Matrix Methods for Linear Systems of Differential Equations

1 Matrix Methods for Linear Systems of Differential EquationsWe now present an application of Matrix Methods to Linear Systems of Differential Equations . We shallfollow the development given in Chapter 9 ofFundamentals of Differential Equations and BoundaryValue Problemsby Nagle, Saff, Snider, third of MatricesIf we allow the entriesaij t in ann nmatrixA t to be functions of the variablet,thenA t is amatrix function of t. Similarly if the entriesxi t of a vectorx t are functions oft,thenx t is avectorfunction of t. A matrixA t is said to becontinuous at t0if eachaij t is continuous t isdifferentiable at t0if eachaij t is differentiable att0and we writedAdt t0 A t0 aij t0 n nAlso abA t dt abaij t dtn nWe have the following differentiation formulas for matricesddt CA CdAdt,Ca constant matrixddt A B dAdt dBdtddt AB AdBdt BdAdtIn the last formula the order in which the matrices are written is important, since Matrix multiplicationneed not be Systems in Normal FormAsystemofnlinear Differential Equations is innormal formif it is expressed asx t A t x t f t 1 wherex t andf t aren 1 column vectors andA t aij t n system is calledhomogeneousiff t 0.

2 Otherwise it is the1elements ofAare constants, the system is said to haveconstant note that a linearnthorder Differential equationy n t pn 1 t y n 1 p0 t y g t 2 can be rewritten as a first order system in normal form using the substitutionx1 t y t ,x2 t y t ,..,xn t y n 1 t 2. 5 Thenx1 t y t x2 t x2 t y t x3 t xn 1 t y n 1 t xn t xn t y n t pn 1 t y n 1 p0 t y g t From 2. 5 we can write this last equation asxn t p0 t x1 t pn 1 t xn t g t Thus the Differential equation 2 can be put in the form 1 withx t x1x2 xn,f t 00 g t andA 010 00001 00 000 01 p0 t p1 t p2 t pn 2 t pn 1 t Theinitial value problemfor the normal system 1 is the problem of finding a Differential vectorfunctionx t that satisfies the system on an intervalIand also satisfies theinitial condition x t0 x0,wheret0is a given point ofIandx0is a given constant :Convert the initial value problemy 3y 2y 0y 0 1y 0 3into an initial value problem for a system in normal.

3 Y 3y t y t x2 t y t x1 x2x2 3x2 2x1 Thusx t 01 2 3x1x2 00We also have the initial conditionx 0 x1 0 x2 0 13 1(Existence and Uniqueness)SupposeA t andf t are continuous on an open intervalIthat contains the pointt0. Then, for anychoice of the initial vectorx0there exists a unique solutionx t on the entire intervalIto the initialvalue problemx t A t x t f t ,x t0 x0 Remark: Just as in Ma 221 we may introduce theWronskianofnvectors functions and use it to test forlinear independence. We haveDefinition: TheWronskianof thenvector functionsx1 t col x11,x21,..,xn1 ,..,xn t col x1n,x2n,..,xnn is defined to be the real-valued function3W x1,..,xn t x11 t x12 t x1n t x21 t x22 t x2n t xn1 t xn2 t xnn t One can show that the Wronskian of solutionsx1,..,xntox Axis either identically zero or neverzero on and intervalI.

4 Also, a set ofnsolutionsx1,..,xntox AxonIis linearly independent ifand only if their Wronskian is never zero onI. Thus the Wronskian provides us with an easy test forlinear independence for solutions ofx 2(Representation of Solutions-Homogeneous Case)Letx1,x2,..,xnbenlinearly independent solutions to the homogeneous systemx t A t x t 3 on the intervalI,whereA t is ann nmatrix function continuous onI. Then every solution of 3 onIcan be expressed in the formx t c1x1 t cnxn t 4 wherec1,..,cnare set of solutions x1,..,xn that are linearly independent onIis called afundamental solution setfor 3 . The Linear combination 4 is referred to as the general solution of 3 .Exercise:Verify thate2te2te2t, e t0e t, e te t0is a fundamental solution set for the systemx t 011101110x t 5 011101110, eigenvectors: 110, 101 1,111 24 Considerx3 t e te t 011101110 e te t0 e t e t0 x3 t Remark: The matrixX t e2t e t e te2t0e te2te t0is a fundamental Matrix for the DE 5.

5 The generalsolution of 5 can be written asx t X t c c1e2te2te2t c2 e t0e t c3 e te t0 Remark:IfwedefineanoperatorLbyL x x Axthen this operator ,L c1x1 c2x2 c1L x1 c2L x2 . Thus ifx1andx2arehomogeneous solutions of the homogeneous equationx Axthec1x1 c2x2is also a solution of this equation. Another consequence of this linearity is thesuperposition principlefor Linear Systems . It states that ifxp1andxp2are solutions respectively of thenonhomogeneoussystemsL x g1andL x g2,thenxp1 xp2is a solution ofL x g1 leads toTheorem 3(Representation of Solutions-Nonhomogeneous Case)Letxpbe a particular solution to the nonhomogeneous systemx t A t x t f t 6 on the intervalI,andlet x1,x2,..,xn be a fundamental solution set onIfor the correspondinghomogeneous systemx t A t x t . Then every solution to 6 onIcan be expressed in the formx t xp t c1x1 t cnxn t 7 wherec1.

