Matrix Exponential. Fundamental Matrix Solution. Objective ...

:Solved~xdt=A~xwithann nconstant coe cient ,theunknownis thevectorfunction~x(t) =24x1(t)..xn(t) MatrixExponentialForm:~x(t) =etA~C= ;whereC1; ; theinitialvalueproblemd~xdt=A~x;~x(t0) =~x0is givenby~x(t) =e(t t0)A~x0:De nition(MatrixExponential):For a squarematrixA,etA=1Xk=0tkk!Ak=I+tA+t22!A 2+t33!A3+ :1 Evaluationof MatrixExponentialin theDiagonalizableCase:SupposethatAis diago-nalizable;thatis, thereareaninvertiblematrixPanda diagonalmatrixD=24 n35such thatA=P DP 1. In thiscase,we haveetA=P etDP 1=P24e nt35P 1 2 4 12139 335.(a)EvaluateetA.(b)Findthegeneralsolu tionsofd~xdt=A~x.(c)Solve theinitialvalueprobelmd~xdt=A~x;~x(0)=24 21435:Solution:ThegivenmatrixAis diagonalized:A=P DP 1withP=241 11=2 12 1=211135;D=2410002000 135:Part(a):We haveetA=P etDP 1=241 11=2 12 1=21113524et000e2t000e t352453 1110 6 4235=245et e2t 3e t3et e2t 2e t et+e t 5et+ 2e2t+ 3e t 3et+ 2e2t+ 2e tet e t5et+e2t 6e t3et+e2t 4e t et+ 2e t35:Part(b):Thegeneralsolutionsto thegivensystemare~x(t) =etA24C1C2C335;whereC1; C2; (c):Thesolutionto theinitialvalueproblemis~x(t) =etA24 21435=24 11et+e2t+ 8e t11et 2e2t 8e t 11et e2t+ 16e t35:2 Evaluationof MatrixExponentialUsingFundamentalMatrix: InthecaseAis notdiagonalizable,oneapproach to obtainmatrixexponentialis to ,weuseanotherapproach.

Matrix Exponential. Fundamental Matrix Solution. Objective: Solve d~x dt = A~x with an n n constant coe cient matrix A. Here, the unknown is the vector function ~x(t) =

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  Solutions, Fundamentals, Objectives, Matrix, Exponential, Matrix exponential, Fundamental matrix solution

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