MHB Verify Change of Variables for System of Equations w/ Constant Coefficients

[Dustinsfl]

For the system of equations
$$
\mathbf{x}' = \begin{pmatrix}
a\cos 2t & a\sin 2t\\
a\sin 2t & -a\cos 2t
\end{pmatrix}\mathbf{x},
$$
verify that under the change of variables:
$$
\mathbf{x} = \begin{pmatrix}
\cos 2t & \sin 2t\\
\sin 2t & -\cos 2t
\end{pmatrix}\mathbf{u},
$$
the equations for $u$ become a system with constant coefficients:
$$
\mathbf{u}' = \begin{pmatrix}
a & 1\\
-1 & -a
\end{pmatrix}\mathbf{u}.
$$Here is what I have
$$
\left[\begin{pmatrix}
\cos 2t & \sin 2t\\
\sin 2t & -\cos 2t
\end{pmatrix}\mathbf{u}\right]' = \begin{pmatrix} a & 0\\ 0 & a\end{pmatrix}\mathbf{u}
$$

What do I do about that left side? I don't see how I will get
$$
\mathbf{u}' = \begin{pmatrix}
a & 1\\
-1 & -a
\end{pmatrix}\mathbf{u}
$$

 

[Ackbach]

Hmm. Using the equation for $\mathbf{x}$ as a function of $\mathbf{u}$, we get
$$\dot{\mathbf{x}}=2\begin{bmatrix}- \sin(2t) & \cos(2t)\\ \cos(2t) & \sin(2t)\end{bmatrix}\mathbf{u}+\begin{bmatrix} \cos(2t) & \sin(2t)\\ \sin(2t) &- \cos(2t)\end{bmatrix}\dot{\mathbf{u}}.$$
But we also have that
$$\dot{\mathbf{x}}=\begin{bmatrix} a \cos(2t) & a \sin(2t)\\ a \sin(2t) &- a \cos(2t)\end{bmatrix}\mathbf{x}=a \begin{bmatrix} \cos(2t) & \sin(2t)\\ \sin(2t) &- \cos(2t)\end{bmatrix} \mathbf{x}=a\mathbf{u}.$$
So equating our two expressions for $\dot{\mathbf{x}}$ yields
$$2\begin{bmatrix}- \sin(2t) & \cos(2t)\\ \cos(2t) & \sin(2t)\end{bmatrix}\mathbf{u}+\begin{bmatrix} \cos(2t) & \sin(2t)\\ \sin(2t) &- \cos(2t)\end{bmatrix}\dot{\mathbf{u}}=a\mathbf{u},$$
or
$$\begin{bmatrix} \cos(2t) & \sin(2t)\\ \sin(2t) &- \cos(2t)\end{bmatrix}\dot{\mathbf{u}}=a\mathbf{u}+2\begin{bmatrix} \sin(2t) & - \cos(2t)\\ - \cos(2t) & - \sin(2t)\end{bmatrix}\mathbf{u}
=a \begin{bmatrix}1 &0\\0&1\end{bmatrix} \mathbf{u}+2\begin{bmatrix} \sin(2t) & - \cos(2t)\\ - \cos(2t) & - \sin(2t)\end{bmatrix}\mathbf{u}$$
$$=\begin{bmatrix}a+2 \sin(2t) &-2 \cos(2t)\\ -2 \cos(2t) &a-2 \sin(2t)\end{bmatrix} \mathbf{u}.$$
Now, let's see if we can invert the matrix on the LHS with a left-multiplication:
$$\begin{bmatrix} \cos(2t) & \sin(2t)\\ \sin(2t) &- \cos(2t)\end{bmatrix}\begin{bmatrix} \cos(2t) & \sin(2t)\\ \sin(2t) &- \cos(2t)\end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0 &1\end{bmatrix},$$
as needed. So now we get
$$\dot{\mathbf{u}}=\begin{bmatrix} \cos(2t) & \sin(2t)\\ \sin(2t) &- \cos(2t)\end{bmatrix}\begin{bmatrix}a+2 \sin(2t) &-2 \cos(2t)\\ -2 \cos(2t) &a-2 \sin(2t)\end{bmatrix} \mathbf{u}$$
$$=\begin{bmatrix}a \cos(2t)+2 \cos(2t) \sin(2t) -2 \cos(2t) \sin(2t) & -2 \cos^{2}(2t) +a \sin(2t)-2 \sin^{2}(2t)\\
a \sin(2t)+2 \sin^{2}(2t)+2 \cos^{2}(2t) &-2 \cos(2t) \sin(2t) -a \cos(2t) +2 \cos(2t) \sin(2t)\end{bmatrix}\mathbf{u}$$
$$=\begin{bmatrix}a \cos(2t)& a \sin(2t)-2\\
a \sin(2t)+2 &-a \cos(2t)\end{bmatrix}\mathbf{u}.$$

Unless I've made a mistake somewhere, this does not reduce to what you need. Are you sure you've stated the problem exactly correctly? Even one sign error could drastically change the result.

