Diff equation by diagonalization

[EvLer]

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*e^lambda*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]

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!

 

[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

 

[HallsofIvy]

?? 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= Ce^t. Then the second equation becomes dx/dt= x- Ce^t or dx/dt- x= -Ce^t. That has integrating factor e^-t: multiplying the equation by e^-t give e^-tdx/dt- e^-tx= d(e^-tx)/dt= e^-t(Ce^t)= C. Integrating, e^-tx= Ct+ D so x= Cte^t+ De^t. With x(0)= D= 1 and y(0)= C= 1, we have x(t)= te^t+ e^t, y(t)= e^t.

Putting that back into matrix form that is x(t)= [te^t+ e^t, e^t].

 

[D H]

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]<br /> <br /> where<br /> <br /> \mathbf 1_{2x2} is the 2x2 identity matrix and<br /> <br /> \mathbf B \equiv \bmatrix 0 &amp;amp; 1 \\ 0 &amp;amp; 0 \endbmatrix<br /> <br /> Then<br /> <br /> \mathbf A^2 = \mathbf 1_{2x2} - 2\mathbf B<br /> <br /> and in general<br /> <br /> \mathbf A^n = \mathbf 1_{2x2} - n\mathbf B<br /> <br /> Thus<br /> <br /> \exp(At) =\mathbf 1_{2x2}\sum_{n=0}^\infty \frac{t^n}{n!} - \mathbf B \sum_{n=1}^\infty \frac{t^n}{(n-1)!}<br /> <br /> Simplifying,<br /> <br /> \exp(At) =\mathbf 1_{2x2}\exp t - \mathbf B t\exp t<br /> <br /> The desired solution is<br /> <br /> \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]

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= Ce^t. Then the second equation becomes dx/dt= x- Ce^t or dx/dt- x= -Ce^t. That has integrating factor e^-t: multiplying the equation by e^-t give e^-tdx/dt- e^-tx= d(e^-tx)/dt= e^-t(Ce^t)= C.

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

Integrating, e^-tx= Ct+ D so x= Cte^t+ De^t. With x(0)= D= 1 and y(0)= C= 1, we have x(t)= te^t+ e^t, y(t)= e^t.

Putting that back into matrix form that is x(t)= [te^t+ e^t, e^t].

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

 

[HallsofIvy]

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