightly and try again Set x tx + ηe t t e t η + e t We have to see if a choice f η exists so that x solves 7 The left side of 7 the time derivative of is x x + tx η + e t e t + t e t η + e t The right side is However, η te t + te t t η η Equating and, canceling the common e t and the other common terms, we find + η η η η η e t η e t After rearranging r here r r η η me generally the system f η is r r η η ξ ξ Notice is an iteration of 8 F the current example, 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 and the general solution is x c x t e t + e t + c [ te t + e t, e t ] 7

8 Example 4 Find the solution of the following system of differential equations x 9 x, x 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 The eigenvalues roots are r, The equations f ξ are A nontrivial solution is ξ, ξ, and The system f η is 9 ξ + 9ξ ξ + ξ x η η Any non-trivial solution is sufficient The choice η and η is a solution Therefe, the general solution is [ ] x c + c t + To find the solution, the initial data must be applied At t x c 4 + c We find c 4 and c 4, and the solution is + 4t xt 4 4t 8

9 Homewk Repeated In Problems -5 find the general solution, and the solution if initial data is provided Also sketch a phase ptrait x 4 x x 4 x 8 4 x x 9 x c x c x c 4 x c 5 x + 4t + 4t e t + c [ + c [ + c [ e t + c [ t + t + te t + / /4 /4 te t + ] Answers e t ] ] / e t ] 4 x 5 x x x x 9