Introduction to Systems and General Solutions to Systems

Transcription

1 Introduction to Systems and General Solutions to Systems 1 Introduction to Systems We now turn our attention to systems of first order linear equations We will be reviewing some linear algebra as we go, as much of the theory of these systems depends on the theory of matrices: 1 Systems of Differential Equations 2 Linear Algebra Review 3 Systems as Matrix Equations 4 Higher Order Equations as Systems 2 Systems of Differential Equations Recall that in algebra we are sometimes interested in solving a system of n equations in n unknowns, such as the following form: a 11 x 1 + a 12 x a 1n x n b 1 a 21 x 1 + a 22 x a 2n x n b 2 a n1 x 1 + a n2 x a nn x n b n Here, the a ij and b i are constant coefficients The solution to this system is a set of n values x 1, x 2, x n which satisfy all the equations simultaneously Similarly, a system of n first order linear equations will be a set of n equations involving n functions x 1 t through x n t, as follows: x 1 x 2 x n p 11 tx 1 + p 12 tx p 1n tx n + g 1 t p 21 tx 1 + p 22 tx p 2n tx n + g 2 t p n1 tx 1 + p n2 tx p nn tx n + g n t 1

2 In this case, the p ij t and g i t are coefficient functions, and the solution is a set of n functions x 1 to x n that satisfy all the equations simultaneously As we would expect, we say that the system is homogeneous if all of the g i t are identically zero, and non-homogeneous if some of the g i t are non-zero 3 Linear Algebra Review More details are contained in 41 of the text, but here is a brief review of basic properties of matrices and vectors: 1 Addition, Subtraction, and Scalar Multiplication: Matrices and vectors of the same size can be added or subtracted by adding or subtracting the individual components Multiplication by a scalar simply multiplies each element in a matrix by that scalar Let A Then and B A 2B 2 Transposes: The transpose A T of a matrix A is the matrix obtained by switching the rows and columns of A Let A 3 + i 1 i 2 1 2i 0 Then A T 3 Multiplication: If the A is an l m matrix, and B is m n matrix, then the product C AB is defined and is an l n matrix We find the entry in row i and column j by 2

3 multiplying the entries in row i of A with the corresponding entries in column j of B and adding: n c ij a ik b kj Let A , B k1, and x 2 3 Find: AB x T x xx T 4 Determinants: The determinant of a 2 2 matrix is straightforward: a b c d ad bc Determinants of larger matrices are defined recursively, by cofactor expansion We expand along any row or column of a matrix, multiplying the entry 1 i+j a ij by the determinant of the submatrix formed by deleting the i th row and j th column We sum the results to get the determinant For example, An interesting application of determinants is that the system Ax 0 has a non-zero solution vector x exactly when the determinant A 0 3

4 5 Row Reduction and Inverses: An inverse of a matrix A is a matrix A 1 such that A 1 A AA 1 I, where I is the identity matrix The identity matrix has ones on the main diagonal and zeros elsewhere; when the multiplication is defined, AI IA A for any matrix A To find A 1, put the identity matrix to the right of A like so: a11 a a 21 a and then perform row operations on both until the left matrix is the identity The identity matrix that was on the right will then contain the inverse A 1 The legal row operations are: a Interchange two rows b Multiply an entire row by a constant c Add a multiple of one row to another row Find the inverse of : Alternatively, we can compute the inverse of 2 2 matrices using the formula a b c d 1 1 d b ad bc c a Note that we start with the reciprocal of the determinant It turns out that A is invertible if and only if A 0 6 Derivatives and Antiderivatives: 4

