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[SOLVED] ODE matrix

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
\mathbf{u}' = \begin{pmatrix}
a & 1\\
-1 & -a
\end{pmatrix}\mathbf{u}.
$$
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

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
$$
\mathbf{u}' = \begin{pmatrix}
a & 1\\
-1 & -a
\end{pmatrix}\mathbf{u}.
$$
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.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193

dwsmith

Well-known member
Feb 1, 2012
1,673
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

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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

Indicium Physicus
Staff member
Jan 26, 2012
4,193
The arguments of the trig functions are all $2t$, not $t$. But that's definitely the right idea.
 

dwsmith

Well-known member
Feb 1, 2012
1,673

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
The arguments of the trig functions are all $2t$, not $t$. But that's definitely the right idea.
Hi Ackbach, :)

dwsmith has mentioned(>>here<<) 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

Indicium Physicus
Staff member
Jan 26, 2012
4,193
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.