# Diff equation by diagonalization

1. Apr 2, 2007

### EvLer

1. The problem statement, all variables and given/known data
solve initial value problem for the equation dx/dt = Ax where A =
[1 -1]
[0 1]
x(0) = [1, 1]T

x(t) = S*elambda*t*S -1*x(0)
where S is diagonal matrix, lambda is eigenvalue;

3. The attempt at a solution
I tried to diagonalize it, but I get one eigenvalue =1 mult 2 and I don't have enough eigenvectors to get S!
Am I missing something? and how do i solve it?

Thanks

2. Apr 2, 2007

### D H

Staff Emeritus
The solution to $dx/dt = Ax$ for scalar x is $x=\exp(At)x_0$. This extends to multiple dimensions with $exp(At) = \sum_r A^r t^r/r!$

3. Apr 2, 2007

### EvLer

ok, so you are saying that A does not have to be diagonalizable for a solution to exist?

I am trying to see relations to the diagonalization

4. Apr 3, 2007

### HallsofIvy

Staff Emeritus
?? Every linear equation has solutions! Where did you get the idea that the matrix had to be diagonalizable?

Diagonalizable makes it easier but when a matrix is not diagonalizable you can still get the "Jordan Normal form".

The simplest way to solve this equation is two treat it as two equations:
dx/dt= x- y, dy/dt= y, since they are "partially disconnected": you can solve for y immediately: from dy/dt= y, y= Cet. Then the second equation becomes dx/dt= x- Cet or dx/dt- x= -Cet. That has integrating factor e-t: multiplying the equation by e-t give e-tdx/dt- e-tx= d(e-tx)/dt= e-t(Cet)= C. Integrating, e-tx= Ct+ D so x= Ctet+ Det. With x(0)= D= 1 and y(0)= C= 1, we have x(t)= tet+ et, y(t)= et.

Putting that back into matrix form that is x(t)= [tet+ et, et].

5. Apr 3, 2007

### D H

Staff Emeritus
This is a linear equation of the form $d\vec x/dt = \mathbf A \vec x$. If A and x were scalars, the solution would obviously be $x=\exp(At)x(0)$. This is true for matrices as well:

$$\vec x \exp(\mathbf At) \vec x(0)$$

where $\exp(At)$ is the matrix exponential

$$\exp(At) = \sum_{n=0}^\infty \frac{A^n t^n}{n!}$$

Writing

$$\mathbf A = \mathbf 1_{2x2} - \mathbf B[/itex] where [tex]\mathbf 1_{2x2}$$ is the $2x2$ identity matrix and

$$\mathbf B \equiv \bmatrix 0 & 1 \\ 0 & 0 \endbmatrix$$

Then

$$\mathbf A^2 = \mathbf 1_{2x2} - 2\mathbf B$$

and in general

$$\mathbf A^n = \mathbf 1_{2x2} - n\mathbf B$$

Thus

$$\exp(At) =\mathbf 1_{2x2}\sum_{n=0}^\infty \frac{t^n}{n!} - \mathbf B \sum_{n=1}^\infty \frac{t^n}{(n-1)!}$$

Simplifying,

$$\exp(At) =\mathbf 1_{2x2}\exp t - \mathbf B t\exp t$$

The desired solution is

$$\vec x(t) =(\mathbf 1_{2x2}\exp t - \mathbf B t\exp t)\bmatrix 1 \\ 1\endbmatrix = \bmatrix 1-t\\1\endbmatrix \exp t$$

6. Apr 3, 2007

### D H

Staff Emeritus
There is a sign error here. That should be d(e-tx)/dt = -C

With the correct sign, integrating yields e-tx= D-Ct, so x= (D-Ct)et. With x(0)=y(0)=1, C=D=1, so x(t)= et-tet, y(t)= et. In matrix form,
$$\vec x(t) = \bmatrix e^t-te^t \\ e^t\endbmatrix$$

7. Apr 3, 2007

### HallsofIvy

Staff Emeritus
Thanks for the correction. One of these days, I really need to learn arithmetic!

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook