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We consider:

$$y''' - 2y'' + 2y' - y = 0$$

The real solution to this equation is:

$$y(x) = c_3e^{x} + c_2e^{x/2}sin\left(\frac{\sqrt{3}x}{2}\right) + c_1e^{x/2}cos\left(\frac{\sqrt{3}x}{2}\right)$$

How do we now represent it as a fundamental- system/matrix ?

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    $\begingroup$ Hint: Write your DEQ as a System of First Order Equations, find eigenvalues / eigenvectors and proceed in the usual way. $\endgroup$ – Moo Feb 10 '18 at 13:46
  • $\begingroup$ @Moo Well, how do I get from one single DEG to a System of equations? $\endgroup$ – Leroy Feb 10 '18 at 17:37
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Write your DEQ as a System of First Order Equations, find eigenvalues / eigenvectors and proceed in the usual way.

We have $$y''' - 2y'' + 2y' - y = 0$$

To write it as a system of first order equations we let $x_1 = y$, so

$$\begin{align} x_1 ' &= y' = x_2 \\ x_2' &= y'' = x_3 \\ x_3' &= y''' = 2y'' - 2 y' + y = 2 x_3 - 2 x_2 + x_1 \end{align}$$

In matrix form, we have

$$X' = Ax = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -2 & 2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$$

Can you proceed?

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  • $\begingroup$ Isn't that a way too complicated method? Since it is just a "one-dimensional system of ODEs" why isn't the real fundamental system just the one-dimensional solution? $\endgroup$ – Leroy Feb 12 '18 at 9:51
  • $\begingroup$ Look at the solution of the ODE and compare to the solution of the system of first order equations. What do you notice? $\endgroup$ – Moo Feb 12 '18 at 13:05
  • $\begingroup$ $y_2$ and $y_3$ are linearly dependent and the second component of the eigenvectors of $y_{2,3}$ look similar ... but it's just not the same ... $\endgroup$ – Leroy Feb 12 '18 at 13:14
  • $\begingroup$ Substitute $y_3$ into the original ODE. Also, I have always seen this defined as a matrix, for example: ocw.mit.edu/courses/mathematics/… and people.math.gatech.edu/~xchen/teach/ode/ExpMatrix.pdf $\endgroup$ – Moo Feb 12 '18 at 13:27
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Proceeding from

$$X' = Ax = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -2 & 2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$$

we find the eigenvalues $\lambda_1 = 1, \lambda_2 = \frac{1}{2}\left(1 + i\sqrt{3}\right), \lambda_3 = \frac{1}{2}\left(1 - i\sqrt{3}\right)$, which correspond to the following eigenvectors: $$v_1 = \begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}, v_2 = \begin{pmatrix}\frac{1}{2}\left(-1 - i\sqrt{3}\right)\\ \frac{1}{2}\left(1 - i\sqrt{3}\right)\\ 1\end{pmatrix}, v_3 = \begin{pmatrix}\frac{1}{2}\left(-1 + i\sqrt{3}\right)\\ \frac{1}{2}\left(1 + i\sqrt{3}\right)\\ 1\end{pmatrix}$$

The Real Fundamental System is then:

$$y_1(x) = e^x\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix} \\y_2(x) = e^{\frac{1}{2}x}\begin{pmatrix}\frac{1}{2}(-\cos(\frac{\sqrt{3}x}{2})+\sqrt{3}\sin(\frac{\sqrt{3}x}{2}))\\ \frac{1}{2}(\cos(\frac{\sqrt{3}x}{2})+\sqrt{3}\sin(\frac{\sqrt{3}x}{2}))\\ 1\end{pmatrix} \\y_3(x) = e^{\frac{1}{2}x}\begin{pmatrix}\frac{1}{2}(-\cos(\frac{\sqrt{3}x}{2})+\sqrt{3}\sin(\frac{\sqrt{3}x}{2}))\\ \frac{1}{2}(\cos(\frac{\sqrt{3}x}{2})+\sqrt{3}\sin(\frac{\sqrt{3}x}{2}))\\ 1\end{pmatrix}$$

Tell me if there's something wrong or another and better representation for the Real Fundamental System.

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  • $\begingroup$ That is what I would have done. Maybe those expressions can be simplified. $\endgroup$ – Moo Feb 10 '18 at 21:21

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