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Solve du/dt=Pu when P is a projection

  • #1

Homework Statement



Solve du/dt=Pu when P is a projection.

[1/2 1/2 = du/dt with
1/2 1/2]

[5 = u(0).
3]

Part of u(0) increases exponentially while the nullspace part stays fixed.

Homework Equations



du/dt = Au with u=u(0) at t=0

The Attempt at a Solution



I'm stuck with this problem. I was thinking that u(t)=eλt*x.

Would du/dt be considered matrix A??? Thanks!
 

Answers and Replies

  • #2
33,273
4,981


Homework Statement



Solve du/dt=Pu when P is a projection.

[1/2 1/2 = du/dt with
1/2 1/2]

[5 = u(0).
3]

Part of u(0) increases exponentially while the nullspace part stays fixed.

Homework Equations



du/dt = Au with u=u(0) at t=0

The Attempt at a Solution



I'm stuck with this problem. I was thinking that u(t)=eλt*x.

Would du/dt be considered matrix A??? Thanks!
I don't understand much of what you have written.
Is this P?
[1/2 1/2]
[1/2 1/2]

If so, it's not a projection matrix - it is just 1/2 I, the identity matrix.

du/dt is a two-dimensional vector, whose components are du1/dt and du2/dt.

Your system of differential equations, not written in matrix form would look something like this:

du1/dt = a11u1 + a12u2
du2/dt = a21u1 + a22u2

This could be written in vector/matrix form as du/dt = Au.
 
  • #3


That's where I'm confused at.

[1/2 1/2 = du/dt. So I guess this means that the matrix equals Pu.
1/2 1/2]'

I'm not sure what to make of the matrix.
 
  • #4
33,273
4,981


That doesn't make any sense to me. What is the exact wording of the problem?
 
  • #5
I like Serena
Homework Helper
6,577
176


Hi tatianaiistb! :smile:

[1/2 1/2]
[1/2 1/2]
is the matrix for the orthogonal projection on (1,1).
(Sorry Mark :redface:)

So with u(t)=(x(t), y(t)) the problem reads:
[tex]\begin{align}\dot x ={1 \over 2}x + {1 \over 2}y \\ \dot y ={1 \over 2}x + {1 \over 2}y\end{align}[/tex]
with
[tex]\begin{align}x_0 = 5 \\ y_0 = 3\end{align}[/tex]

Can you solve that?
 
  • #6


Thank you both... I think I figured out the problem. I greatly appreciate it!
 
  • #7
33,273
4,981


[1/2 1/2]
[1/2 1/2]
is the matrix for the orthogonal projection on (1,1).
(Sorry Mark :redface:)
No problem and no need for apology. I wasn't able to make heads or tails out of what she was trying to do, so I'm glad you jumped in.
 

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