Solution to Matrix Differential Equation

[WolfOfTheSteps]

How do I write the form of the solution to this equation:

<br /> <br /> \dot{\vec{x}}(t) =<br /> \left [ \begin{array}{cc}<br /> a_{11}(t) &amp; a_{12}(t) \\<br /> a_{21}(t) &amp; a_{22}(t)<br /> \end{array} \right ] \vec{x}(t)<br /> <br />


I just need to be able to write x₁(t) and x₂(t) so I can do the rest of the problem I'm working on. Getting this would just be a small step in my solution, but I am very rusty with my differential equations! :(

Initially, I thought to write:

<br /> x_1(t) = \int_{t_0}^{t}x_1(\tau)a_{11}(\tau) + x_2(\tau)a_{12}(\tau)d\tau <br />


<br /> x_2(t) = \int_{t_0}^{t}x_1(\tau)a_{21}(\tau) + x_2(\tau)a_{22}(\tau)d\tau <br />

But that has the solutions with dependence on x₁(t) and x₂(t). That's not the way to write it, is it?

Thanks.

 

[Mark44]

How do I write the form of the solution to this equation:

<br /> <br /> \dot{\vec{x}}(t) =<br /> \left [ \begin{array}{cc}<br /> a_{11}(t) &amp; a_{12}(t) \\<br /> a_{21}(t) &amp; a_{22}(t)<br /> \end{array} \right ] \vec{x}(t)<br /> <br />


I just need to be able to write x₁(t) and x₂(t) so I can do the rest of the problem I'm working on. Getting this would just be a small step in my solution, but I am very rusty with my differential equations! :(

Initially, I thought to write:

<br /> x_1(t) = \int_{t_0}^{t}x_1(\tau)a_{11}(\tau) + x_2(\tau)a_{12}(\tau)d\tau <br />


<br /> x_2(t) = \int_{t_0}^{t}x_1(\tau)a_{21}(\tau) + x_2(\tau)a_{22}(\tau)d\tau <br />

But that has the solutions with dependence on x₁(t) and x₂(t). That's not the way to write it, is it?

Thanks.

Let's look at your system of DEs this way:
<br /> \dot{\vec{x}}(t) = A \vec{x}(t)<br />

What you'd like is a system that looks like this:
<br /> \dot{\vec{x}}(t) = D \vec{x}(t)<br />
where D is a diagonal matrix.

This will untangle things so that you have x1'(t) = d11 x1(t) and x2'(t) = d22 x2(t).

These are easy to solve, since each one involves only a single variable.

Getting the matrix D is the hard part, though, since doing this involves changing to a different basis (for R2 in your case). Without going into too many details, you'll want to find a matrix D that is similar to your original matrix A, which I'm assuming is invertible. Similarity is precisely defined this way: If A ~ D, then for some invertible matrix P, AP = PD.

Equivalently, P^{-1}AP = P^{-1}PD = D.

You will need to come up with a matrix P whose columns are the new basis, and a matrix P^{-1}, the inverse of P.

To wind this up, the columns of P are the eigenvectors of matrix A, and it turns out that the diagonal entries of D are the eigenvalues of A.

I hope I've given you enough to at least get you started searching for the things to learn more about. Solving a system of DEs, even the simplest possible system in two variables requires a significant amount of understanding in linear algebra.

Mark

 

[WolfOfTheSteps]

Let's look at your system of DEs this way:
<br /> \dot{\vec{x}}(t) = A \vec{x}(t)<br />

What you'd like is a system that looks like this:
<br /> \dot{\vec{x}}(t) = D \vec{x}(t)<br />
where D is a diagonal matrix.

This will untangle things so that you have x1'(t) = d11 x1(t) and x2'(t) = d22 x2(t).

These are easy to solve, since each one involves only a single variable.

Getting the matrix D is the hard part, though, since doing this involves changing to a different basis (for R2 in your case). Without going into too many details, you'll want to find a matrix D that is similar to your original matrix A, which I'm assuming is invertible. Similarity is precisely defined this way: If A ~ D, then for some invertible matrix P, AP = PD.

Equivalently, P^{-1}AP = P^{-1}PD = D.

You will need to come up with a matrix P whose columns are the new basis, and a matrix P^{-1}, the inverse of P.

To wind this up, the columns of P are the eigenvectors of matrix A, and it turns out that the diagonal entries of D are the eigenvalues of A.

I hope I've given you enough to at least get you started searching for the things to learn more about. Solving a system of DEs, even the simplest possible system in two variables requires a significant amount of understanding in linear algebra.

Mark

Yes, my linear algebra is a few years out of service, but it is not non-existent. You have refreshed my memory. Thanks.