2
$\begingroup$

I have the system

$$Y' = \begin{pmatrix} -4 & 4 \\ -8 & 4 \end{pmatrix}Y$$ Using DSolve

   DSolve[{x'[t] == -4x[t]+4 y[t], y'[t] == -8x[t] + 4y[t]},{x,y},t]

Produces

$$x = c_2 \sin (4 t)+c_1 (\cos (4 t)-\sin (4 t)) \\y = c_2 (\sin (4 t)+\cos (4 t))-2 c_1 \sin (4 t)$$

Finding the Fundamental Matrix, $e^{A t}$, using MatrixExp as

   MatrixExp[{{-4t,4t},{-8t,4t}}]

Produces

$$\left( \begin{array}{cc} \cos (4 t)-\sin (4 t) & \sin (4 t) \\ -2 \sin (4 t) & \sin (4 t)+\cos (4 t) \\ \end{array} \right)$$

Notice how that perfectly aligns with the DSolve result.

If I use Eigensystem

    Eigensystem[{{-4, 4}, {-8, 4}}]

This produces

$$ \lambda_i = \pm 4i, v_1 = (1 \pm i, 2) $$

That is a correct result, but I am wondering if we can coax the eigenvectors to be a in a form that can produce the previous two results from the imaginary to real conversion using Euler's Formula.

$$e^{4 i t} = (\cos 4 t + i \sin 4t)\begin{pmatrix} 1 - i \\ 2 \end{pmatrix}$$

Is there a way to peer in to how Mathematica calculated the previous results and make the eigenvectors (since they are not unique) be in a form that I can use this expansion to generate either of the previous two forms?

$\endgroup$
2
$\begingroup$

The Idea is, that you have to introduce constants A, B for every individual term too. You'll then recognize the solution to be obtainable, if you set A=Conjugate[B]. Afterwards its just getting the right expression for the constants to match the exact same solution as the one from above.

As is shown in this example:

(
(A + B*I)*Exp[4 I*t]*{1 - I, 2} +
(A - B*I)*Exp[-4 I*t]*{1 + I, 2}
) /. {A -> C[2]/4, B -> C[1]/2 - C[2]/4} // ExpToTrig // Simplify

{C[1] Cos[4 t] + (-C[1] + C[2]) Sin[4 t], C[2] Cos[4 t] + (-2 C[1] + C[2]) Sin[4 t]}

$\endgroup$
2
  • $\begingroup$ Interesting. Is there some way to automate finding $A$ and $B$? $\endgroup$ – Moo Apr 20 at 3:31
  • $\begingroup$ Well, I found it by solving the resulting expression with the DSolve result for A and B. To automated that might be hard since there is no real target to optimize it to. $\endgroup$ – Julien Kluge Apr 20 at 9:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.