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My Problem is this given System of differential equations.

$$y_{1}^{\prime}=5y_{1}+2y_{2} \\ y_{2}^{\prime}=-2y_{1}+y_{2}$$

I am looking for the solution. According to one of my earlier Questions, I tried the method on my own. Now i fear the solution could be wrong. (especially the eigenvectors)

My Approach was: again, i analyze, it must be a ordinary, linear System of equations, with both being of first-order. Than i built the corresponding Matrix as follows:

$$\underbrace{\pmatrix{ y_1^{\prime} \\ y_2^{\prime}}}_{\large{ {\vec y^{\prime}}}} = \underbrace{\pmatrix{5 & 2 \\ -2 & 1}}_{\large{\mathbf A}}\underbrace{\pmatrix{y_1\\y_2}}_{\large{\vec y}}$$

that's why:

$$\vec y^{\prime} = \pmatrix{5 & 2 \\ -2 & 1}\vec y$$

Then I determined the eigenvalues:

they are $r_1 = 3$ and $r_2=3$

Knowing them, I can build the corresponding eigenvectors:

they are $\vec v_1 = \pmatrix{ -1 \\ +1}$ and $\vec v_2 = \pmatrix{ 0 \\ 0}$

Now i plug into the equation:

$$\vec{x} = c_1e^{r_1t}\vec{v_1}+c_2e^{r_2t}\vec{v_2} \\ \vec{x} = c_1e^{3t}\pmatrix{-1 \\ 1}+c_2e^{3t}\pmatrix{0 \\ 0}$$

this lead to my result:

$$y_1 = -c_1e^{3t} + 0c_2e^{3t}\\ y_2 = c_1e^{3t} + 0c_2e^{3t} \\ \\ y_1 = -c_1e^{3t}\\ y_2 = c_1e^{3t}$$

But I doubt it's correct. My suspect are the eigenvectors, I really don't know if they are correct. And this could have lead to a wrong solution.

P.S.: Edits were made to improve language and latex

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    $\begingroup$ This obviously wrong since you can't take the zero vector as an eigenvector. $\endgroup$
    – user63181
    Jul 2, 2013 at 7:51
  • $\begingroup$ You don't need the zero eigenvector. Look at my answer here. $\endgroup$
    – user403781
    Jan 22, 2017 at 7:17

3 Answers 3

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Hints:

You calculated the double eigenvalue correctly, but you cannot have a zero eigenvector as that is not allowed.

You should have done:

$[A - \lambda I]v_1 = 0$ to find the first eigenvector and then $[A- \lambda I]v_2 = v_1$ to find a second generalized eigenvector. You end up with:

  • $\lambda_1 = 3, ~v_1 = (-1, 1)$
  • $\lambda_2 = 3, ~v_2 = (-\dfrac{1}{2}, 0)$

Since we have a repeated eigenvalue, our solution will be written as:

$$Y(t) = c_1 e^{\lambda_1 t} v_1 + c_2 e^{\lambda_2 t}(v_1t + v_2) = c_1e^{3t} \begin{bmatrix}-1\\1\end{bmatrix} + c_2 e^{3t}\left(t\begin{bmatrix}-1\\1\end{bmatrix} +\begin{bmatrix}-\dfrac{1}{2}\\0\end{bmatrix}\right)$$

I would recommend reviewing this in your book as it is critical to understand.

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  • $\begingroup$ Nice work, and recommendation! ;-) $\endgroup$
    – amWhy
    Jul 3, 2013 at 0:10
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Hint:$$X '(t)=AX(t)$$ if you have repeated eigen value like $ c $ and $v $ is eigen vector correspond to $c $ then general solution is $$\large{X(t)=e^{ct}v}$$ and here solution is:$$\large{ X(t) = ce^{3t}\pmatrix{-1 \\ 1}}$$

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Found this one sitting around on my laptop, and realized I'd never posted it; better late than never, I guess! Here goes:

Not to put too fine a point on it, but Maisim Hedyelloo's answer does not tell the whole story. True, the eigenvalues of the matrix $A$ are the repeated pair 3, 3; this follows from the fact that the characteristic polynomial of $A$ is

$\det(A - \lambda I) = \det(\begin{bmatrix} 5 - \lambda & 2 \\ -2 & 1 - \lambda \end{bmatrix}) = (5 - \lambda)(1 - \lambda) + 4 = \lambda^2 - 6\lambda + 9= (\lambda - 3)^2$,

which easily seen to have root $\lambda = 3$ of multiplicity $2$. Now consider the matrix $N$ defined by the equation

$2N = A - \lambda I = A - 3I = \begin{bmatrix} 2 & 2 \\ -2 & -2 \end{bmatrix} = 2 \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix}$,

so that

$N = \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix}$.