5 If we have a matrix containing functions, such as A e 2t 2t cost 3 then we define the derivative of the matrix to be the derivative of the component functions: da dt 2t 2 sint 0 Find the derivative: d dt t 2 7 sint t + 2 5e t Of course this means we can find antiderivatives of matrix functions just by taking antiderivatives of each component function, and then adding a matrix of constant functions 4 Systems as Matrix Equations With the conventions we have adopted above, we can rewrite a system of first order linear differential equations x 1 x 2 x n p 11 tx 1 + p 12 tx p 1n tx n + g 1 t p 21 tx 1 + p 22 tx p 2n tx n + g 2 t p n1 tx 1 + p n2 tx p nn tx n + g n t as a single differential equation using matrices: x Ptx + gt Here we have x 1 x 2 x, Pt x n p 11 t p 12 t p 1n t p 21 t p 22 t p 2n t p n1 t p n2 t p nn t 5, gt g 1 t g 2 t g n t

6 We can easily confirm that xt e is a solution to the homogeneous equation x x since x 3e 6e and e 3e 6e However, it may not yet be clear how to come up with such a solution Of course, if we have n first order equations, we should be able to solve for n initial conditions, x 1 t 0, x 2 t 0, x n t 0 This suggests that we need to have n constants c 1 to c n in our general solution As in the previous example, it is not hard to confirm that xt c 1 e + c 2 e 2t e 2t 2c1 e + c 2 e 2t c 1 e + c 2 e 2t is a solution to the homogeneous equation x x, since x 6c1 e + 2c 2 e 2t 3c 1 e + 2c 2 e 2t and we also have c1 e + c 2 e 2t c 1 e + c 2 e 2t 6c1 e + 2c 2 e 2t 3c 1 e + 2c 2 e 2t Check! Can we then require x and x 2 0 1, for example? Plugging in t 0 gives 2c 1 + c 2 3 c 1 + c 2 1 6

7 This is the same as the matrix equation c1 c or the augmented matrix Solving this system gives our needed c 1 and c 2 : 41 Existence and Uniqueness We mention the existence and uniqueness result for first order linear systems of equations: Theorem 41: Consider the initial value problem y t Ptyt + gt, yt 0 y 0, where y and g are n 1 vector functions, and Pt is an n n matrix function Let the n 2 components of Pt be continuous on the interval a, b, and let t 0 be in a, b The the initial value problem has a unique solution that exists on the entire interval a < t < b Our goal therefore must be to find this unique solution We will begin establishing what we need during the next class 5 Higher Order Equations as Systems We will consider how higher order equations can be rewritten as first order systems One reason why we are interested in systems of first order equations is because any n th order linear ordinary differential equation can be transformed into a system of n first order linear equations If you have an n th order equation in the variable y, simply set x 1 yt, x 2 y t,,x n y n 1 and use the fact that x 1 x 2, x 2 x 3, and so forth to set up your system Convert y + y y + t into a system of two first order equations 7

8 We set x 1 y, and x 2 y Then we have x 1 y x 2, and since y + y y + t, we have y y + y + t, or x 2 x 2 + x 1 + t So we have the system of equations x 1 x 2 x 2 + x 1 + t We see that this system is linear, but non-homogeneous We can of course write the system as a matrix equation: [ ] [ ] x 1 t 0 1 x1 t x 2 t x 2 t x 2 This means that the theory of higher order differential equations can be rewritten in terms of systems of first order equations One reason this is useful is that we can convert numerical methods such as Euler s method for use on systems of first order equations Thus, if we can rewrite higher order equations, we will be able to use numerical methods to approximate their solutions Let s try one more example Convert y y + 2y, y0 4, y 0 1, y 0 2 into a first order system with appropriate initial conditions [ 0 t ] y 1 t yt y 2 t y 3 t y 1 t y 2t y 2t y 3 t y y 2 0 y General Solutions to Systems We continue on systems of equations We will then consider the form of the general solution and introduce the Wronskian for systems 1 General Solutions 2 The Wronskian 3 Linear Independence 4 Fundamental Matrices 8