 

[Dustinsfl]

There was a typo:
$$
\mathbf{x} = \begin{pmatrix}
\cos t & -\sin t\\
\sin t & \cos t
\end{pmatrix}\mathbf{u}
$$

But now I have
$$
a\begin{pmatrix}
\cos t & \sin t\\
\sin t & -\cos 3t
\end{pmatrix}u =
\begin{pmatrix} -\sin t & -\cos t\\
\cos t & -\sin t
\end{pmatrix}u+
\begin{pmatrix}
\cos t & -\sin t\\
\sin t & \cos t
\end{pmatrix}\dot{u}
$$

 

[Ackbach]

There was a typo:
$$
\mathbf{x} = \begin{pmatrix}
\cos t & -\sin t\\
\sin t & \cos t
\end{pmatrix}\mathbf{u}
$$

But now I have
$$
a\begin{pmatrix}
\cos t & \sin t\\
\sin t & -\cos 3t
\end{pmatrix}u =
\begin{pmatrix} -\sin t & -\cos t\\
\cos t & -\sin t
\end{pmatrix}u+
\begin{pmatrix}
\cos t & -\sin t\\
\sin t & \cos t
\end{pmatrix}\dot{u}
$$

Couple comments:

1. I don't think you should have a $3$ in the argument of one of your trig functions.
2. Multiply the whole equation on the left by the inverse of the matrix multiplying the $\dot{\mathbf{u}}$ and see if something nice doesn't happen.
3. Just to be clear, what is the original DE, and what is the suggested substitution?

 

[Dustinsfl]

Couple comments:

1. I don't think you should have a $3$ in the argument of one of your trig functions.
2. Multiply the whole equation on the left by the inverse of the matrix multiplying the $\dot{\mathbf{u}}$ and see if something nice doesn't happen.
3. Just to be clear, what is the original DE, and what is the suggested substitution?

The original is in the first post. I have a by following the trig identities.

 

[Opalg]

There was a typo:
$$
\mathbf{x} = \begin{pmatrix}
\cos t & -\sin t\\
\sin t & \cos t
\end{pmatrix}\mathbf{u}
$$

But now I have
$$
a\begin{pmatrix}
\cos t & \sin t\\
\sin t & -\cos 3t
\end{pmatrix}u =
\begin{pmatrix} -\sin t & -\cos t\\
\cos t & -\sin t
\end{pmatrix}u+
\begin{pmatrix}
\cos t & -\sin t\\
\sin t & \cos t
\end{pmatrix}\dot{u}
$$

Not wanting to get too personal about this, but some people seem to make typos and/or careless arithmetic mistakes practically every time they write something. In mathematics it's really essential to check everything carefully. And if things look like they are going in the wrong direction, go back and double-check all the previous work.

The equation $
a\begin{pmatrix}
\cos t & \sin t\\
\sin t & -\cos 3t
\end{pmatrix}\mathbf{u} =
\begin{pmatrix} -\sin t & -\cos t\\
\cos t & -\sin t
\end{pmatrix}\mathbf{u}+
\begin{pmatrix}
\cos t & -\sin t\\
\sin t & \cos t
\end{pmatrix}\dot{\mathbf{u}}
$ shows that you are heading in exactly the right direction to solve this problem. But the 3 in that first matrix looks really out of place. So go back and double-check every calculation that led to that matrix. (The 3 is not the only thing wrong in that matrix. I think that there are also some wrong signs in some of the other entries.) When you have done that, write the equation in the form $$ \begin{pmatrix}\cos t & -\sin t \\ \sin t & \cos t \end{pmatrix}\dot{\mathbf{u}} = a\begin{pmatrix}? & ? \\ ? & ? \end{pmatrix}\mathbf{u} - \begin{pmatrix} -\sin t & -\cos t \\ \cos t & -\sin t \end{pmatrix}\mathbf{u}.$$ Then you can see why Ackbach's hint 2 is so helpful: if you multiply both sides of the equation on the left by the inverse of $ \begin{pmatrix}\cos t & -\sin t \\ \sin t & \cos t \end{pmatrix}$, you will get an equation starting $\dot{\mathbf{u}} = \ldots$, which is exactly what you want.

 

[Ackbach]

The substitution
$$\mathbf{x}=\begin{bmatrix} \cos(2t) &-\sin(2t)\\ \sin(2t) & \cos(2t)\end{bmatrix}\mathbf{u}$$
renders the original DE in the OP equivalent to
$$\dot{\mathbf{u}}=\begin{bmatrix} a &2\\ -2 &-a\end{bmatrix}\mathbf{u}.$$
Use the procedure I have outlined above.