# Matrix equation

1. Sep 3, 2007

### Physics_wiz

$$\frac{\partial}{\partial t} \phi(t, to) = A(t)\phi(t,to)$$

Show that if

$$\int_{to}^t {A(\lambda) d\lambda}$$ and $$A(t)$$ commute, then the unique solution of $$\phi(t, to)$$ to the above equation is:

$$\phi(t, to) = \exp[\int_{to}^t {A(\lambda) d\lambda}]$$

The professor said it was very simple, but I can't even get any expression with the integral of A(t) and A(t) in the same equation. I tried plugging the answer in and checking it, because I think that's what he said to do...

$$\frac{\partial}{\partial t}\exp[\int_{to}^t {A(\lambda) d\lambda}] = A(t)\exp[\int_{to}^t {A(\lambda) d\lambda}]$$

$$\frac{\partial}{\partial t}[\int_{to}^t {A(\lambda) d\lambda}]\exp[\int_{to}^t {A(\lambda) d\lambda}] = A(t)\exp[\int_{to}^t {A(\lambda) d\lambda}]$$

Using Euler's formula to evaluate the partial of the integral, I get:

$$A(t)\exp[\int_{to}^t {A(\lambda) d\lambda} = A(t)\exp[\int_{to}^t {A(\lambda) d\lambda}$$

Note: $$\phi$$ and $$A$$ are matrices.

2. Sep 3, 2007

### D H

Staff Emeritus
You are almost done, why go any further? In particular, look at the right-hand side. Your answer is right there.

However, how do you justify going from the left-hand side to the right-hand side? (Hint: you need to use that $\int_{to}^t {A(\lambda) d\lambda}$ and $A(t)$ commute.)

3. Sep 3, 2007

### Physics_wiz

Errr I don't see it. Sorry if this is stupid, but I've thought about it a lot and I don't see it.

4. Sep 3, 2007

### Fredrik

Staff Emeritus
That was my first thought too, but I don't think that was a part of his calculation. He's just stating the equality that he wants to show.

Wiz, you are trying to evaluate an expression of the form

$$\frac{d}{dt}e^{B(t)}$$

and you can't be sure that this is

$$B'(t)e^{B(t)}$$

unless [B'(t),B(t)]=0. Do you understand why?

What you need to do is to show that B'(t)=A(t). I wouldn't use any fancy shmancy rules of integration. I would just use the definition of a derivative.

When you have done this, you have shown that $\phi(t,t_0)$ is a solution (for each $t_0$). Apparently you're also supposed to show that there are no other solutions.

Last edited: Sep 3, 2007
5. Sep 3, 2007

### Physics_wiz

What do you mean by that? Both B'(t) and B(t) = 0?

I don't see how this solves the problem.

I was thinking that I should get something along the lines of an equation with A multiplied by its integral on one side, and the integral of A multiplied by A on the other side and I would say that the two sides are equal only if A and its integral commute.

Last edited: Sep 3, 2007
6. Sep 3, 2007

### Fredrik

Staff Emeritus
Yes. I kept editing my post for a while and you replied while that mistake was still in there.

You would probably see this if you try to prove that

$$\frac{d}{dt}e^{B(t)}=B'(t)e^{B(t)}$$

using the power series definition of the exponential. You won't be able to without using [B',B]=0 (which by the way is the same as [A,integral of A]=0).

I don't see why you don't. The identity above, with B'(t) replaced with A(t) (and of course B(t) equal to the integral), is your differential equation, so if you can obtain that result, you have proved that $\phi(t,t_0)$ is a solution.

I haven't really thought about alternative solutions, but I don't think there's any easier way to do this than what I described.

7. Sep 3, 2007

### Fredrik

Staff Emeritus
Nope. [X,Y] is the commutator of the matrices X and Y, which is defined as XY-YX, so [X,Y]=0 means that X and Y commute.

8. Sep 3, 2007

### Physics_wiz

Ahh Yes, I see it now. Thank you so much.

9. Sep 3, 2007

### D H

Staff Emeritus
First part of post deleted. I type too slow.

Aside: Isn't [A,integral of A] a stronger statement than [A,A']=0? [A,A'] will be zero so long as [A,integral of A] is a constant of time. The zero matrix is a special case of a constant of time.

Last edited: Sep 3, 2007