Matrix Methods for Linear Systems of Differential …

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.

2 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; 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.

3 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 :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.

4 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. Also, a set ofnsolutionsx1.

5 ,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.

6 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 . 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 .

7 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,..,cnare : The Linear combination ofxp,x1,..,xnwritten in 7 with arbitrary constantsc1,..,cniscalled thegeneral solutionof 6.

8 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.

9 , 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.

10 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.