# System of first order ODE

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• Nov 19th 2010, 12:57 AM
LHeiner
System of first order ODE
Hey everybody

how can i calculate the Solution $\in \mathbb{R}$ of

$x'=\begin{pmatrix}
a & 0 & b\\
0& b& 0\\
-b& 0& a
\end{pmatrix} x, \ x \ \in \mathbb{R}^3$
?

Thank you very much!
• Nov 19th 2010, 01:02 AM
Ackbeet
What ideas have you had so far?
• Nov 19th 2010, 02:52 AM
LHeiner
I tried to figure out the eigenvalues and eigenvectors but i got stuck here:
$
det(A-\lambda E)=det\begin{pmatrix}
a-\lambda & 0 & b\\
0 & b-\lambda &0 \\
-b& 0 & a-\lambda
\end{pmatrix}=(a^2-2a\lambda+\lambda^2)(b-\lambda)-(-b^3+b^2\lambda)=a^2b-a\lambda-2ab\lambda+2a\lambda^2-\lambda^3+b^3$

$
a^2b-a\lambda-2ab\lambda+2a\lambda^2-\lambda^3+b^3=0$
• Nov 19th 2010, 04:49 AM
Ackbeet
Good approach - to find the eigenvalues and eigenvectors. I think you may have skipped too many steps in taking your determinant. I get

$\det\begin{bmatrix}
a-\lambda & 0 & b\\
0 & b-\lambda &0 \\
-b& 0 & a-\lambda
\end{bmatrix}=(a-\lambda)(b-\lambda)(a-\lambda)+b(0-(-b)(b-\lambda))$

$=(a-\lambda)^{2}(b-\lambda)+b^{2}(b-\lambda)=(b-\lambda)\left[(a-\lambda)^{2}+b^{2}\right]=0.$

This is not the same thing as what you got, if you multiply it all out (sometimes it pays not to do that!). So, what are the eigenvalues?
• Nov 19th 2010, 06:14 AM
LHeiner
so $(b-\lambda)[(a-\lambda)^2+b^2]=0 \Rightarrow \lambda_1=b, \ \
\lambda_2=a-ib, \ \ \lambda_3=a+ib$

since I'm looking for a real solution I take $\lambda_1=b$ which gives me the eigenvector $(0,1,0)^T$.

So the solution is: $x(t)=c_1e^bt(0,1,0)^T$? is this correct?

If so how do i find linear subspaces for which $t \rightarrow \infty$ or $t \rightarrow -\infty$ converges to x=0?
• Nov 19th 2010, 07:18 AM
Ackbeet
I don't think you can ignore the complex eigenvalues - don't worry, I think they'll turn into sines and cosines. You have correctly found the eigenvector for the real eigenvalue. What are the eigenvectors for the complex eigenvalues?
• Nov 19th 2010, 08:20 AM
LHeiner
for $\lambda_2=a-ib \Rightarrow(i,0,1)^T$ and for $\lambda_3=a+ib \rightarrow (-i,0,1)^T$

and now?
• Nov 19th 2010, 08:28 AM
Ackbeet
So, if you call

$A=\begin{bmatrix}
a & 0 & b\\
0 & b &0 \\
-b& 0 & a
\end{bmatrix},$

then what is the diagonalization of $A?$
• Nov 19th 2010, 08:48 AM
LHeiner
Is it:

$\begin{bmatrix}
\lambda_1 & 0 & 0\\
0 & \lambda_2 & 0 \\
0& 0 & \lambda_3
\end{bmatrix},$

But how does this help me in solving the ODE?

In other cases i simply took the eigenvalues and eigenvectors (if complex, split it into sin and cos) and solved the DE with the exponential function (like in the one dimensional case: $x'=ax \Rightarrow x(t)=c_1*e^{at}$
• Nov 19th 2010, 09:05 AM
Ackbeet
Yes, you've correctly identified the diagonal matrix $D$ such that there is an invertible matrix $P$ such that $A=PDP^{-1}.$

The reason you need to diagonalize $A$ is because the solution to the system $\mathbf{x}'=A\mathbf{x}$, as you've hinted at, is the following:

$\mathbf{x}(t)=e^{At}\mathbf{x}_{0}.$

So you have to compute this matrix exponential $e^{At},$ and the best and easiest way to do that is to diagonalize $A.$ Why? Because it's ridiculously easy to compute arbitrary powers of a diagonal matrix (just raise the elements on the diagonal to the desired power!). Also,

$A^{2}=(PDP^{-1})(PDP^{-1})=PDDP^{-1}=PD^{2}P^{-1},$
$A^{3}=(PDP^{-1})(PDP^{-1})(PDP^{-1})=PDDDP^{-1}=PD^{3}P^{-1}.$

In general, you have

$A^{k}=PD^{k}P^{-1}.$

Then you just invoke the Taylor series expansion of the exponential, and you find that

$e^{At}=Pe^{Dt}P^{-1}.$

The RHS there is easy to compute: $e^{Dt}$ is just the matrix with the elements $(e^{Dt})_{ij}=\delta_{ij}(e^{D_{ij}t}).$ That is, you just exponentiate the elements on the main diagonal.

So I would say that diagonalization, when it can be done, is at the heart of solving linear systems of ODE's with constant coefficients.

So, what remains to be done is this: construct your invertible matrix $P,$ and then compute $Pe^{Dt}P^{-1}$, and then compute $Pe^{Dt}P^{-1}\mathbf{x}_{0},$ and you're done.

Make sense?
• Nov 19th 2010, 09:28 AM
Jester
Just a comment. As the middle row of the matrix is (0,b,0), it might be a whole lot easier to simply solve

$\dot{x} = ax + bz,$

$\dot{y} = by,$

$\dot{z} = -b x + az$.

For $b \ne 0$ then $z =\dfrac{\dot{x} - ax}{b}$ giving

$\ddot{x} - 2 a \dot{x} +(a^2+b^2)x = 0$.
• Nov 20th 2010, 01:09 AM
LHeiner
So i tried the way Danny sugested it and got the solution:

$x(t)=e^{at}(c_1sin(bt)+c_2cos(bt))$
$y(t)=c_3 e^{bt}$
$z(t)=-e^{at}(c_4sin(bt)-c_5cos(bt))$

is this correct?
• Nov 20th 2010, 02:36 AM
Ackbeet
Well, why not compute the derivative, and see if it satisfies the original system? Here's one of the really great things about differential equations: as long as you're decent at differentiation (which is quite straight-forward, really), you can check your own answers very easily.

What do you get?
• Nov 20th 2010, 05:10 AM
LHeiner
well, if i didnt make any mistakes the solution should be correct!

can you give me one last hint how i get the linear subspaces which $t \rightarrow \infty$ or $t \rightarrow -\infty$ converges to x=0
• Nov 20th 2010, 06:14 AM
Ackbeet
Quote:

well, if i didnt make any mistakes the solution should be correct!
True, but you don't know ahead of time if you made any mistakes or not. I recommend computing the LHS (the derivatives), and then computing the RHS and show they are equal.

As for the linear subspaces, could you please state your question a little more carefully? Perhaps rephrase it a bit? I'm not following what you're asking, though I have an inkling.
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