MHB Real and Complex Solutions for Linear System with 2x2 Matrix

[Dustinsfl]

$$
\mathbf{u}' = \begin{pmatrix}
a & 1\\
-1 & -a
\end{pmatrix}\mathbf{u}.
$$
The characteristic equation is
$$
\lambda^2 -a^2 + 1 = 0\iff \lambda = \sqrt{a^2 - 1}.
$$
The eigenvalues are complex when $-1 < a < 1$ and real otherwise.
Let's consider the real case first.
We have that $(a - \lambda)u_1 + u_2 = 0\iff u_2 = (\lambda - a)u_1$.
Then
$$
u = \begin{pmatrix}
1\\
\lambda - a
\end{pmatrix}.
$$
The solution for the real case is
$$
\mathbf{u} = A\exp\left[t\sqrt{a^2 - 1}\right]\begin{pmatrix}
1\\
\sqrt{a^2 - 1} - a
\end{pmatrix}
+ B\exp\left[-t\sqrt{a^2 - 1}\right]\begin{pmatrix}
1\\
-\sqrt{a^2 - 1} - a
\end{pmatrix}.
$$
Next, let's look at the complex case.
We have $(a - \lambda)u_1 + u_2 = 0\iff u_2 = (\lambda - a)u_1$.
Then
$$
u = \begin{pmatrix}
1\\
\lambda - a
\end{pmatrix}.
$$
The solution for the complex case is
$$
\mathbf{u} = A\exp\left[ti\sqrt{1 - a^2}\right]\begin{pmatrix}
1\\
i\sqrt{1 - a^2} - a
\end{pmatrix} +
B\exp\left[-ti\sqrt{1 - a^2}\right]\begin{pmatrix}
1\\
-i\sqrt{1 - a^2} - a
\end{pmatrix}.
$$

Correct?

 

[Sudharaka]

$$
\mathbf{u}' = \begin{pmatrix}
a & 1\\
-1 & -a
\end{pmatrix}\mathbf{u}.
$$
The characteristic equation is
$$
\lambda^2 -a^2 + 1 = 0\iff \lambda = {\color{red}\pm}\sqrt{a^2 - 1}.
$$
The eigenvalues are complex when $-1 < a < 1$ and real otherwise.
Let's consider the real case first.
We have that $(a - \lambda)u_1 + u_2 = 0\iff u_2 = (\lambda - a)u_1$.
Then
$$
u = \begin{pmatrix}
1\\
\lambda - a
\end{pmatrix}.
$$
The solution for the real case is
$$
\mathbf{u} = A\exp\left[t\sqrt{a^2 - 1}\right]\begin{pmatrix}
1\\
\sqrt{a^2 - 1} - a
\end{pmatrix}
+ B\exp\left[-t\sqrt{a^2 - 1}\right]\begin{pmatrix}
1\\
-\sqrt{a^2 - 1} - a
\end{pmatrix}.
$$
Next, let's look at the complex case.
We have $(a - \lambda)u_1 + u_2 = 0\iff u_2 = (\lambda - a)u_1$.
Then
$$
u = \begin{pmatrix}
1\\
\lambda - a
\end{pmatrix}.
$$
The solution for the complex case is
$$
\mathbf{u} = A\exp\left[ti\sqrt{1 - a^2}\right]\begin{pmatrix}
1\\
i\sqrt{1 - a^2} - a
\end{pmatrix} +
B\exp\left[-ti\sqrt{1 - a^2}\right]\begin{pmatrix}
1\\
-i\sqrt{1 - a^2} - a
\end{pmatrix}.
$$

Correct?

Hi dwsmith, :)

I see no errors in your solution. But it is unnecessary to consider the real and complex cases separately. You can give the solution as,

\[\mathbf{u} = A\exp\left[t\sqrt{a^2 - 1}\right]\begin{pmatrix}1\\ \sqrt{a^2 - 1} - a\end{pmatrix}+ B\exp\left[-t\sqrt{a^2 - 1}\right]\begin{pmatrix}1\\-\sqrt{a^2 - 1} - a\end{pmatrix}\mbox{ for }a\in\Re\]

Kind Regards,
Sudharaka.

 

[Dustinsfl]

To expand on this now, referring to http://www.mathhelpboards.com/f10/change-variables-1871/#post8706.

How can I take u back to x with the u solution?

 

[Ackbach]

To expand on this now, referring to http://www.mathhelpboards.com/f10/change-variables-1871/#post8706.

How can I take u back to x with the u solution?

Just multiply $\mathbf{u}$ by the necessary rotation matrix to get $\mathbf{x}$.

 

[Dustinsfl]

Just multiply $\mathbf{u}$ by the necessary rotation matrix to get $\mathbf{x}$.

So it will be $xu$ where u is the Ay_1+By_2?

 

[Sudharaka]

Just multiply $\mathbf{u}$ by the necessary rotation matrix to get $\mathbf{x}$.

What Ackbach meant was,

\begin{eqnarray}

\mathbf{x} &=& \begin{pmatrix} \cos t & -\sin t\\ \sin t & \cos t \end{pmatrix}\mathbf{u}\\

&=&\begin{pmatrix} \cos t & -\sin t\\ \sin t & \cos t \end{pmatrix}\left[A\exp\left[t\sqrt{a^2 - 1}\right]\begin{pmatrix}1\\ \sqrt{a^2 - 1} - a\end{pmatrix}+ B\exp\left[-t\sqrt{a^2 - 1}\right]\begin{pmatrix}1\\-\sqrt{a^2 - 1} - a\end{pmatrix}\right]

\end{eqnarray}

Kind Regards,
Sudharaka.