9 7 General Solutions We saw last time that we could use a column vector v as a solution to a system of first order differential equations It is easy to see that if v 1 and v 2 are both solutions to the matrix equation y Ay, then so is any linear combination y c 1 v 1 + c 2 v 2, since and y c 1 v 1 + c 2v 2 Ay Ac 1 v 1 + c 2 v 2 c 1 Av 1 + c 2 Av 2 c 1 v 1 + c 2 v 2 from basic properties of matrices and the assumption that v 1 and v 2 are both solutions to the system Thus, we have determined the principle of superposition holds for solutions to homogeneous systems of first order equations See Theorem 42 in the text This leaves us looking for a general solution to a system of equations We know of course that if we have n first order equations, we ought to be able to solve for n initial conditions Generalizing from our experience with second order and higher equations, we will attempt to find n solutions y 1,,y n to the system, and we will guess that c 1 y c n y n might be the general solution Once we have found n solutions y 1,,y n, it will prove useful to construct a matrix Ψt which has columns given by the solutions In other words, if we have solutions y 1,1 y 2,1 y 1,2 y 2,2 y 1,n y 2,n y 1,y 2,,y n y n,1 y n,2 y n,n we will then form the matrix Ψt y 1 y 2 y n y 1,1 y 1,2 y 1,n y 2,1 y 2,2 y 2,n y n,1 y n,2 y n,n Then any linear combination of these solutions can be written easily as y 1,1 y 1,2 y 1,n y 2,1 y 2,2 y 2,n Ψtc y n,1 y n,2 y n,n c 1 c 2 c n c 1y 1 + c 2 y c n y n 9

10 You can verify the equality if you wish It represents a special form of matrix multiplication Consider the system y Confirm that both of the following are solutions: y 1 : 3e y y 2 t : So this means any linear combination of y 1 and y 2 are solutions We form our solution matrix Ψt t 3e and we know we have solutions of the form c 1 y 1 t + c 2 y 2 t, or Ψtc where c is a 2 1 column vector of constants Can we then solve the system above together with the initial condition 1 y0? 1 We must attempt to solve the equation Ψ0c y0, or 2 2 c c 2 This is just a system of equations, which we can solve using row reduction on the augmented matrix

11 So we get c 1 3/4 and c 2 1/4, and our solution is yt [y 1 y 2 ] 3/4 1/ e 1 4 t It is not difficult to confirm both that this satisfies the system and that it satisfies the initial conditions We are of course somewhat cautious We know from solving second order and higher systems that it is not generally enough just to have the right number of solutions to combine, but that the solutions need to be really different from each other, so that we may satisfy whatever initial conditions we wish In the above example, we were able to satisfy our initial conditions How can we determine in general if linear combination of a set of solutions will be able to satisfy any initial conditions, and therefore form a fundamental set of solutions? 8 The Wronskian We have seen above that if we find n solutions y 1,,y n to a system y Pty, then we can attempt to write all solutions in the form c 1 y 1 t+ c n y n t, or equivalently Ψtc, where Ψt [y 1 y n ] and c is an n 1 column vector of constants Then to be able solve for an initial condition yt 0 y 0 is to be able to solve the matrix equation Ψt 0 c y 0 It is possible to solve this uniquely exactly when the matrix Ψt 0 is non-singular, which happens when the determinant Ψt 0 0 Therefore, we define the Wronskian of the vector functions y 1,,y n at point t to be the determinant y 1,1 t y 1,2 t y 1,n t y 2,1 t y 2,2 t y 2,n t Wy 1,,y n t Ψt y n,1 t y n,2 t y n,n t Then it is probably not surprising that the Wronskian identifies fundamental sets of solutions: Theorem 43:p

