LS.6 Solution Matrices - MIT Mathematics

1 Solution MatricesIn the literature, solutions to linear systems often are expressed using square matricesrather than vectors. You need to get used to the before, we state thedefinitions and results for a 2 2 system, but they generalize immediately ton Fundamental return to the system(1)x =A(t)x,with the general Solution (2)x=c1x1(t) +c2x2(t),wherex1andx2are two independent solutions to (1), andc1andc2are arbitrary form the matrix whose columns are the solutionsx1andx2:(3)X(t) = (x1x2) =(x1x2y1y2).Since the solutions are linearly independent, we called them in afundamentalset ofsolutions, and therefore we call the matrix in (3) afundamental matrixfor the system (1).

2 Writing the general Solution using X(t). As a first application ofX(t), we can use itto write the general Solution (2) efficiently. For according to (2), it isx=c1(x1y1)+c2(x2y2)=(x1x2y1y2) (c1c2),which becomes using the fundamental matrix(4)x=X(t)cwherec=(c1c2),(general Solution to (1)).Note that the vectorcmust be written on the right, even though thec s are usuallywritten on the left when they are the coefficients of the the IVP using X(t). We can now write down the Solution to the IVP(5)x =A(t)x,x(t0) = from the general Solution (4), we have to choose thecso that the initial conditionin (6) is satisfied.

3 Substitutingt0into (5) gives us the matrix equation forc:X(t0)c= the determinant|X(t0)|is the value att0of the Wronskian ofx1amdx2, it isnon-zero since the two solutions are linearly independent (Theorem ). Therefore theinverse matrix exists (by ), and the matrix equation above can be solved forc:c=X(t0) 1x0;using the above value ofcin (4), the Solution to the IVP (1) can now be written(6)x=X(t)X(t0) NOTES: LS. LINEAR SYSTEMSNote that when the Solution is written in this form, it s obvious thatx(t0) =x0, ,that the initial condition in (5) is equation for fundamental matricesWe have been saying a rather than the fundamental matrix since the system (1) doesn t have a unique fundamental matrix: thareare many different ways to pick two independent solutions ofx =Axto form the columnsofX.

4 It is therefore useful to have a way of recognizing a fundamental matrix when you seeone. The following theorem is good for this; we ll need it (t)is a fundamental matrix for the system (1) if its determinant|X(t)|is non-zero and it satisfies the matrix equation(7)X =A X ,whereX means that each entry ofXhas been |X| 6 0, its columnsx1andx2are linearly independent, by section writingX= (x1x2),(7) becomes, according to the rules for matrix multiplication,(x 1x 2) =A(x1x2) = (Ax1Ax2),which shows thatx 1=A x1andx 2=Ax2;this last line says thatx1andx2are solutions to the system (1).

5 2. The normalized fundamental there a best choice for fundamental matrix?There are two common choices, each with its advantages. If the ODE system has con-stant coefficients, and its eigenvalues are real and distinct, then a natural choice for thefundamental matrix would be the one whose columns are the normal modes the solutionsof the formxi=~ ie it,i= 1, is another choice however which is suggested by (6) andwhich is particularly usefulin showing how the Solution depends on the initial conditions. Suppose we pickX(t) so that(8)X(t0) =I=(1 00 1).

6 Referring to the definition (3), this means the solutionsx1andx2are picked so(8 )x1(t0) =(10),x2(t0) =(01).Since thexi(t) are uniquely determined by these initial conditions, the fundamental matrixX(t) satisfying (8) is also unique; we give it a unique matrix Xt0(t) satisfying(9) X t0=A Xt0, Xt0(t0) = Solution MATRICES27is called thenormalized fundamental convenience in use, the definition uses Theorem to guarantee Xt0will actuallybe a fundamental matrix; the condition| Xt0(t)| 6= 0 in Theorem is satisfied, since thedefinition implies| Xt0(t0)|= keep the notation simple, we will assume in the rest of thissection thatt0= 0, as italmost always is; then X0is the normalized fundamental matrix.

7 Since X0(0) =I, we getfrom (6) the matrix form for the Solution to an IVP:(10)The Solution to the IVPx =A(t)x,x(0) =x0isx(t) = X0(t) X0. One way is to find the two solutions in (8 ), and use them as thecolumns of X0. This is fine if the two solutions can be determined by not, a simpler method is this: find any fundamental matrixX(t); then(11) X0(t) =X(t)X(0) verify this, we have to see that the matrix on the right of (11) satisfies the two con-ditions in Definition The second is trivial; the first is easy using the rule for matrixdifferentiation:IfM=M(t)andB, Care constant Matrices , then(BM) =BM ,(M C) =M C,from which we see that sinceXis a fundamental matrix,(X(t)X(0) 1) =X(t) X(0) 1=AX(t)X(0) 1=A(X(t)X(0) 1),showing thatX(t)X(0) 1also satisfies the first condition in Definition Example the Solution to the IVP:x =(0 1 1 0)x,x(0) = the system isx =y, y = x, we can find by inspection the fundamentalset of solutions satisfying (8 ).

8 X= costy= sintandx= sinty= by (10) the normalized fundamental matrix at 0 and Solution to the IVP isx= Xx0=(costsint sintcost) (x0y0)=x0(cost sint)+y0(sintcost).Example the normalized fundamental matrix at 0 forx =(1 31 1) time the solutions (8 ) cannot be obtained by inspection, so we use thesecond method. We calculated the normal modes for this sytemat the beginning of ;using them as the columns of a fundamental matrix gives usX(t) =(3e2t e 2te2te 2t). NOTES: LS. LINEAR SYSTEMSU sing (11) and the formula for calculating the inverse matrix given in , we getX(0) =(3 11 1),X(0) 1=14(1 1 1 3),so that X(t) =14(3e2t e 2te2te 2t) (1 1 1 3)=14(3e2t+e2t3e2t 3e 2te2t e 2te2t+ 3e 2t).

9 The Exponential work in the preceding section with fundamental matriceswas valid for any linearhomogeneous square system of ODE s,x =A(t) , if the system hasconstant coefficients, , the matrixAis a constant matrix,the results are usually expressed by using the exponential matrix, which we now that ifxis any real number, then(12)ex= 1 +x+x22!+..+xnn!+..Definition ann nconstant matrixA, theexponential matrixeAis then nmatrix defined by(13)eA=I+A+A22!+..+Ann!+..Each term on the right side of (13) is ann nmatrix; adding up theij-thentry of each of these Matrices gives you an infinite series whose sum is theij-thentry ofeA.

10 (The series always converges.)In the applications, an independent variabletis usually included:(14)eAt=I+A t+A2t22!+..+Antnn!+..This is not a new definition, it s just (13) above applied to the matrixA tin which everyelement ofAhas been multiplied byt, since for example(At)2=At At=A A t2= out (13) and (14) on these two examples; the first is workedout in your book(Example 2, p. 417); the second is easy, since it is not an infinite (a00b),show:eA=(ea00eb);eAt=(eat00ebt)Ex ample (0 10 0),show:eA=(1 10 1);eAt=(1t0 1)What s the point of the exponential matrix?