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

  • #1
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Homework Statement


I need to (analytically) solve a system of coupled second-order ODEs:
(A) [itex]\frac{du}{dt} - fv = \Omega^2x[/itex]
(B) [itex]\frac{dv}{dt} + fu = \Omega^2y[/itex]

where

[itex]u = \frac{dx}{dt}[/itex]
[itex]v = \frac{dy}{dt}[/itex]

subject to the initial conditions [itex]u(t=0) = U[/itex] and [itex]v(t=0) = 0[/itex].

Homework Equations


---

The Attempt at a Solution


(1) I first converted the ODEs to:
[itex]\frac{d^2x}{dt^2} - f\frac{dy}{dt} = \Omega^2x[/itex]
[itex]\frac{d^2y}{dt^2} + f\frac{dx}{dt} = \Omega^2y[/itex]

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

then making the substitution
[itex]\alpha = x + y[/itex]
[itex]\beta = x - y[/itex]

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), [itex]u=\frac{1}{f}(\Omega^2y - \frac{dv}{dt})[/itex]. Plugging into A, we get:
[itex]\frac{1}{f}(\Omega^2\frac{dy}{dt} - \frac{d^2v}{dt^2})-fv=\Omega^2x[/itex]
(...after rearranging...)
[itex]\frac{d^2v}{dt^2} + v(f-\frac{\Omega^2}{f}) = -\Omega^2x[/itex]

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

Does anyone know where to go from here?
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


I need to (analytically) solve a system of coupled second-order ODEs:
(A) [itex]\frac{du}{dt} - fv = \Omega^2x[/itex]
(B) [itex]\frac{dv}{dt} + fu = \Omega^2y[/itex]

where

[itex]u = \frac{dx}{dt}[/itex]
[itex]v = \frac{dy}{dt}[/itex]

subject to the initial conditions [itex]u(t=0) = U[/itex] and [itex]v(t=0) = 0[/itex].

Homework Equations


---

The Attempt at a Solution


(1) I first converted the ODEs to:
[itex]\frac{d^2x}{dt^2} - f\frac{dy}{dt} = \Omega^2x[/itex]
[itex]\frac{d^2y}{dt^2} + f\frac{dx}{dt} = \Omega^2y[/itex]

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

then making the substitution
[itex]\alpha = x + y[/itex]
[itex]\beta = x - y[/itex]

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), [itex]u=\frac{1}{f}(\Omega^2y - \frac{dv}{dt})[/itex]. Plugging into A, we get:
[itex]\frac{1}{f}(\Omega^2\frac{dy}{dt} - \frac{d^2v}{dt^2})-fv=\Omega^2x[/itex]
(...after rearranging...)
[itex]\frac{d^2v}{dt^2} + v(f-\frac{\Omega^2}{f}) = -\Omega^2x[/itex]

Since [itex]v=\frac{dy}{dt}[/itex], this is now a third order coupled ODE (after we do something similar to above for [itex]u[/itex]), 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.
 
  • #3
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0
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?
 
  • #4
pasmith
Homework Helper
1,774
439
If you set [itex]z = x + iy[/itex] then your system reduces to the second-order autonomous linear ODE with constant coefficients[tex]
\frac{d^2 z}{dt^2} + if\frac{dz}{dt} - \Omega^2 z = 0[/tex] subject to [itex]z'(0) = U[/itex]. 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 [tex]z(t) = Ae^{k_1 t} + Be^{k_2 t}[/tex] where [itex]A[/itex] and [itex]B[/itex] are complex constants and [itex]k_1[/itex] and [itex]k_2[/itex] are the roots of [tex]k^2 + ifk - \Omega^2 = 0.[/tex] It is possible to obtain four linearly-independent real-valued solutions from this general solution.
 
  • #5
8
0
If you set [itex]z = x + iy[/itex] then your system reduces to the second-order autonomous linear ODE with constant coefficients[tex]
\frac{d^2 z}{dt^2} + if\frac{dz}{dt} - \Omega^2 z = 0[/tex] subject to [itex]z'(0) = U[/itex]. 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 [tex]z(t) = Ae^{k_1 t} + Be^{k_2 t}[/tex] where [itex]A[/itex] and [itex]B[/itex] are complex constants and [itex]k_1[/itex] and [itex]k_2[/itex] are the roots of [tex]k^2 + ifk - \Omega^2 = 0.[/tex] 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!
 
  • #6
vela
Staff Emeritus
Science Advisor
Homework Helper
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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
8
0
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?
 

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