# Analytically solve second-order coupled ODE (damping term coupled)

1. Oct 5, 2014

### chilge

1. The problem statement, all variables and given/known data
I need to (analytically) solve a system of coupled second-order ODEs:
(A) $\frac{du}{dt} - fv = \Omega^2x$
(B) $\frac{dv}{dt} + fu = \Omega^2y$

where

$u = \frac{dx}{dt}$
$v = \frac{dy}{dt}$

subject to the initial conditions $u(t=0) = U$ and $v(t=0) = 0$.

2. Relevant equations
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3. The attempt at a solution
(1) I first converted the ODEs to:
$\frac{d^2x}{dt^2} - f\frac{dy}{dt} = \Omega^2x$
$\frac{d^2y}{dt^2} + f\frac{dx}{dt} = \Omega^2y$

and then added and subtracted them to get:
$\frac{d^2(x+y)}{dt^2} - f\frac{d(x-y)}{dt} = \Omega^2(x+y)$
$\frac{d^2(x-y)}{dt^2} - f\frac{d(x+y)}{dt} = \Omega^2(x-y)$

then making the substitution
$\alpha = x + y$
$\beta = x - y$

which just leads me to the exact problem I started with.

Since I got stuck here, I tried it a different way...

(2) Making note that, from (B), $u=\frac{1}{f}(\Omega^2y - \frac{dv}{dt})$. Plugging into A, we get:
$\frac{1}{f}(\Omega^2\frac{dy}{dt} - \frac{d^2v}{dt^2})-fv=\Omega^2x$
(...after rearranging...)
$\frac{d^2v}{dt^2} + v(f-\frac{\Omega^2}{f}) = -\Omega^2x$

Since $v=\frac{dy}{dt}$, this is now a third order coupled ODE (after we do something similar to above for $u$), and I don't know how to solve it.

Does anyone know where to go from here?

2. Oct 5, 2014

### Ray Vickson

It looks like the easiest way to solve your system is to use Laplace transforms. Let $X = X(s) = {\cal L}[x(t)](s)$ and $Y = Y(s) = {\cal L}[y(t)](s)$, then get two coupled linear equations for $X(s),Y(s)$. You would need to use standard results for relating the Laplace transforms of $x'(t)$ and $x''(t)$ to that of $x(t)$, etc.

3. Oct 5, 2014

### chilge

Hi Ray,

Thank you for your suggestion! It's been quite awhile since I've seen Laplace transforms and had forgotten about them. Unfortunately I ran into a bit of trouble when trying to find a solution...

${\cal L}\{x(t)\}=X(s)$
${\cal L}\{\frac{dx}{dt}\} = sX(s) - x(0) = sX(s)$ (with the initial condition $x(t=0) = 0$)
${\cal L}\{\frac{du}{dt}\} = s^2X(s) - u(0) = s^2X(s) - U$

and

${\cal L}\{y(t)\}=Y(s)$
${\cal L}\{\frac{dy}{dt}\} = sY(s) - y(0) = sX(s)$ (with the initial condition $y(t=0) = 0$)
${\cal L}\{\frac{dv}{dt}\} = s^2Y(s) - v(0) = s^2Y(s)$

Then, taking the Laplace transforms of the coupled ODEs in the problem statement, we get:
$s^2X(s) + U - fsY(s) = \Omega^2X(s)$
$s^2Y(s) + fsX(s) = \Omega^2Y(s)$

Isolating $X$ from the second equation, we get $X(s) = \frac{Y(s)(\Omega^2 - s^2)}{fs}$

Then, plugging into the first equation,
$s\frac{Y(s)(\Omega^2 - s^2)}{fs} + U - fsY(s) = \Omega^2\frac{Y(s)(\Omega^2 - s^2)}{fs}$

$Y(s)=\frac{-U}{\frac{\Omega^2s - s^3}{f}-fs-\frac{\Omega^4 - s^2\Omega^2}{fs}} = \frac{fUs}{f^2s^2 + s^4 - 2s^2\Omega^2 + \Omega^4}$

I don't see how I can decompose this into partial fractions so that I can get $y(t) = {\cal L}^{-1}\{Y(s)\}$. Have I gone wrong somewhere or is this problem not solvable using Laplace transforms?

4. Oct 5, 2014

### pasmith

If you set $z = x + iy$ then your system reduces to the second-order autonomous linear ODE with constant coefficients$$\frac{d^2 z}{dt^2} + if\frac{dz}{dt} - \Omega^2 z = 0$$ subject to $z'(0) = U$. The standard method of solving such ODEs will still work even though one of the coefficients is not real, so the general solution will be $$z(t) = Ae^{k_1 t} + Be^{k_2 t}$$ where $A$ and $B$ are complex constants and $k_1$ and $k_2$ are the roots of $$k^2 + ifk - \Omega^2 = 0.$$ It is possible to obtain four linearly-independent real-valued solutions from this general solution.

5. Oct 5, 2014

### chilge

Hi pasmith,

Thank you for your suggestion! That did the trick, and I was able to come up with a solution for $x(t)$ and $y(t)$. I forgot to mention in my problem statement that $f=2\Omega$, so the characteristic equation $k^2+ifk-\Omega^2$ has a double root, and the solution is of the form $z(t)=Ae^{rt}+Bte^{rt}$.

My solutions were
$x(t) = Utcos(\Omega t)$
$y(t) = -Utsin(\Omega t)$
which satisfy the initial coupled ODEs.

Thank you again for your help!

6. Oct 5, 2014

### vela

Staff Emeritus
Another possible approach: You have
\begin{eqnarray*}
\dot{x} &= u \\
\dot{y} &= v \\
\dot{u} &= fv + \Omega^2 x \\
\dot{v} &= -fu + \Omega^2 y
\end{eqnarray*} which you can solve using matrix methods.

7. Oct 6, 2014

### chilge

Hi vela, I only know how to solve the matrix problem numerically. Do you know how I can get an analytical solution with matrix methods?