1
$\begingroup$

I tried to solve a set of diff equations representing a

$$M\ddot X=KX$$

$M=\begin{bmatrix}m & 0\\0 & m\end{bmatrix}$

$X=\begin{bmatrix}x_1' \\x_2'\end{bmatrix}$ (note that $x_1'=x_1-x_{1,eq}$ and $x_2'=x_2-x_{2,eq}$; this was done just to remove some constants from the equations)

$K=\begin{bmatrix}(8a^2b+k) & -k\\-k & (8a^2b+k)\end{bmatrix}$

I then assumed a solution of the form $x_1'(t) = A_1 e^{i\omega t}$ and $x_2'(t) = A_2 e^{i\omega t}$. I then plugged this into the set of differential equations and reduced to:

B$\begin{bmatrix}A_1\\A_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$

where $B=\begin{bmatrix}(-m\omega^2-8a^2b-k) & k\\k & (-m\omega^2-8a^2b-k)\end{bmatrix}$

But when I do $det(B)=0$ (with $\omega$ as the eigenvalues), I get these as eigenvalues for $\omega$:

$$\omega_1 = -\frac{2ia\sqrt{2}\sqrt{b}}{\sqrt{m}}$$ $$\omega_2=\frac{2ia\sqrt{2}\sqrt{b}}{\sqrt{m}}$$ $$\omega_3=-\frac{\sqrt{2}\sqrt{-4a^2b-k}}{\sqrt{m}}$$ $$\omega_4=\frac{\sqrt{2}\sqrt{-4a^2b-k}}{\sqrt{m}}$$

These don't seem real-valued. These equations are supposed to represent a double well potential with a particle inside of each. Furthermore, the 2 particles are connected by a string. $k$ is the spring constant, $a,b$ are parameters of the potential function $=U(x)=b(x^2-a^2)^2$

$\endgroup$
  • $\begingroup$ The calcultions are ok, so, if you don't have what you expected, it's time to revise the premises. I'd be interesting to know where $k$, $a$ and $b$ come from. $\endgroup$ – Rafa Budría Mar 21 '17 at 17:39
  • $\begingroup$ How did you end up with four eigenvalues for a $2\times2$ matrix? $\endgroup$ – amd Mar 21 '17 at 19:26
  • $\begingroup$ Are you sure about the equation? For something oscillating $M\ddot X=-KX$ with $M,K$ positive definite would be more logical and correspond to the energy/Hamiltionian function $E(X,\dot X)=\frac12(\dot X^TM\dot X+X^TKX)$ which has bounded level sets in phase space. $\endgroup$ – LutzL Mar 22 '17 at 17:44
1
$\begingroup$

When I develop the system I get

$\begin{cases} m\ddot x_1'=(8a^2b+k)x_1'-kx_2'=8a^2bx_1'+k(x_1'-x_2') \\ m\ddot x_2'=-kx_1'+(8a^2b+k)x_2'=8a^2bx_2'-k(x_1'-x_2') \\ \end{cases}\tag{E}$

So it is tempting to introduce two new variables $\begin{cases} u=x_1'+x_2' \\ v=x_1'-x_2' \end{cases}\iff\begin{cases} x_1'=\frac{u+v}{2} \\ x_2'=\frac{u-v}{2} \end{cases}$

By adding and substracting the lines of $(E)$ we now have :

$\begin{cases} m\ddot u=8a^2bu\\ m\ddot v=8a^2bv+2kv \end{cases}\iff \begin{cases} \ddot u-(\omega_1)^2u=0\\ \ddot v-(\omega_2)^2v=0 \end{cases}\quad\text{with}\quad\begin{cases} \omega_1=2a\sqrt{\frac{2b}{m}}\\ \omega_2=\sqrt{\omega_1^2+\frac{2k}{m}} \end{cases}$

I assumed all constants $a,b,m,k$ are positive from your physical description of the problem.

The solution in $(u,v)$ is :

$\begin{cases} u=U_1\;e^{\omega_1 t}+U_2\;e^{-\omega_1 t} \\ v=V_1\;e^{\omega_2 t}+V_2\;e^{-\omega_2 t} \end{cases}$

You have hyperbolic solutions, this is why when you assume $e^{i\omega t}$ form you get complex values for $\omega$.

Plus $x_1,x_2$ are sums of these solutions, and you assumed only a single pulsation while you need two.

Depending of your initial conditions, it might be possible that the solutions reduce to $x_i=A\cosh(\omega_1t+\phi)\pm B\cosh(\omega_2t+\psi)$ (possibly with $\phi=\psi=0$). This is when $U1$ and $U_2$ have same sign, else it reduces to $\sinh$.

$\endgroup$
  • $\begingroup$ Ah, yes I have $\ddot x_1(0)=\ddot x_2(0)=0$. $\endgroup$ – loltospoon Mar 21 '17 at 11:30
  • $\begingroup$ But then I thought that if we don't have real-valued $\omega $'s then this cannot describe real motion? Note - this is a physics problem. $\endgroup$ – loltospoon Mar 21 '17 at 11:31
  • 1
    $\begingroup$ But it is real valued, instead of cos,sin, you get cosh,sinh. It is just that the motion is not a wave. Does it have necessarily to be bounded ? Maybe you assumed little movements to get your equation, in this case the problem would have significance only for a short period of time. (i.e. $t$ small). On the other hand, if $b<0$ then we would have cos, sin. $\endgroup$ – zwim Mar 21 '17 at 11:45
  • $\begingroup$ Wait ok really novice questions: are your $w_i$'s above the eigenvalues (vibrational frequencies)? What are your normal modes? I'm supposed to set this up in the form $M\ddot X = KX$, solve for the eigenvalues, and then the eigenvectors because this gives me the vibrational frequencies and the normal modes. $\endgroup$ – loltospoon Mar 21 '17 at 12:26

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.