Hey everybody
how can i calculate the Solution $\displaystyle \in \mathbb{R}$ of
$\displaystyle x'=\begin{pmatrix}
a & 0 & b\\
0& b& 0\\
-b& 0& a
\end{pmatrix} x, \ x \ \in \mathbb{R}^3$ ?
Thank you very much!
I tried to figure out the eigenvalues and eigenvectors but i got stuck here:
$\displaystyle
det(A-\lambda E)=det\begin{pmatrix}
a-\lambda & 0 & b\\
0 & b-\lambda &0 \\
-b& 0 & a-\lambda
\end{pmatrix}=(a^2-2a\lambda+\lambda^2)(b-\lambda)-(-b^3+b^2\lambda)=a^2b-a\lambda-2ab\lambda+2a\lambda^2-\lambda^3+b^3$
$\displaystyle
a^2b-a\lambda-2ab\lambda+2a\lambda^2-\lambda^3+b^3=0$
Good approach - to find the eigenvalues and eigenvectors. I think you may have skipped too many steps in taking your determinant. I get
$\displaystyle \det\begin{bmatrix}
a-\lambda & 0 & b\\
0 & b-\lambda &0 \\
-b& 0 & a-\lambda
\end{bmatrix}=(a-\lambda)(b-\lambda)(a-\lambda)+b(0-(-b)(b-\lambda))$
$\displaystyle =(a-\lambda)^{2}(b-\lambda)+b^{2}(b-\lambda)=(b-\lambda)\left[(a-\lambda)^{2}+b^{2}\right]=0.$
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?
so $\displaystyle (b-\lambda)[(a-\lambda)^2+b^2]=0 \Rightarrow \lambda_1=b, \ \
\lambda_2=a-ib, \ \ \lambda_3=a+ib$
since I'm looking for a real solution I take $\displaystyle \lambda_1=b$ which gives me the eigenvector $\displaystyle (0,1,0)^T$.
So the solution is: $\displaystyle x(t)=c_1e^bt(0,1,0)^T$? is this correct?
If so how do i find linear subspaces for which $\displaystyle t \rightarrow \infty$ or $\displaystyle t \rightarrow -\infty$ converges to x=0?
Is it:
$\displaystyle \begin{bmatrix}
\lambda_1 & 0 & 0\\
0 & \lambda_2 & 0 \\
0& 0 & \lambda_3
\end{bmatrix},$
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: $\displaystyle x'=ax \Rightarrow x(t)=c_1*e^{at}$
Yes, you've correctly identified the diagonal matrix $\displaystyle D$ such that there is an invertible matrix $\displaystyle P$ such that $\displaystyle A=PDP^{-1}.$
The reason you need to diagonalize $\displaystyle A$ is because the solution to the system $\displaystyle \mathbf{x}'=A\mathbf{x}$, as you've hinted at, is the following:
$\displaystyle \mathbf{x}(t)=e^{At}\mathbf{x}_{0}.$
So you have to compute this matrix exponential $\displaystyle e^{At},$ and the best and easiest way to do that is to diagonalize $\displaystyle A.$ 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,
$\displaystyle A^{2}=(PDP^{-1})(PDP^{-1})=PDDP^{-1}=PD^{2}P^{-1},$
$\displaystyle A^{3}=(PDP^{-1})(PDP^{-1})(PDP^{-1})=PDDDP^{-1}=PD^{3}P^{-1}.$
In general, you have
$\displaystyle A^{k}=PD^{k}P^{-1}.$
Then you just invoke the Taylor series expansion of the exponential, and you find that
$\displaystyle e^{At}=Pe^{Dt}P^{-1}.$
The RHS there is easy to compute: $\displaystyle e^{Dt}$ is just the matrix with the elements $\displaystyle (e^{Dt})_{ij}=\delta_{ij}(e^{D_{ij}t}).$ 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 $\displaystyle P,$ and then compute $\displaystyle Pe^{Dt}P^{-1}$, and then compute $\displaystyle Pe^{Dt}P^{-1}\mathbf{x}_{0},$ and you're done.
Make sense?
Just a comment. As the middle row of the matrix is (0,b,0), it might be a whole lot easier to simply solve
$\displaystyle \dot{x} = ax + bz,$
$\displaystyle \dot{y} = by,$
$\displaystyle \dot{z} = -b x + az$.
For $\displaystyle b \ne 0$ then $\displaystyle z =\dfrac{\dot{x} - ax}{b}$ giving
$\displaystyle \ddot{x} - 2 a \dot{x} +(a^2+b^2)x = 0$.
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?
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.well, if i didnt make any mistakes the solution should be correct!
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.