A simple calculation reveals that

$N^2 = 0$,

i.e., $N$ is nilpotent of degree $2$; hence $2N$ is as well: $(2N)^2 = 4N^2 = 0$. Next, we recall that for any initial vector $y(0) = y_0$, the solution to the ODE is $y(t) = e^{At}y_0$; and, based upon what we have done already, the matrix exponential $e^{At}$ is easily had. Note that, by the above, $A = 3I + 2N$, so that

$e^{At} = e^{(3I + 2N)t}$.

At this point, the critical step is to observe that

$e^{(3I + 2N)t} = e^{3It}e^{2Nt}$;

this equation between the matrix exponentials on either side holds by virtue of the fact that the two matrices $3I$ and $2N$ commute; that is $(3I)(2N) = (2N)(3I)$, or, if you prefer the bracket notation, $[3I, 2N] = 0$. Under these circumstances, it can be shown that the rule $e^{X + Y} = e^Xe^Y$ holds for matrices exactly as it does for ordinary real or complex numbers. In the present case of course the relation $(3I)(2N) = (2N)(3I)$ or $[3I, 2N] = 0$ is trivially obvious, since one of the matrices in question is a scalar multiple of the identity matrix $I$; but the result is quite general. A full discussion and proof may be found in Hirsch, Smale, and Devaney's Differential Equations, Dynamical Systems, and an Introduction to Chaos, Second Edition, Elsevier Academic Press, 2004, pp. 123-130. In any event we can now conclude that

$e^{At} = e^{3It}e^{2Nt}$,

and since it is evident that $e^{3It} = e^{3t}I$, we further have

$e^{At} = e^{3t}Ie^{2Nt} = e^{3t}e^{2Nt}$

so all we really need to do is evaluate $e^{2Nt}$; but this is easy: since $N^2 = 0$, the power series expansion of $e^{2Nt}$ is truncated after the first-degree term, hence we have

$e^{2Nt} = I + 2Nt$,

whence

$e^{At} = e^{3t}e^{2Nt} = e^{3t}(I + 2Nt)$.

It is now a relatively simple matter to write down the solution which passes through the point $y_0 = (y_{01}, y_{02})^T$ at the time $t = 0$. Since

$I + 2Nt = \begin{bmatrix} 1 + 2t & 2t \\ -2t & 1 - 2t \end{bmatrix}$,

we have

$y(t) = e^{At}y_0 = e^{3t}(I + 2Nt)y_0 = e^{3t}\begin{pmatrix}y_{01} + 2t(y_{01} + y_{02}) \\ y_{02} - 2t(y_{01} + y_{02})\end{pmatrix}$

$= e^{3t}\begin{pmatrix}y_{01} \\y_{02}\end{pmatrix} + (y_{01} + y_{02})te^{3t}\begin{pmatrix} 2 \\ -2 \end{pmatrix}$;

it is easy to verify, by a direct and elementary computation, that such $y(t)$ satisfies $y'(t) = Ay(t)$ with $y(0) = y_0$. Examining this equation, we observe that in the event that $y_{01} + y_{02} = 0$, i.e. $y_{01} = -y_{02}$ the solution $y(t) = e^{3t}(y_{01}, y_{02})^T = e^{3t}(y_{01}, -y_{01})^T = y_{01}e^{3t}(1, -1)^T$ is in fact of the general form suggested by Maisam Hedyelloo in his answer; but this is indeed a very special solution, lying as it does in the unique, one-dimensional eigenspace of $A$ generated by the eigenvector $(1, -1)^T$. In general we should expect $y_{01} + y_{02} \ne 0$, in which case the solution contains a (vector) term proportional to $te^{3t}$. We can re-write $y(t)$, thus:

