Find the general solution of dX/dt = AX(t) for the given 3x3 matrix A.

[glacier302]

Homework Statement

Find the general solution of dX/dt = AX(t) with the initial condition X(0) = (a1,a2,a3), where A = [0 1 0, 0 0 1, -1 1 1]. (Here, a comma signifies the end of a row).

Homework Equations

The exponential of A is e^A = ∑A^k/k! from k = 0 to k = ∞.
The solution of dX/dt = AX(t) is given by X(t) = e^(tA)*X(0).

The Attempt at a Solution

I know that the first thing I have to is either find the diagonal matrix similar to A if A is diagonalizable, or a Jordan form of A (which is an upper triangular matrix) if A is not diagonalizable. Since the only eigenvalues of
A are -1 and 1 and the dimensions of both eigenspaces are 1, A is not diagonalizable. So I find the Jordan form of A: (Q^-1)AQ = [-1 0 0, 0 1 1, 0 0 1] where Q = [1 1 0, -1 1 1, 1 1 2]and Q^-1 = [1/4 -1/2 1/4,
3/4 1/2 -1/4, -1/2 0 1/2].

Now I'm stuck. I know how to find the solution if A is diagonalizable, because then we have
(Q^-1)AQ = B for some diagonal matrix B so A = QB(Q^-1) and e^tA = Q(e^tB)(Q^-1), and since B is diagonal e^tB is easy to calculate. But how do I calculate e^tB if B is only a Jordan matrix (hence upper triangular), not diagonal?

Any help would be much appreciated : )

 

[HallsofIvy]

I, personally, wouldn't use the exponential form.

Your equation is of the form dX/dt= AX and there exist a matrix, p, such that
p^{-1}Ap= \begin{bmatrix}-1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}

Since p is a constant matrix we can multiply both sides of the equation by p^{-1} to get d(p^{-1}X)/dt= p^{-1}AX= p^{-1}A(pp^{-1})X= (p^{-1}Ap)(p^{-1}X). Now, if we let Y= p^{-1}X, that becomes
\frac{d\begin{bmatrix}y_1\\ y_2\\ y_3\end{bmatrix}}{dt}= \begin{bmatrix}-1 & 0 & 0\\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}y_1 \\ y_2 \\ y_3\end{bmatrix}
which can be written as the three equations
\frac{dy_1}{dt}= -y_1
\frac{dy_2}{dt}= y_2+ y_3
\frac{dy_3}{dt}= y_3

The first and third equations are easy to solve and, once you know y_3 the second equation is easy to solve. Of course, once you have found Y, X= pY.

But, yes, you could do it by finding a matrix exponential. Use the Taylor's series:
e^{Bt}= I+ Bt+ (B/2)t^2+ (B/6)t^3+ ...
Now, look at the powers of B:
B^2= \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{bmatrix}
B^3= \begin{bmatrix}-1 & 0 & 0 \\ 0 & 1 & 3 \\ 0 & 0 & 1\end{bmatrix}
B^4= \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 4\\ 0 & 0 & 1\end{bmatrix}
etc. Get the idea?

 

[glacier302]

Hello,

Thank you very much for your help. I see now that using matrix exponentials is not the easiest way to solve this problem.

Thanks again! : )

 

[HallsofIvy]

But using the eigenvalues and eigenvectors is!

And the matrix exponential is not all that difficult. As I tried to indicate, for
B= \begin{bmatrix}-1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}
B^n= \begin{bmatrix}(-1)^n & 0 & 0 \\ 0 & 1 & n \\ 0 & 0 & 1\end{bmatrix}

So
(Bt)^n= \begin{bmatrix}(-t)^n & 0 & 0 \\ 0 & t^n & nt^n \\ 0 & 0 & t^n\end{bmatrix}
\frac{(Bt)^n}{n!}= \begin{bmatrix}\frac{(-t)^n}{n!} & 0 & 0 \\ 0 & \frac{t^n}{n!} & \frac{nt^n}{n!} \\ 0 & 0 & \frac{t^n}{n!}\end{bmatrix}
= \begin{bmatrix}\frac{(-t)^n}{n!} & 0 & 0 \\ 0 & \frac{t^n}{n!} & \frac{t^{n-1}}{(n-1)!}t \\ 0 & 0 & \frac{t^n}{n!}\end{bmatrix}

so that
e^{Bt}= \sum \frac{(BT)^n}{n!}= \begin{bmatrix}\sum\frac{(-t)^n}{n!} & 0 & 0 \\ 0 & \sum\frac{t^n}{n!} & \sum\frac{t^{n-1}}{(n-1)!}t \\ 0 & 0 & \sum\frac{t^n}{n!}\end{bmatrix}

Note that we can take a factor of t out of that "odd" sum and that
\sum_{n=1}^\infty \frac{t^{n-1}}{(n-1)!}
is the same as
\sum_{i= 0}^\infty \frac{t^i}{i!}= e^t
by taking i= n-1.

That is,
e^{Bt}= \begin{bmatrix}e^{-t} & 0 & 0 \\ 0 & e^t & te^t \\ 0 & 0 & e^t\end{bmatrix}