Let $x(t)$ and $y(t)$ be measures of happiness for the husband and wife, respectively. Negative values indicate unhappiness. Let $x_0$ and $y_0$ be the "natural disposition" of the husband and wife, respectively. This is how happy they would be if they were single. During marriage, the couple develops a style of interaction that is called "validating." A model of their marriage dynamics is:

$$\frac{dx}{dt} = r_1(x_0 - x) + a_1y$$ $$\frac{dy}{dt} = r_2(y_0 - y) + a_2x$$

where $a_1$ measures how easily the husband is influenced by the wife's emotions, and $a_2$ is the corresponding quantity for his wife:

$$0 < \frac{a_1}{r_1} < 1$$ $$0 < \frac{a_2}{r_2} < 1$$

a) Find out where the marriage is heading.

b) A marriage is termed "regulated" and is low risk if the long-term happiness of each spouse is enhanced by the marital interaction. Otherwise it is called "unregulated." Give reasons for the following:

Is the marriage "regulated" if each of the spouses is naturally happy $(x_0 > 0, y_0 > 0)$? What if $x_0 < 0, y_0 < 0$?


closed as off-topic by Thomas, Yiorgos S. Smyrlis, Lost1, TooTone, Andrew D. Hwang Feb 27 '14 at 23:44

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This problem is essentially adapted from Steven Strogatz's book "Nonlinear Dynamics and Chaos"; although this set of ODE's is linear. In his book, its the "Romeo and Juliet" problem.

You want to rewrite this set of odes into linear form $\frac{d\bf x}{dt} = \bf{ Ax} + b$.

Then, since you want to see where the marriage is headed, you want to see where the marriage winds up at $t = \infty$, the steady state.

To do that, set the derivative equal to zero, and solve the resulting linear system for $\mathbf{x}^* = [x^* \space y^*]^T$, the equilibrium happiness values of hubby and wifey.

This should be enough to get you the end. The equilibrium values of x and y will tell you whether this is a happy couple, or two divorce attorneys are about to become even richer.

  • $\begingroup$ The equilibrium values only tell you where it occurs, but not the type. Using phase plane analysis, by inspecting the eigenvalues of the coefficient matrix you can tell the type of equilibrium point that you have. $\endgroup$ – Daryl Nov 13 '12 at 7:03

Well, the easy way would be to just plug some numbers in for those unknown constants and run the system through any old numerical solver (we all love RK4) and plot the solution x vs.t and y vs. t and see what are they doing. The messier thing (but a more complete solution) would be to just solve the linear system symbolically getting the exact solutions leaving the unknown parameters in there and see how the solution depends on them. This is just a first order linear coupled system so you can definitely solve it but it is really REALLY messy. You can also do something qualitative in the middle like phase space analysis.


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