Fundamental Matrix for the System of Odes

Home , Diagonal matrix, Identity matrix, Matrix exponential, Wronskian

Fundamental Matrices, Matrix Exp & Repeated Eigenvalues – Sections 7.7 & 7.8

Given fundamental solutions G GGdx = G x1,..., xn of the ODE Ax dt we put them in an nxn matrix Ψ=()G G , (txx )1 ,..., n with each of the solution vectors being a column.

We call Ψ(t) a fundamental matrix for the system of ODEs.

Example. G dx ⎛⎞12G = ⎜⎟x dt ⎝⎠21 − λ ⎛⎞12 2 det⎜⎟=− (1λ ) − 4 ⎝⎠21− λ

=−−=−λλ2 23(3)(1) λ λ +

⎛⎞12 So the eigenvalues of the matrix A=⎜⎟ ⎝⎠21

in our ODE are λ=3,-1.

The corresponding eigenvectors are found by solving (A-λI)v=0 using Gaussian elimination.

We find that ⎛⎞1 the eigenvector for eigenvalue 3 is: v = ⎜⎟ ⎝⎠1 ⎛⎞1 w = ⎜⎟ the eigenvector for eigenvalue -1 is: ⎝⎠−1

So the corresponding solution vectors for our ODE 11 system are G ==3tt⎛⎞G − ⎛ ⎞ ue12⎜⎟, ue ⎜ ⎟ 11− ⎝⎠ ⎝ ⎠

Our fundamental matrix is: − GG ⎛⎞ee3tt Ψ=()tuu() = 12⎜⎟3tt− ⎝⎠ee−

The general solution is

c GGG=+ =Ψ⎛⎞1 yt() cu11 cu 2 2 () t⎜⎟ ⎝⎠c2

If you need to find c1,c2 satisfying some initial

condition like y(t0)=b, then you need to solve cb G =Ψ⎛⎞⎛⎞11 = yt()00 () t ⎜⎟⎜⎟ ⎝⎠⎝⎠cb22 ⎛⎞cb ⎛⎞ 11=Ψ()t −1 ⎜⎟0 ⎜⎟ ⎝⎠cb22 ⎝⎠

The inverse of the fundamental matrix exists because the Wronskian is not 0.

Matrix Exponential

Recall that the Taylor series for exp(x) is

∞ n x ==−⋅x ennn∑ ,!(1)21." n=0 n! The series converges for all real numbers x.

It also converges for all complex numbers x and for all nxn matrices A. ∞ AAAn 23 exp(AIA ) ==++++∑ " n=0 n!2!3!

So exp(0)=I=identity matrix.

You can legally differentiate these Taylor series term by term. This says for scalar x and nxn matrix A, we have:

∞∞n dxAexp( ) 1 dxA() n− ==∑∑xnn1 A dxnn==00 n!! dx n ∞ 1 n−1 == AxAAxA∑ () exp( ) = ()n −1! n 1

And exp(0)=I, the identity matrix.

So we have a solution to the system of ODEs given by G dy G G G ==Ayy,(0); b dx G G yxAb= exp( )

When a matrix is diagonal D, it is easy to compute exp(D).

⎛⎞a 0 exp⎜⎟ ⎝⎠0 b ⎛⎞⎛⎞10a 0 11⎛⎞⎛⎞aa2300 =++ + +" ⎜⎟⎜⎟⎜⎟⎜⎟23 ⎝⎠⎝⎠01 0b 2!00bb 3! ⎝⎠⎝⎠ ⎛⎞a2 10++a +" ⎜⎟⎛⎞ea 0 ==⎜⎟2! 2 ⎜⎟b ⎜⎟b ⎝⎠0 e ⎜⎟01++b ⎝⎠2! When an nxn matrixGG A has n linearly independent eigenvectors vv1,..., n , corresponding to the eigenvalues λλwe can write 1,..., n

− ATDT= 1, λ ⎛⎞1 " 0 GG ⎜⎟ Tvv==()... & D #%# 1 n ⎜⎟ ⎜⎟0 " λ G ⎝⎠G n The reason is = λ Avvj jj.

−−11= One has exp(TDT ) T exp( D ) T

From this, it is easy to compute the solutions to the G system of ODEs dy G G G ==Ayy,(0); b dx G G yxAb= exp( )

ATDT= −1,

λ ⎛⎞1 " 0 GG ⎜⎟ Tvv==()... & D #%# 1 n ⎜⎟ ⎜⎟λ ⎝⎠0 " n

Q(x)=exp(xD) and the fundamental matrix is Ψ(x)=TQ(x), where D is the diagonal matrix of eigenvalues of A and T is the matrix coming from the corresponding eigenvectors in the same order.

exp(xA) is a fundamental matrix for our ODE

Repeated Eigenvalues

When an nxn matrix A has repeated eigenvalues it may not have n linearly independent eigenvectors. In that case it won’t be diagonalizable and it is said to be deficient.

Example.

⎛⎞11 A = ⎜⎟ ⎝⎠01 11− λ ⎛⎞= det⎜⎟ 0 ⎝⎠01− λ

The roots of this are both 1.

Gaussian elimination solves (A-I)x=0 ⎛⎞⎛⎞11− 1 0 1 AI−=⎜⎟⎜⎟ = ⎝⎠⎝⎠01100−

Solutions have x2=0, x1 arbitrary. So we have only 1 linearly independent eigenvector v 1 G ==⎛⎞1 ⎛⎞ v ⎜⎟⎜⎟ v2 0 ⎝⎠⎝⎠

This gives us one solution to the ode

G 1 ⎛⎞ex dy ===GGx ⎛⎞ Ay;e y ⎜⎟ ⎜⎟ dx ⎝⎠0 ⎝⎠0

How to find a 2nd linearly independent solution?

Our first idea is just to multiply this by x, but that will not be linearly independent as it is a multiple of the eigenvector v.

Matrix Exp solves the problem.

AB=BA implies exp(A+B)=expAexpB

So our fundamental matrix is exp(xA)

⎛⎞01 ⎜⎟ A=I+N, N=⎝⎠00

⎛⎞1 x 2 ⎜⎟ N =0 implies exp(xN)=I+xN=⎝⎠01

Ψ(t)=exp(x(I+N))=exp(xI)exp(xN) ⎛⎞exex x ⎜⎟x =⎝⎠0 e

G xxGG Another way: write =+ uxevew2 for some constant vector w to be determined.

G x dwdxe()G dex G =+vw dx dx dx GG =+()exevewxx + x

G dw x xxG G =+()exevew + dx GG =+Axev()xx ew

xxGG =+xev eAw

Here we used the fact that Av=v.

So now equate coefficients of x and 1 xxGG x xx G G () exevewxeveAw++=+ GG G ⎪⎧()vw+= Aw ⇔ ⎨ GG ⎩⎪xv= xv GG We need to vAIw=−(); solve for w ⎛⎞011 −= Gauss(|) A I v ⎜⎟ ⎝⎠000 Solution is

x2=1, x1 arbitrary.

Putting this all together: 10 GGG=+=xx x⎛⎞ + x ⎛⎞ uxevewxee2 ⎜⎟ ⎜⎟ ⎝⎠01 ⎝⎠

G ⎛⎞(1)xe+ x u = 2 ⎜⎟x ⎝⎠e Check that this gives the same fundamental matrix as exp did.