# Diff equation by diagonalization

## Homework Statement

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;

## 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

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 $exp(At) = \sum_r A^r t^r/r!$

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

HallsofIvy
Homework Helper
?? 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].

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

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

D H
Staff Emeritus
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.

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

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,
$$\vec x(t) = \bmatrix e^t-te^t \\ e^t\endbmatrix$$

HallsofIvy