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

[chilge]

Homework Statement

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.

Homework Equations

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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?

 

[Ray Vickson]

Homework Statement

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.

Homework Equations

---

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?

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.

 

[chilge]

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.

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,
$s\frac{Y(s)(\Omega^2 - s^2)}{fs} + U - fsY(s) = \Omega^2\frac{Y(s)(\Omega^2 - s^2)}{fs}$

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

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?

 

[pasmith]

If you set z = x + iy then your system reduces to the second-order autonomous linear ODE with constant coefficients<br /> \frac{d^2 z}{dt^2} + if\frac{dz}{dt} - \Omega^2 z = 0 subject to z&#039;(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.

 

[chilge]

If you set z = x + iy then your system reduces to the second-order autonomous linear ODE with constant coefficients<br /> \frac{d^2 z}{dt^2} + if\frac{dz}{dt} - \Omega^2 z = 0 subject to z&#039;(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.

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!

 

[vela]

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.

 

[chilge]

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.

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?