Question

find the general solution of the differential system u'=Au where the matrix A is given by
[0 1 1
1 0 1
1 1 0]

Answer 1

\[\left|\begin{matrix}0 - \lambda & 1 & 1 \\ 1 & 0 -\lambda &1 \\ 1 & 1 & 0 - \lambda\end{matrix}\right| = \vec 0\]
leading to \[-\lambda \left| \begin{matrix} -\lambda & 1 \\ 1 & -\lambda\end{matrix}\right| - \left| \begin{matrix} 1 & 1 \\ 1 & -\lambda\end{matrix}\right| +\left| \begin{matrix} 1 & -\lambda \\ 1 & 1 \end{matrix}\right| = 0 \]

from there, it is :

\(-\lambda^3 +\lambda +\lambda + 1 + 1 + \lambda = -\lambda^3 + 3\lambda +2 = 0 \)

\(\lambda = -1\)

by long division you have \((\lambda + 1) ( \lambda - 2) (\lambda + 1)\)

repeated roots

does that make sense?

Answer 2

i have the final answer but i do not know hoe it comes

Answer 3

[e^2x e^-x 0 [c1
e^2x 0 e^-x c2
e^2x - e^-x - e^-x] c3]

Answer 4

I don't know the _kosher_ way of doing this problem anymore, but I know how to do it.

If you allow yourself to believe this is legitimate:

\[\frac{du}{dt} = Au\]\[\frac{du}{u} = Adt\]\[\ln u = At+k\]\[u=ce^{At}\]
we can calculate the matrix exponential with the taylor series after we diagonalize it:

\[u = cP^{-1}e^{Dt}P\]

Although I have a feeling that's not too helpful. :|

Answer 5

from the thread i'm not sure what you already know; but the basic idea behind this is that **if** we have \(\mathbf {\dot x }= A \mathbf{ x}\), where, say, column vector \(\mathbf x(t) = \left[\begin{matrix}x_1(t) \\ x_2(t) \\x_3(t)\end{matrix}\right]\),

and A is a 3x3 matrix of constants [it doesn't have to be 3, it can be anything but you are looking at 3]

**and** if there exists a solution \(\mathbf{x} = \mathbf{v}e^{\lambda t}\), where \(\mathbf v = \left[\begin{matrix}c_1 \\ c_2 \\c_3\end{matrix}\right]\), ie a vector of constants

then we can see that \(\mathbf{\dot x} = \lambda \mathbf{v}e^{\lambda t}\) so \(A\mathbf{v}e^{\lambda t} = \lambda \mathbf{v}e^{\lambda t}\) or \(A \mathbf v = \lambda \mathbf v\), ie it is an eigenvector problem.

the final bit is that we **know** there will be such a solution if we are looking at a system of linear homogeneous constant coefficient first order DE's.

taking a simple example \(\left[\begin{matrix} \dot x \\ \dot y\end{matrix}\right] = \left[\begin{matrix}3 & 2 \\ 1 & 4\end{matrix}\right] \left[\begin{matrix} x \\ y\end{matrix}\right]\)

we can write these as \(\dot x = 3x + 2y\) and \(\dot y = x + 4y\) and from the second one we know that \(\dot x = \ddot y - 4 \dot y \) so \(\ddot y - 4 \dot y = 3(\dot y - 4y) + 2y \) or \(\ddot y - 7 \dot y +10y = 0\).

the auxiliary eqn for that 2nd order constant coefficient linear DE is \(\lambda^2 - 7 \lambda + 10 = 0\) or \((\lambda - 5)(\lambda - 2)= 0\). the general solution is \(y(t) = A e^{5t} + B e^ {2t}\). from \(\dot y = x + 4y\) you will find that \(x(t) = Ae^{5t} - 2B e^{2t}\)

and this is where we notice something. the eigen values for that matrix are the solutions to \((3-\lambda)(4-\lambda) - 1(2) = 0\) or \(\lambda^2 - 7 \lambda + 10 = 0\) [!!]

the eigenvectors are \(\left[\begin{matrix}1 \\ 1\end{matrix}\right]\) and \(\left[\begin{matrix}-2 \\ 1\end{matrix}\right]\)

the solution is \(\left[\begin{matrix}x \\ y\end{matrix}\right] = A \left[\begin{matrix}1 \\ 1\end{matrix}\right] e^{5t} + B \left[\begin{matrix}-2 \\ 1\end{matrix}\right] e^{2t}\)

if we apply this logic to the problem you have presented here, you are solving \(y'''-3y'-2y=0\)

http://www.wolframalpha.com/input/?i=y%27%27%27-3y%27-2y%3D0

....because we can reverse engineer the 3rd order DE from the eigenvalues as \( (D+1)(D+1)(D-2)y = 0\) where \(D = \frac{d}{dt}\), the differential operator. and you will notice the 3rd row of your matrix in the solution offered by Wolfram .

you may well know all of that already. if so, mea culpa.


if not, it is a big mouthful, and you have repeating roots which create the same issues in system solutions as they do in the method of undetermined coefficients, which is what all this piggy-backs off. to my mind at least..... we can follow the repeated roots idea next, if you like

FYI, i'm not actually very good with this stuff, just very interested....in the whole idea. i hope this has added something.