Matrix Exponential. Fundamental Matrix Solution. …

1 :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.

2 We have alreadylearnedhow to solve theinitialvalueproblemd~xdt=A~x;~x(0)=~x 0:We shallcomparethesolutionformulawith~x(t) =etA~x0to knowthegeneralsolutionsofd~x=dt=A~xareof thefollowingstructure:~x(t) =C1~x1(t) + +Cn~xn(t);where~x1(t); ;~xn(t) arenlinearlyindependent ~x(t) = [~x1(t) ~xn(t) ]~Cwith~C= :For theinitialvalueproblem,~Cis determinedby theinitialcondition[~x1(0) ~xn(0)]~C=~x0=)~C= [~x1(0) ~xn(0)] 1~x0:Thus,thesolutionof theinitialvalueproblemis givenby~x(t) = [~x1(t) ~xn(t) ] [~x1(0) ~xn(0)] 1~x0:Comparingthiswith~x(t) =etA~x0, we obtainetA= [~x1(t) ~xn(t) ] [~x1(0) ~xn(0)] 1:In thismethod of evaluatingetA, thematrixM(t) = [~x1(t) ~xn(t) ] plays ,etA=M(t)M(0) nition(FundamentalMatrixSolution):If~x1( t); ;~xn(t) arenlinearlyindependentsolutionsof thendimensionalhomogeneouslinearsystemd~ x=dt=A~x, we callM(t) = [~x1(t) ~xn(t) ]afundamentalmatrixsolutionof thesystem.

3 (Remark1:ThematrixfunctionM(t) satis estheequationM0(t) =AM(t).Moreover,M(t) is aninvertiblematrixforeveryt. Thesetwo propertiescharacterizefundamentalmatrixs olutions.)(Remark2: Givena linearsystem, ,whenwe make any choiceof a fundamentalmatrixsolutionM(t) andcomputeM(t)M(0) 1,we always getthesameresult.) 7 9935 3 3 :We rstsolved~x=dt=A~x. We obtain~x(t) =C1e t243 1135+C2e2t24 11035+C3e2t2410135:Thisgives a fundamentalmatrixsolution:M(t) =243e t e2te2t e te2t0e t0e2t35:ThematrixexponentialisetA=M(t)M( 0) 1=243e t e2te2t e te2t0e t0e2t35243 11 11010135 1=243e t 2e2t3e t 3e2t 3e t+ 3e2t e t+ 2e2t e t+ 2e2te t e2te t e2te t e2t e t+ 2e2t35 5 8423 2614 :We rstsolved~x=dt=A~x. We obtain~x(t) =C1e t243 1135+C2e 3t24 21135+C3e 3t24 1 2t1=2 +tt35:Thisgives a fundamentalmatrixsolution:M(t) =243e t 2e 3t e 3t 2te 3t e te 3t12e 3t+te 3te te 3tte 3t35:ThematrixexponentialisetA=M(t)M(0) 1=243e t 2e 3t e 3t 2te 3t e te 3t12e 3t+te 3te te 3tte 3t35243 2 1 111=211035 1=243e t 2e 3t 8te 3t6e t 6e 3t 20te 3t4te 3t e t+e 3t+ 4te 3t 2e t+ 3e 3t+ 10te 3t 2te 3te t e 3t+ 4te 3t2e t 2e 3t+ 10te 3te 3t 2te 3t35 9 545.

4 Solution1:(UseDiagonalization)Solvingdet (A I) = 0, we obtaintheeigenvaluesofA: 1= 7 + 4i; 2= 7 1= 7 + 4i: areobtainedby solving[A (7 + 4i)I]~v= 0:~v=v2 12+i1 :( )Eigenvectorsfor 2= 7 4i: arecomplexconjugateof thevectorsin ( ).ThematrixAis now diagonalized:A=P DP 1withP= 12+i12 i11 ;D= 7 + 4i007 4i :We haveetA=P etDP 1= 12+i12 i11 e(7+4i)t00e(7 4i)t 12i12+14i12i12 14i =24(12 14i)e(7+4i)t+ (12+14i)e(7 4i)t58ie(7+4i)t 58ie(7 4i)t 12ie(7+4i)t+12ie(7 4i)t(12+14i)e(7+4i)t+ (12 14i)e(7 4i)t35:Solution2:(Usefundamentalsolution sandcomplexexpfunctions)A fundamentalmatrixsolutioncanbe obtainedfromtheeigenvaluesandeigenvector s:M(t) = (12+i)e(7+4i)t(12 i)e(7 4i)te(7+4i)te(7 4i)t :ThematrixexponentialisetA=M(t)M(0) 1= (12+i)e(7+4i)t(12 i)e(7 4i)te(7+4i)te(7 4i)t 12+i12 i11 1=24(12 14i)e(7+4i)t+ (12+14i)e(7 4i)t58ie(7+4i)t 58ie(7 4i)t 12ie(7+4i)t+12ie(7 4i)t(12+14i)e(7+4i)t+ (12 14i)e(7 4i)t35:Solution3:(Usefundamentalsolution sandavoidcomplexexpfunctions)A fundamentalmatrixsolutioncanbe obtainedfromtheeigenvaluesandeigenvector s.