$y(t) = e^{3t}\begin{pmatrix}y_{01} \\y_{02}\end{pmatrix} + (y_{01} + y_{02})te^{3t}\begin{pmatrix} 2 \\ -2 \end{pmatrix}$ $= e^{3t}\begin{pmatrix} y_{01} \\ -y_{01}\end{pmatrix} + e^{3t}\begin{pmatrix} 0 \\ y_{01} + y_{02}\end{pmatrix} + (y_{01} + y_{02})te^{3t}\begin{pmatrix} 2 \\ -2 \end{pmatrix}$ $= y_{01}e^{3t}\begin{pmatrix} 1 \\ -1\end{pmatrix} + (y_{01} + y_{02})e^{3t}\begin{pmatrix} 0 \\ 1\end{pmatrix} + 2(y_{01} + y_{02})te^{3t}\begin{pmatrix} 1 \\ -1 \end{pmatrix}$ $= (y_{01}+ 2(y_{01} + y_{02})t)e^{3t}\begin{pmatrix} 1 \\ -1\end{pmatrix} + 2(y_{01} + y_{02})e^{3t}\begin{pmatrix} 0 \\ \frac{1}{2}\end{pmatrix}$,

which exhibits the complete solution in terms of a component lying in the unique eigenspace of $A$, spanned by $(1, -1)^T$, and a component $2e^{3t}(y_{01} + y_{02})(0, \frac{1}{2})^T$ which lies in the one-dimensional subspace spanned by the generalized eigenvector $(0, \frac{1}{2})^T$. The fact that Amzoti's choice of generalized eigenvector, $(-\frac{1}{2}, 0)^T$, is different that the present one, $(0, \frac{1}{2})^T$, yet still yields a valid solution is, I believe, explained by the fact that there is a certain latitude in the choice of the generalized eigenvector $v_2$ corresponding to the eigenvector $v_1$ via the formula $(A - \lambda)v_2 = v_1$, for if we add a scalar multiple of $v_1$ to $v_2$ we still obtain a generalized eigenvector satisfying the equation $(A - \lambda)v = v_1$: taking $v = v_2 + \alpha v_1$ we see that

$(A - \lambda)(v_2 + \alpha v_1) = (A - \lambda)v_2 + \alpha(A - \lambda)v_1= v_1$

by virtue of the fact that $(A - \lambda)v_1 = 0$; if we now observe that $(\frac{1}{2}, 0)^T = (0, \frac{1}{2})^T + \frac{1}{2}(1, -1)^T$ we see that the apparent discrepancy between Amzoti's choice of generalized eigenvector and mine is resolved. (Note: Amzoti chooses $(-1, 1)^T$ for the eigenvector and $(-\frac{1}{2}, 0)^T$ for the generalized eigenvector, which is consistent with our choice of eigenvector $(1, -1)$ since eigenvectors are, shall we say, scale invariant; that is, $Av = \lambda v$ if and only if $A(\alpha v) = \lambda (\alpha v)$ for $\alpha \ne 0$.)

In the development I have given here, I have used the concepts of nilpotence and the nilpotent part $2N = A - \lambda I$ of the matrix $A$ in lieu of the more conventional methodology which exploits generalized eigenvectors; such an approach makes the direct calculation of $\exp{At}$ simple, as has been seen above. It is in fact the nilpotent part of $A$, $2N$, which gives rise the generalized eigenvectors since $(A - \lambda)v_2 = 2Nv_2 = v_1$, and we have seen that it is the nilpotent part which gives rise to the terms of the functional form $te^{3t}$ in our solution. Such terms may also be derived via generalized eigenvectors directly, as is well-known, and has been mentioned by Amzoti in his/her answer. All of this machinery of course generalizes to higher dimensions; but the discussion of such matters would take us even further afield, and I have, for the moment, written enough. I hope these words are graced to bring with them a little of what Henry Jones, Sr., sought in, if I am not mistaken, Raiders of the Lost Ark, and that is: illumination.

Fiat Lux!

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