12 Let y 1 t,y 2 t,,y n t be a set of n solutions to the order n system y Pty, a < t < b, where the matrix function Pt is continuous on a, b Let Wt represent the Wronskian of this set of solutions If there is a point t 0 in a, b where Wt 0 0, then y 1 t,y 2 t,,y n t form a fundamental set of solutions to the equation In previous example we showed that 3e and t were solutions to the given differential equation Show that they form a fundamental set of solutions We check the Wronskian: t 10e2t 6e 2t 4e 2t 0 3e We see that Wt 0 for any value, and so these form a fundamental set of solutions Notice in the above example, our Wronskian was never zero This is not unusual In parallel with the case for second order and higher equations, the Wronskian of a set of n solutions to an order n equation is either always zero or always non-zero on the interval where a unique solution is guaranteed to exist Finally, we note that as we saw in example above, solving for initial conditions once we have a fundamental set of solutions is fairly straightforward: just solve the system Ψt 0 c y0 for the constants c 1, c 2,, c n We continue on systems of equations We will discuss linear independence and fundamental matrices 9 Linear Independence Recall that we say that a set of vectors x 1, x 2,, x n are linearly independent if there is no collection of n constants c 1, c 2,, c n not all zero for which c 1 x 1 + c 2 x c n x n 0 12

13 If such a collection of constants exists, then we say the x i are linearly dependent How can we tell if a set of vectors is linearly independent? If we have n vectors of x 1, x 2,, x n of length n, they are linearly independent if and only if the only solution to the system Xc x 1 x 2 x n c 1 c 2 c n x 1,1 x 1,2 x 1,n x 2,1 x 2,2 x 2,n x n,1 x n,2 x n,n c 1 c 2 c n is the zero vector Therefore, if the matrix X above is non-singular, the vectors are linearly independent If not, they are dependent Determine whether the vectors 0 x 1 2 2, x , and x 3 are linearly independent or not We see quickly that they are linearly dependent, since the determinant of X x 1 x 2 x 3 is zero: This means that there are constants c 1, c 2, and c 3 not all zero such that c 1 x 1 + c 2 x 2 + c 3 x 3 0 If we wished to find the constants, it would not be difficult; just solve the system Xc 0 using row reduction Of course, we see that all of this means the Wronskian being non-zero assures that we have a linearly independent set of solutions to a differential equation These are of course required to get a general solution 10 The Fundamental Matrix Given a set of n solutions, y 1,,y n, to a linear system of n first order equations, we have already seen how to form a solution matrix y 1,1 y 1,n Ψ y 1,,y n y n,1 y n,n 13

14 If the n solutions form a fundamental set of solutions in other words, if the y i are linearly independent solutions, then we call Ψ a fundamental matrix for the system We have already seen that any solution to the system y Pty must have the form Ψtc where Ψt is our fundamental matrix and c is a column vector of constants: Theorem 45: p 234 Let y 1 t,y 2 t,,y n t be a fundamental set of solutions of y Pty, a < t < b, where the n n matrix function Pt is continuous on a, b Let Ψt [y 1 t,y 2 t,,y n t] denote the n n matrix function formed from the fundamental set Let ŷ 1 t,ŷ 2 t,,ŷ n t be any other set of n solutions of the differential equation, and let ˆΨt [ŷ 1 t,ŷ 2 t,,ŷ n t] denote the n n matrix formed from this other set of solutions Then 1 There is a unique n n constant matrix C such that ˆΨt ΨtC, a < t < b, 2 Moreover, ŷ 1 t,ŷ 2 t,,ŷ n t is also a fundamental set of solutions if and only if the determinant of C is non-zero It is interesting to note and not difficult to prove see p 234 in the text that the fundamental matrix is itself a solution to the matrix differential equation Ψ t PtΨt We saw before that the two functions 3e and t were solutions to y We also proved by checking the Wronskian that these functions formed a fundamental set of solutions So a fundamental matrix for the system above is Ψt t 3e 14 y

15 We note that Ψ satisfies the matrix equation Ψ t PtΨt, since Ψ t 6e t 9e while PtΨt t 3e 6e t 9e also Find the map from the fundamental matrix for these solutions to the alternative solution matrix e t e We need to solve t 3e ˆΨ c11 c e t 3 2 e c 21 c 22 e t e 5 2 e t 3 2 e We see for example that we need c 11 + t c 21 e t, so we will require c 11 0 and c 21 1/2 Continuing in this way, we get the equation and determine that our matrix C is 0 1/2 1/2 0 Theorem 45 therefore assures us the new solution matrix ˆΨ is also a fundamental matrix, since C 1/4 0 15