5 M(t) = e7t 12cos4t sin4t e7t cos4t+12sin4t e7tcos4te7tsin4t :ThematrixexponentialisetA=M(t)M(0) 1= e7t 12cos4t sin4t e7t cos4t+12sin4t e7tcos4te7tsin4t 12110 1= e7tcos4t+12e7tsin4t 54e7tsin4te7tsin4te7tcos4t 12e7tsin4t :5 APPENDIX:CommonMistakesSincetheyearsof freshmencalculus,we alllovedtheexponentialfunctionexwithscal arvariablex. Therearetonsof simpleandbeautifulformulasforthescalarfu nctionex. Thematrixexponentialis, however,a quitedi erent needto be a littlecarefulin I have seenpeoplemake: et(A+B)=etAetB: Thesolutionsofd~xdt=A(t)~xare~x(t) =etA(t)~C. Thesolutionsofd~xdt=A(t)~xare~x(t) =eRt0A(s)ds~C. eB(t) 0=B0(t)eB(t).Alltheabove fourstatements areWRONG!(1)It is truethatet(A+B)=etAetBifAB=BA. Butin general,et(A+B)6= :ForA= 1 1 11 ; B= 1100 ; A+B= 00 11 ;we haveetAetB=12 1 +e2t1 e2t1 e2t1 +e2t e t1 e t01 =12 e t+et2 e t ete t et2 e t+et ;butet(A+B)= 101 etet :(2)We learnedhow to solved~xdt=A~xwhereAis a constant nogeneralsolutionmethod for( )d~xdt=A(t)~xwhereA(t) is particular,~x(t) =etA(t)~C:does notsolve ( ).

6 Thisevenfailsin :Thesolutionsof thescalarequationdxdt= (sint)xis givenbyx(t) =e costC, (3)Likewise,~x(t) =eRt0A(s)ds~Cis alsoawrongsolutionformulafor( ). Thisformulais onlyvalidforscalarequations, ,whenthespacedimensionis :Considerd~xdt= 102t 1 ~x;~x(0)= 10 whereA(t) = 102t 1 :Thesolutionof thisinitialvalueproblemis~x(t) = et(t 12)et+12e t :Thefunctionexp Zt0A(s)ds 10 does notgive ,exp Zt0A(s)ds 10 = exp t0t2 t 10 = et012tet 12te te t 10 = et12tet 12te t :(4)We dohave:(eb(t))0=b0(t)eb(t)holdsforscalar functionsb(t), and(etA)0=AetA=etAAforconstant ,in general,fornonconstant matrixfunctionB(t),(eB(t))0is neitherB0(t)eB(t)noreB(t)B0(t).Example:L etB(t) = t0t2 t . We haveeB(t)= et012tet 12te te t ;andhence eB(t) 0= et0t+12et+t 12e t e t ;B0(t)eB(t)= 102t 1 et012tet 12te te t = et032tet+12te t e t ;eB(t)B0(t) = et012tet 12te te t 102t 1 = et012tet+32te t e t :Allthreematrixfunctions(eB(t))0,B0(t)eB (t)andeB(t)B0(t) aredi erent fromeach [1]EvaluateetAforA= 5 311.

7 [2](a)EvaluateetAforA= 412 38 .(b)Solved~xdt=A~x;~x(0)= 5 1 .[3]EvaluateetAforA=24 14 2 340 31335.[4]EvaluateetAforA=2454 2 12 94 12 8335.[5](a)EvaluateetAforA=2497 3 16 125 8 5235.(b)Solved~xdt=A~x;~x(0)=2411135.[6] EvaluateetAforA= 5 421 .[7]EvaluateetAforA=24 1102 320 :[1]etA=26432e4t 12e2t 32e4t+32e2t12e4t 12e2t 12e4t+32e2t375[2](a)etA= e2t 6te2t12te2t 3te2te2t+ 6te2t (b)~x(t) =etA 5 1 = 5e2t 42te2t e2t 21te2t [3]etA=243et 2e2t 5et+ 6e2t e3t3et 4e2t+e3t3et 3e2t 5et+ 9e2t 3e3t3et 6e2t+ 3e3t3et 3e2t 5et+ 9e2t 4e3t3et 6e2t+ 4e3t35[4]etA=243et 2e t2et 2e t et+e t 6et+ 6e t 4et+ 5e t2et 2e t 6et+ 6e t 4et+ 4e t2et e t35[5](a)etA=243et 2e t+ 4te t2et 2e t+ 3te t et+e t te t 6et+ 6e t 4te t 4et+ 5e t 3te t2et 2e t+te t 6et+ 6e t+ 4te t 4et+ 4e t+ 3te t2et e t te t35(b)~x(t) =etA2411135=244et 3e t+ 6te t 8et+ 9e t 6te t 8et+ 9e t+ 6te t35[6]etA=e3t cos2t+ sin2t 2 sin2tsin2tcos2t sin2t ,or,equivalently,etA=12 (1 i)e(3+2i)t+ (1 +i)e(3 2i)t2ie(3+2i)t 2ie(3 2i)t ie(3+2i)t+ie(3 2i)t(1 +i)e(3+2i)t+ (1 i)e(3 2i)

8 T [7]etA=15242e 3t+ 3 cost+ sint 2e 3t+ 2 cost sinte 3t cost+ 3 sint 4e 3t+ 4 cost 2 sint4e 3t+ cost 3 sint 2e 3t+ 2 cost+ 4 sint 2e 3t+ 2 cost 6 sint2e 3t 2 cost 4 sint e 3t+ 6 cost+ 2 sint359