6 ,cnare : The Linear combination ofxp,x1,..,xnwritten in 7 with arbitrary constantsc1,..,cniscalled thegeneral solutionof 6 . We may express this solution asx xp Xc,whereXis afundamental Matrix for the homogeneous system andcis an arbitrary constant Normal determine a general solution to then nhomogeneous systemx a fundamental solution set x1,..,xn that consists ofnlinearlyindependent solutions to the homogeneous the Linear combinationx Xc c1x1 cnxnwherec col c1,..,cn is any constant vector andX x1,..,xn is thefundamental Matrix , to obtain a general determine a general solution of to the nonhomogeneous systemx Ax a particular solutionxpto the nonhomogeneous the sum of the particular solution and the general solutionXc c1x1 cnxnto the corresponding homogeneous system in part 1,x xp Xc xp c1x1 cnxnto obtain a general Linear Systems with Constant CoefficientsConsider now the systemx t Ax t 8 whereAis a (real)constant n 4 Suppose then nconstant matrixAhasnlinearly independent eigenvectorsu1,u2.

7 , eigenvalue corresponding to er1tu1,er2tu2,..,erntun 9 is a fundamental solution set on , for the homogeneous systemx Ax. Hence the generalsolution ofx Axisx t c1er1tu1 cnerntunwherec1,..,cnare arbitrary : The eigenvalues may be real or complex and need not be riuiwe haveddt eritui rieritui eritAui A eritui so each element of the set 9 is a solution of the system 8 . Also the Wronskian of these solutions isW t det er1tu1,..,erntun e r1 r2 rn tdet u1,..,un 0since the eigenvectors are linearly a general solution ofx 54 10x54 10, eigenvectors: 11 1, 41 4 Thusx t c1et 11 c2e4t 41 Thus the solution isx1 t c1et 4c2e4tx2 t c1et c2e4t SNB gives the following strange looking result:x1 5x1 4x2x2 x1, Exact solution is:x1 t 13C1et 43C1e4t 43C2e4t 43C2etx2 t 13C1e4t 13C1et 43C2et 13C2e4tThis is correct and is equivalent to ,ifweletc1 13C1 43C2andc2 13C1 , it is a most cumbersome form of the :Nagle and Saff page 535 #23.

8 Find a fundamental Matrix for the systemx t 211 10 10 100310007x t Solution:211 10 10 100310007, eigenvectors:1 300 1,1000 2, 1128 7,1010 3 Hence the four linearly independent solutions aree t1 300,e2t1000,e7t 1128,e3t1010 Therefore a fundamental Matrix ise te2t e7te3t 3e t0e7t0002e7te3t008e7t0We know that if a Matrix hasndistinct eigenvalues, then the eigenvectors associated with theseeigenvalues are linearly independent. HenceCorollaryIf then nconstant matrixAhasndistinct eigenvaluesr1,..,rnanduiis an eigenvector associatedwithrithen er1tu1,..,erntun is a fundamental solution set for the homogeneous systemx the initial value problemx t 12 110 14 45x t x 0 100 Solution:12 110 14 45, eigenvectors: 112 1, 214 2, 114 3 Thusx t c1et 112 c2e2t 214 c3e3t 114 c1et 2c2e2t c3e3tc1et c2e2t c3e3t2c1et 4c2e2t 4c3e3tWe definex t via t c1et 2c2e2t c3e3tc1et c2e2t c3e3t2c1et 4c2e2t 4c3e3tso thatx 0 c1 2c2 c3c1 c2 c32c1 4c2 4c3 100 Thus we form 1 2 1 111102440, row echelon form:100 0010 1001 1.

9 Hencec1 0,c2 1,c3 1. Then the solution isx t e2t 214 e3t 114 Complex Eigenvalues9We now discuss how one solves the systemx t Ax t in the case whereAis a real Matrix and the eigenvalues are complex. We shall show how to obtain tworeal vector solutions of the system . Recall that ifr1 i is a solution of the equation thatdetermines the eigenvalues, namely,p det A rI 0thenr2 i is also a solution of this equation, and hence is an eigenvalue. Recall thatr2is calledthe complex conjugate ofr1andr1 a ib,whereaandbare real vectors, be an eigenvector corresponding tor1. Then it is nothard to see thatz a ibis an eigenvector corresponding A r1I z 0then taking the conjugate of this equation and noting that sinceAandIare real matrices thenA AandI I A r1I z A r1I z A r2I z 0sozis an eigenvector corresponding tor2.

10 Therefore the vectorsw1 t e i t a ib andw2 t e i t a ib are two linearly independent vector solutions of . However, they are not real. To get real solutionswe proceed as follows: Sincee i t e t cos t isin t thenw1 t e i t a ib e t cos ta sin tb i sin ta cos tb Thereforew1 t x1 t ix2 t wherex1 t e t cos ta sin tb x2 t e t sin ta cos tb Sincew1 t is a solution of ,thenw1 t Aw1 t sox1 t ix2 t Ax1 t iAx2 t Equating real and imaginary parts of this last equation leads to the real equationsx1 t Ax1 t x2 t Ax2 t 10so thatx1 t andx2 t are real vector solutions of corresponding to the eigenvalues i .Note that we can get the two expressions above forx1 t andx2 t by taking the real and imaginaryparts ofw1 t ExampleFind the general solution of the initial value problemx t 3 12 1x t x 2 01 Solution:We first find the eigenvalues and eigenvectors of the Matrix .