# Solving this set of differential equations doesn't yield real eigenvalues

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 $=U(x)=b(x^2-a^2)^2$

• 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. – Rafa Budría Mar 21 '17 at 17:39
• How did you end up with four eigenvalues for a $2\times2$ matrix? – amd Mar 21 '17 at 19:26
• 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. – LutzL Mar 22 '17 at 17:44

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 $\sinh$.

• Ah, yes I have $\ddot x_1(0)=\ddot x_2(0)=0$. – loltospoon Mar 21 '17 at 11:30
• 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. – loltospoon Mar 21 '17 at 11:31
• 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. – zwim Mar 21 '17 at 11:45
• 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. – loltospoon Mar 21 '17 at 12:26