 

[Dustinsfl]

What Ackbach meant was,

\begin{eqnarray}

\mathbf{x} &=& \begin{pmatrix} \cos t & -\sin t\\ \sin t & \cos t \end{pmatrix}\mathbf{u}\\

&=&\begin{pmatrix} \cos t & -\sin t\\ \sin t & \cos t \end{pmatrix}\left[A\exp\left[t\sqrt{a^2 - 1}\right]\begin{pmatrix}1\\ \sqrt{a^2 - 1} - a\end{pmatrix}+ B\exp\left[-t\sqrt{a^2 - 1}\right]\begin{pmatrix}1\\-\sqrt{a^2 - 1} - a\end{pmatrix}\right]

\end{eqnarray}

Kind Regards,
Sudharaka.

So the solution is
\begin{alignat*}{3}
\mathbf{x} & = & \begin{pmatrix}
\cos t & -\sin t\\
\sin t & \cos t
\end{pmatrix}\left[A\exp\left[t\sqrt{a^2 - 1}\right]\begin{pmatrix}
1\\
\sqrt{a^2 - 1} - a
\end{pmatrix}
+ B\exp\left[-t\sqrt{a^2 - 1}\right]\begin{pmatrix}
1\\
-\sqrt{a^2 - 1} - a
\end{pmatrix}\right]\\
& = & A\exp\left[t\sqrt{a^2 - 1}\right]\begin{pmatrix}
\cos t - \sin t\left(\sqrt{a^2 - 1} - a\right)\\
\sin t + \cos t\left(\sqrt{a^2 - 1} - a\right)
\end{pmatrix}\\
& & + B\exp\left[-t\sqrt{a^2 - 1}\right]\begin{pmatrix}
\cos t + \sin t\left(\sqrt{a^2 - 1} + a\right)\\
\sin t - \cos t\left(\sqrt{a^2 - 1} + a\right)
\end{pmatrix}
\end{alignat*}

 

[Sudharaka]

So the solution is
\begin{alignat*}{3}
\mathbf{x} & = & \begin{pmatrix}
\cos t & -\sin t\\
\sin t & \cos t
\end{pmatrix}\left[A\exp\left[t\sqrt{a^2 - 1}\right]\begin{pmatrix}
1\\
\sqrt{a^2 - 1} - a
\end{pmatrix}
+ B\exp\left[-t\sqrt{a^2 - 1}\right]\begin{pmatrix}
1\\
-\sqrt{a^2 - 1} - a
\end{pmatrix}\right]\\
& = & A\exp\left[t\sqrt{a^2 - 1}\right]\begin{pmatrix}
\cos t - \sin t\left(\sqrt{a^2 - 1} - a\right)\\
\sin t + \cos t\left(\sqrt{a^2 - 1} - a\right)
\end{pmatrix}\\
& & + B\exp\left[-t\sqrt{a^2 - 1}\right]\begin{pmatrix}
\cos t + \sin t\left(\sqrt{a^2 - 1} + a\right)\\
\sin t - \cos t\left(\sqrt{a^2 - 1} + a\right)
\end{pmatrix}
\end{alignat*}

Correct. (Yes)

 

[Ackbach]

The arguments of the trig functions are all $2t$, not $t$. But that's definitely the right idea.

 

[Dustinsfl]

The arguments of the trig functions are all $2t$, not $t$. But that's definitely the right idea.

What?

 

[Sudharaka]

The arguments of the trig functions are all $2t$, not $t$. But that's definitely the right idea.

Hi Ackbach, :)

dwsmith has mentioned(http://www.mathhelpboards.com/f10/change-variables-1871/#post8662) that he had made a typo and the transformation should be,

\[\mathbf{x} = \begin{pmatrix} \cos t & -\sin t\\ \sin t & \cos t \end{pmatrix}\mathbf{u}\]

I was referring to this in my previous post.

Kind Regards,
Sudharaka.

 

[Ackbach]

All I know is this: assuming the original DE is
$$\dot{\mathbf{x}}=a\begin{bmatrix} \cos(2t) & \sin(2t)\\ \sin(2t) & -\cos(2t)\end{bmatrix}\mathbf{x},$$
the substitution
$$\mathbf{x}=\begin{bmatrix} \cos(2t) &-\sin(2t)\\ \sin(2t) & \cos(2t)\end{bmatrix}\mathbf{u}$$
reduces the DE to
$$\dot{\mathbf{u}}=\begin{bmatrix} a &2\\ -2 &-a\end{bmatrix}\mathbf{u}.$$
So you solve this new DE in $\mathbf{u}$ using the usual eigenvalue method, and then multiply by the rotation matrix
$$\begin{bmatrix} \cos(2t) &-\sin(2t)\\ \sin(2t) & \cos(2t)\end{bmatrix}$$
to get the final result.

Whatever substitution you make must have the same arguments for the trig functions as are in the original DE, or else you won't get the nice cancellations you need to get a constant-coefficients equivalent DE in $\mathbf{u}$. If the original DE had simply $t$ as the argument for the trig functions, then so must the substitution.