the ode problem(linear system)
- Thread starter polomnbv
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I want to ask a question
Find the general sol. of x'=Ax with
A=
[2 1 3]
[0 2-1]
[0 0 2]
x(0)=
[1]
[2]
[1]
I know that must be found the eigenvalues,but this is a up-triangle matrix.
So I can't solve it.
Please help me
The eigenvalues of a triangular matrix are the diagonal entries.
I know this,so the lamda=2
then
[x1] [x1]
A [x2]=2[x2]
[x3] [x3]
→{2x1+x2+x3=2x1}
{x2-x3=2x2}
{2x3=2x3}
→x2=-x3,then x1=?
Your second equation is wrong. I get x3=0, x2=0 and x1 is free.
HallsofIvy
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You could just ignore the matrix part. If we let \(\displaystyle x= \begin{bmatrix}x \\ y \\ z\end{bmatrix}\), this equation is the same as the "partially uncoupled" system x'= 2x+y+ 3z, y'= 2y- z, z'= 2z. It is "partially" uncoupled because we have the single equation z'= 2z which has, of course, the solution \(\displaystyle z(t)= C_1e^{2t}\). We can then rewrite the second equation as \(\displaystyle y'= 2y- C_1e^{2t}\), a "non-homogeneous" linear equation. The solution to the homogeneous equation is again \(\displaystyle e^{2t}\) so we seek a specific solution to the entire equation of the form \(\displaystyle y(t)= Ate^{2t}. Then y'= Ae^{2t}+ 2Ate^{2t}= 2Ate^{2t}- C_1e^{2t}\). The "\(\displaystyle 2Ate^{2t}\)" terms cancel, leaving \(\displaystyle Ae^{2t}=-C_1e^{-2t}\) so that \(\displaystyle A= -C_1\) which gives \(\displaystyle y(t)= C_2te^{2t}- C_1e^{2t}\). Putting that into the first equation, \(\displaystyle x'= 2x+ C_2te^{2t}- C_1e^{2t}+ 3C_1e^{2t}\). That is also a non-homogeneous linear equation to solve for x. Yes, once you have found x, y, and z, with the three constants, \(\displaystyle C_1, C_2, C_3\), use the initial condition to determine values for \(\displaystyle C_1, C_2, C_3\).
Last edited:
HallsofIvy
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If you really want to do it in terms of matrices, yes, "2" is a triple eigenvalue.
To find an eigenvector, we have first
\(\displaystyle \begin{bmatrix}2 & 1 & 3 \\ 0 & 2 & -1\\ 0 & 0 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}2x+ y+ 3z \\ 2y- z \\ 2z\end{bmatrix}= \begin{bmatrix}2x \\ 2y \\ 2z\end{bmatrix}\)
which gives the three equations 2x+ y+ 3z= 2x, 2y- z= 2y, 2z= 2z. The last equation is, of course, true for any z so we are left with two equations in terms of three variables. The first is y+ 3z= 0 and the second, -z= 0. Since z= 0, y= -3z= 0. There is no equation left for x so x can be anything- any eigenvector, is of the form <x, 0, 0>= x<1, 0, 0>. Now, that's only a single vector so the "geometric multiplicity" of eigenvalue 2 is only 1 while its "algebraic multiplicity" is 3.
We can't "diagonalize" this matrix, the best we can do is write it in "Jordan Normal form". That is, if, in addition to the single eigenvector, we find two "generalized eigenvectors", and form a matrix B with them as columns, we can write \(\displaystyle B^{-1}AB\) as \(\displaystyle \begin{bmatrix}2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2\end{bmatrix}\). But, in fact, that is not really any simpler to solve than the method I showed in the previous post so, frankly, I wouldn't bother.
To find an eigenvector, we have first
\(\displaystyle \begin{bmatrix}2 & 1 & 3 \\ 0 & 2 & -1\\ 0 & 0 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}2x+ y+ 3z \\ 2y- z \\ 2z\end{bmatrix}= \begin{bmatrix}2x \\ 2y \\ 2z\end{bmatrix}\)
which gives the three equations 2x+ y+ 3z= 2x, 2y- z= 2y, 2z= 2z. The last equation is, of course, true for any z so we are left with two equations in terms of three variables. The first is y+ 3z= 0 and the second, -z= 0. Since z= 0, y= -3z= 0. There is no equation left for x so x can be anything- any eigenvector, is of the form <x, 0, 0>= x<1, 0, 0>. Now, that's only a single vector so the "geometric multiplicity" of eigenvalue 2 is only 1 while its "algebraic multiplicity" is 3.
We can't "diagonalize" this matrix, the best we can do is write it in "Jordan Normal form". That is, if, in addition to the single eigenvector, we find two "generalized eigenvectors", and form a matrix B with them as columns, we can write \(\displaystyle B^{-1}AB\) as \(\displaystyle \begin{bmatrix}2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2\end{bmatrix}\). But, in fact, that is not really any simpler to solve than the method I showed in the previous post so, frankly, I wouldn't bother.