1

$$ \frac{dx}{dt}=\left(\begin{matrix}1 & 9 & 9 \\0 & 19 & 18 \\ 0 & 9 & 10 \end{matrix} \right)\left(\begin{matrix}x_1\\x_2\\x_3 \end{matrix}\right)$$

My try:

We assume it has the solution of the form $x=\alpha e^{\lambda t}$

Setting the Determinant of the system to zeroes.

I have found three eigenvalues,

$$\lambda_1=1,\lambda_2=10, \lambda_3=19 $$

For $\lambda_1=1$

$$\left(\begin{matrix}\alpha_1 + 9\alpha_2 + 9\alpha_3 \\ 19\alpha_2 +18\alpha_3 \\ 9\alpha_2 + 10\alpha_3 \end{matrix} \right)=\left(\begin{matrix}\alpha_1\\ \alpha_2\\ \alpha_3 \end{matrix}\right)$$

Solving it we have,

$\alpha_1$ is independent and $\alpha_2=-\alpha_3$

For $\lambda_2=10$

$$\left(\begin{matrix}\alpha_1 + 9\alpha_2 + 9\alpha_3 \\ 19\alpha_2 +18\alpha_3 \\ 9\alpha_2 + 10\alpha_3 \end{matrix} \right)=\left(\begin{matrix}10\alpha_1\\ 10\alpha_2\\ 10\alpha_3 \end{matrix}\right)$$

$\alpha_1=\alpha_2+\alpha_3$ and $\alpha_2=-2\alpha_3$ and $\alpha_2=0$

For $\lambda_3=19$

I run into similar problems

I know that it is wrong, can someone help me in this matter?

Moo
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Crazy
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    your eigen values seem to be incorrect because the determinant is $28$ whereas the product of the eigen values found by you is $190$. I think you have mistakenly though of this as a diagonal matrix. Based on your matrix, the eigen values should be $1,1,28$. – Anurag A Sep 09 '17 at 18:49
  • Thanks now I spotted it. – Crazy Sep 09 '17 at 18:51
  • Presumably you are asking how to solve the system of linear (constant coefficient) first-order differential equations? You launch into "my try" without quite saying what the problem is. Are you aware of the solution in terms of matrix exponentials? – hardmath Sep 09 '17 at 18:56
  • Yes.$$\left(\begin{matrix}a_{11}&a_{12}&a_{13}\ a_{21}&a_{22}&a_{23} \a_{31}&a_{32}&a_{33}\end{matrix}\right)\left(\begin{matrix}\alpha_1\\alpha_2\ \alpha_3\end{matrix}\right)=\lambda\left(\begin{matrix}\alpha_1\\alpha_2\ \alpha_3\end{matrix}\right)$$ Or simply, $$|A-\lambda \mathrm{I}|=0$$ where$ \mathrm{I}$ is the $n \times n$ identity matrix. – Crazy Sep 09 '17 at 19:01

1 Answers1

1

We have

$$X'= \left(\begin{matrix}x'_1\\x'_2\\x'_3 \end{matrix}\right) = \left(\begin{matrix}1 & 9 & 9 \\0 & 19 & 18 \\ 0 & 9 & 10 \end{matrix} \right)\left(\begin{matrix}x_1\\x_2\\x_3 \end{matrix}\right)$$

We find eigenvalues using the characteristic polynomial by solving $|A - \lambda I| = 0$ giving $-(\lambda -28) (\lambda -1)^2 = 0 $, so

$$\lambda_{1, 2} = 1, \lambda_3 = 28$$

Notice the repeated eigenvalue, so we may or may not have a deficient matrix. Are you familiar with algebraic and geometric multiplicities?

To find the eigenvectors, for each eigenvalue we solve $[A - \lambda_i I]v_i = 0$.

For $\lambda_{1,2} = 1$

$$RREF ~[A - I]v_{1,2} = \left(\begin{matrix}0 & 1 & 1 \\0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right)v_{1,2} = 0$$

It turns out that we can find two linearly independent eigenvectors for this eigenvalue as

$$v_1 = (1, 0, 0), v_2 = (0, -1, 1)$$

For $\lambda_3 = 28$

$$RREF ~[A - 28I]v_{3} = \left(\begin{matrix}1 & 0 & -1 \\0 & 1 & -2 \\ 0 & 0 & 0 \end{matrix} \right)v_{3} = 0$$

$$v_3 = (1, 2, 1)$$

Because we were able to find three linearly independent eigenvectors, we can write the solution as

$$X(t) = (c_1 v_1 + c_2 v_2)e^t + c_3 v_3 e^{28 t} = \left(c_1 \left(\begin{matrix}1\\ 0 \\ 0 \end{matrix}\right) + c_2 \left(\begin{matrix}0\\ -1 \\ 1 \end{matrix}\right)\right)e^t + c_3 \left(\begin{matrix}1\\ 2 \\ 1 \end{matrix}\right)e^{28t}$$

It is worth noting that there are many ways to solve such problems, for example see Systems of differential equations or Nineteen Dubious Ways to Compute the Exponential of a Matrix.

Moo
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  • Additional question for $3 \times 3$ matrix.
    1. Suppose that the multiplicity of the eigenvalues are $m=2$. The first solution will be $x=\alpha_1 e^{\lambda t}$. When do we know the second solution is not $x=\alpha_2 e^{\lambda t}$ but rather $x=(\alpha t+\beta) e^{\lambda t}$?
    2. Assume that the multiplicity is $3$ for all the roots of the equation.

    When do we find $\alpha=k_1\alpha_1+k_2 \alpha_2$. as the third solution?

     – Crazy Sep 10 '17 at 05:04
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    @Crazy: I think it would be instructive to work through this 2 x 2 and then some 3x3 examples, for example http://www-users.math.umn.edu/~leif0020/Spring2011/Repeated_eigenvalues.pdf and https://math.stackexchange.com/questions/985958/on-a-linear-3-times-3-system-of-differential-equations-with-repeated-eigenvalu – Moo Sep 10 '17 at 12:49
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    @Crazy: I should have also mentioned a method called chaining and Jordan Form - which is where this is all heading, as in these sets of excellent notes: http://mathcs.holycross.edu/~spl/old_courses/304_fall_2008/handouts/jordan.pdf and http://www.maths.tcd.ie/~vdots/teaching/files/MA1111+1212-0809/JordanExample.pdf and http://www.ms.uky.edu/~lee/amspekulin/jordan_canonical_form.pdf – Moo Sep 10 '17 at 12:57