Diff equation by diagonalization

  • Thread starter EvLer
  • Start date
  • #1
458
0

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
 

Answers and Replies

  • #2
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
685
The solution to [itex]dx/dt = Ax[/itex] for scalar x is [itex]x=\exp(At)x_0[/itex]. This extends to multiple dimensions with [itex]exp(At) = \sum_r A^r t^r/r![/itex]
 
  • #3
458
0
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
HallsofIvy
Science Advisor
Homework Helper
41,833
962
?? 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
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
685
This is a linear equation of the form [itex]d\vec x/dt = \mathbf A \vec x[/itex]. If A and x were scalars, the solution would obviously be [itex]x=\exp(At)x(0)[/itex]. This is true for matrices as well:

[tex]\vec x \exp(\mathbf At) \vec x(0)[/tex]

where [itex]\exp(At)[/itex] is the matrix exponential

[tex]\exp(At) = \sum_{n=0}^\infty \frac{A^n t^n}{n!}[/tex]

Writing

[tex]\mathbf A = \mathbf 1_{2x2} - \mathbf B[/itex]

where

[tex]\mathbf 1_{2x2}[/tex] is the [itex]2x2[/itex] identity matrix and

[tex]\mathbf B \equiv \bmatrix 0 & 1 \\ 0 & 0 \endbmatrix[/tex]

Then

[tex]\mathbf A^2 = \mathbf 1_{2x2} - 2\mathbf B[/tex]

and in general

[tex]\mathbf A^n = \mathbf 1_{2x2} - n\mathbf B[/tex]

Thus

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

Simplifying,

[tex]\exp(At) =\mathbf 1_{2x2}\exp t - \mathbf B t\exp t[/tex]

The desired solution is

[tex]\vec x(t) =(\mathbf 1_{2x2}\exp t - \mathbf B t\exp t)\bmatrix 1 \\ 1\endbmatrix = \bmatrix 1-t\\1\endbmatrix \exp t[/tex]
 
  • #6
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
685
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,
[tex]\vec x(t) = \bmatrix e^t-te^t \\ e^t\endbmatrix[/tex]
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
41,833
962
Thanks for the correction. One of these days, I really need to learn arithmetic!
 

Related Threads on Diff equation by diagonalization

Replies
6
Views
627
Replies
12
Views
937
Replies
3
Views
2K
Replies
4
Views
1K
Replies
7
Views
4K
Replies
0
Views
1K
Replies
0
Views
3K
Replies
1
Views
1K
Replies
5
Views
904
  • Last Post
Replies
1
Views
1K
Top